AIRCRAFT FLIGHT DYNAMICS AND CONTROL - Ira A....

AIRCRAFT FLIGHT DYNAMICSAND CONTROL

August 30, 2008

Dennis S. BernsteinDepartment of Aerospace EngineeringThe University of MichiganAnn Arbor, MI [email protected]

C o p y r i g h t 2 0 0 8

Contents

Chapter 1. Review of Kinematics 3

1.1 Points, Particles, and Bodies 3

1.2 Physical and Mathematical Vectors 3

1.3 Dot Product 5

1.4 Angle Vector and Cross Product 6

1.5 Frames 8

1.6 Signed Angles 11

1.7 Particle-Fixed Frames 11

1.8 Physical Matrices 12

1.9 Physical Cross Product 15

1.10 Rotation and Orientation Matrices 18

1.11 Euler’s Theorem 24

1.12 Quaternions 29

1.13 Euler Angles 29

1.14 Frame Derivatives 33

1.15 Momentum 34

1.16 Angular Momentum 35

1.17 Angular Velocity Vector 35

1.18 Euler Angle Derivatives 40

1.19 Transport Theorem 41

1.20 Double Transport Theorem 43

1.21 Summation of Angular Velocities 44

1.22 Solid and Open Frame Dots 45

1.23 Problems 45

Chapter 2. Aircraft Kinematics 49

2.1 Frames Used in Aircraft Dynamics 49

2.2 Velocity Vector 52

2.3 Rotational Kinematics 57

2.4 Vector Derivatives in Rotating Frames 59

2.5 Cross Product 60

2.6 Problems 61

Chapter 3. Review of Dynamics 65

iv CONTENTS

3.1 Newton’s First Law for Particles 65

3.2 Newton’s Second Law for Particles 66

3.3 Forces and Moments 66

3.4 Change in Angular Momentum 68

3.5 Momentum and Angular Momentum Relative to Centerof Mass 74

3.6 Continuum Bodies 77

3.7 Properties of the Inertia Matrix 78

3.8 Problems 79

Chapter 4. Aircraft Dynamics 81

4.1 Flight Dynamics and Control 81

4.2 History of Aircraft Stability and Control 81

4.3 Aerodynamic Forces 81

4.4 Translational Momentum Equations 83

4.5 Rotational Momentum Equations 87

4.6 Aircraft Equations of Motion Resolved in FAC 89

4.7 Problems 90

Chapter 5. Linearization 93

5.1 Taylor series 93

5.2 Alternative Linearization Procedure 94

5.3 Trigonometric Functions 94

5.4 Steady Flight 95

5.5 Linearization of the Aircraft Kinematics 96

5.6 Linearization of the Aircraft Dynamics in FAC 97

5.7 Linearization of the Aircraft Dynamics in FSf98

5.8 Problems 101

Chapter 6. Static Stability and Stability Derivatives 105

6.1 Control Surface Deflections 105

6.2 Aerodynamic Force Coefficients 105

6.3 Linearization of Forces in FSf108

6.4 Aerodynamic Moment Coefficients 115

6.5 Linearization of Moments in FSf124

6.6 Effect of Adverse Control Derivatives 126

6.7 Problems 127

Chapter 7. Linearized Equations of Motion 129

7.1 Longitudinal Equations of Motion 129

7.2 Linearized Longitudinal Equations and Transfer Functions 131

7.3 Lateral Equations of Motion 133

7.4 Linearized Lateral Equations and Transfer Functions 135

7.5 Problems 139

CONTENTS v

Chapter 8. Analysis of Flight Modes 141

8.1 Eigenflight 141

8.2 Phugoid Mode 142

8.3 Short Period Mode 145

8.4 Phugoid Approximation 145

8.5 Short Period Approximation 147

8.6 Problems 148

Chapter 9. Control Concepts 151

9.1 Problems 151

Chapter 10. Control of Aircraft 155

10.1 Problems 155

Appendix A. Mathematical Background 157

A.1 Vectors and Matrices 157

A.2 Complex Numbers, Vectors and Matrices 159

A.3 Eigenvalues and Eigenvectors 162

A.4 Single-Degree-of-Freedom Systems 165

A.5 Matrix Differential Equations 167

A.6 Eigensolutions 169

A.7 State Space Form 170

A.8 Linear Systems with Forcing 171

A.9 Standard Input Signals 171

A.10 Laplace Transforms 174

A.11 Solution of ODE’s 176

A.12 Initial Value and Slope Theorems 178

A.13 Final Value Theorem 178

A.14 Laplace Transforms of State Space Models 179

A.15 Pole Locations and Response 181

A.16 Stability 184

A.17 Routh Stability Criterion 186

A.18 Problems 187

Appendix B. Frequency Response 195

B.1 Sinusoidal Gain and Phase Shift 195

B.2 Phase Angle as a Delay or Advance 195

B.3 Frequency Response Law for Linear Systems 198

B.4 Frequency Response Plots for Linear Systems Analysis 199

B.5 Circuits and Filters 201

B.6 Bode Plot 201

B.7 Magnitude Crossover Frequency 202

B.8 Phase of G(ω) 202

B.9 Poles at s = p 203

vi CONTENTS

B.10 Phase Angle 204

B.11 Damped Oscillator 204

B.12 Problems 205

Appendix C. MATLAB Operations 209

C.1 atan2 209

C.2 expm 209

C.3 rlocus 209

C.4 bode 210

Appendix D. Dimensions and Units 211

D.1 Mass and Force 211

D.2 Force, Impulse, and Momentum 211

Bibliography 213

Acknowledgments

I wish to thank all of those who contributed to this effort. Mark Ardema,Robert Fuentes, Anouck Girard, Don Greenwood, Harris McClamroch, andMalcolm Shuster read and commented on portions of the manuscript. SuhailAkhtar, Julie Bellerose, Haoyun Fu, Ashwani Padthe, and Jin Yan con-tributed to the transformation of lecture material to LATEX and assistedin checking and deriving many of the results.

I wish to thank the Aerospace Engineering Department of the Univer-sity of Michigan for helping to support this project.

Dennis Bernstein

Chapter One

Review of Kinematics

1.1 Points, Particles, and Bodies

A point has zero size and zero mass. A particle has zero size andpositive mass. Points and particles have location (position or displacement)relative to other points and particles.

A reference point is a point with respect to which the locations of otherpoints are determined.

Points and particles can have translational motion relative to otherpoints and particles. Translational motion includes velocity and accelera-tion. Points and particles cannot rotate.

For example, the point or particle x has a location relative to the pointor particle y. Likewise, the point or particle x has a velocity and accelerationrelative to the point or particle y.

A body is a finite collection of rigidly interconnected particles. A bodythus has positive size and positive mass. A body can translate and rotate.A body has position relative to points and particles as well as rotation(orientation) relative to other bodies. A body has rotational motion relativeto other bodies. Rotational motion includes angular velocity and angularacceleration.

1.2 Physical and Mathematical Vectors

A physical vector (as distinct from a mathematical vector, which is acolumn of numbers) is an abstract quantity having a tip and a tail and thusmagnitude and direction. A physical vector is not a physical object, andthus it is not located anywhere, although we can envision its tail located atan arbitrary location for convenience. A physical vector is denoted with aharpoon or hat over the symbol denoting the physical quantity. For example,

4 CHAPTER 1

f is a physical vector representing a force applied to a particle in a body,

whiler x/y is the physical vector representing the position of the point x

relative to the point y. We can envision the tip ofr x/y at x and its tail

at y. However,r x/y has no fixed location. A physical vector may have

dimensions or it may be dimensionless. The magnitude ofx is denoted by

|x |. There are 11 distinct types of physical vectors that arise in dynamics,namely:

i) Dimensionless. A dimensionless vector has no physical units associatedwith it. A unit dimensionless vector is written as ı. Three mutuallyorthogonal unit dimensionless vectors comprise a frame.

ii) Angle. The angle from the nonzero physical vectorx to the nonzero

physical vectory is represented by the physical vector

θy /

x, which

is perpendicular to the plane containingx and

y . The direction of

θy /

x

is determined by the right hand rule with fingers curled from

x to

y . The magnitude of

θy /

x

is the number of radians betweenx

andy .

iii) Position. The position of the point y relative to the point x is written

asr y/x.

iv) Velocity. The velocity of the point y relative to the point x with respect

to the frame FA is written asv y/x/A.

v) Acceleration. The acceleration of the point y relative to the point x

with respect to the frame FA is written asa y/x/A.

vi) Momentum. The momentum of the particle y relative to the point x

with respect to the frame FA is written asp y/x/A. The momentum

of the body B relative to the point x with respect to the frame FA is

written aspB/x/A.

vii) Force. A force applied to the particle x is written asf x. The net force

applied to the body B is written asf B.

viii) Angular velocity. The angular velocity of the frame FB relative to the

frame FA is written asωB/A.

ix) Angular acceleration. The angular acceleration of the frame FB rela-

tive to the frame FA with respect to the frame FC is written asαB/A/C.

REVIEW OF KINEMATICS 5

x) Angular momentum. The angular momentum of the particle x relative

to the point w with respect to the frame FA is written asHx/w/A. The

angular momentum of the body B relative to the point w with respect

to the frame FA is written asHB/w/A.

xi) Moment. A moment applied to a particle x relative to the point y is

written asMx/y. The net moment applied to the body B relative to

the point y is written asMB/y.

If x, y, and z are points, thenr z/x =

r z/y +

r y/x. (1.2.1)

Likewise,v z/x/A =

v z/y/A +

v y/x/A (1.2.2)

anda z/x/A =

a z/y/A +

a y/x/A. (1.2.3)

Note thatr x/y = −

r y/x, (1.2.4)

v x/y/A = −

v y/x/A, (1.2.5)

andax/y/A = −

a y/x/A. (1.2.6)

A physical vector can be multiplied by a real scalar, as in 3f x or

−6f x. A physical vector

r x/y(t) can be a function of time.

1.3 Dot Product

Let θ ∈ [0, π] denote the angle between two nonzero physical vectorsx and

y . The dot product

x ·y between

x and

y is defined by

x ·y = |x ||y | cos θ. (1.3.1)

Therefore,

θ = arccos

x ·y|x ||y |

∈ [0, π]. (1.3.2)

6 CHAPTER 1

Translation Rotation

Kinematics Point Frame

Dynamics Particle Body

Figure 1.2.1Conceptual roadmap for kinematics and dynamics. When mass is irrelevant, a particle iseffectively a point. Furthermore, when mass distribution is irrelevant, a body is viewed

as a body-fixed frame.

The physical vectorsx and

y are mutually orthogonal if

x ·y = 0, that is,

if θ = π/2.

1.4 Angle Vector and Cross Product

Letx and

y be physical vectors, and let θ

y /x

∈ [0, π] be the angle

betweenx and

y . The unit angle vector θ

y /x

fromx to

y is the dimension-

less unit vector orthogonal to bothx and

y whose direction is determined

by the right hand rule with the fingers curved fromx to

y and the thumb

pointing in the direction of θy /

x. The angle vector

θy /

x

is defined by

θy /

x

= θy /

xθy /

x. (1.4.1)

See Figure 1.4.2. Note that

θy /

x

= |θy /

x|. (1.4.2)

Ifx and

y are position vectors, then the angle vector

θy /

x

has the dimen-

sions of radians since

length0 = length/length. (1.4.3)


-x

*

y

θθy /

x

Y

Figure 1.4.2

Angle vector

θ y /

x

of magnitude θy /

x

fromx to

y . Note that

θ y /

x

as shown points

out of the page.

The cross product of the physical vectorsx and

y is defined as

x ×

y

= |x ||y |(sin θy /

x)θy /

x. (1.4.4)

Therefore,

|x ×y | = |x ||y | sin θ

y /x, (1.4.5)

θy /

x

=1

|x ×y |

x ×

y , (1.4.6)

y ×

x = −(x ×

y ) = (−x) ×

y =x × (−

y ), (1.4.7)

and

θy /

x

= −θx/

y. (1.4.8)

Fact 1.4.1 Letx and

y be non-parallel physical vectors, and let θ ∈

(0, π) be the angle betweenx and

y . Then

θy /

x

=θy /

x

|x ×y |

x ×

y . (1.4.9)

8 CHAPTER 1

Fact 1.4.2 Letx ,

y , and

z be nonzero physical vectors lying in a

single plane. Then,θz /

x

=θz /

y

+θy /

x. (1.4.10)

1.5 Frames

A frame consists of three unit, dimensionless physical vectors (framevectors) that are mutually orthogonal. Since each frame vector is a physicalvector, the notion of “location” of the frame is meaningless. In addition,since a frame has no location, it cannot translate and thus has no velocity oracceleration. However, it is often useful to associate a reference point with a

6k

9ı

j

Figure 1.5.3A right-handed frame.

frame. When we do this, we call the reference point the origin of the frame,and we draw the frame as if it were located at the reference point, whichmay have nonzero velocity or acceleration.

Letting FA be a frame, we denote its unit vectors by ıA, A, kA. Theframe FA is right handed if the labeling of the frame vectors conforms to

ıA × A = kA.


Henceforth, all frames are right handed. Consequently,

A × kA = ıA

andkA × ıA = A.

See Figure 1.5.3.

Any physical vector can be resolved in any frame. Letx be a physical

vector and let FA be a frame. Thenx∣∣∣A

is the physical vectorx resolved

in FA. In fact,x∣∣∣A

is the mathematical vector defined by

x∣∣∣A

=

ıA ·xA ·xkA ·x

=

x1

x2

x3

, (1.5.1)

where x1, x2, and x3 are the components of the physical vectorx resolved

in FA. Every physical vector is uniquely specified by resolving it in a frame

sincex can be reconstructed from

x∣∣∣A

by means of

x =

[

ıA A kA

] x∣∣∣A

= x1ıA + x2A + x3kA. (1.5.2)

The quantity[

ıA A kA

]is a vectrix since its entries are physical vectors.

Fact 1.5.1 Let FA be a frame and letx and

y be physical vectors.

Thenx =

y (1.5.3)

if and only if

x∣∣∣A

=y∣∣∣A. (1.5.4)

Let FA be a frame and letx and

y be physical vectors, where

x∣∣∣A

=

x1

x2

x3

,y∣∣∣A

=

y1

y2

y3

. (1.5.5)

Then

x ·y =

x∣∣∣

T

A

y∣∣∣A

= x1y1 + x2y2 + x3y3. (1.5.6)

10 CHAPTER 1

6k

6θy /

x

Y

9ı x

y

-θ

j

Figure 1.5.4

The angle vector

θ y /

x

fromx to

y in the ı- plane points in the k direction. Thus, an

angle vector in the k direction corresponds to a counterclockwise rotation in the ı- plane.

Furthermore, the cross product ofx and

y can be represented as

x ×

y = det

ıA A kA

x1 x2 x3

y1 y2 y3

= (x2y3 − x3y2)ıA − (x1y3 − x3y1)A + (x1y2 − x2y1)kA. (1.5.7)

Hence

(x ×

y )∣∣∣A

=x∣∣∣A× y∣∣∣A

=

x2y3 − x3y2

x3y1 − x1y3

x1y2 − x2y1

=

0 −x3 x2

x3 0 −x1

−x2 x1 0

y1

y2

y3

. (1.5.8)

Fact 1.5.2 Letx ,

y ,

z be physical vectors. Then

x × (

y ×

z ) = (x ·z )

y − (

x ·y )

z (1.5.9)


and

(x ×

y ) ×z = (

x ·z )

y − (

y ·z )

x. (1.5.10)

Furthermore,

(x ×

y ) ·z =x · (y ×

z ). (1.5.11)

Finally, let FA be a frame. Then

(x ×

y ) ·z = det[

x∣∣∣A

y∣∣∣A

z∣∣∣A

]

. (1.5.12)

If a frame rotates according to the rotation of a body, then the frameis a body-fixed frame. A body-fixed frame can be painted on a body. Theorigin of a body-fixed frame is usually taken to be a point in the body. Viceversa, the orientation of a body is usually defined by the orientation of abody-fixed frame.

1.6 Signed Angles

A signed angle θ ∈ [−π, π] is used to determine whether an angle vectorcorresponds to a clockwise or counterclockwise rotation about a frame axis.A positive rotation about a frame axis is determined by the right hand rule.See Figure 1.5.4.

1.7 Particle-Fixed Frames

If the origin of a frame coincides with the location of a point or particleand if the orientation of frame depends on the position of the point orparticle, then the frame is a particle-fixed frame. The following particle-fixed frames are used in practice:

i) Cylindrical frame. The frame vectors are radial, transverse, and

vertical, denoted by (er, eθ, k).

ii) Spherical frame. The frame vectors are radial, polar, and azimuthal,denoted by (er, eθ, eφ).

iii) Normal-tangential-binormal frame. The frame vectors are tangen-tial, normal, and binormal, denoted by (et, en, eb).

iv) Local vertical/local horizontal (LVLH) frame. The frame vectorsare denoted by (eLV, eLH, eLN). eLV points toward the center of the

12 CHAPTER 1

Earth,

eLN

=eLV ×

v x/OE/E

|eLV ×v x/OE/E|

, (1.7.1)

wherev x/OE/E is the velocity of the particle relative to the center

of the Earth with respect to the Earth frame, and

eLH= eLN × eLV. (1.7.2)

1.8 Physical Matrices

Letx1, . . . ,

xn and

y 1, . . . ,

y n be physical vectors. Then

M

=

n∑

i=1

x iy i (1.8.1)

is a physical matrix. Physical matrices are also called dyadics or second-ordertensors.

Physical matrices transform physical vectors. Letx ,

y , and

z be

physical vectors, and defineM =

xy . (1.8.2)

ThenM

z = (

xy )

z =

xy ·z = (

y ·z )

x (1.8.3)

and

z

M =

z (

xy ) = (

z ·x)

y . (1.8.4)

Furthermore, letw and

v be physical vectors and define

N

=wv . (1.8.5)

ThenM

N =

M

wv =

(M

w

)

v =

x(

y ·w)

v = (

y ·w)

xv (1.8.6)

andM

Nz = (

xy )(

wv )

z =

x(

y ·w)(

v ·z ) = (

y ·w)(

v ·z )

x. (1.8.7)

Note that (xy )

z and

x(

yz ) are generally different.


Let FA be a frame. Then the physical identity matrix

U is defined by

U = ıAıA + AA + kAkA. (1.8.8)

Fact 1.8.1 For all physical vectorsx ,

Ux =

x. (1.8.9)

Furthermore,

U is independent of the choice of frame in (1.8.8).

Letx and

y be physical vectors, and define

M

=xy . (1.8.10)

ThenM

T

= (xy )T

=yx. (1.8.11)

Furthermore, let

N and

L be physical matrices. Then(N +

L

)T

=

N

T

+

L

T

(1.8.12)

and(N

L

)T

=

L

TN

T

. (1.8.13)

Fact 1.8.2 Letx and

y be physical vectors, and defineM

=xy −

yx. (1.8.14)

ThenM

T

= −M. (1.8.15)

Let FA and FB be frames. Then the physical rotation matrix

RB/A

from FA to FB is defined byRB/A

= ıB ıA + BA + kBkA. (1.8.16)

14 CHAPTER 1

Fact 1.8.3 Let FA and FB be frames. ThenRB/AıA = ıB, (1.8.17)

RB/AA = B, (1.8.18)

RB/AkA = kB. (1.8.19)

Furthermore,

RB/A =

R

T

A/B (1.8.20)

andRB/A

RA/B =

U. (1.8.21)

Fact 1.8.4 Let FA, FB, and FC be frames. ThenRC/A =

RC/B

RB/A. (1.8.22)

Proof The result is immediate.

Fact 1.8.5 Letx and

y be physical vectors, let FA be a frame, and

define

M

=xy . Then

M

∣∣∣∣∣A

= (xy )∣∣∣A

=x∣∣∣A

y∣∣∣

T

A. (1.8.23)

Note that

M

∣∣∣∣∣A

is a 3 × 3 matrix.

The following result is analogous to Fact 1.5.1.

Fact 1.8.6 Let

M and

N be physical matrices. Then

M =

N (1.8.24)

if and only ifM

∣∣∣∣∣A

=

N

∣∣∣∣∣A

. (1.8.25)


Fact 1.8.7 Let FA be a frame, let

M and

N be physical matrices, and

letx and

y be physical vectors. Then

M

T∣∣∣∣∣∣A

=

M

∣∣∣∣∣

T

A

, (1.8.26)

(

M +

N)

∣∣∣∣∣A

=

M

∣∣∣∣∣A

+

N

∣∣∣∣∣A

, (1.8.27)

(

M

x)

∣∣∣∣∣A

=

M

∣∣∣∣∣A

x∣∣∣A, (1.8.28)

(x

M)

∣∣∣∣∣A

=

M

∣∣∣∣∣

T

A

xA, (1.8.29)

(

M

N)

∣∣∣∣∣A

=

M

∣∣∣∣∣A

N

∣∣∣∣∣A

, (1.8.30)

[x(

M

y )]

∣∣∣∣∣A

=x∣∣∣A

y∣∣∣

T

A

M

∣∣∣∣∣

T

A

, (1.8.31)

[(x

M)

y ]

∣∣∣∣∣A

=

M

∣∣∣∣∣

T

A

x∣∣∣A

y∣∣∣

T

A, (1.8.32)

x · (

M

y ) = (

x

M) ·y =

x∣∣∣

T

A

M

∣∣∣∣∣A

y∣∣∣A. (1.8.33)

1.9 Physical Cross Product

Letx be a physical vector. Then, for all physical vectors

y , the

physical cross product matrix

M =

x×

is defined byM

y =

x×y

=x ×

y . (1.9.1)

16 CHAPTER 1

Fact 1.9.1 Letx be a physical vector, let FA be a frame, and define

M

=x×

. ThenM = (ıA ·x)(kAA − AkA) + (A ·x)(ıAkA − kAıA)

+ (kA ·x)(A ıA − ıAA). (1.9.2)

Furthermore,

x×∣∣∣A

=x∣∣∣

×

A. (1.9.3)

Proof Lety be a physical vector and let

x∣∣∣A

=

x1

x2

x3

,y∣∣∣A

=

y1

y2

y3

.

ThenM

∣∣∣∣∣A

y∣∣∣A

= (

M

y )

∣∣∣∣∣A

= (x ×

y )∣∣∣A

=x∣∣∣A× y∣∣∣A

=

0 −x3 x2

x3 0 −x1

−x2 x1 0

y1

y2

y3

.

Therefore,

M

∣∣∣∣∣A

is given by

M

∣∣∣∣∣A

=

0 −kA ·x A ·xkA ·x 0 −ıA ·x−A ·x ıA ·x 0

.

It now follows from Fact 1.8.7 that

M is given by (1.9.2).

The following result shows that the representation forx×

given by(1.8.3) is independent of the choice of frame.

Fact 1.9.2 Let FA and FB be frames, letx be a physical vector, define


M by (1.9.2), and define

N by

N = (ıB ·x)(kBB − BkB) + (B ·x)(ıBkB − kBıB)

+ (kB ·x)(B ıB − ıBB). (1.9.4)

ThenM =

N. (1.9.5)

Proof Lety be a physical vector. Then

(M

y

)∣∣∣∣∣A

=

M

∣∣∣∣∣A

y∣∣∣A

=x∣∣∣

×

A

y∣∣∣A

=(x ×

y)∣∣∣A.

HenceM

y =

x ×

y .

Likewise,Ny =

x ×

y .

Therefore, for all physical vectory ,

M

y =

Ny .

Therefore,M =

N.

Fact 1.9.3 Letx be a physical vector. Then

(x×)T

= −x×

(1.9.6)

and

(x×)2

=xx − |x |2

U. (1.9.7)

Proof The result follows from (1.9.2) and Fact 1.8.2.

18 CHAPTER 1

Fact 1.9.4 Letx and

y be physical vectors. Then

(x ×

y )× =yx −

xy . (1.9.8)

Proof Letz be a physical vector. Using Fact 1.5.2 we have

(x ×

y )×z = (

x ×

y )×z

= −z × (

x ×

y )

= (x ·z )

y − (

y ·z )

x

= (yx)

z − (

xy )

z

= (yx −

xy )

z .

Fact 1.9.5 Letx be a physical vector, and let

R be a physical rotation

matrix. Then

(

Rx)× =

Rx×

R

T

. (1.9.9)

Proof To be added.

1.10 Rotation and Orientation Matrices

Fact 1.10.1 Let FA and FB be frames, and define

RB/A

=

RB/A

∣∣∣∣∣B

. (1.10.1)

Then

RB/A = OA/B, (1.10.2)

where

OA/B

=

ıA · ıB ıA · B ıA · kB

A · ıB A · B A · kB

kA · ıB kA · B kA · kB

. (1.10.3)

Furthermore,RB/A

∣∣∣∣∣A

= RB/A. (1.10.4)

Finally,

RA/B = R−1B/A (1.10.5)


and

OA/B = O−1B/A. (1.10.6)

Proof Let ei denote the ith column of the 3× 3 identity matrix. Notethat

RB/A

∣∣∣∣∣B

= e1 ıA|TB + e2 A|TB + e3 kA

∣∣∣

T

B

=

ıA|TBA|TBkA|TB

=

ıA|TB e1 ıA|TB e2 ıA|TB e3A|TB e1 A|TB e2 A|TB e3

kA

∣∣∣

T

Be1 kA

∣∣∣

T

Be2 kA

∣∣∣

T

Be3

=




= OA/B.

Finally, (1.10.5) follows from (1.8.21).

We can write OA/B in terms of row and column vectrices as

OA/B =

ıAAkA

·[

ıB B kB

]. (1.10.7)

The 3 × 3 matrix RA/B is the rotation from FA to FB. The 3 × 3 matrixOA/B is the orientation of FA relative to FB.

The following result shows that the entries of OA/B are the cosines ofthe angles between pairs of vectors in frames FA and FB. Consequently,OA/B is also called a direction cosine matrix.

Fact 1.10.2 Let FA and FB be frames. Then

OA/B =

cos θıA/ıB cos θıA/B cos θıA/kB

cos θA/ıB cos θA/B cos θA/kB

cos θkA/ıBcos θkA/B

cos θkA/kB

. (1.10.8)

The following result shows that OA/B is an orthogonal matrix.

20 CHAPTER 1


RB/A

∣∣∣∣∣A

=

R

T

A/B

∣∣∣∣∣∣B

=

(RA/B

∣∣∣∣∣B

)T

= OTB/A = OA/B = O

−1B/A. (1.10.9)

The following result relates vectrices corresponding to different frames.

Fact 1.10.4 let FA and FB be frames. Then

ıBBkB

= OB/A

ıAAkA

, (1.10.10)

where

OB/A =

ıB · ıA ıB · A ıB · kA

B · ıA B · A B · kA

kB · ıA kB · A kB · kA

. (1.10.11)

The following identities are extremely useful.


U =

[

ıB B kB

]OB/A

ıAAkA

(1.10.12)

and

RB/A =

[

ıB B kB

]

ıAAkA

=[

ıB B kB

]OA/B

ıBBkB

. (1.10.13)

Proof Using (1.10.10) it follows that

[

ıB B kB

]OB/A

ıAAkA

=[

ıB B kB

]

ıBBkB

= ıB ıB + BB + kBkB

=

U.

Fact 1.10.6 Let FA and FB be frames, and letx be a physical vector.


Thenx∣∣∣B

= OB/Ax∣∣∣A

(1.10.14)

andx∣∣∣B

= RA/Bx∣∣∣A. (1.10.15)

Fact 1.10.7 Let FA and FB be frames, letx be a physical vector, and

lety =

RB/A

x. Then

y∣∣∣A

= RB/Ax∣∣∣A

= R2B/A

x∣∣∣B

(1.10.16)

andy∣∣∣B

= RB/Ax∣∣∣B

=x∣∣∣A. (1.10.17)

The following result shows that OA/B is a proper, orthogonal matrix,that is, a rotation matrix.


det OB/A = 1. (1.10.18)

22 CHAPTER 1

Proof Note that

det OB/A = det[

ıA|B A|B kA|B]

= det[ıA|B A|B ıA|B × A|B

]

= det

x1 y1 x2y3 − x3y2

x2 y2 x3y1 − x1y3

x3 y3 x1y2 − x2y1

= x1(x1y22 − x2y1y2 − x3y1y3 + x1y

23)

− y1(x1x2y2 − x22y1 − x2

3y1 + x1x3y3)

+ (x2y3 − x3y2)2

= x21y

22 − x1x2y1y2 − x1x3y1y3 + x2

1y23

− x1x2y1y2 + x22y

21 + x2

3y21 − x1x3y1y3

+ x22y

23 − 2x2x3y2y3 + x2

3y22

= x21(y

22 + y2

3) + x22(y

21 + y2

3) + x23(y

21 + y2

2)

− 2x1y1(x2y2 + x3y3) − 2x2x3y2y3

= (x21 + x2

2 + x23)(y

21 + y2

2 + y23)

− x21y

21 − x2

2y22 − x2

3y23

− 2x1y1(x2y2 + x3y3) − 2x2x3y2y3

= |ıA|2|A|2 − (x1y1 + x2y2 + x3y3)2

= |ıA|2|A|2 − ıA|TB A|TB= 1.

The following result is a consequence of Fact 1.9.5.

Fact 1.10.9 Let R be a rotation matrix and let x ∈ R3. Then,

(Rx)× = Rx×RT. (1.10.19)

Example 1.10.1 Let FA and FB be frames such that

ıB = −kA, B = A, kB = ıA. (1.10.20)


Therefore,

RB/A rotates FA clockwise by π/2 radians about A. Furthermore,

OB/A =

0 0 −10 1 01 0 0

. (1.10.21)

Finally,

[

ıB B kB

]OB/A

ıAAkA

= −ıBkA + kBıA + BA

= ıA ıA + AA + kAkA

=

U,

which confirms (1.10.12).

Fact 1.10.10 Let

M be a physical matrix. ThenM

∣∣∣∣∣B

= OB/A

M

∣∣∣∣∣A

OA/B. (1.10.22)

Proof WriteM =

n∑

i=1

x iy i.

We thus haveM

∣∣∣∣∣B

=

n∑

i=1

x i

∣∣∣B

y i

∣∣∣

T

B

=n∑

i=1

OB/Ax i

∣∣∣A

(

OB/Ay i

∣∣∣A

)T

= OB/A

n∑

i=1

x i

∣∣∣A

y i

∣∣∣

T

AO

TB/A

= OB/A

M

∣∣∣∣∣A

OA/B.

Fact 1.10.11 Let FA, FB, and FC be frames. Then

OC/A = OC/BOB/A. (1.10.23)

24 CHAPTER 1

Proof Using (1.8.22), (1.10.2), and (1.10.22), we have

OC/A =

RA/C

∣∣∣∣∣C

=

(RA/B

RB/C

)∣∣∣∣∣C

=

RA/B

∣∣∣∣∣C

RB/C

∣∣∣∣∣C

= OC/B

RA/B

∣∣∣∣∣B

OB/COC/B

= OC/BOB/A.

Fact 1.10.12 Let FA and FB be frames, letx be a physical vector,

and define

M and

N by (1.9.2) and (1.9.4), respectively. Then

OA/B

N

∣∣∣∣∣B

=

M

∣∣∣∣∣A

OA/B. (1.10.24)

Note that (1.10.24) is the identity




0 −kB ·x B ·xkB ·x 0 −ıB ·x−B ·x ıB ·x 0

=

0 −kA ·x A ·xkA ·x 0 −ıA ·x−A ·x ıA ·x 0




.

(1.10.25)

1.11 Euler’s Theorem

Fact 1.11.1 Let n be a dimensionless unit-length physical vector, letθ ∈ [0, 2π), and define

Rn(θ)

= (cos θ)

U + (1 − cos θ)nn+ (sin θ)n×. (1.11.1)

Then

Rn(θ) is a physical rotation matrix. Furthermore, for all physical vec-

torsx , the physical vector

y =

Rn(θ)

x is obtained by rotating

x according


to the right hand rule by the angle θ about n. In particular,Rn(θ)n = n. (1.11.2)

Finally,

Rn(−θ) =

R−n(θ) =

R

T

n (θ), (1.11.3)

and thusR−n(−θ) =

Rn(θ). (1.11.4)

Proof Using (1.9.7) we have

Rn(θ)

R

T

n (θ) = (cos θ)2U + 2(cos θ)(1 − cos θ)nn− (cos θ)(sin θ)n×

+ (cos θ)(sin θ)n× + (1 − cos θ)2nn− (sin θ)2(n×)2

= (cos θ)2U + (1 − cos2 θ)nn− (sin θ)2

(

nn−U

)

= (cos2 θ + sin2 θ)

U + (1 − cos2 θ − sin2 θ)nn

=

U.

Next, letx be a physical vector, and write

x = xparn + xperpp, where p is

a unit-length physical vector that is orthogonal to n. We then haveRn(θ)

x = (cos θ)

x + xpar(1 − cos θ)n+ xperp(sin θ)n× p

= xpar(cos θ)n+ xperp(cos θ)p+ xpar(1 − cos θ)n+ xperp(sin θ)n× p

= xparn+ xperp[(cos θ)p+ (sin θ)n× p].

Fact 1.11.2 Define n, θ, and

Rn(θ) as in Fact 1.11.1, and let

S be a

physical rotation matrix. Then,

R

Sn

(θ) =

S

Rn(θ)

S

T

. (1.11.5)

Fact 1.11.3 Letx and

y be nonzero, non-parallel physical vectors

such that |x | = |y |, and let θ ∈ (0, π) denote the angle betweenx and

y .

Then

y =

Rθ

y /x

(θ)x. (1.11.6)

26 CHAPTER 1

Furthermore,y = (cos θ)

x + (sin θ)θ

y /x×x. (1.11.7)

Proof Note that

Rθ

y /x

(θ)x = [(cos θ)

U + (1 − cos θ)θ

y /xθy /

x

+ (sin θ)θ×y /

x]x

= (cos θ)x + (1 − cos θ)θ

y /xθy /

x·x + (sin θ)θ

y /x×x

= (cos θ)x + (sin θ)θ

y /x×x

= (cos θ)x +

sin θ

|x ×y |

(x ×

y ) ×x

= (cos θ)x +

sin θ

|x ×y |

[(x ·x)

y − (

x ·y )

x ]

= (cos θ)x +

sin θ

|x ||y | sin θ[|x |2y − |x ||y |(cos θ)x ]

= (cos θ)x +

1

|x ||y |[|x |2y − |x ||y |(cos θ)x ]

=y .

Note that (1.11.7) can be written as

y =

[

(cos θ)

U + (sin θ)θ×

y /x

]

x. (1.11.8)

However, the physical matrix coefficient ofx in (1.11.8) is not a physical

rotation matrix.

Fact 1.4.2 shows that angle vectors are additive when both angles liein the same plane. The following result considers the general case.

Fact 1.11.4 Letx ,

y , and

z be physical vectors such that |x | = |y | =

|z |, assume thatx and

y are not parallel and that

y and

z are not parallel,

let θ ∈ (0, π) denote the angle betweenx and

y , let φ ∈ (0, π) denote the

angle betweeny and

z , and let ψ ∈ (0, π) denote the angle between

x and

z . Then

Rθ

z /x

(ψ) =

Rθ

z /y

(φ)

Rθ

y /x

(θ). (1.11.9)


The following result is Euler’s theorem.

Fact 1.11.5 Let FA and FB be frames. Then there exist a unit-lengthphysical vector nB/A and θB/A ∈ [0, 2π) such that

RB/A =

RnB/A

(θB/A). (1.11.10)

In particular, nB/A is given by

n×B/A =

RB/A −

RA/B (1.11.11)

and

RB/A = (cos θB/A)I + (1 − cos θB/A)nB/AnTB/A + (sin θB/A)n×B/A, (1.11.12)

where

nB/A= nB/A

∣∣B. (1.11.13)

Finally,

cos θB/A = 12 (tr RB/A − 1) (1.11.14)

and

cos 12θB/A = 1

2

√

1 + tr RB/A. (1.11.15)

Proof To be added.

The following result considers the resolved form of a physical rotationmatrix that rotates physical vectors about a frame axis. Let FA be a frameand let θ ∈ [0, 2π), and define

R1(θ)

=

RıA(θ)

∣∣∣∣∣A

, (1.11.16)

R2(θ)

=

RA(θ)

∣∣∣∣∣A

, (1.11.17)

R3(θ)

=

RkA

(θ)

∣∣∣∣∣A

. (1.11.18)

28 CHAPTER 1

Fact 1.11.6 Let FA be a frame, and let θ ∈ [0, 2π). Then

R3(θ) =

1 0 00 cos θ sin θ0 − sin θ cos θ

, (1.11.19)

R2(θ) =

cos θ 0 − sin θ0 1 0

sin θ 0 cos θ

, (1.11.20)

R3(θ) =

cos θ sin θ 0− sin θ cos θ 0

0 0 1

. (1.11.21)

An alternative way to express a rotation is in terms of the exponential

of a physical matrix. For a physical matrix

M define

exp(

M ) =

U +

M + 1

2

M + 1

3!

M + · · · . (1.11.22)

Fact 1.11.7 Letx and

y be non-parallel physical vectors such that

|x | = |y |, and let θ ∈ (0, π) denote the angle betweenx and

y . Then

y = exp

(

θ

|x ×y |

(x ×

y )×

)

x, (1.11.23)

that is,

y = exp

(θ×

y /

x

)x. (1.11.24)

Furthermore,

exp

(θ×

y /

x

)

=

Rθ

y /x

(θ). (1.11.25)

Fact 1.11.8 Letx ,

y , and

z be physical vectors such that |x | = |y | =

|z |, assume thatx and

y are not parallel and that

y and

z are not parallel,

let θ ∈ (0, π) denote the angle betweenx and

y , let φ ∈ (0, π) denote the

angle betweeny and

z , and let ψ ∈ (0, π) denote the angle between

x and

z . Then

exp

(θ×

z /

x

)

= exp

(θ×

z /

y

)

exp

(θ×

y /

x

)

. (1.11.26)


1.12 Quaternions

Definition 1.12.1 Let FA and FB be frames, and define nB/A and θB/Aas in Fact 1.11.5. Then the quaternion qB/A that transforms FA to FB isdefined by

qB/A=

[ηB/A

εB/A

]

=

[

sin 12θB/A

(cos 12θB/A)nB/A

]

. (1.12.1)

Fact 1.12.1 Let FA and FB be frames. Then,

RB/A = (2η2B/A − 1)I + 2εB/Aε

TB/A + 2ηB/Aε

×

B/A. (1.12.2)

Proof It follows from (1.11.12) that...

Fact 1.12.2 Let FA, FB, and FC be frames. Then,

qC/A =

[ηC/A

εC/A

]

=

[

ηC/BηB/A − εTC/BεB/A

ηB/AεC/B + ηC/BεB/A − εC/B × εB/A

]

. (1.12.3)

Proof See [2, p. 17].

1.13 Euler Angles

A transformation from one reference frame to another can be achievedthrough a sequence of three rotations. Each rotation yields a new referenceframe. The three angles that define the transformations are referred to asEuler angles. The rotation process involves four frames, namely, the initialand final frames as well as two intermediate frames.

There are twelve different Euler-angle sequences depending on the axeschosen. The 3-2-1 and 3-1-3 sequences are the most frequently used, where1, 2, and 3 refer to rotations about ı, , and k axes, respectively, of theoriginal and intermediate frames.

For a 3-2-1 rotation sequence, the Euler angles are denoted by Ψ, Θ,and Φ, and are called yaw, pitch, and roll, respectively. The 3-2-1 sequenceis typically used for aircraft. The 3-2-1 rotation sequence is represented by

FAΨ−→3

FBΘ−→2

FCΦ−→1

FD, (1.13.1)

where the symbol above the arrow represents the rotation angle, and thenumber below the arrow represents the axis about which the rotation is

30 CHAPTER 1

carried out. The matrix

OB/A = RA/B = R3(AΨ→ B) =

cos Ψ sin Ψ 0− sinΨ cos Ψ 0

0 0 1

(1.13.2)

gives the orientation of FB with respect to FA as a function of the rotationangle Ψ, where

R3(AΨ→ B)

=

RkA

(Ψ)

∣∣∣∣∣A

. (1.13.3)

The subscript 3 represents the axis of rotation, while AΨ→ B indicates that

the angle Ψ is measured from ıA to ıB or from A to B as shown in Figure1.13.5. Similarly, the orientation matrices

OC/B = RB/C = R2(BΘ→ C) =

cos Θ 0 − sinΘ0 1 0

sin Θ 0 cos Θ

(1.13.4)

and

OD/C = RC/D = R1(CΦ→ D) =

1 0 00 cos Φ sin Φ0 − sinΦ cos Φ

(1.13.5)

represent the result of rotations from FB to FC and from FC to FD, respec-tively, also specifying the corresponding rotation angles and axes of rotation.

Specifically, R2(BΘ→ C) is the orientation of FC with respect to FB given

by a rotation of FB about B by the angle Θ. In particular,

R2(BΘ→ C) =

RjA(Θ)

∣∣∣∣∣A

(1.13.6)

and

R1(CΦ→ D) =

RiA(Φ)

∣∣∣∣∣A

. (1.13.7)

The three rotations in the 3-2-1 rotation sequence are depicted infigures 1.13.5, 1.13.6, and 1.13.7. Note that

OA/B = RB/A = R3(B−Ψ→ A) = R

T3 (A

Ψ→ B) = RTA/B = O

TB/A. (1.13.8)

Also note that the sign pattern in (1.13.4) is different from the sign patternsin (1.13.2) and (1.13.5).


?ıA ıB

-A

:B

W

kA = kB O Ψ

:Ψ

Figure 1.13.5Rotation from FA to FB.

Using Fact 1.10.11 to combine the 3-2-1 rotation sequence yields

OD/A = OD/COC/BOB/A

= R1(CΦ→ D)R2(B

Θ→ C)R3(AΨ→ B)

= R3,2,1(AΨ,Θ,Φ−→ D)

= RA/D. (1.13.9)

For a 3-1-3 rotation sequence, the Euler angles are denoted by Φ, Θ,and Ψ, which represent precession, nutation, and spin, respectively. Theseangles are used for spacecraft dynamics and spinning tops as well as fororbits. For orbits, the 3-1-3 Euler angles are denoted by Ω, i, and ω, andare called right ascension of the ascending node, inclination, and argumentof periapsis, respectively. The 3-1-3 rotation sequence is represented by

FAΦ−→3

FBΘ−→1

FCΨ−→3

FD. (1.13.10)

32 CHAPTER 1

6

kC

-ıB

zıC

B = C

kB

?Θ

-Θ

Figure 1.13.6Rotation from FB to FC.

?C D

-kC

:kD

W

ıC = ıD O Φ

:Φ

Figure 1.13.7Rotation from FC to FD.


The orientation matrices for the 3-1-3 sequence are given by

OB/A = RA/B = R3(AΦ→ B) =

cos Φ sin Φ 0− sinΦ cos Φ 0

0 0 1

,

OC/B = RB/C = R1(BΘ→ C) =

1 0 00 cos Θ sinΘ0 − sin Θ cos Θ

,

OD/C = RC/D = R3(CΨ→ D) =

cos Ψ sin Ψ 0− sinΨ cos Ψ 0

0 0 1

.

Combining the 3-1-3 sequence yields

OD/A = OD/COC/BOB/A

= R3(CΨ→ D)R1(B

Θ→ C)R3(AΦ→ B)

= R3,1,3(AΦ,Θ,Ψ−→ D)

= RA/D. (1.13.11)

1.14 Frame Derivatives

Let FA be a frame and letr be a vector expressed as

r = r1ıA + r2A + r3kA. (1.14.1)

We can thus write

r∣∣∣A

=

r1r2r3

. (1.14.2)

The derivative ofr with respect to the frame FA is given by

A•

r = r1ıA + r1

A•

ı A + r2A + r2A•

A + r3kA + r3

A•

k A

= r1ıA + r2A + r3kA. (1.14.3)

We thus have

A•

r

∣∣∣∣∣A

=

r1r2r3

, (1.14.4)

that is,

A•

r

∣∣∣∣∣A

=˙(

r∣∣∣A

)

. (1.14.5)

34 CHAPTER 1

Fact 1.14.1 Let FA be a frame, and let x, y, and z be points. Thenv z/x/A =

v z/y/A +

v y/x/A (1.14.6)

anda z/x/A =

a z/y/A +

a y/x/A, (1.14.7)

where

v y/x/A

=A•

r y/x (1.14.8)

and

a y/x/A

=A•

v y/x/A =

A••

r y/x . (1.14.9)

Definition 1.14.1 Let FA be a frame, letx and

y be physical vectors,

and define

M =

xy . Then

A•

M =

A•

x

y +

x

A•

y . (1.14.10)


A•

U= 0 (1.14.11)

and

A•

ıB ıB +A•

B B +A•

kB kB = −(ıBA•

ıB + BA•

B + kB

A•

kB). (1.14.12)


B•

RA/B = −

RA/B

B•

RB/A

RA/B (1.14.13)

andA•

RA/B = −

RA/B

A•

RB/A

RA/B. (1.14.14)

Proof The result follows from (1.8.21).

1.15 Momentum

Let FA be a frame, let x be a particle with mass m, and let y be apoint. Then the momentum px/y/A of x relative to y with respect to FA is


defined by

px/y/A

= mv x/y/A = m

A•

r x/y . (1.15.1)

Furthermore, let B be a body with mass mB, and let c denote the center ofmass of B. Then the momentum pB/y/A of B relative to y with respect toFA is defined by

pB/y/A= mB

v c/y/A = mB

A•

r c/y . (1.15.2)

1.16 Angular Momentum

Let FA be a frame, let x be a particle with mass m, and let w be apoint. Then the angular momentum of x relative to w with respect to FA isdefined by

Hx/w/A

=r x/w ×m

v x/w/A. (1.16.1)

Let FA be a frame, let B be a body composed of l rigidly interconnectedparticles y1, . . . ,ml of mass m1, . . . ,ml, respectively, and let w be a point.Then the angular momentum of B relative to w with respect to FA is definedby

HB/w/A

=

l∑

i=1

Hmi/w/A, (1.16.2)

whereHmi/w/A =

rmi/w ×mi

vmi/w/A. (1.16.3)

1.17 Angular Velocity Vector

Given frames FA and FB, the angular velocity vectorωB/A describes

the time-dependent rotation of FB relative to FA. In particular, the physical

vectorωB/A can be viewed as the instantaneous axis of rotation, where the

rate of rotation is given by |ωB/A| and the direction of rotation is given bythe curled fingers of the right hand with the thumb pointing in the direction

ofωB/A.

Definition 1.17.1 Let FA and FB be frames. Then

ΩB/A

=

RA/B

B•

RB/A . (1.17.1)

36 CHAPTER 1


B•

RB/A =

RB/A

ΩB/A (1.17.2)

andB•

RA/B = −

ΩB/A

RA/B. (1.17.3)

Proof Result (1.17.2) follows from (1.17.1). Result (1.17.3) followsfrom (1.14.13).


ıA·A•

ı B =B•

ı A · ıB, ıA·A•

B =B•

ı A · B, ıA·A•

k B =B•

ı A · kB, (1.17.4)

A·A•

ı B =B•

A · ıB, A·A•

B =B•

A · B, A·A•

k B =B•

A · kB, (1.17.5)

kA·A•

ı B =B•

k A · ıB, kA·A•

B =B•

k A · B, kA·A•

k B =B•

k A · kB. (1.17.6)

Proof Note that

ıB/A =

ıB · ıAıB · AıB · kA

, ıA/B =

ıA · ıBıA · BıA · kB

. (1.17.7)

Therefore

ıA·A•

ı B = ıA|TA ·A•

ı B

∣∣∣∣A

= eT1

·

︷︸︸︷

(ıB|A)

=

·

︷︸︸︷

ıB · ıA

= eT1

·

︷︸︸︷

( ıA|B)

= ıB|TB ·B•

ı A

∣∣∣∣B

= ıA·B•

ı A .



A•

ı B =

ΩB/A ıB, (1.17.8)

A•

B =

ΩB/AB, (1.17.9)

A•

k B =

ΩB/AkB. (1.17.10)

Proof Using Fact 1.17.2 we haveΩB/AıB =

(

ıAB•

ı A + AB•

A + kA

B•

k A

)

ıB

=

(B•

ı A ·ıB)

ıA +

(B•

A ·ıB)

A +

(B•

k A ·ıB)

kA

=

(

ıA·A•

ı B

)

ıA +

(

A·A•

ı B

)

A +

(

kA·A•

ı B

)

kA

=A•

ı B .

Fact 1.17.4 Let FA and FB be frames. ThenΩA/B = ıB

A•

ı B + BA•

B + kB

A•

k B, (1.17.11)

ΩA/B = −

(A•

ı B ıB +A•

B B +A•

k B kB

)

, (1.17.12)

ΩB/A = ıA

B•

ı A + AB•

A + kA

B•

k A, (1.17.13)ΩB/A = −

(B•

ı A ıA +B•

A A +B•

k A kA

)

. (1.17.14)

Furthermore, letx be a physical vector. Then

ΩA/B

x +

x

ΩA/B = 0. (1.17.15)

38 CHAPTER 1

Proof Note that

ΩB/A =

RA/B

B•

RB/A

= (ıAıB + AB + kAkB)(ıBB•

ı A + BB•

A + kB

B•

k A)

= ıAB•

ı A + AB•

A + kA

B•

k A,

which proves (1.17.13). Next, using (1.14.12) yields (1.17.14). Finally,(1.17.15) follows from (1.17.11) and (1.17.12).


Ω

T

B/A = −ΩB/A (1.17.16)

andΩA/B = −

ΩB/A. (1.17.17)

Hence,

ΩA/B =

Ω

T

B/A. (1.17.18)

Proof Using (1.17.13) and (1.17.14) it follows that

Ω

T

B/A = (ıAB•

ı A + AB•

A + kA

B•

kA)T

=B•

ı A ıA +B•

A A +B•

k A kA

= −ΩB/A.

Furthermore, using (1.14.12) and Fact 1.17.4 we have

ΩA/B = ıB

A•

ı B + BA•

B + kB

A•

k B

= −(

A•

ı B ıB +A•

B B +A•

k B kB

)

= −[(

ΩB/AıB

)

ıB +

(ΩB/AB

)

B +

(ΩB/AkB

)

kB

]

= −ΩB/A

U

= −ΩB/A.


Fact 1.17.6 Let FA and FB be frames. Then there exists a physical

vectorωB/A such that

ΩB/A =

ω×

B/A. (1.17.19)

Furthermore,ωB/A = −

ωA/B, (1.17.20)

ΩB/A = −ΩA/B = ΩTA/B = −ΩT

B/A = ω×

B/A, (1.17.21)

and

ωB/A = −RA/BωA/B = −OB/AωA/B, (1.17.22)

where

ΩB/A

=

ΩB/A

∣∣∣∣∣B

, (1.17.23)

ωB/A

=ωB/A

∣∣∣B

=

ω1

ω2

ω3

, (1.17.24)

and

ω1

ω2

ω3

×

=

0 −ω3 ω2

ω3 0 −ω1

−ω2 ω1 0

. (1.17.25)

Furthermore,

ωB/A

∣∣∣A

= OA/BωB/A (1.17.26)

andΩB/A

∣∣∣∣∣A

= OA/BΩB/AOB/A. (1.17.27)

The following result gives Poisson’s equation.


RB/A = RB/Aω×

B/A. (1.17.28)

Furthermore,

OA/B = OA/Bω×

B/A, (1.17.29)

40 CHAPTER 1

and thus

OB/A = −ω×

B/AOB/A. (1.17.30)

Proof Resolving (1.17.2) in FB yields (1.17.28).

The following result relates the derivative of the quaternion to theangular velocity vector.


ηB/A = −12ω

TB/AεB/A, (1.17.31)

εB/A = 12 (ηB/AωB/A − ωB/A × εB/A). (1.17.32)

Furthermore,

ωB/A = 2[

ηB/AI + ε×B/A −εB/A]

qB/A. (1.17.33)

Equations (1.17.31) and (1.17.32) can be written as

qB/A = Q(ωB/A)qB/A, (1.17.34)

where

Q(ωB/A)

= 12

0 −ωTB/A

ωB/A −ω×

B/A

.

Furthermore,

eQ(ωB/A)t = cos(12 |ωB/A|t)I +

2 sin(12 |ωB/A|t)

|ωB/A|Q(ωB/A). (1.17.35)

1.18 Euler Angle Derivatives

The angular velocity vector can be related to the derivatives of theEuler angles. For 3-1-3 Euler angles (see (1.13.10)) we have

ωD/A =

ωD/C +

ωC/B +

ωB/A (1.18.1)

= ΨkC + ΘıB + ΦkA. (1.18.2)


Since kA = kB, ıB = ıC, and kC = kD, resolvingωD/A in FD yields

ωD/A = ΨkD + ΘıC + ΦkB

= ΨkD + Θ[(cos Ψ)ıD − (sin Ψ)D] + Φ[(cos Θ)kC + (sin Θ)C]

= ΨkD + Θ(cos Ψ)ıD − Θ(sin Ψ)D + Φ(cos Θ)kD

+ Φ(sin Θ)[(cos Ψ)D + (sin Ψ)ıD]

= [Θ(cos Ψ) + Φ(sin Θ) sin Ψ]ıD + [Φ(sin Θ) cos Ψ − Θ(sin Ψ)]D

+ [Ψ + Φ(cos Θ)]kD.

Hence

ωD/A

∣∣∣D

=

(sin Θ) sin Ψ cos Ψ 0(sin Θ) cos Ψ − sinΨ 0

cos Θ 0 1

Φ

Θ

Ψ

. (1.18.3)

Note that the matrix in (1.18.3) is independent of Φ.

For 3-2-1 (yaw-pitch-roll) Euler angles (see (1.13.1)) we have

ωD/A =

ωD/C +

ωC/B +

ωB/A (1.18.4)

= ΦıC + ΘB + ΨkA. (1.18.5)

Since ıD = ıC, C = B, and kB = kA, resolvingωD/A in FD yields

ωD/A = ΦıD + ΘC + ΨkB

= ΦıD + Θ[(cos Φ)D − (sin Φ)kD] + Ψ[(cos Θ)kC − (sin Θ)ıC]

= ΦıD + Θ(cos Φ)D − Θ(sin Φ)kD + Ψ(cos Θ)[(cos Φ)kD + (sin Φ)D]

− Ψ(sin Θ)ıD

= [−Ψ(sin Θ) + Φ]ıD + [Ψ(cos Θ) sin Φ + Θ cos Φ]D

+ [Ψ(cos Θ) cos Φ − Θ(sin Φ)]kD.

Hence

ωD/A

∣∣∣D

=

− sin Θ 0 1(cos Θ) sin Φ cos Φ 0(cos Θ) cos Φ − sinΦ 0

Ψ

Θ

Φ

. (1.18.6)

Note that the matrix in (1.18.6) is independent of Ψ.

1.19 Transport Theorem

The following important result is the transport theorem.

Fact 1.19.1 Letx(t) be a physical vector, and consider frames FA

42 CHAPTER 1

and FB. Then,

A•

x (t) =

B•

x (t) +

ωB/A ×

x(t). (1.19.1)

Proof Note that(

B•

x (t) +

ωB/A ×

x(t)

)∣∣∣∣∣A

=B•

x (t)

∣∣∣∣∣A

+ωB/A

∣∣∣A×x(t)|A

= OA/B

B•

x (t)

∣∣∣∣∣B

+ωB/A ×

x(t)|A

= OA/B

(˙

x(t)∣∣∣B

)

+ ΩB/Ax(t)|A

= OA/B

(˙

OB/Ax(t)

∣∣∣A

)

+ ΩB/Ax(t)|A

= OA/BOB/A

(˙

x(t)∣∣∣A

)

+ OA/BOB/Ax(t)

∣∣∣A

+ OA/BOB/Ax(t)

∣∣∣A

=

(˙

x(t)∣∣∣A

)

+(

OA/BOB/A + OA/BOB/A

)x(t)

∣∣∣A

=A•

x (t)

∣∣∣∣∣A

.

The following result is an immediate consequence of the transporttheorem.

Fact 1.19.2 Letω(t) be a physical vector, and consider frames FA

and FB. Then,

A•

ω B/A =

B•

ω B/A . (1.19.2)

We define

αB/A

=A•

ω B/A (1.19.3)

and

αB/A/C

=C•

ω B/A . (1.19.4)

The following result is based on Figure 1.19.8.


Fact 1.19.3 Let FA and FB be frames with origins OA and OB, re-spectively, and let x be a point. Then,

v x/OA/A =

v x/OB/B +

ωB/A ×

r x/OB+vOB/OA/A. (1.19.5)

O

B

A

B

A B

A

x

x/Or

O

x/Or

r /O

O

Figure 1.19.8Geometry for relative motion.

1.20 Double Transport Theorem

Applying the transport theorem twice yields the double transport the-orem.

Fact 1.20.1 Letx(t) be a physical vector, and consider frames FA

and FB. Then,

A••

x (t) =

B••

x (t) + 2

ωB/A ×

B•

x(t)

︸︷︷︸

Coriolis acceleration

+B•

ω B/A ×

x(t)︸︷︷︸

angular-accelerationacceleration

+ωB/A × (

ωB/A ×

x(t))︸︷︷︸

centripetal acceleration

. (1.20.1)

The following specialization of Fact 1.20.1 is based on Figure 1.19.8.

Fact 1.20.2 Let FA and FB be frames with origins OA and OB, re-spectively, and let x be a point. Then,ax/OA/A =

ax/OB/B +

ωB/A ×

v x/OB/B +αB/A ×

r x/OB(1.20.2)

+ωB/A ×

v x/OB/A +ωB/A × (

ωB/A ×

r x/OB) +

aOB/OA/A

=ax/OB/B + 2

ωB/A ×

v x/OB/B +αB/A ×

r x/OB(1.20.3)

+ωB/A × (

ωB/A ×

r x/OB) +

aOB/OA/A.

44 CHAPTER 1

1.21 Summation of Angular Velocities

Fact 1.21.1 Let FA, FB, and FC be frames. Then,ΩC/A =

ΩC/B +

ΩB/A (1.21.1)

andωC/A =

ωC/B +

ωB/A. (1.21.2)

Furthermore,

ΩC/A = ΩC/B + OC/BΩB/AOB/C (1.21.3)

and

ωC/A = ωC/B + OC/BωB/A. (1.21.4)

Proof Using the transport theorem, (1.17.15), and (1.17.17), we have

ΩC/A = ıA

C•

ı A + AC•

A + kA

C•

k A

= ıA

(B•

ı A +

ΩB/C ıA

)

+ A

(B•

A +

ΩB/CA

)

+ kA

(B•

k A +

ΩB/CkA

)

= ıA

(ΩB/CıA

)

+ A

(ΩB/CA

)

+ kA

(ΩB/CkA

)

+ ıAB•

ı A + AB•

A + kA

B•

k A

= ıA

(

ıA

ΩC/B

)

+ A

(

A

ΩC/B

)

+ kA

(

kA

ΩC/B

)

+

ΩB/A

=

ΩC/B +

ΩB/A,

where we used the identity[

ıA

(

ıA

ΩC/B

)]∣∣∣∣∣A

= ıA|A ıA|TAΩC/B

∣∣∣∣∣

T

A

,


and the fact that

ıA|A ıA|TA + A|A A|TA + kA

∣∣∣AkA

∣∣∣

T

A= I3.

Next, to prove (1.21.3), note that

ΩC/A =

ΩC/A

∣∣∣∣∣C

= ΩC/B +

ΩB/A

∣∣∣∣∣C

= ΩC/B + OC/B

ΩB/A

∣∣∣∣∣B

OB/C

= ΩC/B + OC/BΩB/AOB/C.

1.22 Solid and Open Frame Dots

If we are considering only two frames, one of which is a body frame,then we can use slightly simpler notation.

Consider a position vectorr (t), consider a frame FA, let FB be a

body-fixed frame, and denoteA•

r(t) and

B•

r(t) by

•

r(t) and

r(t), respectively.

Furthermore, we letω/• denote

ωB/A. Then,

••

r(t) =

r(t) + 2

ω/• ×

r(t)

︸︷︷︸


+

ω/• ×

r (t)

︸︷︷︸


+ω/• × (

ω/• ×

r (t))

︸︷︷︸


.

(1.22.1)

Hence,

r(t) =

••

r(t) − 2

ω/• ×

r(t)

︸︷︷︸


−

ω/• ×

r (t)

︸︷︷︸


− ω/• × (

ω/• ×

r (t))

︸︷︷︸


.

(1.22.2)

1.23 Problems

Problem 1.23.1 Definex

= 3ıA − 4A,y

= −1ıA + 5A − 2kA,

M

=

xy , and

N

=yx . Then do the following:

46 CHAPTER 1

i) Resolve

M ,

N ,

M

N ,

M

x ,

M

y ,

Nx , and

Ny in FA.

ii) Check that (

M

x)|A =

M |A

x |A.


Symbol Definition

x Point or particle x

B Body B

x Physical vector

|x | Magnitude of vector x

ı, , k Unit vectors

x(t) Vector

x as a function of time

θy /

x

Angle vector

r y/x Position vector

v y/x/A Velocity vector

a y/x/A Acceleration vector

p y/x/A Momentum vector

pB/x/A Momentum vector

f Force vector

ωB/A Angular velocity vector

αB/A Angular acceleration vector

αB/A/C Angular acceleration vector

Table 1.1 Symbols for Chapter 1, part 1.

48 CHAPTER 1

Symbol Definition

ı, , k Cartesian frame

FA Frame A

er, eθ, k Cylindrical frame

er, eθ, eφ Spherical frame

et, en, eb Normal-tangential-binormal frame

eLV, eLH, eLN Local vertical/local horizontal frame

RA/B Rotation from FB to FA

RTA/B Rotation transpose, RB/A

OA/B Orientation of FA with respect to FB

OTA/B Orientation transpose, OB/A

R2(BΘ→ C) Orientation of FC with respect to FB given

by a rotation of FB about B by the angle Θ

3-2-1 rotation: Ψ,Θ,Φ Yaw, pitch, and roll angles

3-1-3 rotation: Φ,Θ,Ψ Precession, nutation, and spin angles

3-1-3 rotation: Ω, i, ω Orbital Euler angles


Chapter Two

Aircraft Kinematics

2.1 Frames Used in Aircraft Dynamics

To describe aircraft flight dynamics we consider four frames, namely,the Earth frame, aircraft frame, stability frame, and wind frame. The air-craft has six degrees of freedom, specifically, three translational degrees offreedom and three rotational degrees of freedom.

Translational Rotational

x: Surge Ψ: Yawy: Sway Θ: Pitchz: Plunge Φ: Roll

Longitudinal Lateral

x: Surge Ψ: YawΘ: Pitch y: Swayz: Plunge Φ: Roll

Table 2.1 Aircraft degrees of freedom using 3-2-1 Euler angles

Since a single degree of freedom is modeled by a second-order differ-ential equation, the dynamics of a six-degree-of-freedom rigid body suchas an aircraft are described by 6 second-order differential equations or 12first-order differential equations. For simplicity, we assume throughout thatthe velocity of the air with respect to the Earth is zero, that is, there is noambient wind.

2.1.1 Earth Frame FE

The Earth frame is assumed to be an inertial frame as explained inChapter 3. The origin OE of the Earth frame is any convenient point onthe Earth. The axes ıE and E are horizontal, while the axis kE pointsdownward.

The acceleration due to gravity is the physical acceleration vectorg = gkE, (2.1.1)

50 CHAPTER 2

?kE

9ıE

jE

OE

Figure 2.1.1The Earth frame FE.

where g = 9.8 m/s2. For a falling particle x,g is given by

g =

ax/OE/E. (2.1.2)

2.1.2 Aircraft Frame FAC

The aircraft frame is fixed to (painted on) the body of the aircraft andhas the origin OAC placed at the center of mass of the aircraft. The frameorigin OAC, along with the frame vectors ıAC and kAC, are chosen to lie inthe aircraft plane of symmetry.

Vectors in the aircraft frame are resolved as

r =r∣∣∣AC

.

We letr AC

=rOAC/OE

(2.1.3)

denote the location of the aircraft with respect to the origin of the Earthframe.

AIRCRAFT KINEMATICS 51

?kAC

ψ

*

9AC

θ

?

jıAC

φ

OAC

Figure 2.1.2The aircraft frame FAC

2.1.3 Stability Frame FS

LetV AC be the velocity of the aircraft relative to OE with respect to

FE, that is,

V AC

=vOAC/OE/E =

E•

r AC =

E•

r OAC/OE

. (2.1.4)

The unit vector ıS is aligned alongV AC,proj as shown in Figure 2.1.3. The

projected velocity vectorV AC,proj is the projection of

V AC onto the ıAC-

kAC plane as shown in Figure 2.1.4, where the slideslip angle β is the angle

from ıS toV AC, and the angle of attack α is the angle from ıS to ıAC. The

wind is in the pilot’s right ear when β is positive.

Vectors in the stability frame are resolved as

r =r∣∣∣S.

52 CHAPTER 2

?kAC

-ıAC

zıSz

V AC,proj

S = AC

kS

6α

-

α

Figure 2.1.3Rotation from the stability frame FS to the aircraft frame FAC.

Note that

OTS/AC = R

T2 (AC

−α→ S) = R2(Sα→ AC) = OAC/S.

2.1.4 Wind Frame FW

In the wind frame, the unit vector ıW points along the velocity vectorV AC as shown in Figure 2.1.4. The stability frame is transformed to the windframe by means of a rotation through β about the kW axis. Consequently,

FEΨ−→3

FE′Θ−→2

FE′′Φ−→1

FAC−α−→2

FSβ−→3

FW. (2.1.5)

Hence,

OTS/W = R

T3 (W

−β→ S) = R3(Sβ→ W) = OW/S.

2.2 Velocity Vector

The velocity of the aircraft with respect to the Earth is represented

byV AC. Letting U , V , and W be the components of

V AC in the aircraft


?S

-ıS - V AC,proj

zV AC

zıW

W

kS = kW

?β

β

Figure 2.1.4

Rotation from the wind frame FW to the stability frame FS and the projection of

V AC

onto ıS.

frame, we haveV AC = UıAC + V AC +WkAC,

and thusV AC resolved in the aircraft frame is

V AC

∣∣∣∣AC

=

UVW

. (2.2.1)

Letting US, VS, and WS be the components ofV AC in the stability frame,

we haveV AC = USıS + VSS +WSkS,

and thus

V AC

∣∣∣∣S

=

US

VS

WS

. (2.2.2)

The airspeed VAC is given by

VAC= |

V AC| =

√

U2 + V 2 +W 2 =√

U2S + V 2

S +W 2S .

54 CHAPTER 2

It can be seen from Figure 2.1.3 that

ıAC = (cosα)ıS − (sinα)kS,

AC = S,

kAC = (sinα)ıS + (cosα)kS,

which can be written as

ıAC

AC

kAC

=

cosα 0 − sinα0 1 0

sinα 0 cosα

︸︷︷︸

OAC/S=R2(Sα→AC)

ıSSkS

. (2.2.3)

Hence,

UVW

= R2(Sα→ AC)

US

VS

WS

. (2.2.4)

Since ıS is aligned alongV AC,proj, it follows that WS = 0. Then,

UVW

=


sinα 0 cosα

US

VS

0

, (2.2.5)

which yields

U = US cosα, (2.2.6)

V = VS, (2.2.7)

W = US sinα. (2.2.8)

Hence

tanα =W

U(2.2.9)

and

US =√

U2 +W 2. (2.2.10)

Furthermore,

sinα =W√

U2 +W 2(2.2.11)

and

cosα =U√

U2 +W 2. (2.2.12)


WkS

UkAC

*ıAC

::

ıS

V AC

6γOα

:

α

O

Θ horizontalS = AC

Figure 2.2.5Flight path angle γ and pitch angle Θ. In this figure, the angle of attack α is positive.

Note that Θ = α + γ.

The reverse of (2.2.4) is

US

VS

WS

= OS/AC

UVW

= R2(AC−α→ S)

UVW

=

cosα 0 sinα0 1 0

− sinα 0 cosα

UVW

. (2.2.13)

We define the flight path angle γ as the angle from the horizontal to ıSas shown in the Figure 2.2.5 and Figure 2.2.6. When the roll angle Φ = 0,the pitch angle of the aircraft is given by

Θ = α+ γ, (2.2.14)

which is the angle from the horizontal to ıAC.

Now, consider the wind frame FW, and recall that the aircraft velocity

56 CHAPTER 2

WkAC

UkS

: ıAC

*

*

ıS

V AC

6γO−α

:

−α

O

Θ horizontalS = AC

Figure 2.2.6Flight path angle γ and pitch angle Θ. In this figure, the angle of attack α is negative.

Note that Θ = α + γ.

V AC is aligned along ıW. From (2.1.5) we know that

V AC

∣∣∣∣W

= OW/S

V AC

∣∣∣∣S

= R3(Sβ→ W)

V AC

∣∣∣∣S

,

which can be written asV AC

∣∣∣∣S

= OS/W

V AC

∣∣∣∣W

= R3(W−β→ S)

V AC

∣∣∣∣W

.

Hence,

US

VS

0

=

cos β − sinβ 0sin β cos β 0

0 0 1

︸︷︷︸

OS/W=R3(W−β→S)

VAC

00

. (2.2.15)

Therefore,

US = VAC cos β (2.2.16)

and

VS = V = VAC sin β. (2.2.17)


Combining (2.2.5) and (2.2.15) we obtain

UVW

=


sinα 0 cosα

cos β − sin β 0sin β cos β 0

0 0 1

VAC

00

=

(cosα) cos β −(cosα) sin β − sinαsin β cos β 0

(sinα) cos β −(sinα) sin β cosα

VAC

00

=

(cosα) cos βsin β

(sinα) cos β

VAC. (2.2.18)

2.3 Rotational Kinematics

We now form a transformation matrix from FE to FAC using the yaw,pitch, and roll angles Ψ, Θ, and Φ as the 3-2-1 Euler angles. Note that yawis a rotation about the kE-axis, pitch is a rotation about the ′E-axis, androll is a rotation about the ı′′E-axis. Transformation from FE to FAC involvestwo intermediate frames, namely, FE′ and FE′′

The orientation matrices corresponding to the three rotations are givenby

OE′/E = R3(EΨ→ E′) =

cos Ψ sin Ψ 0− sin Ψ cos Ψ 0

0 0 1

, (2.3.1)

OE′′/E′ = R2(E′ Θ→ E′′) =

cos Θ 0 − sinΘ0 1 0

sinΘ 0 cos Θ

, (2.3.2)

and

OAC/E′′ = R1(E′′ Φ→ AC) =

1 0 00 cos Φ sin Φ0 − sinΦ cos Φ

.

The overall transformation can be represented as

FEΨ−→3

FE′Θ−→2

FE′′Φ−→1

FAC. (2.3.3)

58 CHAPTER 2

The vectrices of the four reference frames are related by

ıE′

E′

kE′

= R3(EΨ→ E′)

ı11k1

= R3(EΨ→ E′)

ıEEkE

, (2.3.4)

ıE′′

E′′

kE′′

= R2(E′ Θ→ E′′)

ıE′

E′

kE′

, (2.3.5)

and

ıAC

AC

kAC

=

ı44k4

= R1(E′′ Φ→ AC)

ıE′′

E′′

kE′′

. (2.3.6)

Combining (2.3.4), (2.3.5), and (2.3.6) we obtain

ıAC

AC

kAC

= R1(E′′ Φ→ AC)

ıE′′

E′′

kE′′

= R1(E′′ Φ→ AC)R2(E

′ Θ→ E′′)

ıE′

E′

kE′

= R1(E′′ Φ→ AC)R2(E

′ Θ→ E′′)R3(EΨ→ E′)

ıEEkE

= R3,2,1(EΨ,Θ,Φ−→ AC)

ıEEkE

= OAC/E′′OE′′/E′OE′/E

ıEEkE

= OAC/E

ıEEkE

. (2.3.7)

Now consider the aircraft velocity vectorV AC, which can be expressed

in the aircraft frame asV AC = UıAC + V AC +WkAC, (2.3.8)


and in the Earth frame asV AC = UEıE + VEE +WEkE. (2.3.9)

Using (1.13.1) on resolving vectors through a change of frame, we have

V AC

∣∣∣∣AC

= O3,2,1(EΨ,Θ,Φ−→ AC)

V AC

∣∣∣∣E

. (2.3.10)

For angular velocities, using the rotation sequence 3-1-3 (precession-nutation-spin) Euler angles (see 1.18.3), the angular velocity of the aircraftrelative to the Earth frame is defined by

ωAC

=ωAC/E. (2.3.11)

We use the notation

ωAC

∣∣∣AC

=

PQR

. (2.3.12)

Therefore,

ωAC

∣∣∣AC

=ωAC/E′′ +

ωE′′/E′ +

ωE′/E

= ΨkAC + ΘıE′′ + ΦkE′

=

(sin Θ) sin Ψ cos Ψ 0(sin Θ) cos Ψ − sinΨ 0

cos Θ 0 1

Φ

Θ

Ψ

. (2.3.13)

For the sequence 3-2-1 (yaw-pitch-roll) Euler angles, (see 1.18.6), we write

ωAC

∣∣∣AC

= ΦıAC + ΘE′′ + ΨkE′

=

− sin Θ 0 1(cos Θ) sin Φ cos Φ 0(cos Θ) cos Φ − sinΦ 0

Ψ

Θ

Φ

. (2.3.14)

2.4 Vector Derivatives in Rotating Frames

Letr be a position vector. Then the transport theorem given by Fact

1.19.1 implies that

E•

r =

AC•

r +

ωAC ×

r .

60 CHAPTER 2

Note thatE•

ωAC =

AC•

ω AC +

ωAC ×

ωAC =AC•

ω AC,

that is, the angular acceleration of the aircraft relative to the Earth frameis the same as the angular acceleration of the aircraft relative to the aircraftframe. Now consider the linear acceleration

E••

r =

E•

︷︸︸︷

AC•

r +

E•

︷︸︸︷ωAC ×

r

=AC••

r +

αAC/E/E

=AC••

r +

ωAC ×

AC•

r +

E•

ωAC ×

r +ωAC ×

(AC•

r +

ωAC ×

r

)

=AC••

r + 2

ωAC ×

AC•

r

︸︷︷︸

aCor

+E•

ωAC ×

r︸︷︷︸

aang

+ωAC ×

(ωAC ×

r)

︸︷︷︸

acent

, (2.4.1)

where, as in (1.22.1), aCor, aang, and acent are the Coriolis acceleration,angular-acceleration acceleration, and centripetal acceleration, respectively.

2.5 Cross Product

The cross product, introduced in Section 1.3, of

ωAC

∣∣∣AC

=

PQR

(2.5.1)

and

r |AC =

r1r2r3

(2.5.2)

can be represented as

ωAC

∣∣∣AC

×r |AC = det

ıAC AC kAC

P Q Rr1 r2 r3

(2.5.3)

= (Qr3 −Rr2)ıAC − (Pr3 −Rr1)AC + (Pr2 −Qr1)kAC.


Defining

ωAC

∣∣∣

×

AC=

0 −R QR 0 −P−Q P 0

,

we have

(ωAC ×

r )∣∣∣AC

=ωAC

∣∣∣AC

× r∣∣∣AC

=ωAC

∣∣∣

×

AC

r∣∣∣AC

=

0 −R QR 0 −P−Q P 0

r1r2r3

. (2.5.4)

2.6 Problems

Problem 2.6.1 Resolve the velocity vectorV AC =

V AC/E = 72 m/s ıAC − 6.3 m/s AC − 43 m/s kAC

in the stability and wind frames.

Problem 2.6.2 An aircraft is flying with an angle of attack of 14 and

no sideslip. The airspeed |V AC| is 94 m/s. Resolve

V AC in the aircraft frame

and in the stability frame.

Problem 2.6.3 An aircraft is flying with an angle of attack of 10 and

a sideslip angle of −19. The airspeed |V AC| is 80 m/s. Resolve

V AC in the

aircraft, stability, and wind frames.

Problem 2.6.4 Resolve the gravity vectorg symbolically in the air-

craft and stability frames. Then, set Ψ = 22, Θ = 5, Φ = −24, and

α = −17, and compute the components ofg in the aircraft and stability

frames.

Problem 2.6.5 Consider the position vectorr = 6.2 m ıE − 14.1 m E + 65.2 m kE

(m = meter) and assume that the orientation of the aircraft frame withrespect to the Earth frame is given by the yaw, pitch, roll Euler angles

Ψ = 62, Θ = 7, Φ = −12. Then resolver in the aircraft frame. Check

your answer by comparing the magnitude ofr computed from

r resolved

in both frames.

62 CHAPTER 2

Problem 2.6.6 An aircraft is flying with velocityV AC, which is con-

stant with respect to the Earth frame. Its angle of attack is −13, sideslip

angle is 24, and airspeed |V AC| is 85 m/s. The aircraft then rolls about ıAC

by +23 while the velocity vector remains fixed with respect to the Earth

frame. Resolve the velocity vectorV AC in the aircraft frame after the roll

is complete, and determine the new angle of attack and sideslip.

Problem 2.6.7 Write a Matlab program (and include your code) thatimplements the transformation from Euler angle rates to angular velocitycomponents given in class. Next, derive and confirm the reverse transfor-mation

Ψ = (Q sin Φ +R cos Φ) sec Θ,

Θ = Q cos Φ −R sin Φ,

Φ = P +Q(sin Φ) tan Θ +R(cos Φ) tan Θ,

and write a Matlab program that implements it. Finally, let

(Ψ,Θ,Φ) = (14,−29,−52),

(Ψ, Θ, Φ) = (19 deg/sec,−7 deg/sec, 16 deg/sec),

and compute (P,Q,R). Then use the computed values of (P,Q,R) in thereverse transformation and compute (Ψ, Θ, Φ). Verify that you recover theoriginal values of (Ψ, Θ, Φ).


Symbol Definition

FE Earth Frame

ıE, E, kE Earth frame axes

FAC Aircraft frame

ıAC, AC, kAC Aircraft frame axes

V AC Velocity of the aircraft relative to FE,

vOAC/OE/E

β Sideslip angle from ıS to ıW

FS Stability frame

ıS, S, kS Stability frame axes

r Component of a physical vector resolved in FS

FW Wind frame

ıW, W, kW Wind frame axes

U, V,W Components ofV AC resolved in FAC

US, VS,WS Components ofV AC resolved in FS

α Angle of attack angle from ıS to ıAC

γ Flight path angle from the horizontal to ıS

Θ Pitch angle from the horizontal to ıAC

ωAC Angular velocity of aircraft relative to FE,

ωAC/E

P,Q,R Components ofωAC resolved in FAC

aCor Coriolis acceleration

aang Angular-acceleration acceleration

acent Centripetal acceleration

Table 2.2 Symbols for Chapter 2.

Chapter Three

Review of Dynamics

Forces and moments can be applied to particles and bodies. Newton’s lawsrelate forces and moments to changes in momentum and angular momen-tum, respectively. These laws are axioms rather than provable mathematicalresults.

3.1 Newton’s First Law for Particles

We do not define the notion of a force, but rather we accept force asan intuitive notion and use it to define the concept of an inertial frame.

An unforced particle is a particle that has no forces acting on it.

The following result is Newton’s first law.

Fact 3.1.1 There exists a frame FA, called an inertial frame, suchthat, for all unforced particles x and y,

A••

r y/x = 0. (3.1.1)

Note that (3.1.1) can be written asa y/x/A. The origin OA of the

inertial frame FA is often taken to be the location of an unforced particle x.Note, however, that OA plays no role in this definition.

Fact 3.1.2 Let FA be an inertial frame, and let FB be a frame. Then

FB is an inertial frame if and only ifωB/A = 0.

Proof Use the double transport theorem Fact 1.20.1. Sufficiency isimmediate. Necessity is a little harder. Prove this result.

Fact 3.1.3 Let FA and FB be inertial frames. Then, for every physical

66 CHAPTER 3

vectorx(t),

A•

x (t) =

B•

x (t). (3.1.2)

3.2 Newton’s Second Law for Particles

Fact 3.2.1 Let FA be an inertial frame, let y be a particle with mass

m, letf y be a force acting on y, and let x be an unforced particle. Then,

mA••

r y/x =

f y. (3.2.1)

Fact 3.2.2 Let FA and FB be inertial frames, let y be a particle, letf y be a force acting on y, and let x be an unforced particle. Then,

A••

r y/x =

B••

r y/x. (3.2.2)

Proof Use (3.2.1) or double transport with Fact 3.1.2.

3.3 Forces and Moments

Let y be a particle in a body B, let w be a point, and letf y be a force

applied to y. Then the momentMy/w applied to y relative to w is defined

byMy/w

=r y/w ×

f y. (3.3.1)

Note that the point w may or may not be in the body B. See Fig. 3.3.1.

Let B be a body composed of particles m1, . . . ,ml, and letfmi

be a

force applied to mi. Then the net forcefB applied to B is defined by

f B

=

l∑

i=1

fmi

. (3.3.2)

Let B be a body composed of l rigidly interconnected particles m1, . . . ,

ml, let w be a point, and letfmi

be a force applied to mi. Then the net

REVIEW OF DYNAMICS 67

B

y

r

f

w

y/w

y

Figure 3.3.1

Representation of the moment

My/w =r y/w ×

f y relative to the point w due to the

force

f y applied to the particle y in the body B.

momentMB/w applied to B relative to w is defined by

MB/w

=

l∑

i=1

Mmi/w =

l∑

i=1

rmi/w ×

fmi

. (3.3.3)

Then net momentMB/w is a torque if the net force

fB is zero.

Fact 3.3.1 Let B be a body, letf x and

f y be forces applied to particles

x and y in B, respectively, assume thatf y = −

f x, and let w be a point.

Then the net moment applied to B relative to w isMB/w =

r x/y ×

f x. (3.3.4)

68 CHAPTER 3

Proof Note thatMB/w =

r x/w ×

f x +

r y/w × (−

f x)

=r x/w ×

f x +

r w/y ×

f x

=(r x/w +

r w/y

)

×f x

=r x/y ×

f x.

Note that the net force in Fact 3.3.1 is zero, and thus (3.3.4) is a

torque. Furthermore, the net momentMB/w in (3.3.4) is independent of w,

and is equivalent to the moment applied to x relative to y due to the forcef x applied to x. For the case of l particles, we have the following extension.

Fact 3.3.2 Let B be a body, and assume that the net force appliedto the body is zero. Then, the net moment applied to the body due to theforces is independent of the point relative to which the moment is taken.

3.4 Change in Angular Momentum

Let FA be an inertial frame, let B be a body composed of l intercon-nected particles m1, . . . ,ml, and let w be an unforced particle, as shown inFig. 3.4.2. We let mi denote both the name and the mass of the ith particle.Define the angular momentum of mi relative to w with respect to FA as

Hmi/w/A

=rmi/w ×mi

vmi/w/A, (3.4.1)

and, as in (1.16.3), define the angular momentum of B as

HB/w/A

=l∑

i=1

Hmi/w/A. (3.4.2)

Fact 3.4.1 Let FA be an inertial frame, let w be an unforced particle,let B be a body composed of l interconnected particles whose masses are

m1, . . . ,ml, respectively, and letfmi

be an external force applied to mi.Then

A•

HB/w/A =

MB/w, (3.4.3)

where

MB/w

=

l∑

i=1

rmi/w ×

fmi

. (3.4.4)


f

y

i

j

ij

ji

iy /w

m

f

m

f

r

wi

Figure 3.4.2Body B composed of rigidly interconnected particles.

Proof Since mi and mj are rigidly connected, letf ij denote the con-

straint force on mi due to the connection to mj. Note thatf ij = −

f ji and

f ii = 0.

The moment applied to mi relative to w is given by

Mmi/w

=rmi/w ×

fmi

+l∑

j=1

f ij

.

70 CHAPTER 3

The change in angular momentum with respect to FA is thus given by

A•

Hmi/w/A =

A•

r mi/w ×mi

vmi/w/A +

rmi/w ×mi

A•

v mi/w/A

=vmi/w/A ×mi

vmi/w/A +

rmi/w ×mi

A•

v mi/w/A

=rmi/w ×mi

A•

v mi/w/A

=rmi/w ×mi

A••

r mi/w

=rmi/w ×

fmi

+

l∑

j=1

f ij

.

Hence,

A•

Hmi/w/A =

Mmi/w.

Summing over the particles m1, . . . ,ml yields

A•

HB/w/A =

l∑

i=1

A•

Hmi/w/A

=

l∑

i=1

Mmi/w

=

l∑

i=1

rmi/w ×

fmi

+

l∑

i,j=1,...,l

i6=j

rmi/w ×

f ij

=MB/w +

l∑

i,j=1,...,l

i<j

(rmi/w −

rmj/w

)

×f ij

=MB/w +

l∑

i,j=1,...,l

i<j

rmi/mj

×f ij

=MB/w,

where we used the fact thatf ij = −

f ji, as well as the fact that

rmi/mj

is

parallel tof ij.

The following result introduces the physical inertia matrix

I B/w of B

relative to w.


Fact 3.4.2 Let B be a body with body-fixed frame FB, let FA be aframe, and let w be a point that is fixed in B. Then

HB/w/A =

I B/w

ωB/A, (3.4.5)

whereI B/w

=

l∑

i=1

mi

(∣∣∣rmi/w

∣∣∣

2U −

rmi/wrmi/w

)

. (3.4.6)

Furthermore,

I B/w

∣∣∣∣∣B

=

Ixx −Ixy −Ixz−Ixy Iyy −Iyz−Ixz −Iyz Izz

, (3.4.7)

where

Ixx

=

l∑

i=1

mi(y2i + z2

i ), Ixy

=

l∑

i=1

mixiyi,

Iyy

=

l∑

i=1

mi(x2i + z2

i ), Ixz

=

l∑

i=1

mixizi,

Izz

=

l∑

i=1

mi(x2i + y2

i )

︸︷︷︸

moments of inertia

, Iyz

=

l∑

i=1

miyizi

︸︷︷︸

products of inertia

. (3.4.8)

Proof Use (3.4.1) and the transport theorem to obtain

Hmi/w/A =

rmi/w ×mi

(B•

r mi/w +

ωB/A ×

rmi/w

)

,

whereB•

r mi/w is the velocity of the particle mi relative to w with respect to

FB. Since the particle mi is fixed in B relative to w, we haveB•

r mi/w = 0 for

all i = 0, . . . , l, and henceHmi/w/A =

rmi/w ×mi

(ωB/A ×

rmi/w

)

.

Summing over the l particles in the body yields

HB/w/A =

l∑

i=1

Hmi/w/A

=l∑

i=1

rmi/w ×mi

(ωB/A ×

rmi/w

)

.

72 CHAPTER 3

Next using (1.5.9) it follows that

HB/w/A =

l∑

i=1

mirmi/w ×

(ωB/A ×

rmi/w

)

=

l∑

i=1

mi

[(rmi/w ·rmi/w

)ωB/A −

(rmi/w ·ωB/A

)rmi/w

]

=

l∑

i=1

mi

[∣∣∣rmi/w

∣∣∣

2U −

rmi/wrmi/w

]

ωB/A

=

I B/w

ωB/A,

which proves (3.4.5). ResolvingHB/w/A in FB yields

HB/w/A

∣∣∣∣B

=

I B/w

∣∣∣∣∣B

ωB/A

∣∣∣B

=l∑

i=1

mi

(∣∣∣rmi/w

∣∣∣

2I3 −

rmi/w

∣∣∣B

rmi/w

∣∣∣

T

B

)

PQR

=l∑

i=1

mi

x2i + y2

i + z2i 0 0

0 x2i + y2

i + z2i 0

0 0 x2i + y2

i + z2i

−

x2i xiyi xizi

yixi y2i yizi

zixi ziyi z2i

PQR

, (3.4.9)

wherermi/w = xiıB + yiB + zikB

andωB/A = P ıB +QB +RkB.

Hence

I B/w

∣∣∣∣∣B

is given by (3.4.7) and (3.4.8).

Fact 3.4.3 Let FA be an inertial frame, let B be a body, and let FB

be a body-fixed frame. Then,I B/w

B•

ω B/A +

ωB/A ×

I B/w

ωB/A =

MB/w. (3.4.10)


Proof Combining (3.4.3) and (3.4.5) yields

A•

HB/w/A =

A•

︷︸︸︷I B/w

ωB/A =

MB/w, (3.4.11)

where the net momentMB/w on B relative to w is given by (3.4.4). Next,

the transport theorem and (3.4.5) yield

A•

HB/w/A =

B•

HB/w/A +

ωB/A ×

HB/w/A

=

B•

︷︸︸︷I B/w

ωB/A +

ωB/A ×

I B/w

ωB/A. (3.4.12)

Since the locations of the mass particles of B do not change relative to B,we have

B•

︷︸︸︷I B/w

ωB/A =

I B/w

B•

ω B/A . (3.4.13)

Combining (3.4.3), (3.4.12), and (3.4.13), yields (3.4.10).

Fact 3.4.4 Let B be a body, let w be a point, let

I B/w be given by

(3.6.1), let

R be a physical rotation matrix, and let

I

′

B/w be the physical

inertia matrix of

RB. Then

I

′

B/w =

R

I B/w

R

T

. (3.4.14)

Fact 3.4.5 Let B be a body with body-fixed frame FB, let w be a

point, let

I B/w be the physical inertia matrix of B, and let FA be a frame.

ThenI B/w

∣∣∣∣∣A

= OA/B

I B/w

∣∣∣∣∣B

OB/A. (3.4.15)

Proof The result follows from Fact 1.10.10.

74 CHAPTER 3

3.5 Momentum and Angular Momentum Relative toCenter of Mass

Note that the center of mass c of a body B composed of rigidly inter-connected particles m1, . . . ,ml whose masses are m1, . . . ,ml, respectively,satisfies

l∑

i=1

mirmi/c = 0. (3.5.1)

Fact 3.5.1 Let B be a body composed of rigidly interconnected parti-cles m1, . . . ,ml whose masses are m1, . . . ,ml, respectively, let c denote the

center of mass of B, letfmi

be an external force applied to the particle mi,assume that B is subject to gravity, let FA be an inertial frame, and let wbe an unforced particle. Then

A•

p B/w/A=

f B, (3.5.2)

where

fB

=l∑

i=1

(fmi

+mig ). (3.5.3)

Proof Note that

A•

p B/w/A =

l∑

i=1

mi

A•

v mi/w/A

=l∑

i=1

(fmi

+mig )

=fB.

The following result shows that gravity does not contribute to the netmoment on a rigid body relative to its center of mass.

Fact 3.5.2 Let B be a body composed of rigidly interconnected parti-cles m1, . . . ,ml whose masses are m1, . . . ,ml, respectively, let c denote thecenter of mass of B, assume that B is subject to gravity, and, for i = 1, . . . , l,

letfmi

be an additional external force applied to the particle mi. Then

MB/c =

l∑

i=1

rmi/c ×

fmi

. (3.5.4)


Proof Using (3.5.1) it follows that

MB/c =

l∑

i=1

rmi/c × (

fmi

+mig )

=

l∑

i=1

rmi/c ×

fmi

+

l∑

i=1

rmi/c ×mi

g

=

l∑

i=1

rmi/c ×

fmi

+

(l∑

i=1

mirmi/c

)

×g

=

l∑

i=1

rmi/c ×

fmi

.

Fact 3.5.3 Let B be a body composed of rigidly interconnected parti-cles m1, . . . ,ml whose masses are m1, . . . ,ml, respectively, let c be the centerof mass of B, let mB

=∑l

i=1mi be the mass of B, let FA be an inertialframe, let w be an unforced particle, assume that B is subject to gravity, for

i = 1, . . . , l letfmi

be an additional external force applied to the particlemi, and define

Hc/w/A

=r c/w ×mB

A•

r c/w (3.5.5)

and

HB/c/A

=l∑

i=1

(

rmi/c ×mi

A•

r mi/c

)

. (3.5.6)

ThenHB/w/A =

HB/c/A +

Hc/w/A. (3.5.7)

Furthermore,

A•

HB/c/A =

MB/c (3.5.8)

andA•

H c/w/A =

M c/w, (3.5.9)

whereM c/w

=r c/w ×

fB, (3.5.10)

fB

=l∑

i=1

(fmi

+mig ), (3.5.11)

76 CHAPTER 3

and

MB/c

=

l∑

i=1

rmi/c ×

fmi

. (3.5.12)

Proof Since w is an unforced particle, it follows from (3.4.3) that

HB/w/A =

l∑

i=1

(rmi/c +

r c/w

)

×mi

(A•

r mi/c +

A•

r c/w

)

=r c/w ×mB

A•

r c/w +

r c/w ×

l∑

i=1

mi

A•

r mi/c

+

l∑

i=1

rmi/c ×mi

A•

r c/w +

l∑

i=1

rmi/c ×mi

A•

r mi/c . (3.5.13)

Since c is the center of mass of B, it follows from (3.5.1) that the secondand third terms in (3.5.13) are zero. Therefore,

HB/w/A =

r c/w ×mB

A•

r c/w +

l∑

i=1

rmi/c ×mi

A•

r mi/c,

which is equivalent to (3.5.7).

Note thatHc/w/A is the angular momentum of a particle of mass mB

relative to w with respect to FA, whereHB/c/A is the angular momentum

of B relative to its center of mass c with respect to FA.

Next, differentiating (3.5.7) yields

A•

HB/w/A =

A•

H c/w/A +

A•

HB/c/A, (3.5.14)

whereA•

H c/w/A

=r c/w ×mB

A••

r c/w (3.5.15)

andA•

HB/c/A =

l∑

i=1

rmi/c ×mi

A••

r mi/c . (3.5.16)

Now it follows from Fact 3.5.1 that

fB = mB

A••

r c/w, (3.5.17)


wherefB is the net force acting on B. Hence it follows from (3.5.15) and

(3.5.17) that

A•

H c/w/A =

M c/w, (3.5.18)

whereM c/w

=r c/w ×

fB. (3.5.19)

Furthermore, note that

MB/w =

l∑

i=1

rmi/w ×

fmi

=

l∑

i=1

(rmi/c +

r c/w

)

×fmi

=l∑

i=1

rmi/c ×

fmi

+r c/w ×

f B

=l∑

i=1

rmi/c ×

fmi

+M c/w

=MB/c +

M c/w. (3.5.20)

Therefore, it follows from (3.5.20) and (3.5.12) that

MB/c +

M c/w =

MB/w

=

A•

HB/w/A

=

A•

H c/w/A +

A•

HB/c/A

=M c/w +

A•

HB/c/A .

Hence (3.5.8) holds.

3.6 Continuum Bodies

Replacing finite sums with integrals, IB/w is given by the followingresult.

78 CHAPTER 3

Fact 3.6.1 Let B be a body and let w be a point. ThenI B/w =

∫

B

∣∣∣r dm/w

∣∣∣

2U −

r dm/wr dm/w dm. (3.6.1)

Furthermore,

IB/w=

I B/w

∣∣∣∣∣B

=

Ixx −Ixy −Ixz−Iyx Iyy −Iyz−Izx −Izy Izz

, (3.6.2)

where

Ixx=

∫

B

(y2 + z2

)dm, Ixy

=

∫

B

xy dm,

Iyy

=

∫

B

(x2 + z2

)dm, Ixz

=

∫

B

xz dm,

Izz

=

∫

B

(x2 + y2

)dm

︸︷︷︸

moments of inertia

, Iyz

=

∫

B

yz dm

︸︷︷︸

products of inertia

.

To be added: Center of mass for a continuum body

3.7 Properties of the Inertia Matrix

Every rigid body B has an inertia dyadic

I B/w, which can be resolved

in a body-fixed frame FB to obtain

IB/w=

I B/w

∣∣∣∣∣B

=

Ixx −Ixy −Ixz−Ixy Iyy −Iyz−Ixz −Iyz Izz

.

The inertia matrix measures the mass distribution with respect to the chosencoordinate frame. The inertia matrix has the following properties:

• IB/w is symmetric and positive definite, which implies that all of theeigenvalues of the inertia matrix are positive.

• The eigenvalues λ1 ≥ λ2 ≥ λ3 of IB/w satisfy

λ1 < λ2 + λ3,

that is, the eigenvalues satisfy the triangle rule.


• If IB/w is diagonal, that is,

IB/w =

λ1 0 00 λ2 00 0 λ3

, (3.7.1)

then the coordinate axes of FB are the principal axes, and λ1, λ2, λ3

are the principal moments of inertia.

• If the inertia matrix is not diagonal, then there exists a frame withrespect to which the resolved inertia tensor is diagonal. Equivalently,there exists a proper orthogonal matrix S such that SIST is diagonal.

Example 3.7.1 The inertia matrix for a sphere has the form

IB/c =

λ 0 00 λ 00 0 λ

, (3.7.2)

that is, λ1 = λ2 = λ3.

Example 3.7.2 The inertia matrix for a cube is the same as that of asphere.

Example 3.7.3 The inertia matrix of a rectangular prism is of theform

IB/c =

λ1 0 00 λ2 00 0 λ3

, (3.7.3)

where the axes of FB are chosen to be perpendicular to the sides of the box.

Example 3.7.4 Two of the three moments of inertia are equal for acylinder, in particular, λ1 = λ2 > λ3.

3.8 Problems

Problem 3.8.1 Assuming the symmetry of a typical aircraft, showthat 4 of the entries of the inertia tensor resolved in the aircraft frame arezero. Next, write out the entries of the inertia tensor resolved in the stabilityframe in terms of the entries of the inertia tensor resolved in the aircraftframe.

Problem 3.8.2 Explain why the inertia matrix of a homogeneous cubeis independent of the frame you choose for resolving the inertia tensor.

Problem 3.8.3 Consider a rigid body spinning about a principal axis

80 CHAPTER 3

Symbol Definition

c Center of mass of B

dm Mass element of B

I B/w Physical inertia matrix of body B relative to w

λ Eigenvalue of the inertia matrix

HB/w/A Angular momentum vectorMB/y Net moment vector


and unperturbed by any applied moments. Use Euler’s equation to showthat the body will spin indefinitely about the principal axis.

Chapter Four

Aircraft Dynamics

4.1 Flight Dynamics and Control

From a dynamics and control point of view, there are two main re-quirements for an aircraft. An aircraft needs to be 1) stable so that it canmaintain course despite disturbances, and 2) controllable so that it can bemaneuvered. A challenge arises due to the fact that stability and control-lability of an aircraft are conflicting requirements. Forces and moments areneeded for control.

4.2 History of Aircraft Stability and Control

• Cayley - Oar with cruciform tail

• Lanchester - Wing dihedral (gliders).

• Bryon - Analyzed linearized flight.

4.3 Aerodynamic Forces

The net aerodynamic forceFA on the center of mass OAC of the air-

craft is given byFA =

D +

L, (4.3.1)

whereD = −DıW (4.3.2)

andL

= −LkW (4.3.3)

82 CHAPTER 4

are the drag and lift vectors, respectively, andD and L are positive numbers.Hence,

FA = −DıW − LkW. (4.3.4)

Note that the direction of the drag vectorD is opposite to the velocity

vectorV AC, and thus is in the direction of the wind relative to the aircraft.

Furthermore, the lift forceL is normal to

V AC and lies in the plane of

symmetry of the aircraft. We usually ignore the contribution of the controlsurfaces to the net aerodynamic force.

We can writeFA as

FA =

FAx

+FAy

+FAz

,

whereFAx

,FAy

, andFAz

are the frontal, side, and downward forces on the

aircraft in the directions ıS, S, and kS, respectively. Hence,D =

FAx

+FAy

(4.3.5)

andL =

FAz

. (4.3.6)

The frontal forceFAx

is given by

FAx

= FAx

V AC

|V AC|

= FAxıS. (4.3.7)

Hence, FAxis a negative number. Furthermore, the side force

FAy

is givenby

FAy

= FAyAC = FAy

S. (4.3.8)

Note that FAy> 0 denotes a left-to-right force, which corresponds to β < 0.

Finally, since kW = kS, the liftL is given by

L = −LkS = FAz

kS = −LkW = FAzkW. (4.3.9)

Hence, we haveFA = FAx

ıS + FAyS − LkS,

AIRCRAFT DYNAMICS 83

?kS = kW

6

L =

FAz

W

:ıAC

WkAC

FAx

i

D

-ıS

-V AC,proj

K α

qV AC

qıW

β

FAy

Figure 4.3.1

Aerodynamic forces. The vectors ıW and W point obliquely out of the page, while

D

points obliquely into the page.

and thus

FA

∣∣∣∣S

=

FAx

FAy

FAx

=

−D cosβ−D sinβ

−L

= OS/W

FA

∣∣∣∣W

=

cos β − sin β 0sin β cos β 0

0 0 1

−D0−L

.

Note that

D =√

F 2Ax

+ F 2Ay. (4.3.10)

4.4 Translational Momentum Equations

Recall thatr AC

=rOAC/OE

(4.4.1)

denotes the location of the center of mass of the aircraft with respect to theorigin of FE, which is assumed for convenience to be an unforced particle,

and define the velocityV AC of the aircraft relative to the Earth with respect

84 CHAPTER 4

to the Earth frame by

V AC

=vOAC/OE/E =

E•

r AC =

E•

r OAC/OE/E . (4.4.2)

Then,

m

E•

V AC = m

g +

FA +

FT, (4.4.3)

whereFT is the engine thrust force. Using the transport theorem (1.19) to

introduce the derivative with respect to the aircraft frame FAC into (4.4.3)yields

m

AC•

V AC +

ωAC ×

V AC

= mg +

FA +

FT, (4.4.4)

whereωAC =

ωAC/E. Resolving (4.4.4) in the aircraft frame yields

m

AC•

V AC

∣∣∣∣∣∣AC

+ (ωAC ×

V AC)

∣∣∣∣AC

= mg∣∣∣AC

+FA

∣∣∣∣AC

+FT

∣∣∣∣AC

.

(4.4.5)

Note that

ωAC

∣∣∣AC

=

PQR

. (4.4.6)

Resolving the gravity vector in the Earth frame asg = gkE yields

g∣∣∣E

=

00g

.


Transforming to the aircraft frame yields

g∣∣∣AC

= R3,2,1(EΨ,Θ,Φ−→ AC)

00g

= R1(E′′ Φ→ AC)R2(E

′ Θ→ E′′)R3(EΨ→ E′)

00g

=

[1 0 00 cos Φ sinΦ0 − sin Φ cos Φ

] [cos Θ 0 − sin Θ

0 1 0sin Θ 0 cos Θ

] [cos Ψ sinΨ 0− sin Ψ cos Ψ 0

0 0 1

] [00g

]

=

−g sin Θg(sin Φ) cos Θg(cos Φ) cos Θ

. (4.4.7)

Hence, we define

gxgygz

=

−g sin Θg(sin Φ) cos Θg(cos Φ) cos Θ

. (4.4.8)

Notice that the yaw angle Ψ does not appear ing∣∣∣AC

.

Next, consider the aerodynamic forceFA resolved in the stability

frame

FA

∣∣∣∣S

=

FAx

FAy

FAz

=

−D cos β−D sin β

−L

. (4.4.9)

These forces can be expressed in the aircraft frame by noting that

FA

∣∣∣∣AC

= R2(Sα→ AC)


−L

=


sinα 0 cosα


−L

=

−D(cos β) cosα+ L sinα−D sinβ

−D(cos β) sinα− L cosα

=

FAx

FAy

FAz

. (4.4.10)

86 CHAPTER 4

FT

*

rT

cg

V AC

KΦT

Figure 4.4.2Thrust force and engine geometry.

For the thrust force we express

FT

∣∣∣∣AC

=

FTx

FTy

FTz

=

cos ΦT 0 sin ΦT

0 1 0− sin ΦT 0 cos ΦT

FT

00

=

FT cos ΦT

0−FT sinΦT

, (4.4.11)

where FT is the engine force and ΦT is the angle from ıAC to the engineforce direction, shown in Fig. 4.4.2. We assume that the engine thrust hasa zero component in the AC direction.

Now, since

AC•

V AC =

AC•

︷︸︸︷(

UıAC + V AC +WkAC

)

= U ıAC + V AC + W kAC,

we have

AC•

V AC

∣∣∣∣∣∣AC

=

U

V

W

. (4.4.12)


Substituting (4.4.6), (4.4.8), (4.4.10), (4.4.11), and (4.4.12) into (4.4.5) weobtain the surge, sway, and plunge equations

m(

U − V R+WQ)

= mgx + FAx+ FTx

, (4.4.13)

m(

V + UR−WP)

= mgy + FAy, (4.4.14)

m(

W − UQ+ V P)

= mgz + FAz+ FTz

, (4.4.15)

or, equivalently,

m(

U − V R+WQ)

= −mg sin Θ −D(cos β) cosα+ L sinα+ FT cos ΦT,

(4.4.16)

m(

V + UR−WP)

= −D sinβ +mg(sin Φ) cos Θ, (4.4.17)

m(

W − UQ+ V P)

= mg(cos Φ) cos Θ−D(cos β) sinα

− L cosα− FT sin ΦT. (4.4.18)

4.5 Rotational Momentum Equations

LetωAC = P ıAC +QAC +RkAC (4.5.1)

be the angular velocity of the aircraft frame FAC with respect to the Earthframe FE. Let

HAC/E

=HAC/c/E (4.5.2)

denote the angular momentum of the aircraft relative to the center of massc

= OAC of the aircraft with respect to FE. Using (3.4.5) we have

HAC/E

∣∣∣∣AC

=

Hx

Hy

Hz

=

Ixx −Ixy −Ixz−Ixy Ixy −Iyz−Ixz −Iyz Izz

PQR

.

For most aircraft, AC-kAC is a plane of symmetry. Hence

Ixy =

∫

ACxy dm = 0 (4.5.3)

and

Iyz =

∫

ACyz dm = 0. (4.5.4)

88 CHAPTER 4

Consequently,

Hx = IxxP − IxzR, (4.5.5)

Hy = IyyQ, (4.5.6)

Hz = IzzR− IxzP. (4.5.7)

Next, Euler’s equation (3.4.10) for the aircraft is given by

I AC/c

AC•

ω AC +

ωAC ×

I AC/c

ωAC =

MAC/c, (4.5.8)

whereMAC/c

=MA/c +

MT/c (4.5.9)

is the net moment acting on the aircraft relative to c, andMA/c and

MT/c

are the aerodynamic and thrust moments relative to c, respectively. The netaerodynamic moment is produced by the air flow over both the fixed andvariable (control) surfaces of the aircraft.

In the aircraft frame, we have

MAC/c

∣∣∣∣AC

=

LAC

MAC

NAC

=MA/c

∣∣∣∣AC

+MT/c

∣∣∣∣AC

, (4.5.10)

where

MA/c

∣∣∣∣AC

=

LA

MA

NA

and

MT/c

∣∣∣∣AC

=

LT

MT

NT

.

Hence,

LAC

MAC

NAC

=

LA

MA

NA

+

LT

MT

NT

. (4.5.11)

Now resolving Euler’s equation (4.5.8) in the aircraft frame yields(I AC/c

AC•

ω AC

)∣∣∣∣∣AC

+

(

ωAC ×

I AC/c

ωAC

)∣∣∣∣∣AC

=MAC/c

∣∣∣∣AC

, (4.5.12)


which yields

IxxP − IxzR

IyyQ

IzzR− IxzP

+

0 −R QR 0 −P−Q P 0

IxxP − IxzRIyyQ

IzzR− IxzP

=

LAC

MAC

NAC

.

(4.5.13)

Equation (4.5.13) can now be written component-wise as the roll,pitch, and yaw equations

IxxP + (Izz − Iyy)QR− Ixz

(

R+ PQ)

= LAC, (4.5.14)

IyyQ+ (Ixx − Izz)PR+ Ixz(P 2 −R2

)= MAC, (4.5.15)

IzzR+ (Iyy − Ixx)PQ+ Ixz

(

QR− P)

= NAC. (4.5.16)

Equations (4.5.14), (4.5.15), and (4.5.16) are the roll, pitch, and yaw mo-ment equations, respectively. Each moment equation has three terms, eachrepresenting a physical effect. The first term is the angular accelerationterm, the next term is the gyroscopic precession term, and the last term isthe inertia coupling term. For the roll equation,

IxxP︸︷︷︸

angular acceleration

+ (Izz − Iyy)QR︸︷︷︸

gyroscopic precession

− Ixz

(

R+ PQ)

︸︷︷︸

inertial coupling

= LAC.

4.6 Aircraft Equations of Motion Resolved in FAC

Translational kinematics (see (2.2.1))

E•

r

∣∣∣∣∣AC

=VAC

∣∣∣∣AC

=

UVW

.

Rotational kinematics (see (2.3.14) and (2.3.12))

PQR

=

− sinΘ 0 1(cos Θ) sin Φ cos Φ 0(cos Θ) cos Φ − sin Φ 0

Ψ

Θ

Φ

.

Translational momentum (see (4.4.16), (4.4.17), and (4.4.18))

m(

U − V R+WQ)

= −mg sin Θ −D(cos β) cosα+ L sinα+ FT cos ΦT,

(4.6.1)

m(

V + UR−WP)

= −D sin β +mg(sin Φ) cos Θ, (4.6.2)

90 CHAPTER 4

m(

W − UQ+ V P)

= mg(cos Φ) cos Θ−D(cos β) sinα

− L cosα− FT sin ΦT.(4.6.3)

Rotational momentum (see (4.5.14), (4.5.15), and (4.5.16))

IxxP + (Izz − Iyy)QR− Ixz

(

R+ PQ)

= LAC = LA + LT, (4.6.4)

IyyQ+ (Ixx − Izz)PR+ Ixz(P 2 −R2

)= MAC = MA +MT, (4.6.5)

IzzR+ (Iyy − Ixx)PQ+ Ixz

(

QR− P)

= NAC = NA +NT. (4.6.6)

4.7 Problems

Problem 4.7.1 Resolve the aerodynamic force vectorFA symbolically

in the aircraft frame.

Problem 4.7.2 Consider Fig. 4.4.2. Determine

LT

MT

NT

in terms of ΦT and rT.


Symbol Definition

FA Aerodynamic force vector

D Drag force

FAxFrontal force in FS

FAySide force in FS

FAzVertical force in FS

L Lift force

FT Engine thrust force vector

FT Magnitude of the thrust force

ΦT Angle between the thrust force and ıSMAC Total moment vector on the aircraft

LAC Total roll moment on the aircraft

MAC Total pitch moment on the aircraft

NAC Total yaw moment on the aircraft

MA Aerodynamic moment vector

LA Aerodynamic roll moment in FAC

MA Aerodynamic pitch moment in FAC

NA Aerodynamic yaw moment in FAC

MT Thrust moment vector

LT Thrust roll moment in FAC

MT Thrust pitch moment in FAC

NT Thrust yaw moment in FACI AC/c Aircraft inertia tensor


92 CHAPTER 4

Symbol Definition

LAC Total roll moment in FS

MAC Total pitch moment in FS

NAC Total yaw moment in FS

LA Aerodynamic roll moment in FS

MA Aerodynamic pitch moment in FS

NA Aerodynamic yaw moment in FS

LT Thrust roll moment in FS

MT Thrust pitch moment in FS

NT Thrust yaw moment in FS


Chapter Five

Linearization

The goal of linearization is to approximate the nonlinear aircraft equationsof motion with linear equations to facilitate the analysis of flight character-istics in the vicinity of steady flight. The linearized equations are easier toanalyze than the original nonlinear equations and involve stability deriva-tives that can be estimated using computational fluid dynamics (CFD) codesor measured through experiments in a wind tunnel.

5.1 Taylor series

The Taylor series expansion of an infinitely differentiable func-tion f in a neighborhood of a is expressed as

f(x) = f(a) + f ′(a)(x− a) +f ′′(a)

2(x− a)2 +

f (3)(a)

3!(x− a)3 + · · ·

(5.1.1)

In a more compact form, we have

f(x) =

∞∑

i=0

f (i)(a)

i!(x− a)i. (5.1.2)

For a multivariable function, we write

f(x, y) = f(a, b) +∂f

∂x

∣∣∣∣(a,b)

(x− a) +∂f

∂y

∣∣∣∣(a,b)

(y − b)

+1

2

∂2f

∂x2

∣∣∣∣(a,b)

(x− a)2 +∂2f

∂x∂y

∣∣∣∣(a,b)

(x− a)(y − b)

+1

2

∂2f

∂y2

∣∣∣∣(a,b)

(y − b)2 + · · · . (5.1.3)

Truncating the series (5.1.3) to the first power in x and y yields the approx-

94 CHAPTER 5

imation

f(x, y) ≈f(a, b) +∂f

∂x

∣∣∣∣(a,b)

(x− a) +∂f

∂y

∣∣∣∣(a,b)

(y − b). (5.1.4)

Example 5.1.1 Consider the lift model

L(α, α, β, β, δe) = CL(α, α, β, β, δe)pdS,

where δe and pd are the elevator angle and dynamic pressure, respectively.Then

CL(α0 + δα, δα,β0 + δβ, δβ, δe)

≈ CL(α0, 0, 0, 0, 0)

+∂CL∂α

(α0, 0, 0, 0, 0)δα +∂CL∂δα

(α0, 0, 0, 0, 0)δα

+∂CL∂β

(α0, 0, 0, 0, 0)δβ +∂CL

∂β(α0, 0, 0, 0, 0)δβ

+∂CL∂δe

(α0, 0, 0, 0, 0)δe. (5.1.5)

5.2 Alternative Linearization Procedure

The following alternative procedure is sometimes simpler than usingthe Taylor expansion.

• Replace each variable with the sum of its steady-state value and aperturbation. For instance, U = U0 + u.

• Since the product of two small quantities is even smaller, ignore prod-ucts of perturbation variables.

• Subtract the nominal equation to delete all terms involving only thesteady-state values.

5.3 Trigonometric Functions

Using the Taylor expansion, we have

sin(Φ0 + φ) ≈ sinΦ0 + (cos Φ0)φ, (5.3.1)

cos(Φ0 + φ) ≈ cos Φ0 − (sin Φ0)φ. (5.3.2)

LINEARIZATION 95

Hence, for Φ0 = 0, we have the small-angle approximations

sinφ ≈ φ (5.3.3)

and

cosφ ≈ 1. (5.3.4)

Note that (5.3.1) and (5.3.2) can also be obtained from the trigonometricidentities

sin(Φ0 + φ) = (sin Φ0) cos φ+ (cos Φ0) sinφ (5.3.5)

and

cos(Φ0 + φ) = (cos Φ0) cosφ− (sin Φ0) sinφ. (5.3.6)

5.4 Steady Flight

Recall that

V AC =

vOAC/OE/E =

E•

r AC =

E•

r OAC/OE

. (5.4.1)

The conditions for steady flight are

AC•

V AC = 0 (5.4.2)

andAC•

ω AC = 0. (5.4.3)

These conditions mean that the velocity vectorV AC and the angular velocity

vectorωAC are constant with respect to the aircraft frame FAC. Under these

conditions, we denoteV AC and

ωAC by

V AC0

andωAC0

, respectively, where

V AC0

∣∣∣∣AC

=

U0

V0

W0

(5.4.4)

and

ωAC0

∣∣∣AC

=

P0

Q0

R0

. (5.4.5)

The aircraft can fly in several possible steady regimes, including

• Straight-line flight, either climb, level, or descent

96 CHAPTER 5

• Circular flight, which is banked or sideslipping

• Helical flight

Steady flight also includes hover as well as both rotation around andconstant-speed translation along any body axis. For example, rotation

around ıAC along withV AC = VACıAC constitutes steady flight known as a

barrel roll.

5.5 Linearization of the Aircraft Kinematics

We linearize the aircraft equations of motion about straight-line,constant-speed wings-level steady flight. Straight-line constant-speed flightmeans

U0 = constant, β0 = 0, Ψ0 = constant, Θ0 = α0 + γ0, (5.5.1)

while wings-level flight implies

Φ0 = 0, P0 = Q0 = R0 = 0. (5.5.2)

Here α0, β0, and γ0 denote the steady values of the angle of attack, sideslip,and flight path angles, respectively, while Ψ0, Θ0, and Φ0 denote steadyvalues of the 3-2-1 sequence of Euler rotations from FE to FAC. Note thatβ0 = 0 implies V0 = 0.

Next, consider the perturbed angular velocities

ωAC/E

∣∣∣AC

=

PQR

=

P0 + pQ0 + qR0 + r

=

pqr

, (5.5.3)

and the perturbed 3-2-1 Euler angles

ΨΘΦ

=

Ψ0 + ψΘ0 + θΦ0 + φ

=

Ψ0 + ψΘ0 + θφ

. (5.5.4)

It thus follows from (1.18.6) that

pqr

=

− sinΘ0 0 10 1 0

cos Θ0 0 0

ψ

θ

φ

. (5.5.5)

LINEARIZATION 97

?WAC

?kE

O

L

W kS

W

FAz

T

ΦTI*

V AC,proj

ıAC*

K α0:

V AC

: ıSγ0

K

-ıE

*

FAx

Figure 5.5.1Steady flight conditions. All vectors shown lie in the aircraft plane of symmetry.

5.6 Linearization of the Aircraft Dynamics in FAC

Consider the surge equation resolved in FAC given by

m(

U − V R+WQ)

= −mg sinΘ + FAx+ FTx

. (5.6.1)

At nominal flight conditions, (5.6.1) becomes

m(

U0 − V0R0 +W0Q0

)

= −mg sinΘ0 + FAx0+ FTx0

. (5.6.2)

Using (5.5.1), (5.5.2), and (5.6.2) yields

−mg sin Θ0 + FAx0+ FTx0

= 0. (5.6.3)

Now, replace each variable with the sum of its nominal value and a pertur-bation to obtain

m[

U0 + u− (V0 + v)(R0 + r) + (W0 + w)(Q0 + q)]

= −mg sin(Θ0 + θ) + FAx0+ fAx

+ FTx0+ fTx

.

(5.6.4)

98 CHAPTER 5

Substituting the nominal flight conditions (5.5.1) and (5.5.2) into (5.6.4)and using the trigonometric identity (5.3.5) yields

m(U0 + u− V0r −R0v − V0R0 − vr +W0Q0 +W0q +Q0w + wq)

= −mg[(sin Θ0) cos θ + (cos Θ0) sin θ] + FAx0+ fAx

+ FTx0+ fTx

.

(5.6.5)

Neglecting the products of perturbation variables and making small-angleapproximations for θ yields

m(U0 + u− V0r −R0v − V0R0 +W0Q0 +W0q +Q0w)

= −mg(sin Θ0 + θ cos Θ0) + FAx0+ fAx

+ FTx0+ fTx

.

(5.6.6)

Now substracting the nominal equation (5.6.3) from (5.6.6) yields the lin-earized surge equation

mu = −mW0q −mg(cos Θ0)θ + fAx+ fTx

, (5.6.7)

which is linear in terms of the perturbation variables u, q, and θ. Similarly,we have the linearized sway and plunge equations

mv = −mU0r +mW0p+mg(cos Θ0)φ+ fAy(5.6.8)

and

mw = mU0q −mg(sin Θ0)θ + fAz+ fTz

. (5.6.9)

The linearized rotational momentum equations for roll, pitch, and yaware given by

Ixxp− Ixz r = lAC = lA + lT, (5.6.10)

Iyy q = mAC = mA +mT, (5.6.11)

Izz r − Ixz p = nAC = nA + nT. (5.6.12)

5.7 Linearization of the Aircraft Dynamics in FSf

The frozen stability frame FSfis the stability frame (see Section 2.1.3

and figures 2.1.3 and 2.1.4) that corresponds to nominal flight conditions.

When steady flight is perturbed, the stability frame FS changes asV AC

changes. However, the frozen stability frame FSfremains fixed to the aircraft

as a body-fixed frame according to the orientation of the steady velocity

LINEARIZATION 99

vectorV AC0

. Note that

OSf/AC =

cosα0 0 sinα0

0 1 0− sinα0 0 cosα0

, (5.7.1)

V AC0

∣∣∣∣Sf

=

USf0

VSf0

WSf0

=

USf0

VSf0

0

= OSf/AC

U0

V0

W0

, (5.7.2)

and

ωAC0

∣∣∣Sf

=

PSf0

QSf0

RSf0

. (5.7.3)

Note that WSf0= 0 by definition of FSf

. Furthermore, we have

USf0

VSf0

WSf0

=

USf0

VSf0

0

=

cosα0 0 sinα0


U0

V0

W0

. (5.7.4)

For the frozen stability frame it follows from (5.5.5) that

pqr

=

cosα0 0 sinα0


pqr

=

cosα0 0 sinα0


− sin Θ0 0 10 1 0

cos Θ0 0 0

ψ

θ

φ

=

sin γ0 0 cosα0

0 1 0cos γ0 0 − sinα0

ψ

θ

φ

. (5.7.5)

We now linearize about straight-line, constant-speed, wings-levelsteady flight in the frozen stability frame. Straight-line constant-speed flightimplies

USf0= constant, β0 = 0, Ψ0 = constant, Θ0 = α0 + γ0, (5.7.6)

while wings-level flight implies

Φ0 = 0, PSf0= QSf0

= RSf0= 0. (5.7.7)

Note that β0 = 0 implies that VSf0= 0. Linearizing the aircraft equations of

motion resolved in the frozen stability frame with

FAx= FAx0

+ fAx, (5.7.8)

100 CHAPTER 5

FAy= FAy0

+ fAy, (5.7.9)

FAz= FAz0

+ fAz, (5.7.10)

LAC = LAC0+ lAC, (5.7.11)

MAC = MAC0+ mAC, (5.7.12)

NAC = NAC0+ nAC, (5.7.13)

LA = LA0+ lA, (5.7.14)

MA = MA0+ mA, (5.7.15)

NA = NA0+ nA, (5.7.16)

FTx= FTx0

+ fTx, (5.7.17)

FTy= FTy0

+ fTy, (5.7.18)

and

FTz= FTz0

+ fTz(5.7.19)

yields

m ˙u = −mg(cos γ0)θ + fAx+ fTx

, (5.7.20)

m ˙v = −mUSf0r +mg(cos Θ0)φ+ fAy

, (5.7.21)

m ˙w = mUSf0q −mg(sin γ0)θ + fAz

+ fTz. (5.7.22)

Note that (5.7.20)-(5.7.22) have the same form as (5.6.7)-(5.6.9) except thatWSf0

= 0 in (5.6.4)-(5.6.5). The linearized rotational equations of motionare given by

Ixx ˙p− Ixz ˙r = lAC = lA + lT, (5.7.23)

Iyy ˙q = mAC = mA + mT, (5.7.24)

Izz ˙r − Ixz ˙p = nAC = nA + nT. (5.7.25)

LINEARIZATION 101

Also note that

˙u˙v˙w

= OSf0/AC

uvw

= R2(AC−α0→ Sf0)

uvw

=

cosα0 0 sinα0


uvw

. (5.7.26)

Furthermore, the aerodynamic forces are expressed as (4.4.8), while thethrust forces are

FTx

0FTz

=

cosα0 0 sinα0


FT cos ΦT

0−FT sin ΦT

. (5.7.27)

5.8 Problems

Problem 5.8.1 A stunt plane is flying in steady circular flight, wherethe circular flight path is contained in a vertical plane. Sideslip and angle ofattack are zero. At the lowest point on the circle the pilot’s head is closer tothe ground than her feet are. The plane completes one revolution in 83 sec

and the radius of the circle is 4,100 ft. ResolveV AC and

ωAC in the aircraft

frame. Draw a diagram that illustrates your answer.

Problem 5.8.2 Using the translational and rotational dynamics, ex-plain why the Moon is in steady flight around the Earth. Let FM and FE

denote body-fixed Moon and Earth frames, respectively. In other words,

show that

M•

V M/E = 0 and

M•

ω M/E =

E•

ωM/E = 0, where

V M/E

=vOM/OE/E

and

M•

V M/E =

M•

v OM/OE/E . Draw relevant diagrams, and determine the ve-

locity vectorV M/E and the angular velocity vector

ωM/E in the Moon frame.

Finally, compute the magnitude of the centripetal acceleration of the Moonusing Newton’s law of universal gravitation (you will need µE = GmE) andcompare your result to the magnitude of the centripetal acceleration vector.

(Hint: Note thatV M/E =

ωM/E×

rM/E and that the Earth’s accelerationg

due to Earth’s gravity at the Moon’s location is the centripetal accelerationαcent =

ωM/E ×

V M/E of the Moon.)

102 CHAPTER 5

Problem 5.8.3 Consider the rotational kinematics equation

Φ = P +Q(sin Φ)(tan Θ) +R(cos Φ)(tan Θ).

Linearize this equation near the steady (not necessarily zero) values (P0, Q0,R0,Θ0,Φ0). (Hint: It will be helpful to note that (d/dΘ) tan Θ = sec2 Θ.)

Problem 5.8.4 Derive the linearized sway and plunge equations(5.6.8) and (5.6.9).

Problem 5.8.5 Derive the linearized roll, pitch, and yaw equations(5.6.10)-(5.6.12).

Problem 5.8.6 Multiply (5.6.7)-(5.6.9) by OSf/AC to obtain (5.7.20)-(5.7.22).

LINEARIZATION 103

Symbol Definition

fAxAerodynamic force perturbation in FAC along ıAC

fAyAerodynamic force perturbation in FAC along AC

fAzAerodynamic force perturbation in FAC along kAC

fTxThrust force perturbation in FAC along ıAC

fTzThrust force perturbation in FAC along kAC

lAC Total roll moment perturbation on the aircraft

mAC Total pitch moment perturbation on the aircraft

nAC Total yaw moment perturbation on the aircraft

lA Aerodynamic roll moment perturbation in FAC

mA Aerodynamic pitch moment perturbation in FAC

nA Aerodynamic yaw moment perturbation in FAC

lT Thrust roll moment perturbation in FAC

mT Thrust pitch moment perturbation in FAC

nT Thrust yaw moment perturbation in FAC

fAxAerodynamic force perturbation in FSf

along ıSf

fAyAerodynamic force perturbation in FSf

along Sf

fAzAerodynamic force perturbation in FSf

along kSf

fTxThrust force perturbation in FSf

along ıSf

fTzThrust force perturbation in FSf

along kSf

lAC Total moment perturbation in FSf

mAC Total pitch moment perturbation in FSf

nAC Total yaw moment perturbation in FSf

lA Aerodynamic roll moment perturbation in FSf

mA Aerodynamic pitch moment perturbation in FSf

nA Aerodynamic yaw moment perturbation in FSf

lT Thrust roll moment perturbation in FSf

mT Thrust pitch moment perturbation in FSf

nT Thrust yaw moment perturbation in FSf


Chapter Six

Static Stability and Stability Derivatives

6.1 Control Surface Deflections

Rudder δr > 0 means trailing edge left. Causes negative yaw.

Elevator δe > 0 means trailing edge down. Causes negative pitch.

Aileron δa > 0 means right aileron up. Causes positive roll. The aileronsalways move in opposite directions.

Static stability means that the initial motion of the aircraft after aperturbation is such that the magnitude of the perturbation decreases.

6.2 Aerodynamic Force Coefficients

Let

S

= wing area,

b

= wing tip-to-tip distance,

c

= wing mean chord,

ρ = air density,

pd

= dynamic pressure =1

2ρV 2

AC.

We define the aerodynamic coefficients

Cx

= CFAx

=FAx

pdS, (6.2.1)

Cy

= CFAy

=FAy

pdS, (6.2.2)

106 CHAPTER 6

and

CL

= −Cz

= −CFAz

= − FAz

pdS=

L

pdS. (6.2.3)

6.2.1 Lift

From (6.2.3) we have

L(α, α, β, β, δe) = CL(α, α, β, β, δe)pdS.

Using a Taylor expansion about the nominal conditions we have

CL(α0 + δα, δα, β0 + δβ, δβ, δe)

≈ CL(α0, 0, 0, 0, 0) + CLα(α0, 0, 0, 0, 0)δα

+c

2USf0

CLα(α0, 0, 0, 0, 0)δα + CLβ

(α0, 0, 0, 0, 0)δβ

+b

2USf0

CLβ(α0, 0, 0, 0, 0)δβ + CLδe

(α0, 0, 0, 0, 0)δe

= CL0+ CLα0

δα+c

2USf0

CLα0δα + CLβ0

δβ +b

2USf0

CLβ0δβ + CLδe0

δe,

(6.2.4)

where α0 is the nominal angle of attack. It is usually the case that CLα0> 0

and CLδe0> 0. Note that CLα0

and CLβ0are nondimensional due to the

notational convention

CLα0=

∂CL

∂(

cδα2USf0

)

∣∣∣∣∣∣∣0

(6.2.5)

and

CLβ0=

∂CL

∂(

bδβ2USf0

)

∣∣∣∣∣∣∣0

. (6.2.6)

6.2.2 Drag Polar

From (6.2.1) we have

D(α, α, β, β, δe) = CD(α, α, β, β, δe)pdS.

STATIC STABILITY AND STABILITY DERIVATIVES 107

Ignoring the dependence on the elevator δe, we write the drag polar as theparabolic function

CD(α, α, β, β, δe) = CDpar+C2L(α, α, β, β, δe)

πeAR

= CDpar+KC2

L(α, α, β, β, δe), (6.2.7)

where AR = b2

S is the aspect ratio, e is the Oswald efficiency factor, and

K = 1πeAR . Typically, e ≈ 0.8, while e always satisfies e < 1, which accounts

for nonelliptical lift distribution. CDparis the parasitic drag coefficient. For

α = α0, we obtain

CD0

= CD(α0, 0, 0, 0, 0) = CDpar+KC2

L0≥ CDpar

, (6.2.8)

where CL0

= CL(α0, 0, 0, 0) is the lift coefficient at α = α0 and CL0> 0 by

assumption.

Now, consider the effect of the elevator on the drag. The drag coeffi-cient is written as

CD(α, α, β, β, δe) = CDpar+KC2

L(α, α, β, β, δe) + g(δe), (6.2.9)

where g(δe) is the elevator effect. From the Taylor expansion for CD(α0 +δα, δα, β0 + δβ, δβ, δe), we express

CD(α0 + δα, δα, β0 + δβ, δβ, δe)

≈ CD(α0, 0, 0, 0, 0) + CDα(α0, 0, 0, 0, 0)δα

+c

2USf0

CDα(α0, 0, 0, 0, 0)δα + CDβ

(α0, 0, 0, 0, 0)δβ

+b

2USf0

CDβ(α0, 0, 0, 0, 0)δβ + CDδe

(α0, 0, 0)δe

= CD0+ CDα0

δα +c

2USf0

CDα0δα + CDβ0

δβ +b

2USf0

CDβ0δβ + CDδe0

δe.

(6.2.10)

On the other hand, differentiating (6.2.9) with respect to α yields

CDα0= CDα

(α0, 0, 0, 0, 0)

=∂

∂α

[

CDpar+KC2

L(α, α, β, β, δe) + g(δe)]∣∣∣∣α=α0, α=0, β0=0, β0=0, δe=0

= 2KCL(α0, 0, 0, 0, 0)CLα(α0, 0, 0, 0, 0)

= 2KCL0CLα0

. (6.2.11)

Now by differentiating with respect to α, we obtain

CDα0= 2KCL0

CLα0. (6.2.12)

108 CHAPTER 6

Similarly, for β and β we obtain

CDβ0= 2KCL0

CLβ0(6.2.13)

and

CDβ0= 2KCL0

CLβ0. (6.2.14)

Finally,

CDδe0= 2KCL0

CLδe0+ gδe0 . (6.2.15)

Hence, substituting (6.2.11), (6.2.12), (6.2.13), (6.2.14), and (6.2.15) into(6.2.9) yields

CD(α0 + δα, δα,β0 + δβ, δβ, δe)

≈ CD0+ 2KCL0

CLα0δα+ 2KCL0

CLα0δα

+ 2KCL0CLβ0

δβ + 2KCL0CLβ0

δβ

+(2KCL0

CLδe0+ gδe0

)δe. (6.2.16)

6.3 Linearization of Forces in FSf

In Section 5, we obtained the linearized aircraft equations in the frozenstability frame FSf

given by (5.7.20), (5.7.21), and (5.7.22). As in Section5.7, we now look at the perturbed steady flight forces in FSf

. We wantto express the perturbed aerodynamic forces and moments in terms of thestability derivatives.

ForV AC0

we have

V AC0

∣∣∣∣Sf

=

USf0

00

. (6.3.1)

Then perturbingV AC yields

V AC

∣∣∣∣Sf

=

USf0

00

+

uvw

=

USf0+ uvw

, (6.3.2)

which is represented in Fig. 6.3.1. Hence, we express

δα ≈ w

USf0+ u

≈ w

USf0

. (6.3.3)


-V AC0

Iα0

Iδα

I

α

u--

ıSf

*

V AC0

+ ∆V AC

6

∆V AC

: ıAC

?kSf

6

w

Figure 6.3.1Perturbation of VAC0 .

For the perturbed angular velocityωAC

∣∣∣Sf

, we have

ωAC

∣∣∣Sf

=

pQSf0

+ qr

. (6.3.4)

Recall from (4.4.10) that the aerodynamic forces resolved in the sta-bility frame are

FA

∣∣∣∣Sf

=

−D(cosβ) cosα+ L sinα−D sin β

−D(cosβ) sinα− L cosα

=

FAx

FAy

FAz

=

FAx0+ fAx

FAy0+ fAy

FAz0+ fAz

,

(6.3.5)

where fAx, fAy

, and fAzare perturbations to the components of the aerody-

namic force resolved in the frozen stability frame. We can express fAxand

fAzas functions of u, δα, δα, q, and δe, and fAy

as a function of δβ, δβ, p,r, δa, and δr. Using Taylor expansions, we write

fAx(u, δα, δα, q, δe) ≈ ∂fAx

∂u

∣∣∣∣0

u +∂fAx

∂δα

∣∣∣∣0

δα+∂fAx

∂δα

∣∣∣∣0

δα

+∂fAx

∂q

∣∣∣∣0

q +∂fAx

∂δe

∣∣∣∣0

δe, (6.3.6)

fAy(δβ, δβ, p, r, δa, δr) ≈ ∂fAy

∂δβ

∣∣∣∣0

δβ +∂fA

∂δβ

∣∣∣∣0

δβ +∂fAy

∂p

∣∣∣∣0

p

+∂fAy

∂r

∣∣∣∣0

r +∂fAy

∂δa

∣∣∣∣0

δa+∂fAy

∂δr

∣∣∣∣0

δr, (6.3.7)

110 CHAPTER 6

and

fAz(u, δα, δα, q, δe) ≈ ∂fAz

∂u

∣∣∣∣0

u +∂fAz

∂δα

∣∣∣∣0

δα+∂fAz

∂δα

∣∣∣∣0

δα

+∂fAz

∂q

∣∣∣∣0

q +∂fAz

∂δe

∣∣∣∣0

δe. (6.3.8)

In order to nondimensionalize the longitudinal and lateral force partial

derivatives, we use uUSf0

,(

c2USf0

)

δα,(

b2USf0

)

δβ,(

b2USf0

)

p,(

c2USf0

)

q, and(

b2USf0

)

r. Hence (6.3.6) becomes

fAx(u, δα, δα, q, δe) =

∂fAx

∂(

uUSf0

)

∣∣∣∣∣∣∣0

u

USf0

+∂fAx

∂δα

∣∣∣∣0

δα +∂fAx

∂(

cδα2USf0

)

∣∣∣∣∣∣∣0

δαc

2USf0

+∂fAx

∂(

cq2USf0

)

∣∣∣∣∣∣∣0

qc

2USf0

+∂fAx

∂δe

∣∣∣∣0

δe. (6.3.9)

Next, we express the force and moment partials in terms of CD andCL. First note from (4.3.10) that

CD =√

C2x + C2

y . (6.3.10)

Then, we can write

FAx= CxpdS = FAx0

+ fAx, (6.3.11)

where

Cx = CFAx=FAx

pdS= CFAx0

+fAx

pdS= Cx0

+fAx

pdS(6.3.12)

and

pd =1

2ρV 2

AC =1

2ρ[(USf0

+ u)2 + v2 + w2]. (6.3.13)

Using (6.3.12) the first partial in (6.3.9) is given by

∂fAx

∂(

uUSf0

)

∣∣∣∣∣∣∣0

=∂(CxpdS)

∂(

uUSf0

)

∣∣∣∣∣∣∣0

=∂Cx

∂(

uUSf0

)

∣∣∣∣∣∣∣0

pdS + Cx0S

∂pd

∂(

uUSf0

)

∣∣∣∣∣∣∣0

. (6.3.14)


Now, we define

Cxu0

=∂Cx

∂(

uUSf0

)

∣∣∣∣∣∣∣0

(6.3.15)

and

pdu0

=∂pd

∂(

uUSf0

)

∣∣∣∣∣∣∣0

(6.3.16)

so that

∂fAx

∂(

uUSf0

)

∣∣∣∣∣∣∣0

= Cxu0pdS + Cx0

Spdu0. (6.3.17)

To determine Cxu0, we use (6.3.10) to compute

CDu0

=∂CD

∂(

uUSf0

)

∣∣∣∣∣∣∣0

=∂√

C2x + C2

y

∂(

uUSf0

)

∣∣∣∣∣∣∣0

=1

2(C2

x0+ C2

y0)−1/2

2Cx0

∂Cx

∂(

uUSf0

)

∣∣∣∣∣∣∣0

+ 2Cy0∂Cy

∂(

uUSf0

)

∣∣∣∣∣∣∣0

=1

CD0

Cx0

∂Cx

∂(

uUSf0

)

∣∣∣∣∣∣∣0

+ Cy0∂Cy

∂(

uUSf0

)

∣∣∣∣∣∣∣0

=1

CD0

(Cx0

Cxu0+ Cy0Cyu0

), (6.3.18)

where CDu0is the dimensionless speed damping derivative. From (6.3.18),

we obtain

Cxu0=CD0

CDu0− Cy0Cyu0

Cx0

. (6.3.19)

Next, to determine pdu0, first note that

∂pd

∂u

∣∣∣∣0

= ρ(USf0+ u) ≈ ρUSf0

, (6.3.20)

112 CHAPTER 6

and thus, from (6.3.16),

pdu0= ρU2

Sf0= 2pd0

. (6.3.21)

Since, at steady state, Cx0=√

C2D0

− C2y0 , it follows from (6.3.14) and

(6.3.19) that

∂fAx

∂(

uUSf0

)

∣∣∣∣∣∣∣0

=

(CD0

CDu0− Cy0Cyu0

Cx0

+ 2√

C2D0

− C2y0

)

pd0S. (6.3.22)

Now, for the second partial in (6.3.9), we have

∂fAx

∂δα

∣∣∣∣0

=

(CD0

CDα0− Cy0Cyα0

Cx0

+ CL0

)

pd0S. (6.3.23)

For the δα derivative in (6.3.9), we obtain

∂fAx

∂(

cδα2USf0

)

∣∣∣∣∣∣∣0

=

(CD0

CDα0− Cy0Cyα0

Cx0

)

pd0S. (6.3.24)

Finally, for the pitch-rate derivative in (6.3.9), we obtain

∂fAx

∂(

cq2USf0

)

∣∣∣∣∣∣∣0

=

(CD0

CDq0− Cy0Cyq0

Cx0

)

pd0S. (6.3.25)

Substituting (6.3.22), (6.3.23), (6.3.24), and (6.3.25) into (6.3.9) yields

fAx(u, δα, δα, q, δe) ≈

(CD0

CDu0− Cy0Cyu0

Cx0

+ 2√

C2D0

− C2y0

)pd0

S

USf0

u

+

(CD0

CDα0− Cy0Cyα0

Cx0

+ CL0

)

pd0Sδα

+

(CD0

CDα0− Cy0Cyα0

Cx0

)

pd0S

c

2USf0

δα

+

(CD0

CDq0− Cy0Cyq0

Cx0

)

pd0S

c

2USf0

q

+

(CD0

CDδe0− Cy0Cyδe0

Cx0

)

pd0Sδe. (6.3.26)


Finally, if Cy0 ≈ 0, then (6.3.26) becomes

fAx(u, δα, δα, q, δe) ≈

(CDu0

+ 2CD0

) pd0S

USf0

u

+(CDα0

+ CL0

)pd0

Sδα

+CDα0pd0

Sc

2USf0

δα

+CDq0pd0

Sc

2USf0

q

+CDδe0pd0

Sδe. (6.3.27)

Next, from (6.2.2) we have

FAy= CypdS.

The side force is largely due to the vertical tail. Using the Taylor expansion,we have

Cy(δβ, δβ, δa, δr) ≈ Cy0 + Cyβ0δβ +

b

2USf0

Cyβ0δβ + Cyδa0

δa+ Cyδr0δr,

(6.3.28)

where

Cyβ

=∂Cy∂β

(6.3.29)

is the side force coefficient due to the sideslip derivative. Cyβ0< 0 implies

static stability since sideslip to the right induces a force to the left. On theother hand, if δr > 0, the left rudder causes a right force and thus Cyδr0

> 0.For a symmetric aircraft, Cy0 = 0. In nondimensional form, we express

∂fAy

∂δβ

∣∣∣∣0

= Cyβ0pd0

S, (6.3.30)

∂fAy

∂(

δβb2USf0

)

∣∣∣∣∣∣∣0

= Cyβ0pd0

Sbδβ

2USf0

, (6.3.31)

∂fAy

∂(

bp2USf0

)

∣∣∣∣∣∣∣0

= Cyp0pd0

Sbp

2USf0

, (6.3.32)

∂fAy

∂(

br2USf0

)

∣∣∣∣∣∣∣0

= Cyr0pd0

Sbr

2USf0

, (6.3.33)

114 CHAPTER 6

∂fAy

∂δa

∣∣∣∣0

= Cyδa0pd0

S, (6.3.34)

and

∂fAy

∂δr

∣∣∣∣0

= Cyδr0pd0

S. (6.3.35)

Substituting the above expressions into (6.3.7), we obtain

fAy(δβ, δβ, p, r, δa, δr) ≈ Cyβ0

pd0Sδβ + Cyβ0

pd0

Sb

2USf0

δβ

+ Cyp0pd0

Sb

2USf0

p+ Cyr0pd0

Sb

2USf0

r

+ Cyδa0pd0

Sδa+ Cyδr0pd0

Sδr. (6.3.36)

Similarly, we can show that

∂fAz

∂(

uUSf0

)

∣∣∣∣∣∣∣0

= −(CLu0

+ 2CL0

)pd0

S, (6.3.37)

where

CLu0

=∂CL

∂(

uUSf0

)

∣∣∣∣∣∣∣0

, (6.3.38)

which is dimensionless. Now, for the second partial in (6.3.8), we have

∂fAz

∂δα

∣∣∣∣0

= −(CLα0

+ CD0

)pd0

S. (6.3.39)

For the δα stability derivative, we obtain

∂fAz

∂(

cδα2USf0

)

∣∣∣∣∣∣∣0

= −CLα0

pd0S

2USf0

, (6.3.40)

where

CLα0

=∂CL

∂(

cδα2USf0

)

∣∣∣∣∣∣∣0

. (6.3.41)


For the pitch-rate stability derivative, we obtain

∂fAz

∂(

cq2USf0

)

∣∣∣∣∣∣∣0

= −CLq0

pd0S

2USf0

. (6.3.42)

Note that CLq0> 0 due to the angle of attack of the tail. Substituting into

(6.3.8), we have

fAz(u, δα, δα, q, δe) ≈

(CLu0

+ 2CL0

)pd0

Su

USf0

+(CLα0

+CD0

)pd0

Sδα + CLα0pd0

Sδα

(c

2USf0

)

+ CLq0pd0

Sq

(c

2USf0

)

+ CLδe0pd0

Sδe. (6.3.43)

6.4 Aerodynamic Moment Coefficients

For moment coefficients, we have Taylor expansions for the roll coeffi-cient

LA

pdSb= Cl(δβ, δβ, δa, δr)

≈ Cl0 +Clβ0δβ +

b

2USf0

Clβ0δβ +Clδa0

δa+ Clδr0δr, (6.4.1)

the pitch coefficient

MA

pdSc= Cm(α0 + δα, u, δe) ≈ Cm0

+ Cmα0δα+

1

USf0

Cmu0u+ Cmδe0

δe,

(6.4.2)

and the yaw coefficient

NA

pdSb= Cn(δβ, δβ, δr, δa)

≈ Cn0+ Cnβ0

δβ +b

2USf0

Cnβ0δβ + Cnδr0

δr + Cnδa0δa. (6.4.3)

Here, Clβ0, Cmα0

, and Cnβ0are stability derivatives, Clδa0

, Cmδr0, and

Cnδr0are control derivatives, and Clδr0

and Cnδa0are cross control deriva-

tives. Note that

Clβ=∂Cl∂β

, (6.4.4)

116 CHAPTER 6

K ΓUΓ

Figure 6.4.2Dihedral wings with angle Γ . The aircraft is pointing out of the page.

Cmα

=∂Cm∂α

, (6.4.5)

and

Cnβ

=∂Cn∂β

. (6.4.6)

6.4.1 Clβ0

Suppose the aircraft velocityV AC is perturbed by ∆

V AC with δβ > 0.

We need LA < 0 and hence Clβ0< 0 in order to roll away from the sideslip

perturbation and thus reduce the sideslip perturbation in accordance withstatic stability. Note that δα increases as δβ decreases.

Clβ0is the stability derivative corresponding to the roll moment LA

caused by sideslip. It is affected by the fuselage, the wing dihedral, the wingposition on the fuselage, and its sweep angle.

6.4.1.1 Effect of Wing Dihedral on Clβ0

We first consider the wing dihedral with the aircraft sideslipping tothe right, as shown in Fig. 6.4.2.

Note that, if Γ > 0, the wing is dihedral, otherwise it is anhedral. InFig. 6.4.3, Vn1

is the normal component of Vair due to V , and Vn2is the

normal component of Vair due to W . The normal velocityV n of

V air on the


VUΓ

Vn1

−knK Γ

WVn2

Figure 6.4.3Wing normal velocity (front view). The aircraft is sideslipping to its right.

right wing is thenV n = (Vn1

+ Vn2)(−kn), (6.4.7)

and the change ∆Vn in normal velocity ofV air on the right wing due to

dihedral is

∆Vn = (Vn1+ Vn2

−W )(−kn). (6.4.8)

Note that

cosΓ =Vn2

W(6.4.9)

and

sinΓ =Vn1

V. (6.4.10)

Substituting (6.4.9) and (6.4.10) into (6.4.8), we have

∆Vn = (cosΓ )W + (sinΓ )V −W

≈W + V Γ −W

= V Γ. (6.4.11)

From Fig. 6.4.4, we have

V = (tanβ)U ≈ δβU. (6.4.12)

Hence,

tan δα =∆Vn

U≈ ΓV

U≈ δβUΓ

U= δβΓ. (6.4.13)

For small perturbations, we have,

δα ≈ δβΓ. (6.4.14)

118 CHAPTER 6

wing cross section

*V AC,proj + ∆

V n

6

∆V nK δα

: V AC,proj

α0

K

Figure 6.4.4Perturbed angle of attack.

Thus, α0 is perturbed by δα, which effectively increases the angle of attackof the right wing due to sideslipping to the right, as shown in Fig. 6.4.5. Asimilar analysis shows that, on the left wing, the angle of attack is decreasedby δα. This effect results in a negative roll moment. Hence Clβ0

< 0, whichimplies static stability.

6.4.1.2 Effect of Wing Position on Clβ0

Consider first a high wing as in Fig. 6.4.6. The cross flow field velocitydue to sideslip to the right is equal to δβU . This sideslip causes the aircraftto roll to the left, with a roll moment LA < 0. Hence, since positive sideslipinduces a negative roll moment, it follows that Clβ0

< 0. Hence high winghas the same effect as wing dihedral.

Consider now a low wing as in Fig. 6.4.7, and a positive sideslippingto the right, with, again, a flow field velocity δβU . In this case, Clβ0

> 0,and a positive sideslip induces a positive roll moment, while a negativesideslip causes a negative roll moment. Wing dihedral is sometimes used tocounteract this effect.

6.4.1.3 Effect of Wing sweep on Clβ0

As shown in Fig. 6.4.8, the sweep angle of the wings affects the aircraftresponse to sideslip. A sideslip to the right causes increased lift by the rightwing, and thus, causes a negative roll moment.


V

U

V AC

V air

δβ

Figure 6.4.5Effect of sideslip and dihedral.

?

δα < 06

δα > 0

Figure 6.4.6High wing (front view). The aircraft is sideslipping to its right. The flow around the

fuselage perpendicular to the wing produces a negative roll moment.

6

δα > 0

?

δα < 0

Figure 6.4.7Low wing (front view). The aircraft is sideslipping to its right. The flow around the

fuselage perpendicular to the wing produces a positive roll moment.

120 CHAPTER 6

>

>

>

>

>

Vair

Figure 6.4.8Clβ0

is negative due to wing sweep since for δβ > 0 the lift of the right wing is greaterthan the lift of the left wing.

Finally, note that Clβ0is also affected by the horizontal and vertical

tails.

6.4.2 Cmα0

For pitch static stability, we perturb α0 to α0 + δα, and obtain

Cm(α0 + δα) ≈ Cm0+ Cmα0

δα, (6.4.15)

where Cm0

= Cm(α0) and the pitch moment is

MA = MA0+ mA, (6.4.16)

with

MA0= Cm0

pdSc. (6.4.17)

The perturbed moment is

MA = Cm(α0 + δα)pdSc

= Cm0pdSc+ ∆CmpdSc, (6.4.18)

where

∆Cm= Cm(α0 + δα) − Cm(α0)

≈ Cmα0δα. (6.4.19)


?WAC

•c

?kE

O

L

WkS

U kAC

cp •

• ac

xcp

xac

xc

0

ıAC*

K α0:

V AC

: ıS-ıE

9

D

Figure 6.4.9Pitching moment analysis assuming zero sideslip.

Hence, the change in moment is

mA = ∆CmpdSc ≈ Cmα0pdScδα. (6.4.20)

If δα > 0, that is, the nose goes up, then the change in moment is negative.Thus,

∆Cm < 0, (6.4.21)

and hence

Cmα0≈ ∆Cm

δα< 0. (6.4.22)

We link Cmαto the center of mass c. First, referring to Fig. 6.4.9, we

note the following properties:

• ac is the aerodynamic center, which is usually fixed, with xac = c/4.

• The momentMac = MacS (6.4.23)

about ac is independent of α and is generally nonzero.

• cp is the center of pressure, where xcp moves as α changes.

122 CHAPTER 6

• The moment about cp is zero for all α.

We compute the momentM c about the center of mass,

M c = (xc − xcp)

︸︷︷︸

<0

ıAC ×L

= (xc − xac)ıAC ×L

︸︷︷︸

aerodynamic moment

+ (xac − xcp)ıAC ×L

︸︷︷︸

aerodynamic moment

= (xc − xac)L(cosα)S +Mac

= [(xc − xac)L+Mac](cosα)S∼= [(xc − xac)︸︷︷︸

<0

L+Mac︸︷︷︸

<0

]S. (6.4.24)

WritingM c = McS, (6.4.25)

we have

Mc

pdSc= [(xc − xac)L+Mac]

1

pdSc

or

Mc

pdSc=xc − xac

c

L

pdS+Mac

pdSc. (6.4.26)

Therefore, defining

Cmc

=Mc

pdSc, (6.4.27)

we have

Cmc=xc − xac

cCL + Cmac

, (6.4.28)

where the pitch moment coefficient Cmacis independent of α. Since Cmα

=Cmc

, we have

Cmα=xc − xac

cCLα

+ 0. (6.4.29)

At the nominal flight conditions, (6.4.29) becomes

Cmα0=xc − xac

cCLα0

. (6.4.30)

Since xc − xac < 0 we have Cmα0< 0 with static stability as shown. On

the other hand, if xc − xac > 0, then Cmα0> 0 and the aircraft is statically

unstable.

STATIC STABILITY AND STABILITY DERIVATIVES 123Cm

α

C ′

mα

•α′

0

new trimdue to increasedmass at nose

Cmα

•

original trim

α0

Figure 6.4.10Change in pitching moment coefficient due to a change in trim.

Now suppose that additional mass is added at the nose. Then, fromFig. 6.4.10, we can see that C ′

mα< Cmα

< 0. Hence, Cmαbecomes more

negative as xc decreases.

6.4.3 Cnβ0

Now suppose that yaw is perturbed so that the sideslip angle isnonzero. For the yawing moment we have

NA = CnpdSb,

where Cn is the yawing moment coefficient. Using the Taylor expansion, weexpress

Cn(β0 + δβ, δβ, δa, δr)

≈ Cn0+ Cnβ0

δβ +b

2USf0

Cnβ0δβ + Cnδa0

δa+ Cnδr0δr, (6.4.31)

where Cnβ0is the static directional stability derivative, or weather vane sta-

bility derivative defined by

Cnβ

=∂Cn∂β

, (6.4.32)

124 CHAPTER 6

Cnδa0is the adverse aileron-yaw effect, usually negative, and Cnδr0

is therudder control derivative. For a symmetric airfoil, Cn0

= 0.

Cnβ0> 0 implies static stability since sideslip to the right causes a yaw

moment to the right.

Suppose δβ > 0. In terms of static stability, rolling away from sideslipso that δβ decreases is statically stable. Furthermore, yawing into sideslipso that δβ decreases is statically stable.

Consider a perturbation δα of α0 yielding the angle of attack α0 + δα.Then, |δα| decreases to zero if Cmα0

< 0.

Now let β0 = 0. For a small perturbation δβ, the aircraft rolls awayfrom δβ due to Clβ0

< 0.

Hence, due to Cmα0, the aircraft returns to its original pitch angle, but

Clβ0causes the aircraft to roll and ultimately attain a new heading. Hence,

the aircraft is asymptotically stable in pitch and semistable in heading.

6.5 Linearization of Moments in FSf

We now look at the perturbed aerodynamic moments in the frozenstability frame FSf

. We can express the pitch-moment perturbation mAC asa function of u, δα, δα, q, and δe. Using a Taylor expansion, we write

mA(u, δα, δα, q, δe) ≈ ∂mA

∂u

∣∣∣∣0

u+∂mA

∂δα

∣∣∣∣0

δα +∂mA

∂δα

∣∣∣∣0

δα

+∂mA

∂q

∣∣∣∣0

q +∂mA

∂δe

∣∣∣∣0

δe. (6.5.1)

Likewise, the lateral moments lA and nA can be expressed as functions ofδβ, δβ, p, r, δa, and δr.

In order to nondimensionalize the longitudinal and lateral moment

partial derivatives, we use uUSf0

,(

c2USf0

)

δα,(

b2USf0

)

δβ,(

b2USf0

)

p,(

c2USf0

)

q,

and(

b2USf0

)

r.

At steady state, we can show that

∂mA

∂(

uUSf0

)

∣∣∣∣∣∣∣0

=(Cmu0

+ 2Cm0

)pd0

Sc, (6.5.2)


where

Cmu

=∂Cm

∂(

uUSf0

) . (6.5.3)

Note that Cmu0, which is the Mach tuck derivative, changes with

V AC. Cm0

is the aerodynamic trim moment to counter the thrust moment. Note thatwhen u increases, xac shifts toward the back of the plane. From (6.4.29),Cmα

< 0, which means that the nose goes down. Hence, the pitch momentmust be negative, which results in Cm < 0 at USf0

+ u, and thus Cmu0< 0.

Now, differentiating with respect to δα, we have

∂mA

∂δα

∣∣∣∣0

= Cmα0pd0

Sc, (6.5.4)

where

Cmα0

=∂Cm∂δα

∣∣∣∣0

. (6.5.5)

For the δα derivative, we obtain

∂mA

∂(

cδα2USf0

)

∣∣∣∣∣∣∣0

=Cmα0

pd0Sc

2USf0

, (6.5.6)

where

Cmα0

=∂Cm

∂(

cδα2USf0

)

∣∣∣∣∣∣∣0

. (6.5.7)

For the pitch-rate derivative, we obtain

∂mA

∂(

qc2USf0

)

∣∣∣∣∣∣∣0

=Cmq0

pd0Sc

2USf0

, (6.5.8)

where the pitch damping derivative is defined by

Cmq0

=∂Cm

∂(

cq2USf0

)

∣∣∣∣∣∣∣0

(6.5.9)

and where Cmq0< 0. Finally, for the δe derivative, we obtain

∂mA

∂δe

∣∣∣∣0

= Cmδe0pd0

Sc. (6.5.10)

126 CHAPTER 6

Finally, using (6.5.2), (6.5.4), (6.5.6), (6.5.8), and (6.5.10), (6.5.1) becomes

mA(u, δα, δα, q, δe) ≈(Cmu0

+ 2Cm0

)pd0

Sc

USf0

u+ Cmα0pd0

Scδα

+ Cmα0pd0

Sc2

2USf0

δα +Cmq0pd0

Sc2

2USf0

q

+ Cmδe0pd0

Scδe. (6.5.11)

6.6 Effect of Adverse Control Derivatives

6.6.1 Adverse Aileron-Yaw

As shown in Fig. 6.6.11, since the left wing has increased lift due toaileron down, it also has increased drag. Hence, the aircraft turns right butsideslips to the left, which explains the “adverse” terminology. To cancelthis moment, we can use the rudder δr. From the yaw moment equation(6.4.31), we have

Cnδr0δr + Cnδa0

δa = 0, (6.6.1)

or,

δr =−Cnδa0

Cnδr0

δa. (6.6.2)

Note that Cnδa0and Cnδr0

do not change sign, whereas δa and δrchange sign as these control surfaces move. The adverse aileron-yaw momentis due to Cnδa0

< 0.

6.6.2 Adverse Rudder-Roll

To cancel the adverse rudder-roll moment, we use the aileron as shownin Fig. 6.6.12. From (6.4.1), we write

Clδr0δr + Clδa0

δa = 0. (6.6.3)

Solving for the aileron δa, we obtain

δa =−Clδr0

Clδa0

δr. (6.6.4)

Note that Cnδa0and Cnδr0

do not change sign, whereas δa and δrchange sign as these control surfaces move. However, the sign of Clδr0

de-pends on the angle of attack. To see this, it is helpful to view the rudder as


down

roll r i g h t

less lift, less drag

upδa > 0

yaw left

more lift, more drag

Figure 6.6.11Adverse aileron-yaw.

u α α q δeSurge CD ±Pitch Cm

Plunge CL

Table 6.1 Longitudinal stability derivatives and control.

producing a forcef δr on the vertical tail along the AC axis in the direction

opposite to the rudder deflection. Now, consider the use of left rudder sothat δr > 0 and the yaw moment is negative. For high α, that is, for largepositive α, the roll moment is negative. Hence the roll due to rudder isconsistent with the yaw motion, and thus is proverse. In this case, Clδr0

isnegative. On the other hand, for low α, that is, large negative α, the rollmoment is positive. Hence the roll due to rudder opposes the yaw motion,and thus is adverse. In this case, Clδr0

is positive.

6.7 Problems

Problem 6.7.1 Consider the expression (6.6.2), which shows how toset the rudder to cancel adverse aileron-yaw. Then draw two diagrams (onefor right rudder and one for left rudder) and check the signs of all terms in

128 CHAPTER 6

AC

α

αlowV

AC

high

VAC

i

δr > 0

fδr

Figure 6.6.12Adverse and proverse rudder-roll.

δβ δβ p r δa δeRoll Cl < 0Sway Cy < 0Yaw Cn

Table 6.2 Lateral stability derivatives and control.

the equation to confirm that they are all correct.

Problem 6.7.2 Consider the expression (6.6.4), which shows how toset the ailerons to cancel adverse rudder-roll. Then draw two diagrams (onefor rudder right and one for rudder left) and check the signs of all terms inthe equation to confirm that they are all correct.

Problem 6.7.3 Determine the linearized expression for the frozen sta-bility frame roll-moment perturbation lA.

Chapter Seven

Linearized Equations of Motion

7.1 Longitudinal Equations of Motion

We now incorporate the stability derivatives within the linearizedequations of motion (5.7.20)-(5.7.25). Note that we divide by the mass mand inertia Iyy to solve for linear and angular acceleration. For the linearizedlongitudinal equations, we obtain

˙u =pd0

S

mUSf0

(CD0

CDu0− Cy0Cyu0

Cx0

+ 2√

C2D0

− C2y0

)

︸︷︷︸

Xu0

u

+pd0

S

mUSf0

(CFTxu0

+ 2CFTx0

)

︸︷︷︸

XFTu0

u+pd0

S

m

(CD0

CDα0−Cy0Cyα0

Cx0

+CL0

)

︸︷︷︸

Xα0

δα

+pd0

Sc

2mUSf0

(CD0

CDα0− Cy0Cyα0

Cx0

)

︸︷︷︸

Xα0

δα +pd0

Sc

2mUSf0

(CD0

CDq0− Cy0Cyq0

Cx0

)

︸︷︷︸

Xq0

q

− g(cos γ0)θ +pd0

S

m

(CD0

CDδe0− Cy0Cyδe0

Cx0

)

︸︷︷︸

Xδe0

δe, (7.1.1)

130 CHAPTER 7

˙w =pd0

S

mUSf0

(CLu0+ 2CL0

)

︸︷︷︸

Zu0

u+pd0

S

mUSf0

(CFTzu0

+ 2CFTz0

)

︸︷︷︸

ZFTw0

u

+pd0

S

m(CLα0

+ CD0)

︸︷︷︸

Zα0

δα +pd0

Sc

2mUSf0

CLα0

︸︷︷︸

Zα0

δα

+ USf0q +

pd0Sc

2mUSf0

CLq0

︸︷︷︸

Zq0

q − g(sin γ0)θ +pd0

S

mCLδe0

︸︷︷︸

Zδe0

δe, (7.1.2)

and

˙q =pd0

Sc

IyyUSf0

(Cmu0+ 2Cm0

)

︸︷︷︸

Mu0

u+pd0

Sc

IyyUSf0

(CmFTu0

+ 2CmFT0

)

︸︷︷︸

MFTu0

u

+pd0

Sc

IyyCmα0

︸︷︷︸

Mα0

δα +pd0

Sc

IyyCmFTα0

︸︷︷︸

MFTα0

δα+pd0

Sc2

2IyyUSf0

Cmα0

︸︷︷︸

Mα0

δα

+pd0

Sc2

2IyyUSf0

Cmq0

︸︷︷︸

Mq0

q +pd0

Sc

IyyCmδe0

︸︷︷︸

Mδe0

δe. (7.1.3)

Hence, we have the linearized surge, plunge, and pitch-rate equations

˙u =(

Xu0+XFTu0

)

u+Xα0δα +Xα0

δα+Xq0 q − g(cos γ0)θ +Xδe0δe,

(7.1.4)

˙w = (Zu0+ ZFTw0

)u+ Zα0δα+ Zα0

δα +(USf0

+ Zq0)q

− g(sin γ0)θ + Zδe0δe, (7.1.5)

and

˙q =(

Mu0+MFTu0

)

u+(

Mα0+MFTα0

)

δα+Mα0δα +Mq0 q +Mδe0δe.

(7.1.6)

Note that

w ≈ USf0δα

and

˙w ≈ USf0δα.

LINEARIZED EQUATIONS OF MOTION 131

Hence, in terms of δα, (7.1.5) can be written as

δα =Zu0

+ ZFTw0

USf0

u+Zα0

USf0

δα +Zα0

USf0

δα+

(USf0

+ Zq0USf0

)

q

− g(sin γ0)

USf0

θ +Zδe0USf0

δe. (7.1.7)

7.2 Linearized Longitudinal Equations and TransferFunctions

We can write the linearized longitudinal equations (7.1.4), (7.1.6), and(7.1.7) in state space form as

˙uδα˙q

θ

=

Xu0+XFTu0

Xα00 −g cos γ0

Zu0+ZFTw0

USf0−Zα0

Zα0

USf0−Zα0

USf0+Zq0

USf0−Zα0

−g sinγ0USf0

−Zα0

Mu0+MFTu0

Mα0+MFTα0

Mq0 0

0 0 1 0

uδαqθ

+

Xδe0

Zδe0

USf0−Zα0

Mδe0

0

δe. (7.2.1)

Taking the Laplace transform of the longitudinal equations, we obtain

sû = (Xu0+XFTu0

)û+Xα0δ ˆα− g(cos γ0)θ +Xδe0δe, (7.2.2)

s(USf0− Zα0

)δα = (Zu0+ ZFTw0

)û+ Zα0δα+ [(USf0

+ Zq0)s− g sin γ0]θ

+ Zδe0δe, (7.2.3)

and

s2θ = (Mu0+MFTu0

)û+ (Mα0+MFTα0

)δα +Mq0sθ +Mδe0δe. (7.2.4)

132 CHAPTER 7

Stability

ParameterDefinition Units

Xu0

pd0S

mUSf0

(CD0

CDu0−Cy0

Cyu0

Cx0

+ 2√

C2D0

− C2y0

)

1/sec

XFTu0

pd0S

mUSf0

(CFTxu0

+ 2CFTx0

) 1/sec

Xα0

pd0S

m

(CD0

CDα0−Cy0

Cyα0

Cx0

+ CL0

)

ft/sec2-rad

Xα0

pd0Sc

mUSf0

(CD0

CDα0−Cy0

Cyα0

Cx0

)

ft − sec/rad

Xq0pd0

ScmUSf0

(CD0

CDq0−Cy0

Cyq0

Cx0

)

1/sec

Xδe0pd0

Sm

(CD0

CDδe0−Cy0

Cyδe0

Cx0

)

ft/sec2-rad

Zu0

pd0S

mUSf0

(CLu0+ 2CL0

) 1/sec

Zα0

pd0S

m (CLα0+ 2CD0

) ft/sec2-rad

Zα0

pd0Sc

2mUSf0

CLα0ft/sec-rad

Zq0pd0

Sc2mUSf0

CLq0ft/sec-rad

Zδe0pd0

Sm CLδe0

ft/sec2-rad

ZFTw0

pd0S

mUSf0

(CFTzu0

+ 2CFTz0

) 1/sec

Mu0

pd0Sc

IyyUSf0

(Cmu0+ 2Cm0

) rad/ft-sec

MFTu0

pd0Sc

IyyUSf0

(CmFTu0

+ 2CmFT0

) 1/ft-sec

Mα0

pd0Sc

IyyCmα0

1/sec2

MFTα0

pd0Sc

IyyCmFTα0

1/sec2

Mα0

pd0Sc2

2IyyUSf0

Cmα01/sec

Mq0pd0

Sc2

2IyyUSf0

Cmq01/sec

Mδe0pd0

Sc

IyyCmδe0

1/sec2

Table 7.1 Longitudinal stability parameters.


Thus,

s −

(

Xu0+ XFTu0

)

−Xα0g cos γ0

−(Zu0+ ZFTw0

)(

USf0− Zα0

)

s − Zα0−

(

USf0+ Zq0

)

s + g sin γ0

−

(

Mu0+ MFTu0

)

−

(

Mα0+ MFTα0

)

s2− Mq0s

[ˆu

δα

θ

]

=

Xδe0

(USf0− Zα0

)Zδe0Mδe0

δe. (7.2.5)

Inverting the 3 × 3 matrix coefficient in (7.2.5) yields

Gu/δe(s) =ˆu(s)

δe(s)=

Aus3 +Bus

2 + Cus+Du

s4 + Es3 + Fs2 +Gs+H, (7.2.6)

Gδα/δe(s) =δα(s)

δe(s)=Aαs

3 +Bαs2 + Cαs+Dα

s4 + Es3 + Fs2 +Gs+H, (7.2.7)

and

Gθ/δe(s) =θ(s)

δe(s)=

Bθs2 + Cθs+Dθ

s4 + Es3 + Fs2 +Gs+H. (7.2.8)

Note that the numerator in (7.2.8) is second order, see Problem 7.5.1.

The denominator p(s) = s4 +Es3 + Fs2 +Gs+H can be factored as

p(s) = (s2 + 2ζspωn,sps+ ω2n,sp)(s

2 + 2ζphωn,phs+ ω2n,ph). (7.2.9)

Although p(s) cannot be factored symbolically, it can always be factorednumerically. It has four roots given by two complex conjugate pairs thatdepend on the flight condition, the mass distribution, and the airplane ge-ometry.

7.3 Lateral Equations of Motion

Similarly, we obtain the lateral equations of motion

˙v =pd0

S

mCyβ0

︸︷︷︸

Yβ0

δβ +pd0

Sb

2mUSf0

Cyβ0

︸︷︷︸

Yβ0

δβ +pd0

Sb

2mUSf0

Cyp0

︸︷︷︸

Yp0

p

+pd0

Sb

2mUSf0

Cyr0

︸︷︷︸

Yr0

r − USf0r + g(cos Θ0)φ+

pd0S

mCyδa0

︸︷︷︸

Yδa0

δa+pd0

S

mCyδr0

︸︷︷︸

Yδr0

δr,

(7.3.1)

134 CHAPTER 7

˙p =IxzIzz

˙r +

pd0Sb

IxxClβ0

︸︷︷︸

Lβ0

+pd0

S

IzzClFTβ0

︸︷︷︸

LFTβ0

δβ +pd0

Sb2

2IxxUSf0

Clp0

︸︷︷︸

Lp0

p

+pd0

Sb2

2IxxUSf0

Clr0

︸︷︷︸

Lr0

r +pd0

Sb

IxxClδa0

︸︷︷︸

Lδa0

δa +pd0

Sb

IxxClδr0

︸︷︷︸

Lδr0

δr, (7.3.2)

and

˙r =IxzIzz

˙p+

pd0Sb

IzzCnβ0

︸︷︷︸

Nβ0

+pd0

Sb

IzzCnFTβ0

︸︷︷︸

NFTβ0

δβ +pd0

Sb2

2IzzUSf0

Cnp0

︸︷︷︸

Np0

p

+pd0

Sb2

2IzzUSf0

Cnr0

︸︷︷︸

Nr0

r +pd0

Sb

IzzCnδa0

︸︷︷︸

Nδa0

δa+pd0

Sb

IzzCnδr0

︸︷︷︸

Nδr0

δr. (7.3.3)

Using stability parameter notation, we can write (7.3.1)-(7.3.3) as

˙v = Yβ0δβ + Yβ0

δβ + Yp0p− USf0r + Yr0 r + g(cos Θ0)φ+ Yδa0

δa+ Yδr0δr,

(7.3.4)

˙p =IxzIzz

˙r + (Lβ0+ LFTβ0

)δβ + +Lp0 p+ Lr0 r + Lδa0δa+ Lδr0δr, (7.3.5)

and

˙r =IxzIzz

˙p+Nβ0δβ +NFTβ0

δβ +Np0 p+Nr0 r +Nδa0δa+Nδr0δr. (7.3.6)

Using

δβ ≈ v

USf0

, (7.3.7)

thus,

v ≈ δβUSf0(7.3.8)

and

v ≈ δβUSf0. (7.3.9)


Then we can rewrite (7.3.4) as

δβ =Yβ0

USf0

δβ +Yβ0

USf0

δβ +Yp0USf0

p− r +Yr0USf0

r +g(cos Θ0)

USf0

φ

+Yδa0

USf0

δa+Yδr0USf0

δr. (7.3.10)

7.4 Linearized Lateral Equations and TransferFunctions

The lateral linearized equations were given by (7.3.4), (7.3.5), and(7.3.6). The linearized perturbation states are the sideslip, roll rate, yawrate, roll angle, yaw angle and sway. There are two control inputs, δa andδr. For the linearized lateral equations, we obtain

δβ˙p˙r˙φ

= R

Yβ0

USf0−Yβ0

Yp0

USf0−Yβ0

Yr0−USf0

USf0−Yβ0

g cos Θ0

USf0−Yβ0

Lβ0+ LFTβ0

Lp0 Lr0 0

Nβ0+NFTβ0

Np0 Nr0 0

0 0 1 0

δβprφ

+R

Yδa0Yδr0

Lδa0Lδr0

Nδa0Nδr0

0 0

[δaδr

]

. (7.4.1)

where

R

=

1 0 0 0

0 1 − Ixz

Izz0

0 − Ixz

Izz1 0

0 0 0 1

−1

=

1 0 0 0

0 I2zz

I2zz−I2xz

Ixz Izz

I2zz−I2xz

0

0 Ixz Izz

I2zz−I2xz

I2zz

I2zz−I2xz

0

0 0 0 1

. (7.4.2)

We now use the small angle approximation for the derivatives of the

136 CHAPTER 7

Stability parameter Definition Units

Yβ0

pd0S

m Cyβ0ft/sec2-rad

Yβ0

pd0Sb

2mUSf0

Cyβ0ft/sec-rad

Yp0pd0

Sb2mUSf0

Cyp0ft/sec-rad

Yr0pd0

Sb2mUSf0

Cyr0ft/sec-rad

Yδa0

pd0S

m CDδa0ft/sec2-rad

Yδr0pd0

Sm CDδr0

ft/sec2-rad

Lβ0

pd0Sb

IxxClβ0

1/sec2

Lp0pd0

Sb2

2IxxUSf0

Clp01/sec

Lr0pd0

Sb2

2IzzUSf0

Cnp01/sec

Lδa0

pd0Sb

IxxClδa0

1/sec2

Lδr0pd0

Sb

IxxClδr0

1/sec2

LFTβ0

pd0S

IzzClFTβ0

1/sec2

Nβ0

pd0Sb

IzzCnβ0

1/sec2

NFTβ0

pd0Sb

IzzCnFTβ0

1/sec2

Np0pd0

Sb2

2IzzUSf0

Cnp01/sec

Nr0pd0

Sb2

2IzzUSf0

Cnr01/sec

Nδa0

pd0Sb

IzzCnδa0

1/sec2

Nδr0pd0

Sb

IzzCnδr0

1/sec2

Table 7.2 Lateral stability parameters.


3-2-1 Euler-angle perturbations given by (5.7.5) to obtain

ψ

θ

φ

=

sin γ0 0 cosα0

0 1 0cos γ0 0 − sinα0

−1

pqr

=1

cos Θ0

sinα0 0 cosα0

0 cos Θ0 0cos γ0 0 − sin γ0

pqr

=

(sinα0)p+(cosα0)rcos Θ0

q

(cos γ0)p−(sin γ0)rcos Θ0

. (7.4.3)

Hence, the derivatives of the Euler-angle perturbations are given in termsof the angular velocity vector perturbations in the frozen stability frame by

ψ =(sinα0)p+ (cosα0)r

cos Θ0, (7.4.4)

θ = q, (7.4.5)

φ =(cos γ0)p− (sin γ0)r

cos Θ0. (7.4.6)

Taking Laplace transforms of the linearized lateral equations yields

(USf0− Yβ0

)sδβ = Yβ0δβ + Yp0s

ˆφ+ (Yr0 − USf0)s ˆψ + g(cos Θ0)

ˆφ

− (USf0− Yβ0

)Yδa0δa+ (USf0

− Yβ0)Yδr0δr, (7.4.7)

s2 ˆφ =IxzIzz

s2 ˆψ + (Lβ0+ LFTβ0

)δβ + Lp0sˆφ+ Lr0s

ˆψ + Lδa0δa+ Lδr0δr,

(7.4.8)

s2 ˆψ =IxzIzz

s2 ˆφ+ (Nβ0+NFTβ0

)δβ +Np0sˆφ+Nr0s

ˆψ +Nδa0δa+Nδr0δr.

(7.4.9)

Then, using these approximations and taking the Laplace transform,

138 CHAPTER 7

the linearized equations become

s(USf0− Yβ0

) − Yβ0−(sYp0 + g cos Θ0) s(USf0

− Yr0)

−Lβ0− LFTβ0

s2 − sLp0 −(s2 Ixz

Ixx+ sLr0)

−Nβ0−NFTβ0

−(s2 Ixz

Izz+ sNp0) s2 − sNr0

β(s)

φ(s)

ψ(s)

=

(USf0− Yβ0

)Yδa0(USf0

− Yβ0)Yδr0

Lδa0Lδr0

Nδa0Nδr0

[δa(s)δr(s)

]

. (7.4.10)

Inverting (7.4.10) yields

Gβ/δa(s) =β(s)

δa(s)=Aβ,δas

3 +Bβ,δas2 + Cβ,δas+Dβ,δa

s4 + E′s3 + F ′s2 +G′s+H ′, (7.4.11)

Gφ/δa(s) =φ(s)

δa(s)=

Bφ,δas2 + Cφ,δas+Dφ,δa

s4 + E′s3 + F ′s2 +G′s+H ′, (7.4.12)

and

Gψ/δa(s) =ψ(s)

δa(s)=

Bψ,δas2 + Cψ,δas+Dψ,δa

s4 + E′s3 + F ′s2 +G′s+H ′. (7.4.13)

For the rudder perturbation, (7.4.10) yields

Gβ/δr(s) =β(s)

δr(s)=Aβ,δrs

3 +Bβ,δrs2 + Cβ,δrs+Dβ,δr

s4 + E′s3 + F ′s2 +G′s+H ′, (7.4.14)

Gφ/δr(s) =φ(s)

δr(s)=

Bφ,δrs2 + Cφ,δrs+Dφ,δr

s4 +E′s3 + F ′s2 +G′s+H ′, (7.4.15)

and

Gψ/δr(s) =ψ(s)

δr(s)=Aψ,δrs

3 +Bψ,δrs2 + Cψ,δrs+Dψ,δr

s(s4 + E′s3 + F ′s2 +G′s+H ′). (7.4.16)

The characteristic equation is written as

p(s) = (s2 + 2ζDrωn,Dr + ω2n,Dr)

(

s+1

τr

)(

s+1

τs

)

= 0. (7.4.17)

From solving (7.4.17), we find the Dutch roll mode, which involves sideslip,roll, yaw oscillations, the roll mode, and the spiral mode. Note that thespiral mode is a slow roll and a slow yaw, and it is stable for Clβ0

<< 0.The roll mode is unstable for large α and τr and has little response to δa.


7.4.1 Yaw Stability

Consider the heading perturbation ψ. Since the transfer function(7.4.16) has a pole at 0, it is not asymptotically stable, but rather is semistable.The integrator component of (7.4.16) due to the pole at 0 shows that an im-pulsive change in δr produces a steady-state effect on the heading. Thistype of stability is consistent with the fact that the aircraft has no naturalmechanism to return its heading to the original value after a sideslip or rollperturbation, thus leading to a new steady-state heading angle.

7.5 Problems

Problem 7.5.1 Show that the numerator of the transfer function in(7.2.8) from δe to θ is second order. Hint: Show CB = 0 and Section A.14.

Problem 7.5.2 Determine Au, Aα, and Bθ in terms of the stabilityparameters. Hint: See Section A.14.

Problem 7.5.3 Use (7.4.16) to determine the steady-state heading-angle perturbation from an impulse rudder deflection.

Chapter Eight

Analysis of Flight Modes

8.1 Eigenflight

Eigenflight is the motion of an aircraft near nominal flight as given byan eigensolution of the linearized dynamics. For details see Section A.3.

Consider the linear time invariant system

x(t) = Ax(t) (8.1.1)

with initial condition

x(0) = x0.

Let the solution be of the form

x(t) = Re eλtv, (8.1.2)

which satisfies (8.1.1) with the initial condition

x(0) = Re v = x0. (8.1.3)

Substituting (8.1.2) into (8.1.1) we obtain the eigenvalue-eigenvector equa-tion

Av = λv. (8.1.4)

Note that λ may be complex, in which case the associated eigenvector v mayalso be complex. Now write

λ = −ζωn ± ωn

√

1 − ζ2 = σ + ωd (8.1.5)

and

v = vR + vI, (8.1.6)

and substitute λ and v into (8.1.2). We obtain

x(t) = Re e(σ+ωd)t(vR + vI)

= Re eσt[cos(ωdt) + sin(ωdt)](vR + vI)

= eσt[vR cos(ωdt) − vI sin(ωdt)]. (8.1.7)

142 CHAPTER 8

Alternatively, we can represent each component of v in polar form

v =

r1eφ1

r2eφ2

r3eφ3

r4eφ4

. (8.1.8)

The component-wise magnitude and angle of v are given by

|v| =

r1r2r3r4

, ∡v =

φ1

φ2

φ3

φ4

. (8.1.9)

Substituting v into (8.1.2) yields

x(t) = Re e(σ+ωd)t

r1eφ1

r2eφ2

r3eφ3

r4eφ4

= Re

r1eσte(ωdt+φ1)

r2eσte(ωdt+φ2)

r3eσte(ωdt+φ3)

r4eσte(ωdt+φ4)

, (8.1.10)

where ri cosφi is the initial value of the ith component, eσt is the decayenvelope, and φi is the phase shift of the ith component.

8.2 Phugoid Mode

Solving for the complex roots λph and its conjugate λph in (7.2.9),the root with the lower frequency ωn,ph is the phugoid mode. The phugoidmode is associated with low damping, that is, ζph << 1, and a long periodTph = 2π

ωn,phof oscillation, typically 30 to 120 seconds.

We now show that the phugoid mode is a roller coaster oscillation,which trades kinetic and potential energy. In (7.2.6) and (7.2.8), θ and uare oscillatory and δα ≈ 0, which implies that α is approximately constant.

Consider a 747-100 aircraft flying at 40,000 ft at Mach 0.8. The aircraftstate is x = Ax, where

x =

uδαqθ

,

ANALYSIS OF FLIGHT MODES 143

0.5 1−0.5−1

0.5

1

−0.5

−1

phugoid

shortperiod

Figure 8.2.1Phugoid and short period eigenvalues.

and

A =

−2.02e(−2) 7.88 −6.5e(−1) −3.22e(+1)−2.54e(−4) −1.02 9.05e(−1) 07.95e(−11) −2.5 −1.39 0

0 0 1 0

.

Solving for the eigenvalues of A we can express

λsp, λsp = −ζspωn,sp ± ωn,sp

√

1 − ζ2sp

and

λph, λph = −ζphωn,ph ± ωn,ph

√

1 − ζ2ph.

We find λph = −0.0087 ± 0.074 for the mode phugoid and λsp = −1.204 ±1.49 for the short period, which are represented in Fig. 8.2.1. Therefore,ωd,ph = 0.074 rad/sec, ωn,ph = 0.0745 rad/sec, ζph = 0.117, ωd,sp = 1.49rad/sec, ωn,sp = 1.92 rad/sec, and ζsp = 0.628. These results are summa-rized in Table 8.1

By scaling the pitch perturbation, the phugoid eigenvector is given by

vph =

−1 m/sec−9.6e(−5) ± 5.0e(−7) rad

1.7e(−4) ± 8.4e(−6) rad/sec−3.8e(−4) ± 2.3e(−3) rad

. (8.2.1)

144 CHAPTER 8

Mode T T1/2 ωn ζ %

phugoid 84.34 sec 79.67 sec 0.0745 rad/sec 0.117 5%

short period 3.27 sec 0.57 sec 1.92 rad/sec 0.628 38%

Table 8.1 Eigenflight mode characteristics.

The component-wise magnitude of vph is

|vph| =

1 m/sec9.6e(−5) rad

1.7e(−4) rad/sec2.33e(−3) rad

(8.2.2)

with phase

∠vph =

0

0.30

2.83

80.62

. (8.2.3)

Using (8.1.10) the eigensolution is written as

x(t) = eσtRe eωd,pht

r1eφ1

r2eφ2

r3eφ3

r4eφ4

= eσt

r1 cos(ωd,pht+ φ1)r2 cos(ωd,pht+ φ2)r3 cos(ωd,pht+ φ3)r4 cos(ωd,pht+ φ4)

. (8.2.4)

It follows from (8.2.2) and (8.2.3) that

θ(t) = 2.33e(−3) cos(ωd,pht+ 1.407)eσt rad. (8.2.5)

Then, we have

u(t) = USf0cos(ωd,pht+ 0)e0.0087t m/sec, (8.2.6)

δα(t) = 9.6e(−5) cos(ωdpht+ 0.005)e0.0087t rad ≈ constant, (8.2.7)

and

θ(t) = 2.33e(−3)ωd,ph sin(ωd,pht+ 1.407)e0.0087t rad/sec. (8.2.8)


The pitch angle is

Θ = α+ γ (8.2.9)

with nominal values

Θ0 = α0 + γ0. (8.2.10)

Then,

θ = δα + δγ, (8.2.11)

where 8.2.7 shows that δα is small compared to θ. Note that α is relativelyconstant although α0 can be very different from zero.

8.3 Short Period Mode

The short period mode is associated with the higher frequency eigen-values in (7.2.9), λsp, and its conjugate λsp. The short period mode isunderdamped with high ζsp but always less than 1, high frequency ωn,sp,and short period period Tsp = 2π

ωn,sp, with Tsp typically about 2 sec.

Solving for λsp and λsp, we obtain

λsp, λsp = −ζspωn,sp ± ωn,sp

√

1 − ζ2sp. (8.3.1)

From numerics, the eigensolution characteristics are that u is constant whileθ and δα oscillate.

8.4 Phugoid Approximation

To approximate the phugoid mode, set δα = 0, γ0 = 0, Zα0= 0,

Zα0= 0, and ZFTw0

= 0. Then (7.2.5) becomes

[

s−(

Xu0+XFTu0

)

−g−Zu0

USf0s

][ˆuδα

]

=

[0

USf0Zδe0

]

δe. (8.4.1)

The characteristic equation then becomes

p(s) = s2 −(

Xu0+XFTu0

)

s− Zu0g

USf0

= 0 (8.4.2)

146 CHAPTER 8

and hence,

ωn,ph ≈√

−Zu0g

USf0

=

[( −gUSf0

) −pd0S(CLu0

+ 2CL0)

mUSf0

]1/2

≈√

gpd0S(CLu0

+ 2CL0)

mU2Sf0

. (8.4.3)

Since CLu0<< CL0

and CL0= mg

pd0S , we obtain

ωn,ph ≈√

gpd0S

mU2Sf0

(2mg

pd0S

)

≈√

2g2

U2Sf0

≈√

2g

USf0

. (8.4.4)

8.4.1 Phugoid Damping

For the phugoid damping we write

ζph ≈−Xu0

+XFTu0

2ωn,ph. (8.4.5)

Table 7.1 implies

Xu0=

−(CDu0+ 2CD0

)pd0S

mUSf0

(8.4.6)

and

XFTu0

=(CFTxu0

+ 2CFTx0

)pd0S

mUSf0

. (8.4.7)

Let set CFTxu0

= 0 and CFTx0

= 0. Then,

ζph =(CDu0

+ 2CD0)pd0

S

2mUSf0ωn,ph

=(CDu0

+ 2CD0)pd0

S

2√

2mg

=(CDu0

+ 2CD0)

2√

2

1

CL0

. (8.4.8)

At low speed, CDu0= 0, and (8.4.8) becomes

CD0√2CL0

, (8.4.9)

which is proportional to 1L/D . At high L/D, ζph is low.


8.5 Short Period Approximation

To approximate the short period mode, set u = 0 in (7.2.5), whichyields

[ (USf0

− Zα0

)s− Zα0

−(USf0

+ Zq0)s+ g sin γ0

−(

Mα0+MFTα0

)

s2 −Mq0s

][δα

θ

]

=

[Zδe0Mδe0

]

δe. (8.5.1)

We now assume that γ0 = 0, Zα0= 0, Zq0 = 0, and MFTα0

= 0. Then,

(8.5.1) becomes[USf0

s− Zα0−USf0

s−Mα0

s2 −Mq0s

] [δα

θ

]

=

[Zδe0Mδe0

]

δe. (8.5.2)

Then, the characteristic equation is obtained from

(s2 −Mq0s)(USf0

s− Zα0

)− USf0

sMα0= 0, (8.5.3)

or,

USf0s3 +

(−Mq0USf0

− Zα0

)s2 +

(Mq0Zα0

−Mα0USf0

)s = 0. (8.5.4)

Dividing by s and USf0, we obtain the quadratic equation

p(s) = s2 −(

Mq0 +Zα0

USf0

)

s+

(Mq0Zα0

USf0

−Mα0

)

= 0. (8.5.5)

Hence, the transfer functions (7.2.7) and (7.2.8) can be written as

δα(s)

δe(s)=Zδe0s+Mδe0USf0

−Mq0Zδe0USf0

p(s)(8.5.6)

and

θ(s)

δe(s)=USf0

Mδe0s+Mα0Zδe0 − Zα0

Mδe0

USf0sp(s)

. (8.5.7)

From (8.5.5), we express

ωn,sp ≈√

Mq0Zα0

USf0

−Mα0(8.5.8)

and

ζsp = −Mq0 +

Zα0

USf0

2ωn,sp. (8.5.9)

148 CHAPTER 8

Note that we need Mα0< 0 for static stability. For dynamic stability we

need

Mα0<Mq0Zα0

USf0

(8.5.10)

and

Mq0 +Zα0

USf0

< 0. (8.5.11)

Now, if the cg of the aircraft is forward enough, ωn,sp becomes

ωn,sp ≈√

−Mα0≈√

−Cmα0pd0

Sc

Iyy. (8.5.12)

Note that ωn,sp increases with −Cmα0and pd0

, and decreases with Iyy. Wealso note that the main component of ζsp isMq0 which is the damping deriva-

tive,Zα0

Mα0

is determined by other aspects and Mα0is due to the horizontal

tail size with Mα0≈ 1

3Mq0 .

8.6 Problems

Problem 8.6.1 The longitudinal dynamics of a business jet have thecharacteristic polynomial

p(s) = s4 + 2.01s3 + 8.05s2 + 0.085s + 0.068.

Check stability using Routh, identify the short period and phugoid roots,determine the damping ratio and natural frequencies of the roots, and com-pute the time for each response to decay by 50%.

Problem 8.6.2 The longitudinal dynamics of an F-104 fighter flyingstraight and level are modeled by the 4th-order system x = Ax, where

x =[

u δα θ θ]T

and the dynamics matrix is given by

A =

−2.02e(−2) 7.88 −3.22e(+1) −6.5e(−1)−2.54e(−4) −1.02 0 9.05e(−1)

0 0 0 17.95e(−11) −2.5 0 −1.39

.

Here, “e” is the matlab notation for powers of 10. Note the order of states,which is slightly different from the order used in class. Then do the following:

i) Compute the phugoid and short period eigenvalues. Plot thesepoles as four ×’s in the s-plane.


ii) Compute the damping ratio, undamped natural frequency, dampednatural frequency, undamped period, and time to 50% decay foreach mode.

Problem 8.6.3 For the F-104:

i) Compute the eigenvector for the phugoid mode, and normalize theθ component to 1.

ii) Convert the phugoid eigenvector into its magnitude and angle com-ponents.

Problem 8.6.4 Simulate the phugoid eigenflight mode of the F-104using the eigensolution x(t) = Re eAtvph, where vph is the phugoid eigenvec-tor. Assume that the initial condition for θ is given by θ(0) = 7.9 degrees.Then use Matlab to plot all four states for 3 full “periods” of the dampedmotion.

Chapter Nine

Control Concepts

9.1 Problems

Problem 9.1.1 Consider the asymptotically stable loop transfer func-tion

L(s) =−1

s2 + 2s+ 3.

Show that the closed-loop transfer function S(s) = 1/(1 + L(s)) is asymp-totically stable, and use Matlab to plot the magnitude and phase Bode plotsfor S. Then, write a Matlab program to show numerically that

∫∞

0ln |S(ω)| dω = 0.

Explain this result in terms of the “balance” between attenuation and am-plification.

Problem 9.1.2 Consider the unstable loop transfer function

L(s) =4

(s− 1)(s + 2).

Show that the closed-loop transfer function S(s) is asymptotically stable,and use Matlab to plot the Bode plots for S. Then, write a Matlab programto show numerically that

∫∞

0ln |S(ω)| dω = π.

Discuss the “balance” between attenuation and amplification.

Problem 9.1.3 Consider the second-order transfer function

G(s) =1

s2 + 3s− 2

controlled by the PI controller

K(s) = KP +KI

s

152 CHAPTER 9

in a servo loop. Determine the values of KP and KI that render the closed-loop system stable by sketching the region in the KP ,KI plane of stabilizinggains.

Problem 9.1.4 Consider the type II control system with plant

G(s) =1

s(s+ 1)

and PI controller

K(s) = KP +KI

s

in a servo loop. Choose values of KP and KI to obtain zero steady stateerror for a ramp command. Use Simulink to simulate the system and chooseKP and KI for good rise time and reasonable overshoot for a step command.Plot the error response for a ramp command, and verify numerically andanalytically that the ramp command error converges to zero.

Problem 9.1.5 Sketch the root loci for the following loop transferfunctions by using the root locus rules given in class and then check yoursketches using Matlab’s root locus function:

i) L(s) = (s+1)(s+2)(s−1)(s−2)(s−3) .

ii) L(s) = (s+2)2

(s+1)(s2+1) .

Problem 9.1.6 Consider the loop transfer function

L(s) =2.5(s + 100)

(s+ 1)2.

Sketch the gain and phase Bode plots of L(s). Use your plot to indicate themagnitude crossover frequency ωmco, the phase crossover frequency ωpco, thegain margin, and the phase margin.

Problem 9.1.7 Consider the damped rigid body plant

G(s) =1

s(s+ 1).

i) Assume unity feedback so that L(s) = G(s). Sketch the Nyquistplot and determine the gain and phase margins.

ii) Instead of unity feedback, consider the lead controller Gc(s) =k(s + 2)/(s + 20) so that L(s) = Gc(s)G(s). For k = 1, use Mat-lab to determine whether the lead provided by this lead controllerincreases the phase margin.

CONTROL CONCEPTS 153

iii) Draw the root locus in terms of k.

iv) Choose k so that the complex conjugate poles have damping ratioζ =

√2/2.

v) For the value of k that you chose, determine the steady state errorfor the unit ramp input 1/s2.

(Hint: You can solve the problem directly by equating the product of (s −a)(s2 + 2ζωns + ω2

n) with the cubic obtained from the closed loop transferfunction with k as the unknown parameter. Then you can get a cubicequation in a or ωn.)

Chapter Ten

Control of Aircraft

10.1 Problems

Problem 10.1.1 At Mach .6, an experimental aircraft has slightlyunstable short period eigenvalues 0.04±0.12 and stable phugoid eigenvalues−4.3 ± 5.7. Wind tunnel testing reveals that the elevator-to-pitch transferfunction has one zero located at −0.2. For proportional control, sketch theroot locus, determine the center and asymptotes, and discuss the stabilityof the closed-loop longitudinal dynamics for high values of k.

Problem 10.1.2 An experimental aircraft has unstable phugoid eigen-values 0.04 ± 0.12 and stable short period eigenvalues −4.3 ± 5.7. Windtunnel testing reveals that, at Mach .6, the elevator-to-pitch transfer func-tion has three real zeros located at −0.3,−1.7,−8.4. Sketch the root locusand discuss the stability of the closed-loop longitudinal dynamics using aconstant feedback controller. Indicate the point on the root locus at whichall poles have at least

√2/2 damping and explain why this minimum value

of damping is desirable.

Problem 10.1.3 The lateral dynamics of an experimental aircraft aremodeled by the transfer function 1/(τs+1), where τ > 0 is a time constant.For this transfer function a basic servo loop is closed with the integral con-troller KI = s.

i) Determine the values of KI for which the closed-loop system isasymptotically stable.

ii) For which values of KI is the steady-state error to a unit-sloperamp command less than 0.025?

iii) For a value of KI such that the closed-loop system is asymptoti-cally stable, determine the amplitude of the harmonic steady-stateresponse to the command r(t) = r0 cos(ωt).

Problem 10.1.4 At a given Mach number, the open-loop longitudinal

156 CHAPTER 10

dynamics of an experimental aircraft are given by L(s) = 1/s(s+ 1)2.

i) Sketch the Bode plot of L (magnitude and phase plots).

ii) Determine the phase crossover frequency !pco and the gain marginin dB of the closed-loop system and illustrate them on the Bodeplot. (Hint: log10 2 = 0.3)

iii) Sketch the Nyquist plot of L.

iv) Indicate the phase crossover frequency !pco and the gain marginon the Nyquist plot. Be sure that what you show on the Nyquistplot is consistent with the Bode plot.

Problem 10.1.5 At a given Mach number, the open-loop longitudinaldynamics of an unstable experimental aircraft are given by

L(s) =4(s + 10)

(s− 1)(s − 2).

i) Sketch the Bode plot of L (magnitude and phase plots).

ii) Sketch the Nyquist plot of L.

iii) Apply the Nyquist test to this system and use it to assess closed-loop stability.

Appendix A

Mathematical Background

A.1 Vectors and Matrices

A mathematical vector is a column of scalars

x =

x1

x2...xn

∈ R

n.

The transpose of x is the row vector

xT =[x1 x2 · · · xn

].

The dot product of two vectors x, y ∈ Rn is given by

xTy = x1y1 + x2y2 + · · · + xnyn ∈ R,

where

y =

y1

y2...yn

∈ R

n.

A matrix has the form

A =

a11 a12 · · · a1m

a21 a22 · · · a2m...

. . ....

an1 an2 · · · anm

∈ R

n×m.

Letting

y =

y1...yn

= Ax,

158 APPENDIX A

then, for i = 1, ..., n,

yi = rowi(A)x =[ai1 ai2 · · · aim

]x.

Letting

B =

b11 b12 · · · b1lb21 b22 · · · b2l...

. . ....

bm1 bm2 · · · bml

∈ R

m×l,

the product of A and B is given by

AB = A[

col1(B) · · · coll(B)]

=[Acol1(B) · · · Acoll(B)

]∈ R

n×l.

A square matrix has the form A ∈ Rn×n. The identity matrix is given by

I =

1 0 · · · 00 1 · · · 0...

. . ....

0 0 · · · 1

∈ R

n×n.

Suppose A, B ∈ Rn×n and AB = I. Then B = A−1, that is, B is the inverse

of A. A has an inverse if and only if det A 6= 0.

The determinant is given by

det

[a bc d

]

= ad− bc

and

det

a b cd e fg h i

= adet

[e fh i

]

− bdet

[d fg i

]

+ cdet

[d eg h

]

.

Inverses are given by[a bc d

]−1

=1

det

[a bc d

]

[d −b−c a

]

MATHEMATICAL BACKGROUND 159

and

a b cd e fg h i

−1

=1

det

[a b c

d e f

g h i

]

det

[e f

h i

]

− det

[d f

g i

]

det

[d e

g h

]

− det

[b c

h i

]

det

[a c

g i

]

− det

[a b

g h

]

det

[b c

e f

]

−det

[a c

d f

]

det

[a b

d e

]

T

A.1 Existence and Uniqueness of Solutions

Let A ∈ Rn×n and y ∈ R

n and consider

Ax = b. (A.1)

We wish to determine whether x ∈ Rn satisfying (A.1) exists, and, if so,

whether the solution x is unique. It is helpful to consider the special case

Ax = 0. (A.2)

Fact A.1.1 x = 0 is a solution of (A.2).

Fact A.1.2 x = 0 is the unique solution of (A.2) if and only if det A 6=0.

Fact A.1.3 If det A = 0 then (A.2) has at least one solution, namely,x = 0.

Note A(αx) = 0 for all α. Hence, (A.2) has either one solution oran infinite number of solutions. (A.2) may also have linearly independentsolutions.

A.2 Special matrices

For A ∈ Rn×n, Table A.1 presents several types of special matrices.

A.2 Complex Numbers, Vectors and Matrices

A complex number z ∈ C is written as

z = x+ y,

160 APPENDIX A

Name Property Example

Symmetric AT = A

a b

b c

Skew Symmetric AT = −A

0 b

−b 0

Orthogonal A−1 = AT

1 0

0 1

Nilpotent A2 = 0

1 1

−1 −1

Idempotent A2 = A 12

1 1

1 1

Nonsingular det A 6= 0

1 2

3 4

Table A.1 Special matrices.


where =√−1 and x and y are real numbers.

Let z1 = x1 + y1 ∈ C and z2 = x2 + y2 ∈ C, where x1, x2, y1, y2

∈ R.

Definition A.2.1 Complex addition is defined by

z1 + z2 = x1 + x2 + (y1 + y2).

Definition A.2.2 Complex multiplication

z1z2 = (x1 + y1)(x2 + y2)

= x1x2 − y1y2 + (x1y2 + x2y1).

Definition A.2.3 Complex conjugation is

z1 = x1 − y1.

Definition A.2.4 Complex division is defined as

z1z2

=z1z2z2z2

=x1x2 + y1y2 + (x2y1 − x1y2)

x22 + y2

2.

Definition A.2.5 The magnitude |z| of the complex number z is definedas

|z1| =√z1z1 =

√

x12 + y1

2.

The complex number z = x+ y is written in polar form as

z = |z|eθ = |z|(cos θ + sin θ),

where θ = atan2(y, x) (see Appendix (C.1)). Therefore, for all z 6= 0 and allα ∈ R,

zα = |z|α [(cosαθ) + (sinαθ)] .

A complex vector is a vector of complex numbers. We write

z =

x1 + y1

x2 + y2...

xn + yn

∈ C

n.

Similarly, a complex matrix is a matrix of complex numbers. We write

A =

x11 + y11 · · · x1m + y1m...

...xn1 + yn1 · · · xnm + ynm

∈ C

n×m.

162 APPENDIX A

Define

z =

x1 − y1

x2 − y2...

xn − yn

∈ C

n

and

A =

x11 − y11 · · · x1m − y1m...

...xn1 − yn1 · · · xnm − ynm

∈ C

n×m.

Vector and matrix addition are given by entry-wise addition. Define theconjugate transpose of a vector z ∈ C

n by

z∗ = zT,

and a matrix A ∈ Cn×m by

A∗ = AT = AT,

as well as the magnitude of a complex vector z ∈ Cn by

|z| =√zTz =

√z∗z.

Then the dot product of complex vectors z1, z2 ∈ Cn is given by

z1∗z2 = z1

Tz2 ∈ C.

A.3 Eigenvalues and Eigenvectors

Let A ∈ Rn×n, x ∈ C

n, λ ∈ C, and

Ax = λx, (A.1)

where x is nonzero. Then λ is an eigenvalue of A and x is a correspondingeigenvector. Eigenvalues are related to modal frequencies, while eigenvectorsare related to mode shapes.

Fact A.3.1 (A.1) holds if and only if (λI −A)x = 0.

Fact A.3.2 (A.1) has a nonzero solution x if and only if det(λI−A) = 0.

Example A.3.1 Let

A =

[1 00 2

]

Then

(λI −A)x = 0


implies that[λ− 1 0

0 λ− 2

] [x1

x2

]

=

[00

]

.

Therefore,

(λ− 1)x1 = 0 and (λ− 2)x2 = 0. (A.2)

Since x 6= 0, we must have either x1 6= 0 or x2 6= 0, and, correspond-ingly, either λ = 1 or λ = 2. If λ1 = 1, then

x =

[x1

0

]

, where x1 6= 0, (A.3)

while, if λ2 = 2, then

x =

[0x2

]

, where x2 6= 0. (A.4)

Example A.3.2 Let

A =

[0 −11 0

]

.

Then

(λI −A)x = 0

implies that[

λ 1−1 λ

] [x1

x2

]

=

[00

]

.

Therefore,

λx1 + x2 = 0,

−x1 + λx2 = 0,

which implies that λ = or λ = −. If λ = , then

x =

[1−

]

,

(A.5)

while, if λ = −, then

x =

[1

]

.

Definition A.3.1 The characteristic polynomial of A is

p(s)

= det(sI −A). (A.6)

164 APPENDIX A

Definition A.3.2 The characteristic equation of A is

p(λ) = 0. (A.7)

Note that s represents an arbitrary complex number in (A.6), whereass = λ in (A.7) denotes a root of p.

Fact A.3.3 A has n eigenvalues, which are not necessarily distinct.

Fact A.3.4 If A is diagonal, then the diagonal entries of A are theeigenvalues of A.

Fact A.3.5

detA =n∏

i=1

λi.

Fact A.3.6

trA =

n∑

i=1

Aii.

Fact A.3.7 If λ is an eigenvalue of A then λ2 is an eigenvalue of A2.

Proof . Note that

Av = λv.

Therefore,

A2v = λAv = λ2v.

Fact A.3.8 If A is nonsingular and λ is an eigenvalue of A, then 1/λis an eigenvalue of A−1.

Proof . Note that

λv = Av.

Therefore,

λA−1v = A−1Av,

and thus,

A−1v =1

λv.

Fact A.3.9 If A is symmetric then every eigenvalue of A is real.


Fact A.3.10 If A is skew symmetric then every eigenvalue of A isimaginary.

Definition A.3.3 xTAx is a quadratic form.

Definition A.3.4 Let A ∈ Rn×n. Then, A is positive semidefinite

(PSD) if xTAx ≥ 0 for all x ∈ Rn. Furthermore, A is positive definite (PD)

if xTAx > 0 for all nonzero x ∈ Rn.

Fact A.3.11 Let A ∈ Rn×n. Then, A is PSD if and only if every

eigenvalue of A is nonnegative.

Fact A.3.12 A is PD if and only if every eigenvalue of A is positive.

A.4 Single-Degree-of-Freedom Systems

m

-q

-f

c

k

Figure A.1The damped oscillator

We study the dynamics of the spring-mass-damper system shown inFigure A.1, where q(t) is the displacement of the mass m. Note that q(t) > 0if and only if the spring extends from its relaxed configuration, whereasq(t) < 0 if and only if the spring compresses from its relaxed configuration.Likewise, q(t) > 0 if and only if the dashpot pulls to the left with force−cq(t), whereas q(t) < 0 if and only if the dashpot pulls to the right withforce −cq(t).

Applying Newton’s second law (see Chapter 3) to the mass we have

mq(t) = ftotal(t) = f(t) − kq(t) − cq(t).

Dividing through by m yields

q(t) +c

mq(t) +

k

mq(t) =

1

mf(t).

166 APPENDIX A

Next define the natural frequency

ωn

=

√

k

m

and the damping ratio

ζ

=c

2√mk

.

Then

q(t) + 2ζωnq(t) + ω2nq(t) =

1

mf(t).

Undamped Rigid Body (URB) (k = 0, c = 0)

m

-q

-f

Figure A.2Undamped rigid body

q(t) =1

mf(t) (A.1)

Damped Rigid Body (DRB) (k = 0)

m

-q

-f

c

Figure A.3Damped rigid body


q(t) +c

mq(t) =

1

mf(t) (A.2)

Undamped Oscillator (UO) (c = 0)

m

-q

-f

k

Figure A.4Undamped oscillator

q(t) +k

mq(t) =

1

mf(t). (A.3)

m

-q

-f

c

k

Figure A.5Damped oscillator

Damped Oscillator (DO)

q(t) +c

mq(t) +

k

mq(t) =

f(t)

m. (A.4)

A.5 Matrix Differential Equations

Consider the scalar differential equation

x(t) = ax(t), x(0) = x0. (A.1)

168 APPENDIX A

Then

x(t) = eatx0

is a solution of (A.1), as can easily be verified by substitution. Since (A.1)is a linear ODE, it has a unique solution.

Now consider the matrix differential equation

x(t) = Ax(t), x(0) = x0, (A.2)

where x ∈ ℜn and A ∈ ℜn×n. Then

x(t) = eAtx0 (A.3)

is the unique solution. To verify that (A.3) is a solution of (A.2) considerthe Taylor series expansion of eAt given by

eAt = I + tA+1

2t2A2 +

1

3!t3A3 + · · · .

Then

d(eAt)

dt= 0 +A+ tA2 +

1

2t2A3 + · · ·

= A

(

I + tA+1

2t2A2 + · · ·

)

= AeAt.

Example A.5.1 Let

A =

[1 00 2

]

.

From Example A.3.1 we know that the eigenvalues are λ1 = 1 and λ2 = 2.The matrix exponential is given by

eAt = I + t

[1 00 2

]

+1

2t2[

1 00 4

]

+ · · ·

=

[

1 + t+ 12t

2 + · · · 0

0 1 + 2t+ 12 t

2(4) + · · ·

]

=

[

et 0

0 e2t

]

. (A.4)

Notice that eAt involves eλt.

Example A.5.2 Let

A =

[0 ω−ω 0

]

.


From Example A.3.1 we know that λ1 = ω or λ2 = −ω. The matrixexponential is given by

eAt = I + ωt

[0 1−1 0

]

+1

2ω2t2

[−1 00 −1

]

+1

3!ω3t3

[0 −11 0

]

+ · · ·

=

[1 − 1

2ω2t2 + · · · ωt− 1

6ω3t3 + · · ·

−ωt+ 16ω

3t3 − · · · 1 − 12ω

2t2 + · · ·

]

=

[cosωt sinωt− sinωt cosωt

]

. (A.5)

A.6 Eigensolutions

Fact A.6.1 Let λ be an eigenvalue of A ∈ Rn×n and let v be an

eigenvector associated with λ. Furthermore, let

x(t) = Ax(t), x(0) = v. (A.1)

Then

x(t) = Re[eλtv] (A.2)

is a solution of (A.1).

Proof . Let v = y + z, where y, z ∈ Rn. Then,

x(t) =Re[eλtv]

=Re[e(σ+ω)t(y + z)]

=Re[eσt(cosωt+ sinωt)(y + z)]

= eσt(y cosωt− z sinωt).

Hence,

x(t) = eσt[(σ cosωt− ω sinωt)y − (σ sinωt+ ω cosωt)z].

Also

Ax(t) =ARe[eλtv]

= Re[eλtAv]

= Re[eλtλv]

= Re[eσt(cosωt+ sinωt)(σ + ω)(y + z)]

= eσt[(σ cosωt− ω sinωt)y − (σ sinωt+ ω cosωt)z].

Therefore,

x(t) = Ax(t), x(0) = Re v,

which shows that x(t) = Re[eλtv] is a solution of x(t) = Ax(t).

170 APPENDIX A

As an alternative proof, we do not decompose the complex variablesλ and v into their real and complex parts. Then,

x(t) =d

dtRe[eλtv] = Re

[d

dt(eλtv)

]

= Re[eλtλv]

= Re[eλtAv] = ARe[eλtv] = Ax(t), (A.3)

which, along with x(0) = Re[eλv] = Re v, shows that (A.3) satisfies thedifferential equation.

Definition A.6.1 The solution (A.2) of (A.1) is an eigensolution.

A.7 State Space Form

Consider the nth-order differential equation

dnq(t)

dtn+ an−1

dn−1q(t)

dtn−1+ · · · + a1

dq(t)

dt+ a0q(t) = b0u(t) (A.1)

and define the state vector

x

=

x1

x2...

xn−1

xn

=

qdqdt...

dn−2qdtn−2

dn−1qdtn−1

. (A.2)

In state space form the differential equation (A.1) can be written as

x1

x2...

xn−1

xn

=

0 1 0 · · · 00 0 1 · · · 0...

. . .

0 0 · · · 0 1−a0 −a1 · · · −an−1 −an

x1

x2...

xn−1

xn

+

00...0b0

u(t).

(A.3)

Equation (A.3) is the state equation

x(t) = Ax(t) +Bu(t),

while the output equation is of the form

y(t) = Cx(t) +Du(t).

Example A.7.1 Consider the differential equation for the damped os-cillator

q(t) = − c

mq(t) − k

mq(t) +

f(t)

m.


Define the state variables x1(t)

= q(t) and x2(t)

= q(t) and assume that q(t)is the output. Then

[x1(t)x2(t)

]

=

[0 1

− km − c

m

] [x1(t)x2(t)

]

+

[01m

]

f(t),

y(t) =[

1 0][x1(t)x2(t)

]

. (A.4)

A.8 Linear Systems with Forcing

Consider the linear state space system

x(t) = Ax(t) +Bu(t), x(0) = x0, (A.1)

y(t) = Cx(t) +Du(t). (A.2)

Then, the solution x(t) of (A.1) is given by

x(t) = eAtx0 +

∫ t

0eA(t−τ)Bu(τ) dτ (A.3)

and thus

y(t) = CeAtx0︸︷︷︸

free response

+

∫ t

0CeA(t−τ)Bu(τ) dτ +Du(t)

︸︷︷︸

forced response

.

Define the impulse response function

H(t)

= CeAtBx0 + δ(t)D.

Then

y(t) = CeAtx0 +

∫ t

0H(t− τ)u(τ) dτ

︸︷︷︸

convolution

.

A.9 Standard Input Signals

Definition A.9.1 δ(t) is an impulse function with the property

δ(t)

=

0, t 6= 0,∞, t = 0,

and∫ b

aδ(t) dt = 1, a ≤ 0 < b. (A.1)

172 APPENDIX A

Note that δ(t) is a right-sided impulse (see Figure A.7). From (A.1) it followsthat [δ(t)] = 1/time. The delta function has the sifting property

∫ b

ag(t)δ(t − t0) dt = g(t0), a ≤ t0 < b. (A.2)

6

0 t

Figure A.6The delta function δ(t)

0 ε

1/ε

t

Figure A.7Approximate right-sided delta function

A force impulse at time t0 has the form

f(t) = f0δ(t− t0), (A.3)

where the dimensions [f0] of f0 are momentum.

An impulse at t = 0 imparts a nonzero initial velocity but zero dis-placement. The forced response with an impulse is thus equivalent to aparticular free response.


6

0 t0 t

Figure A.8The delta function δ(t − t0)

Definition A.9.2 The function 1(t) is a unit step function, where

1(t)

=

0, t < 0,1, t ≥ 0.

0

1

t

Figure A.9The unit step function 1(t)

Definition A.9.3 The function

f(t) = f0t1(t) (A.4)

is a ramp.

Definition A.9.4 The function

f(t) = f0 sin (ωt+ φ) (A.5)

is a sinusoid.

174 APPENDIX A

0 t0 t

Figure A.10The unit step function 1(t − t0)

0 t

Figure A.11The ramp function

A.10 Laplace Transforms

q(s)

= Lq(t)

=

∫∞

0e−stq(t) dt. (A.1)

Note that [q(s)] = time× [q(t)] and [s] = 1/time. From (A.1) it follows that

Lq(t) =

∫∞

0e−stq(t) dt

= e−stq(t)∣∣∞

t=0+ s

∫∞

0e−stq(t) dt

= sq(s) − q(0).

Similarly,

Lq(t) = s2q(s) − sq(0) − q(0).


Note that

Lδ(t) =

∫∞

0e−stδ(t) dt = e0 = 1.

L1(t) =

∫∞

0e−st1(t) dt =

∫∞

0e−st dt = −1

se−st

∣∣∣∣

∞

0

=1

s.

L

∫ t

0q(τ) dτ

=1

sLq(t).

Leat

=1

s+ a.

L sin(ωt) = L

1

2

[eωt − e−ωt

]

=1

2

[1

s− ω− 1

s+ ω

]

=1

2

2ω

s2 + ω2

=ω

s2 + ω2. (A.2)

L cos(ωt) = L

1

ω

d

dtcos(ωt)

=1

ωs

ω

s2 + ω2

=s

s2 + ω2. (A.3)

Note that

L cos(0t) =s

s2 + 02=

1

s= L 1(t) .

J.1 Time delay

The Laplace transform of a time delay is given by

Ly(t− τ) = e−τsy(s). (A.4)

176 APPENDIX A

J.2 s-shift

An s-shift in the Laplace transform is represented in the time domainas

Le−aty(t) = y(s+ a). (A.5)

Hence

L−1y(s+ a) = e−aty(t) (A.6)

and

L−1y(s− a) = eaty(t). (A.7)

Example A.10.1

Le−at sin(ωt) =ω

(s+ a)2 + ω2.

J.3 Time multiplication

Lty(t) = −y′(s). (A.8)

Hence,

L−1y′(s) = −ty(t),

which is the dual of

Ly(t) = sy(s) − y(0). (A.9)

A.11 Solution of ODE’s

Taking the Laplace transform of

q(t) + q(t) = 0

yields

Lq(t) + Lq(t) = 0.

Therefore,

s2q(s) − sq(0) + q(0) + q(s) = 0

and

q(s) =sq(0)

s2 + 1+

q(0)

s2 + 1.


Hence,

q(t) = q(0) cos(t) + q(0) sin(t).

Example A.11.1 Consider the differential equation

q(t) = − c

mq(t) − k

mq(t) +

f(t)

m.

Taking the Laplace transform yields

q(s) =mq(0)s + cq(0) +mq(0)

ms2 + cs+ k︸︷︷︸

free response

+1

ms2 + cs+ k︸︷︷︸

forced response

f(s).

Definition A.11.1 A zero of a rational function is a root of the numer-ator of the rational function. A pole of a rational function is a root of thedenominator of the rational function.

Example A.11.2 Consider the forced response of the damped rigidbody

q(t) + 2q(t) = u(t).

If u(t) = δ(t), then the displacement is given by

q(s) =1

s2 + 2s.

Define the velocity v(t) = q(t). Then

v(t) + 2v(t) = u(t).

Now, if u(t) = 1(t), then the velocity is given by

v(s) =1

s(s+ 2).

Therefore, the displacement impulse response is identical to the velocityresponse.

K.1 Partial Fractions

Example A.11.3 Consider

y(s) =(s+ 2)(s + 4)

s(s+ 1)(s + 3).

Use partial fractions to show

y(t) =8

31(t) − 3

2e−t − 1

6e−3t.

178 APPENDIX A

A.12 Initial Value and Slope Theorems

The following result is the initial value theorem.

Fact A.12.1

y(0+) = lims→∞

sy(s)

The following result is the initial slope theorem.

Corollary A.12.1

y(0+) = lims→∞

s[sy(s) − y(0)]

Example A.12.1 If y(t) = 1(t), then

y(0+) = lims→∞

s1

s= 1.

If y(t) = cos(ωt), then

y(0+) = lims→∞

ss

s2 + ω2= 1.

Example A.12.2 Consider the unit-slope ramp

y(t) = t.

Then

y(0+) = lims→∞

s[sy(s) − y(0)]

= lims→∞

s(s1

s2− 0)

= 1.

A.13 Final Value Theorem

We say that limt→∞ y(t) exists if it is a number. Note that ∞ is nota number.

Case I y(t) remains bounded but limt→∞ y(t) does not exist. For example,y(t) = sin(ωt).

Case II y(t) is not bounded and thus limt→∞ y(t) does not exist. For ex-ample, y(t) = et sin t.

Case III y(t) is not bounded and limt→∞ y(t) does not exist but is infinite.For example, y(t) = et. In this case, limt→∞ y(t) = ∞.


Case IV limt→∞ y(t) exists.

The following result is the final value theorem.

Theorem A.1. Every pole of y(s) is either in the open left half planeor is zero if and only if

limt→∞

y(t) = lims→0+

sy(s). (A.1)

In addition, limt→∞ y(t) exists if and only if s = 0 is not a repeated pole ofy(s).

The cases in Theorem A.1. are legal. For all other cases the final valuetheorem gives the wrong answer.

Example A.13.1 If y(t) = 1(t), then

limt→∞

y(t) = lims→0+

s1

s= 1.

If y(t) = −t1(t), then

limt→∞

y(t) = lims→0+

s1

s2= −∞.

If y(t) = e−t, then

limt→∞

y(t) = lims→0+

s1

s+ 1= 0.

If y(t) = et, then

limt→∞

y(t) = lims→0+

s1

s− 1= −1,

which is incorrect. Note that s → 0+ is needed in the second example toobtain the correct sign of ∞ when there is an odd number of repeated polesat s = 0.

A.14 Laplace Transforms of State Space Models

Consider

x(t) = Ax(t) +Bu(t),

y(t) = Cx(t) +Du(t).

Then

sx(s) − x(0) = Ax(s) +Bx(s),

y(s) = Cx(s) +Du(s).

180 APPENDIX A

Hence,

y(s) = C(sI −A)−1x(0)︸︷︷︸

free response

+

G(s)︷︸︸︷[C(sI −A)−1B +D

]u(s)

︸︷︷︸

forced response

.

G(s) thus has the realization

G(s) ∼[A BC D

]

.

We write

G(s) = C(sI −A)−1B +D

=1

sC

(

I − 1

sA

)−1

B +D

=1

sC

[

I +1

sA+

1

s2A2 + · · ·

]

B +D

= D +1

sCB +

1

s2CAB + · · ·

= H0 +1

sH1 +

1

s2H2 + · · · , (A.1)

where H0, H1, and H2 are Markov parameters. Note that the expansion(A.1) of G(s) converges if |s| >> 1.

G(s) is a ratio of polynomials in s, for example,

G(s) =as3 + · · ·s3 + · · · . (A.2)

Taking the limit of G(s) we obtain

a = lims→∞

G(s) = D. (A.3)

In the case D = 0, G(s) is of the form

G(s) =as2 + · · ·s3 + · · · , (A.4)

and thus

a = lims→∞

sG(s) = CB. (A.5)

Furthermore, if D = 0 and CB = 0, then G(s) is of the form

G(s) =as+ · · ·s3 + · · · , (A.6)


and thus

a = lims→∞

s2G(s) = CAB. (A.7)

Example A.14.1 Consider the differential equation for the dampedoscillator

q(t) = − c

mq(t) − k

mq(t) +

f(t)

m.

Define the state variables x1(t)

= q(t) and x2(t)

= q(t) and assume that thevelocity q(t) is the output. Then

[x1(t)x2(t)

]

=

[0 1

− km − c

m

] [x1(t)x2(t)

]

+

[01m

]

f(t),

y(t) =[

0 1][x1(t)x2(t)

]

,

and

G(s) =[

0 1]([

s 00 s

]

−[

0 1

− km − c

m

])−1 [

01m

]

=s

ms2 + cs+ k.

Every pole of G(s) is an eigenvalue of A. In addition, in most casesevery eigenvalue of A is a pole of G(s). As an exception, consider theDRB with velocity output y(t) = v(t) = q(t). Then the eigenvalues of

A =

[0 10 − c

m

]

are λ1 = 0 and λ2 = − cm . Also,

y(s) = v(s) =s

s2 + cms

=1

s+ cm

. (A.8)

The pole at 0 is canceled by the zero at 0 since the position state is unob-servable by the velocity measurement.

A.15 Pole Locations and Response

182A

PP

EN

DIX

A

Open Left Half Plane (OLHP) Imaginary Axis (IA) Open Right Half Plane (ORHP)

not repeated ce−at sin(ωt)

repeated cte−at sin(ωt)

not repeated sin(ωt)

repeated t sin(ωt)

not repeated ceat sin(ωt)

repeated cteat sin(ωt)

not repeated ce−at

repeated cte−at

not repeated c

repeated ct

not repeated ceat

repeated cteat

not repeated ce−at sin(ωt)

repeated cte−at sin(ωt)

not repeated sin(ωt)

repeated t sin(ωt)

not repeated ceat sin(ωt)

repeated cteat sin(ωt)


Consider the transfer function

G(s) =1

s2 + 2ζωns+ ω2n

. (A.1)

If ζ ≥ 1, then the roots of the characteristic equation are

λ1, λ2 = −ζωn ± ωn

√

ζ2 − 1.

If ζ > 1, then the roots are real and distinct, which is the overdamped case.If ζ is large, then λ1 ≈ −ωn/2ζ and λ2 ≈ −2ζωn. Thus λ2 is the faster poleand λ1 is the slower pole. For ζ = 1 the roots are repeated and given byλ = −ζωn, which is the critically damped case. For 0 < ζ < 1 the roots arecomplex, that is,

λ1, λ2 = −ζωn ± ωn

√

1 − ζ2,

which is the underdamped case. Defining the damped natural frequency

ωd = ωn

√

1 − ζ2, (A.2)

we have

λ1, λ2 = −ζωn ± ωd. (A.3)

Finally, if ζ = 0, then the roots are imaginary, which is the undamped case.

I

×

−ζωn

ωn

√

1 − ζ2

θ

ωn

Figure A.12Relationship between pole location, damping, and natural frequency

Notice that in the undamped, underdamped, and critically dampedcases we have

|λ| = ωn

and

ζ = sin θ.

Therefore, the distance from the pole to the origin determines the naturalfrequency, while the angle it subtends at the origin from the imaginary axis

184 APPENDIX A

Property Name Free Response

ζ = 0 undamped sinusoid

0 < ζ < 1 underdamped decaying sinusoid

ζ = 1 critically damped decaying exponentials

ζ > 1 overdamped decaying exponentials

Table A.2 Damping and free response.

determines the damping. Furthermore, the impulse response of the system(A.1) is given by

y(t) = e−ζωnt cos(ωn

√

1 − ζ2t+ φ).

Therefore the real part of the root determines the rate of decay, and theimaginary part determines the frequency of oscillation.

We summarize the various free responses in Table A.2.

A.16 Stability

Stability concerns the free response only. However, stability can alsobe determined from G(s).

Unstable

• x(t) → ∞ as t→ ∞ for at least one x(0).

• At least one entry of eAt is unbounded.

• Not Lyapunov stable.

• At least one pole of G(s) is in the ORHP or is repeated on theimaginary axis.

• Example: Undamped rigid body (URB).

Lyapunov Stability (LS)

• For all x(0), x(t) is bounded.

• Every entry of eAt is bounded.

• Each eigenvalue of A is in the OLHP or is not repeated on theimaginary axis.


• Each pole of G(s) is in the OLHP or is not repeated on theimaginary axis.

• Example: UO.

Semistability (SS)

• For all x(0), limt→∞ x(t) exists.

• Each eigenvalue of A is in the OLHP or is not repeated at theorigin.

• limt→∞ eAt exists.

• Each pole of G(s) is in the OLHP or is not repeated at the origin.

• Example: DRB.

Asymptotic Stability (AS)

• For all x(0), x(t) → 0 as t→ ∞.

• Each eigenvalue of A is in the OLHP.

• limt→∞ eAt = 0.

• Each pole of G(s) is in the OLHP.

• Example: DO.

Fact A.16.1 AS =⇒ SS =⇒ LS

AS

SSLSUS

Figure A.13Stability Venn Diagram

Definition A.16.1 The transfer function G(s) is bounded-input,bounded-output (BIBO) stable if, for every bounded input signal u(t), theoutput y(t) of G(s) is bounded.

186 APPENDIX A

Example A.16.1 Consider the undamped rigid body

v(s) =1

msf(s)

with velocity output and force step input f(t) = f01(t). Then v(t) is un-bounded. Mathematically, v(t) is the integral of a step function. Physically,the velocity under nonzero constant forcing increases without bound. Hence

G(s) =1

ms

is not BIBO stable.

Example A.16.2 Consider the undamped oscillator with position out-put and sinusoidal force input f(t) = sinωnt. The output y(t) has repeatedpoles on the imaginary axis. The corresponding time-domain response,which is an oscillation with a linear envelope, is known as resonance. Hence

G(s) =1

s2 + ω2n

(A.1)

is not BIBO stable.

Fact A.16.2 G(s) is BIBO stable if and only if G(s) is asymptoticallystable.

A.17 Routh Stability Criterion

Fact A.17.1 Suppose that all of the roots of the polynomial

p(s) = sn + an−1sn−1 + · · · + a1s+ a0

are in the open left half plane. Then a0, . . . , an−1 are positive.

Fact A.17.2 Suppose that n = 2. Then both roots of

p(s) = s2 + a1s+ a0

are in the open left half plane if and only if a0 and a1 are positive.

Fact A.17.3 Suppose n = 3. Then all three roots of

p(s) = s3 + a2s2 + a1s+ a0

are in the open left half plane if and only if a0, a1, a2 are positive and

a0 < a1a2. (A.1)

Fact A.17.4 Suppose n = 4. Then all four roots of

p(s) = s4 + a3s3 + a2s

2 + a1s+ a0


are in the open left half plane if and only if a0, a1, a2, a3 are positive and

a0a23 + a2

1 < a1a2a3. (A.2)

A.18 Problems

Problem A.18.1 Consider the 3 × 3 matrix

A =

3 0 24 3 57 2 6

.

Manually compute the determinant and inverse of A. Then compute thesequantities using Matlab to check your answer.

Problem A.18.2 Use Matlab to compute the eigenvalues of the matrixA in Problem A.18.1. Then use Matlab to show that the determinant is theproduct of the eigenvalues, and show that the trace (the sum of the diagonalentries) is the sum of the eigenvalues. Furthermore, use Matlab to computethe eigenvalues of A2 and A−1, and discuss how they are related to theeigenvalues of A.

Problem A.18.3 Using the “rand” command in Matlab, form a ran-dom 4 × 2 matrix A. Then compute the 4 × 4 matrix AAT. Using Matlabto compute the eigenvalues of AAT, check whether this matrix is positivesemidefinite. Then, show mathematically (by hand, not using Matlab) thatxTAATx ≥ 0 for all vectors x. (Hint: Define z = ATx.)

Problem A.18.4 Let A be an n×n matrix and let p(s) = det(sI −A)be the characteristic polynomial of A. The Cayley-Hamilton theorem statesthat p(A) = 0. Check this fact by obtaining the characteristic polynomialp(s) for the matrix

A =

[0 1

−a0 −a1

]

,

and then showing that p(A) = 0. Repeat these steps for

A =

0 1 00 0 1

−a0 −a1 −a2

.

Do all of these symbolic calculations by hand.

Problem A.18.5 The eigenvalues of a matrix are the roots of its char-acteristic polynomial. Consider the 3 × 3 matrix in Problem A.18.4 witha0 = −2, a1 = 5, and a2 = 3, and use Matlab to show numerically that theroots of p(s) are indeed the eigenvalues of A. Use roots(p) and eig(A) for

188 APPENDIX A

your computations.

Problem A.18.6 Show that the matrix

A =

[cos θ sin θ− sin θ cos θ

]

is orthogonal. Also, check whether the matrix

B =

[cos θ sin θsin θ − cos θ

]

is orthogonal. Determine by hand (not by Matlab) the determinants andeigenvalues of these matrices. Finally, choose several values of θ and applythese matrices to the vector [1 1]T. Discuss how the resulting vectorscompare to the original vector in terms of their length and direction. Youcan use a calculator but do not use Matlab.

Problem A.18.7 Use the dot product to compute the angle betweenthe vectors [3 2]T and [−2 4]T.

Problem A.18.8 Solve the differential equation

x(t) = ax(t) + b

analytically by evaluating the convolution integral expression given by (A.3)in Section A.8. Under what conditions does limt→∞ x(t) exist? When thelimit exists, determine the limiting value. How could you guess the limitingvalue without solving the equation?

Problem A.18.9 Using the solution to Problem A.18.8, write downthe solution to the scalar ordinary differential equation

x(t) = −2x(t) + 8.

This equation represents the step response of a linear system, where theconstant 8 is the value of the step input. Plot the solutions for the ini-tial conditions x(0) = 5 and x(0) = −4 in the same figure. Be sure tolabel all axes of your figure and give it an appropriate caption. Determinelimt→∞ x(t) from the plot and compare that numerical value with the ana-lytical limit. Explain how the limiting value of x(t) depends on the constantsin the problem, namely, the coefficient of x(t), the step value, and the initialcondition.

Problem A.18.10 Consider x = Ax, and let λ = σ+ ω be a complexvalue of A with associated complex eigenvector

v =

[31

]

+

[14

]

.

Let ω = 1 and σ = −1, 0, 1 (three different values), and plot the eigensolution


x(t) = [x1(t) x2(t)]T = Re(eλtv) in the x1, x2 plane for each value of σ.

Explain how the properties of the eigensolution depend on σ.

Problem A.18.11 Let λ, λ = −ζωn ± ωn

√

1 − ζ2 denote a complexconjugate pair of underdamped eigenvalues. Show that

ωn = |λ|, (A.1)

ζ = −λ+ λ

ωn, (A.2)

and

ωd =λ− λ

2. (A.3)

Problem A.18.12 Let λ1, λ2 = −ζωn ± ωn

√

ζ2 − 1 denote a pair ofoverdamped eigenvalues. Show that

ωn =√

λ1λ2 (A.4)

and

ζ = − 1

ωn. (A.5)

Problem A.18.13 Consider the damped oscillator (DO) with acceler-ation output. Write this system in A,B,C,D form.

Problem A.18.14 Use the initial value theorem to determine the initialvalue of each of the following functions: y(t) = t, y(t) = t+ 1, y(t) = sin 2t,and y(t) = cos 2t.

Problem A.18.15 Use the initial value theorem to determine the initialslopes of the functions in the previous problem.

Problem A.18.16 Use Laplace transforms to analytically determinethe total response of v + 2v = sin 5t for an arbitrary initial conditionv(0). Then show that you can choose a special initial condition v(0) sothat the total response is exactly sinusoidal, that is, there is no transient(non-sinusoidal) component of the solution. Finally, confirm your answer byusing ODE45 to simulate the system with this special initial condition aswell as another initial condition.

Problem A.18.17 A motor with constant applied torque is modeledas the damped rigid body Jθ + cθ = τ01(t), where J is the load inertia, cis the viscous damping coefficient, and τ0 is the moment. The initial angleθ(0) and initial angular rate θ(0) are zero. Use Laplace transforms and thefinal value theorem to determine the terminal angular rate limt→∞ θ(t). Also

190 APPENDIX A

compute the same limit by using the time-domain solution obtained fromLaplace transforms.

Problem A.18.18 For

y(s) =1

s(s2 + s+ 1),

use partial fractions to show that

y(t) = 1(t) − e−1

2t cos

(√3

2t

)

− 1√3e−

1

2t sin

(√3

2t

)

.

Problem A.18.19 Consider an object falling under the force of gravity.Ignore drag and model the motion as an undamped rigid body. Use Laplacetransforms to express the displacement q(t) as a function of t, q(0), q(0),and g.

Problem A.18.20 A body with mass M is falling under the force ofgravity. Atmospheric drag is modeled by a dashpot coefficient D, so thatthe system is modeled as a damped rigid body. Assuming initial velocityv0, use Laplace transforms to find the velocity v(t) for t > 0. Then use thefinal value theorem to compute the terminal velocity limt→∞ v(t).

Problem A.18.21 Consider the damped rigid body with position out-put, and write it in state space form x = Ax. Then determine eAt by us-ing the fact that the Laplace transform of the matrix exponential eAt is(sI − A)−1. Next, determine limt→∞ eAt by using the expression for eAt aswell as by applying the final value theorem to each entry of (sI −A)−1. Usethis result to determine the free response of the damped rigid body withinitial displacement q(0), initial velocity q(0), and output y(t) given by themass position. What kind of matrix is limt→∞ eAt?

Problem A.18.22 For each of the transfer functions G(s) below withinput u and output y, determine whether the use of the final value theoremis legal, and, if so, use it to determine the limit of the output y(t) as t→ ∞.Explain why or why not the use of the final value theorem is legal in eachcase.

5a) G(s) =−5

s(s+ 7)2, u(t) = 3e−2t.

5b) G(s) =5

s− 3, u(t) = 71(t) − 3δ(t − 4.2).

5c) G(s) =−4

s2 + 2.4s + .3, u(t) = 6t2.


Problem A.18.23 Consider the damped rigid body with ramp forceinput u(t) = t and velocity output. Use Laplace transforms and partialfractions to determine the forced response for t ≥ 0.

Problem A.18.24 An engineer has shown that the output response ofa new airframe developed for a UAV application is given by

y(t) = 2e−6t.

Use the initial value to determine y(t). (Note the two dots.) Check youranswer by computing y(t) directly. Note: You must use the initial valuetheorem. (Hint: I showed in class that

Ly(t) = s2y(s) − sy(0) − y(0).)

Problem A.18.25 For both systems described below (described by atransfer function and input) and without using a calculator or computer,determine whether the use of the final value theorem is legal, and, if so, usethe final value theorem to determine the limit of the output y(t) as t→ ∞.Explain why or why not the use of the final value theorem is legal in eachcase. If the limit exists, use Matlab to plot the response and confirm youranswer. (Hint: For plotting consider the Matlab functions IMPULSE andLSIM.)

a) G(s) =s2 − 1

s(3s3 + 2s2 + 4s + 5), u(t) = 2δ(t).

b) G(s) =1

s(s2 + 3s + 1), u(t) = −2e−t sin 2t.

Problem A.18.26 For the systems described below and without usinga calculator or computer, determine whether the use of the final value the-orem is legal, and, if so, use the final value theorem to determine the limitof the output y(t) as t → ∞. Explain why or why not the use of the finalvalue theorem is legal in each case. If the limit exists, use Matlab to plot

192 APPENDIX A

the response in order to confirm your answer.

a) The free response of the damped rigid body with initial displace-ment q(0), initial velocity q(0), and output y(t) given by the massposition. Obtain the limit symbolically. Then use Matlab to sim-ulate the system with the numerical values m = 3 kg, c = 4 kg/s,q(0) = 1 m, and q(0) = −2 m/s.

b) The forced response of a system whose transfer function is

G(s) =4s2 − 12s− 16

2s5 + 2s4 + 4s3 + 2s2 + s

with the impulsive input u(t) = 3δ(t− 1).

Problem A.18.27 Flight testing of a new aircraft reveals that it hasa pair of underdamped complex conjugate poles. Testing reveals that thetime to 50% decay is T seconds, while analytical modeling shows that theimaginary parts of the poles are ±ωd, where ωd > 0. Derive an expressionfor the damping ratio ζ in terms of T and ωd. Show that your expression forζ satisfies 0 < ζ < 1.

Problem A.18.28 Flight testing of a new aircraft reveals that it be-haves like an overdamped oscillator. Analytical models are used to deter-mine the value of ωn. Measurements show that the time to 50% decay is Tseconds. Derive an expression for the damping ratio ζ in terms of T and ωn.Finally, show that your expression for ζ satisfies ζ > 1.

Problem A.18.29 Without taking inverse Laplace transforms, sketchthe step response of the transfer function

G(s) =4s− 3

s2 + .8s + 4.

Be sure to qualitatively capture the direction of the step response for smallpositive time t as well as the settling behavior for large time t. What featuresdoes your sketch illustrate? Then use Matlab to plot the step response.

Problem A.18.30 Consider the eigenvalues of the matrix

A =

[0 1

−k/m −c/m

]

for the undamped (c = 0) and damped (c > 0) oscillators. Let k = 2.5 kg/s2

and m = 7 m. Plot the locations of the eigenvalues as ×’s in the complexplane for a range of values of c. Choose a range that includes undamped,


underdamped, critically damped, and overdamped cases. For each value ofc, plot a “×” in the complex plane.

Problem A.18.31 Show analytically that the poles of the undamped,underdamped, and critically damped oscillators satisfy |λ1| = |λ2| = ωn.Furthermore, show that the poles of the overdamped oscillator satisfy λ1λ2 =ω2

n. Explain the meaning of these results in terms of the plot you made inthe previous problem.

Problem A.18.32 Consider the second-order state space (A,B,C,D)system with

A =

[0 1−2 −3

]

, B =

[01

]

, C =[

1 1], D = 0.

Determine the transfer function G(s) for this system. Show that G(s) isactually a first-order transfer function due to pole-zero cancelation. Finally,determine a first-order state space (A, B, C, D) system whose transfer func-tion is the same G. (Remark: Given a transfer function G we are usuallyinterested in a realization (A,B,C,D) of G of the lowest possible order, thatis, where the size of A is as small as possible.)

Appendix B

Frequency Response

B.1 Sinusoidal Gain and Phase Shift

Assume all sinusoids below have the same frequency. The sinusoids inFig. B.1 have different amplitudes but the same phase. In Fig. B.2, thesinusoids have the same amplitude but different phases.

B.2 Phase Angle as a Delay or Advance

Consider

f(t) = sin(ωt)

and

g(t) = sin(ωt+ φ),

where φ ∈ [−π, π] is the phase shift. To solve for the time shift, we solvefor ∆t from g(t) = 0. We obtain (see Fig. B.3)

sin(ω∆t+ φ) = 0

or,

∆t = −φω.

We thus have

g(t) = sin(ωt − ω∆t)

= sin(ω(t− ∆t)).

If φ > 0, then ∆t < 0, and thus g(t) is advanced relative to f(t). On theother hand, if φ < 0, then ∆t > 0, and thus g(t) is delayed relative to f(t).

Now consider

f(t) = α sin(ωt+ φ) (B.1)

196 APPENDIX B

0 5 10 15 20−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

x

y

Figure B.1Sinusoids with same frequency but different amplitudes.

0 5 10 15 20−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

x

y

Figure B.2Sinusoids with same amplitude but different phases.

FREQUENCY RESPONSE 197

−4 −2 0 2 4 6 8−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

t

g

sin φ

∆ =− φ/ωt

Figure B.3Phase shift as a time delay.

and

g(t) = α sin(ωt + φ)

as shown in Fig. B.4. Now suppose that 0 < φ− φ ≤ 180. In this case, wesay that g(t) leads f(t) or f(t) lags g(t). Note that leading by more than180 is equivalent to lagging by less than 180.

0 5 10 15 20−1.5

−1

−0.5

0

0.5

1

1.5

x

y

g(t) f(t)

Figure B.4Sinusoids g(t) and f(t), where g(t) leads f(t).

We can express f(t) as the phasor

f(t) = Im αe(ωt+φ), (B.2)

which is shown in Fig. B.5. Leading and lagging are represented in Fig.B.6.

198 APPENDIX B

(ω + φ)

f(t)

Re

Im

ω + φ

eα

t

j t

Figure B.5Phasor representation.

B.3 Frequency Response Law for Linear Systems

Suppose that a sinusoidal input (such as a forcing) is applied to anasymptotically stable linear system. Then the output (response) of the sys-tem approaches sinusoidal motion whose frequency is the same as the inputfrequency. The limiting sinusoidal motion is called the harmonic steady stateresponse. Note, however, that the response does not have a “limit” in theusual mathematical sense since it does not approach a constant value.

The transient behavior of the system before its response reaches har-monic steady state depends on the poles and zeros of the system as well asthe initial conditions of the internal states.

The ratio of the amplitude of the harmonic steady state response to theamplitude of the sinusoidal input is equal to the magnitude of the transferfunction evaluated at the input frequency ω, that is, |G(ω)|, while the phaseshift of the harmonic steady-state response relative to the phase of the inputis given by the phase angle of the transfer function at the input frequency,that is, atan2(Im G(ω),Re G(ω)). Note that the harmonic steady-stateresponse leads the input if 0 < ∠G(ω) < 180, whereas the harmonic steady-state response lags the output if −180 < ∠G(ω) < 0

We thus have

u(t) = u0 sin(ωt + φ) (B.1)


lags

Re

Im

φ

φ^

leads

Figure B.6Phasor representation of leading and lagging sinusoids.

and

yhss(t) = |g(ω)|u0 sin(ωt+ φ+ ∠G(ω)).

B.4 Frequency Response Plots for Linear SystemsAnalysis

Assume that G(s) is asymptotically stable. As seen previously, if theinput u(t) is sinusoidal, that is,

u(t) = U cos(ωt), (B.1)

where U is the amplitude and ω is the frequency of the sinusoidal input,then the harmonic steady-state output is

yhss(t) = M(ω)U(ω) cos(ωt + φ(ω)), (B.2)

where M(ω) is the gain and φ is a phase angle, which satisfy

M(ω)eφ(ω) = G(ω). (B.3)

That is,

M(ω) = |G(ω)| (B.4)

and

φ(ω) = ∠G(ω). (B.5)

Plots of the gain M versus input frequency ω and of the phase angle

200 APPENDIX B

φ versus input frequency ω are the Bode frequency response plots. Theplot of gain versus frequency is the magnitude plot, while the plot of phaseangle versus frequency is the phase angle plot. It is convenient to expressthese plots using base 10 logarithmic scales for the gain and frequency. Inparticular, the gain is often scaled 20log10M , referred to as decibels (dB).

The Matlab command for plotting the Bode frequency response isBode(A,B,C,D), where A, B, C, D define the state equations for the lin-ear system with the desired input variable and the desired output variable.This function can also be used in the form Bode(n,d), where the vectors nand d represent the numerator and denominator polynomials of the transferfunction for the linear system with specified input and output variables.

The Bode frequency response plot can be constructed for all trans-fer functions G(s) whether or not G(s) is asymptotically stable, semistable,Lyapunov stable, or unstable. However, the gain and phase shift character-istics of G(s) have a harmonic steady-state interpretation only in the casethat G(s) is asymptotically stable.

Properties of the Bode frequency response plots include:

• The gain |G(ω)| is finite if and only if ω is not a pole of G.

• The DC gain is the gain at zero frequency; the DC phase angle iseither zero degrees or −180 degrees. In the Bode plot the DC gain isevident from the low frequency part of the magnitude plot.

• For a linear system with more poles than zeros, that is, n > m, thegain goes to zero as the frequency tends to infinity; in the Bode plot,the magnitude plot tend to negative infinity dB as the frequency tendsto infinity.

• Each local maximum corresponds to a pair of complex eigenvalues withdamping ratio less than

√2/2.

• The total change in the phase angle from low frequency to high fre-quency is given by (n−m)90, where n−m is the excess of poles overzeros.

• The frequency ωmco at which the gain is unity (or, equivalently, zerodB) is the magnitude crossover frequency. The phase margin is definedas

PM = 180 −G(ωmco). (B.6)

• The frequency ωpco at which the phase angle is 180 degrees is the phasecrossover frequency. The gain margin is defined as

GM = −20 log |G(ωpco)|. (B.7)


B.5 Circuits and Filters

A lowpass filter reduces noise at high frequency. From the impedancelaw, we have

V = IZ = ZI

or

I =V

Z.

Here Z = Z(s) is a transfer function.

Example B.5.1 Consider

Vin − Vout = RI

and

Vout =I

Cs=Vin − Vout

RCs. (B.1)

Solving for Vout in (B.1, we obtain(

1 +1

RCs

)

Vout =1

RCsVin

or

Vout =1

RCs+ 1Vin =

1RC

s+ 1RC

Vin.

Now, the current I is

I =Cs

RCs+ 1Vin.

Hence, at high frequency,

I =Cω

RCω + 1Vin(ω) ≈ 1

RVin(ω).

B.6 Bode Plot

Consider

G(s) =1

ms.

With s = ω, G(s) becomes

G(ω) =1

mω=

−mω

.

202 APPENDIX B

Hence

|G(ω)| =1

mω.

We plot 20log |G(ω)| versus log ω. We write

y = 20log |G(ω)|

= 20log

∣∣∣∣

1

mω

∣∣∣∣

= −20log |mω|= −20log |m| − 20log |ω|= −20x+ b, (B.1)

where x = log |m| and b = −20log |ω|.

B.7 Magnitude Crossover Frequency

Suppose |G(ωmco)| = 1 sec/kg. We use ωmco for magnitude crossover.Then

1

mωmco= 1

sec

kg−rad, (B.1)

ωmco =1

m

rad

sec. (B.2)

B.8 Phase of G(ω)

Consider

G(ω) =1

mω=

−mω

= − 1

mω. (B.1)

The magnitude of G(ω) is

|G(ω)| =1

mω(B.2)

with phase angle

∠G(ω) = −90. (B.3)

Figure B.7 represents the phase of G(ω) on the unit circle while Fig. B.8represented G(ω) on the Bode phase angle plot.


o

Im G(j )

Re G(j )ω

ω

270

mω1( )G(j ) = −j ω

−90

o

Figure B.7Phase angle representation on the unit circle.

−90

ω

ωG(j )

log

o

Figure B.8Phase angle representation on the Bode phase plot.

B.9 Poles at s = p

Consider

G(s) =α

s− p. (B.1)

Substituting s = ω, we obtain

G(ω) =α

ω − p. (B.2)

Note that if p < 0, then G is asymptotically stable, whereas, if p > 0, thenG is unstable. The magnitude of G(ω) is

|G(ω)| =|α|

√

ω2 + p2. (B.3)

204 APPENDIX B

For ω << p,

|G(ω)| =|α||p| (B.4)

while, for ω >> p,

|G(ω)| =|α||ω| . (B.5)

Now, for ω = |p|, the magnitude of G(ω) becomes

|G(|p|)| =|α|√

2p2=

√2

2

|α||p| . (B.6)

Hence, ∆dB is found from

∆dB = 20 log |G(|p|)| − 20 log |G(0)|

= 20 log

(√2

2

|α||p|

)

− 20 log

( |α||p|

)

= 20 log

(√2

2

)

= 20 log 2−1/2 = −10 log 2 = −3 dB. (B.7)

B.10 Phase Angle

Consider

z = a+ b (B.1)

as represented on Fig. B.9. To find the phase angle θ, we can write

tan θ =Im z

Re z. (B.2)

Consequently,

∠G(ω) = atan2(Im G(ω),Re G(ω)). (B.3)

The function atan2(a,b) is discussed in Appendix C.

B.11 Damped Oscillator

Consider

q(s) = G(s)f (s), (B.1)


z = a + jb

Im

Re

z = θ

Figure B.9Phase angle representation in the complex plane.

where

G(s) =1

ms2 + cs+ k

=1/m

s2 + 2ζωns+ ω2n

. (B.2)

Substituting s = ω, yields

G(ω) =1/m

−ω2 + 2ζωn(ω) + ω2n

=1/m

ω2n − ω2 + (2ζωnω)

=1/m[ω2

n − ω2 − (2ζωnω)]

(ω2n − ω2)2 + (2ζωnω)2

. (B.3)

B.12 Problems

Problem B.12.1 Consider the transfer function

G(s) =1

s2 + 2ζωns+ ω2n

with ωn = 1 rad/sec. Use Matlab to plot the magnitude and phase Bodeplots of this function for ω from .01 to 100 for ζ = .1, .3, .5, .7, .9. Put allplots in the same figure.

Problem B.12.2 Consider the transfer function

G(s) =1

s2 + 2ζωns+ ω2n

.

206 APPENDIX B

Assume that the system is underdamped, that is, 0 < ζ < 1. Use calculusto determine the resonance frequency ωr at which the magnitude |G(ω)|is maximized. Check whether your answer agrees with the figure from theprevious problem (B.12.1). In addition, determine the range of values of ζfor which the magnitude of the transfer function is never greater than itsDC, that is, ω = 0, value.

Problem B.12.3 Sketch Bode magnitude and phase plots AND theNyquist plot by hand for each of the following transfer functions. You canuse Matlab to print an “empty” log-log grid for your sketch. Be sure that therange of ω is large enough to include all important features of your plots.Explain how each plot was constructed. Check your sketches by plottingwith the Matlab Bode and Nyquist functions. Question: What is strangeabout the Nyquist plots that Matlab draws?

i) G(s) = s+2s+10 (lead).

ii) G(s) = s+10s+2 (lag).

iii) G(s) = s−2s+10 .

iv) G(s) = s−2s+2 (allpass).

v) G(s) = ss+2 (washout).

Problem B.12.4 For each of the following transfer functions, sketchthe Bode magnitude and phase plots by hand. You can use Matlab to printan “empty” log-log grid for your sketch. Be sure that the range of ω is largeenough to include all important features of your plots. Explain how eachplot was constructed. Check your sketches by plotting with Matlab.

i) G(s) = s2

(s+1)2(s+10)2 (rooftop).

ii) G(s) = (s+1)(s+100)s(s+5)(s+10) .

Problem B.12.5 Suppose that a sensor with first-order dynamics hasa time constant of .031 sec and a DC (that is, zero-frequency) magnitudeof 9.2. (Hint: G(s) = α

Ts+1 .) Now suppose that the input to the sensor is4.6 volts corrupted by (which means “added to”) 60 Hz electrical noise withamplitude 3.3 mV. Describe the sensor output after a large amount of timepasses by giving the amplitude and phase of all of its harmonic components.

Problem B.12.6 Suppose that a voltage amplifier with first-order dy-namics G(s) = α

Ts+1 has a time constant of T = 0.037 sec and DC gainα = 43.1. A sinusoidal input signal with amplitude of 2.8 volts yields, after


an initial transient, a sinusoidal response with amplitude 76.2 volts. Whatwere the frequencies of the input sinusoid and the output sinusoid?

Problem B.12.7 A lag filter has a pole at -2, a zero at -6, and a DCgain of 10. At the frequency 4 rad/sec, what is the magnitude of the filter indB and what is the phase angle of the filter in degrees? (Use just a calculatorfor this problem.)

Appendix C

MATLAB Operations

C.1 atan2

The MATLAB command atan2(y, x) computes the angle θ ∈ [−π, π]of the complex number x+ y.

Example C.1.1 For x = 1 and y = 1,

θ = atan2(y, x) = atan2(1, 1) =π

4.

For x = −1 and y = −1,

θ = atan2(−1,−1) =5π

4

. However, note that, in both cases, tan θ = 1.

C.2 expm

The MATLAB command expm(A) computes the exponential eA of thematrix A.

C.3 rlocus

The MATLAB command rlocus(n′, d′) if used for plotting the rootlocus of a transfer function where n′ and d′ are vectors that represent thepolynomials that define the numerator and the denominator of a character-istic polynomial. Specifically,

n′(s) = (s− z′1) · · · (s− z′m) (C.1)

is a polynomial of degree m and

d′(s) = (s − p′1) · · · (s− p′n) (C.2)

is a polynomial of degree n, with m ≤ n.

210 APPENDIX C

C.4 bode

The Matlab command for plotting the Bode frequency response isbode(A,B,C,D), where A, B, C, D define the state equations for the linearsystem with the desired input variable and the desired output variable. Thefunction can also be used in the form bode(n′, d′) where n′ and d′ are definedas in Section C.3.

Appendix D

Dimensions and Units

D.1 Mass and Force

For pound force,

1 lb = 1 slug-ft/sec2 = 4.4 N, (D.1)

where N denotes Newton. On the surface of the Earth, 1 kg weighs 9.8 N,while 1 slug weighs 32.2 lb. Conversion between slugs and kilograms is

1 slug = 14.59 kg. (D.2)

D.2 Force, Impulse, and Momentum

The impulse δ(t) has units

[δ(t)] = 1/sec. (D.1)

For the force impulse f(t) = f0δt, the coefficient f0 has the units

[f0] = [f(t)]/[δ(t)] = kg-m/sec (D.2)

which are the units of momentum. Specific impulse is given by f0/m whichhas the units

[f0/m] = m/sec, (D.3)

that is velocity.

Bibliography

[1] D. T. Greenwood, Advanced Dynamics, Cambridge University Press,Cambridge, 2003.

[2] P. C. Hughes, Spacecraft Attitude Dynamics, Wiley, New York, 1986.

[3] T. R. Yechout, Introduction to Aircraft Flight Mechanics, AIAA Edu-cation Series, VA, 2003.

[4] J. Roskam, Airplane Flight Dynamics and Automatic Flight Controls,DARCorporation, Lawrence, Kansas, 2001.

[5] R. C. Nelson, Flight Stability and Automatic Control, McGraw-Hill, 2ndEdition, 1997.

[6] R. F. Stengel, Flight Dynamics, Princeton University Press, 2004

[7] B. L. Stevens and F. L. Lewis, Aircraft Control and Simulations, Wiley-Interscience, 2nd Ed, 2003.

[8] D. G. Hull, Fundamentals of Airplane Flight Mechanics, Springer, 2007.

AIRCRAFT FLIGHT DYNAMICS AND CONTROL - Ira A....

Documents

Transcript of AIRCRAFT FLIGHT DYNAMICS AND CONTROL - Ira A....