Summary Dynamics & Control

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Dynamics & Control of Mechanical Systems [4DB00] Summary

Wouter Kuijpers

February 13, 2014

This document gives an overview of the theory of Dynamics & Control of Mechanical Systems[4DB00].

Contents

I Control Systems 2

1 Laplace Transformation 21.1 System Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 The Laplace Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3 Signals in the Laplace domain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.4 Frequently Used Systems’ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2 Block Diagrams 7

3 Bode Diagrams 8

4 Controller Design 104.1 Complex Plane: Poles and Zeroes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104.2 Stability Criteria of Routh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114.3 Stability Criteria of Nyquist . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114.4 Margins . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124.5 Loop Shaping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134.6 Design for Performance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144.7 Limits to Performance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

II Lagrangian Mechanics 15

III Appendix 20

5 Detailed Lagrangian Mechanics 20

1



Part I

Control Systems

1 Laplace Transformation

1.1 System Example

Suppose we monitor a Water Tank’s height h (See Figure 1), when it is filled by ϕin and waterleaves through ϕout. The flow of water leaving the Water Tank is restricted by R; the tank has asurface of Ah and a volume V . This system can be described using a differential equation, for thiswe monitor the content of buffers:

∆V = Ahh = ϕin − ϕout

Figure 1: A Water Tank.

Furthermore with knowledge of Physics we can rewrite ϕout to:

Ah

˙h = ϕin −

1

R h(t)

h = 1

Ah

ϕin − 1

Rh(t)

h = 1

Ahϕin − 1

AhRh(t)

We now have captured the behavior of the Water Tank in aDifferential Equation. But what can we do with it? We can seehow the system reacts to several different inputs ϕin, or how thesystem reacts differently when a system parameter is changed. Buthow do we solve this Differential Equation; there exist two ways

to do this:

Numerical: Calculating the content of the buffer at verysmall intervals of time. This can be easily done using MATLABSimulink; see Figure 2 where the differential equation above is implemented 1:

Figure 2: A MATLAB Simulink simuation in which the differential equation is implemented.

1hdot = h

2



Analytically: One example of a Analytic approach is solving the Differential Equation. No-tice that we can rewrite the Differential Equation to:

h = ah(t) + bϕin(t) a = − 1

AhR, b =

1

Ah

To find the Homogeneous Solution of this differential equation we set the ϕin(t) = 0, h(0) to h0.This means that we observe the behavior when no input is applied; Homogeneous Solution. ThisHomogeneous Solution or Natural Response describes how the system dissipates or acquiresenergy. An example of a Natural Response is a systems’ response under a starting position notequal to 0. The form or nature of this response is dependent only on the system, not on the input.So in our search for the Homogeneous Solution we have the following equation:

h(t) = ah(t)

Which has the following solution:

h(t) = h0eλt

→ h(t) = λh0eλt so a = λ =

− 1

AhR

The solution of this differential equation states that the h will have a e-power behavior, which isreferred to as First-Order Behavior. The larger λ the faster the system responds, the faster itis in Steady-State. λ = a = 1

AhR; which states that the system becomes faster when R and Ah

become smaller.

With the expression above we can also make a statement about the Stability of a system. Wehave derived the following formula for h:

h(t) = λh0eλt

This is the Homogeneous Solution or Natural Response of the system. For this a system tobe stable the Natural Response eventually has to approach zero. This means that at some pointh(t) = 0. This states that eλt should approach a constant value at t →∞, this implies that λ < 0.A system is Stable when λ < 0.

1.2 The Laplace Transform

In the previous subsection we have looked at a system and we have observed a Numerical Methodto find the solution. Also we have derived a Analytic Solution this included solving a differentialequation. Solving more difficult differential equations is hard and time-consuming.

Therefore we introduce the Laplace Transform, which is a mathematical transformation whichallows function represented in the time domain to be represented with the Laplace operator. The

Laplace Transform of a function represented in the time domain x(t) is given by:

L{x(t)} = X (s) =

∞0

x(t)e−stdt

But why would we use another operator to make life even more difficult? Now we enter thederivative of the function above:

L{x(t)} =

∞0

dy

dx(t)e−stdt

3



After integration by parts we get the following expression:

L{x(t)} = sX (s)− x(0)

The expression on the right is in the Laplace domain, this shows that differentiating a function inthe time domain results in a multiplication in the Laplace domain.

In the previous example we have already transferred a differential equation from the time domainto the Laplace domain. Suppose we have a Mass-Spring-Damper system suspended from a roof.The input to this system is the roof. This results in the following differential equation:

my = c(u− y) + k(u− y)

my + cy + ky = cu + ku

Now we take the Laplace transform of both sides of the equation:

L{my + cy + ky} = L{cu + ku}

As the Laplace transform is a Linear Operation we can also write:

L{my}+ L{cy}+ L{ky} = L{cu}+ L{ku}

Y (s)

ms2 + cs + k

= U (s) (cs + k)

The equation above describes the relation between the input of the system U (s) and the output:Y (s). The Transfer Function, H (s), also describes this relation in a way that it can be used inBlock Diagrams:

H (s) = Y (s)

U (s) =

cs + k

ms2 + cs + k

Note that H (s) is complex number, as s is a complex number. This complex number has a Modulus

and a Phase this means that the system may affect the Amplitude (Modulus) and/or the Phaseof the input signal; which results in the output signal.

General Definition Laplace Transformation Suppose the following differential equationis transformed to the Laplace domain:

any(n)(t) + an−1y(n−1)(t) + . . . + a0y(t) = bmu(m)(t) + bm−1u(m−1)(t) + . . . + b0u(t)

This a Linear Differential Equation because it does not contain: addition of a constant, squares,product of signals or other functions. To transfer this Linear Differential Equation to theLaplace domain we assume for a moment that the start condition is 0.

ans(n) + an−1s(n−1) + . . . + a0

Y (s) =

bms(m) + bm−1s(m−1) + . . . + b0

U (s)

To use the behavior described by this differential equation in a Block Diagram we use the followingformula:

Y (s) = H (s)U (s) with H (s) = ans(n) + an−1s(n−1) + . . . + a0

bms(m) + bm−1s(m−1) + . . . + b0

4



1.3 Signals in the Laplace domain

In the previous subsection the concept of the Laplace transformation was discussed. With thistransformation it is possible to transform differential equation to equation in the Laplace domain,which are easier to solve. This however also requires writing the signals in the Laplace domain,U (s) in the last example is the Laplace Transform of the input.

The following equation can be used to represent multiple signals by adjusting parameters:

u(t) = u(t) cos(ωt) , with u = ueλt , with λ < 0

u(t) = ueλt cos(ωt)

With Euler’s Formula 2 we can rewrite this equation to:

u(t) = ueλt cos(ωt) = ueλtRe

e jωt

= Re

ueλte jωt

= Re

uest

, with s = λ + jω

With this we can make four different signals:

1. Constant Value: If s = 0 + 0 j the signal’s Exponential Contribution is 0 and there is nooscillation. Hence u is a Constant Value.

2. Exponential Signal: If s = a +0 j only the Exponential Part remains, there is no oscillation.Hence ueλt is what remains: a Exponential Signal.

3. Oscillating Signal: If s = 0 + bj only the Oscillating Part remains, there is no ExponentialSignal. Hence ue jωt is what remains: a Oscillating Signal.

4. Out-damping Signal: A combination of the two above signals is the Exponentially Out-damping Signal.

With Oscillating Signals it is possible to find the Frequency Response Function of a system.

This is done with Oscillating Signals (λ = 0):

est = eλte jωt = e jωt = cos(ωt) + j sin(ωt)

For Oscillating Signals s = j ω, this can be substituted in the Transfer Function of the system wefind the Frequency Response Function:

H ( jω) = an( jω)(n) + an−1( jω)(n−1) + . . . + a0

bm( jω)(m) + bm−1( jω)(m−1) + . . . + b0

Note that the above transfer function still is a complex number; it has a Modulus and Phase. Incase of a Transfer Function this Modulus will be referred to as the Amplification. Calculating, and

plotting the Amplification and Phase of the Transfer Function at different values for ω will resultin a Bode Plot.

2eix = cos(x) + i sin(x)

5



1.4 Frequently Used Systems’

In this section we will develop the Laplace domain representatives of some frequently used systems:

1. Integrator: y = K u(t) L−→ sY (s) = K U (s) → H (s) = K

s

2. Differentiator: y(t) = K u(t) L−→ Y (s) = K sU (s) → H (s) = K s

3. First Order System: τ 0 y(t) + y0 = K u(t) L−→ (τ 0s + 1)Y (s) = K U (s) → H (s) = K

τ 0s+1

4. Second Order System: y(t) + 2βωn y(t) + ω2ny(t) = K ω2

nu(t) L−→ s2 + 2βωns + ω2

n

Y (s) =

Kω2nU (s) → H (s) =

Kω2n

s2 + 2βωns + ω2n

In the equation for the Second Order System ωn represents the Resonance Frequency of thesystem. The β represents the Relative Damping of the system. We will now observe an ex-ample in which these parameters will be determined. Suppose a standard Seconds Order Systemimplemented with a Mass, Spring and Damper:

mx + cx + Kx = K ∗ F (t) L−→ ms2 + cs + k

X (s) = K ∗ F (s)

H (s) = K

ms2 + cs + k =

K m

s2 + cm

s + km

The general equation for a Second Order System:

H (s) = Kω2

n

s2 + 2βωns + ω2n

ω

2

n =

k

m → ωn = k

m

2βωn = 2β

k

m =

c

m → β =

1

2

c

m

m

k =

1

2

c√ km

6



2 Block Diagrams

In Section 1.2 the concept of the Transfer Function was presented. An input signal X (s) can bemultiplied by a systems behavior (H (s))to yield the output signal Y (s):

Y (s) = X (s)H (s)

This allows us to draw block diagrams as presented in Figure 3. With the Transfer Functions

Figure 3: A block diagram in which a disturbance is included.

of the used blocks given in the s-domain it is easy to find the Transfer Function of the TotalSystem. The system in Figure 3 can also be displayed as a single block with input r and outputx. Because r will be the input to this block, we neglect F d for a moment. The derivation of theTransfer Function for this Total System Block starts at the output:

x = (F + 0)H = F H = eC H = (r − x)CH → x = rCH − xCH

x + xCH = rCH → x(1 + CH ) = rCH → H tot(s) = x(s)

r(s) =

HC

1 + CH

With this approach we can also research the Sensitivity of the Error with respect to the reference;again start at the (desired) output:

e = r − x = r − (F + 0)H = r − eCH → e = r − eCH

e + eCH = r → S (s) = e(s)

r(s) =

1

1 + CH

We can also find the Influence of the Disturbance on the Error signal, for this case we assumer = 0:

e = 0− x = −(F + F d)H = −eCH − F dH → e = −eCH − F dH

e + eCH = −F dH → e(1 + CH ) = −F dH → P (s) = F d(s)

e(s) =

−H

1 + CH

In similar way we can find the equations completely describing e and x in this block diagram:

e = 1

1 + HC ∗ r +

−H

1 + HC ∗ F d

x = HC

1 + HC ∗ r +

H

1 + HC ∗ F d

In Control Systems we aim for High Gain Feedback if we increase C we will see that e =0 ∗ r + 0 ∗ F d = 0 and x = 1 ∗ r + 0 ∗ F d = r, these are the results we would like to yield whencontrolling a system.

7



3 Bode Diagrams

It was already elaborated that Transfer Functions are complex numbers, complex numbers havea Modulus and a Phase. The Frequency Response Function depicts the Modulus andthe Phase of the system as a function of the frequency of the input signal. For this FrequencyResponse Function we use an pure sine as input s = jω. The Modulus (M (ω)) and Argument(α(ω)) can be calculated with:

H ( jω) = T 1T 2N 1N 2

|H ( jω)| = M (ω) = |T 1( jω)||T 2( jω)||N 1( jω)||N 2( jω)| |T 1( jω)| =

Re(T 1)2 + Im(T 1)2

ΛH ( jω) = α(ω) = ΛT 1 + ΛT 2 − ΛN 1 − ΛN 2 ΛT 1 = arctan

Im(T 1)

Re(T 1)

Bode Diagrams can be easily made with the help of MATLAB, but here we will present a man-ual derivation of the Bode Diagram. We will focus on determining the Asymptotes of the BodeDiagram. We will observe one frequency lower- and one larger than the Cutoff-Frequency. At

the Cutoff-Frequency the Imaginary Part and the Real Part are equally of magnitude. Someexample derivations are given below.

Usually Bode Diagrams have a logarithmic scale x-axis or ω-axis. Note that the values on thisaxis are ω and not log(ω). The values on the Modulus-diagrams’ y-axis, |H ( jω)|-axis, are inDecibels [dB]. To transfer A to its value on the Decibel-scale we use:

X [dB] = 20 ∗10 log(A)

The following formula is easier to solve; note however that the Decibel-value is still X .

A 1 10 108 0.001 2 0.5√

2 12

√ 2

X 0 20 160 −60 6 −6 3 −3

First-Order System

H (s) = 1

τ s + 1 → H ( jω) =

1

jωτ + 1

ω |H ( jω)| 20 ∗ log(|H ( jω)|) arg(H ( jω)

ωτ 1 →ω

1

τ

√ 12√

02+12 = 1

1 = 1 20

∗log(1) = 0 arg(1) = 0◦

ωτ 1 →ω 1

τ

√ 12√

(ωτ )2+02 = 1

ωτ

20 ∗ log 1ωτ

=

−20 ∗ log(τ )− 20 ∗ log(ω)Slope −20dB for ω 1

τ

arg( 1 jωτ

) = − jωτ

= −90◦

ωτ = 1 →ω = 1

τ

11+1 = 1√

2 = 1

2

√ 2 20 ∗ log(12

√ 2) = 1

arg( 11+1 j ) =

− arctan (1/1) = −45◦

8



Integrator

H (s) = 1

s → H ( jω) =

1

jω


ω 1√ 12√ (ω)2

= 1ω

20 ∗ log1ω

= −20 ∗ log(ω)

Slope −20dB for ω 1 arg( 1

jω) = −1

ω = −90◦

ω = 1√ 12√ 12

= 11 = 1 20 ∗ log(1) = 0 arg(1) = 0◦

Double Integrator

H (s) = 1

s2 → H ( jω) =

1

( jω)2


ω 1√ 12√

((ω)2)2 = 1

ω2

20∗log 1ω2

= −40∗log(ω)

Slope −40dB for ω 1 arg( 1

( jω)2 ) = −1ω2 = −180◦

ω = 1√ 12

(√ 12)2

= 11 = 1 20 ∗ log(1) = 0 arg(1) = 0◦

Proportional-Differential (PD) Controller

H (s) = ds + k → H ( jω) = djω + k


dω k →ω k

d

√ k2 = k 20 ∗ log(k) arg(k) = 0◦

dω k →ω

k

d (dω)2 = dω

20 ∗ log(dω) =log(d) + 1 ∗ log(ω) Slope

+20dB for ω

k

d

arg(dωj) = 90◦

dω = k →ω = k

d

d k

d

2+ k2 =

√ 2k2 =

k√

2 20 ∗ log(k

√ 2)

arg(kj + k) =arg(k( j + 1)) = 45◦

9



4 Controller Design

4.1 Complex Plane: Poles and Zeroes

In Section 1.2 we derived a general Laplace domain equation for an n-order differential equation:

H (s) = ans

(n)

+ an−1s

(n

−1)

+ . . . + a0bms(m) + bm−1s(m−1) + . . . + b0

In the above equation both numerator and denominator are polynomials in s, polynomials can alsobe written in ”root-form”:

H (s) = ans(n) + an−1s(n−1) + . . . + a0

bms(m) + bm−1s(m−1) + . . . + b0= K

(s− z1)(s − z2) . . . (s− zm)

(s− p1)(s− p2) . . . (s− pn) = K

i=1...m(s − zi)i=1...n(s− pi)

The Zeros of a Transfer Function are given by the roots of numerator:

N (s) = 0 → lims→zi

H (s) = 0 → zi for i = 1, . . . , m

The Poles of a Transfer Function are given by the roots of denominator:

D(s) = 0 → lims→ pi

H (s) = ∞→ pi for i = 1, . . . , n

The poles determine if a system is stable or not. A System is Stable when the response to animpulse fades . A System is Instable when the response to an impulse grows without bounds.The relation between Poles of a system and the Stability of a system is:

1. Re( pi) < 0 for every pole of the system, then the system is Stable.

2. Re( pi) > 0 for one of the poles of the system, then the system is Instable.

3. Re( pi)≤

0 for every pole of the system, then the system is Neutral Stable.

Figure 4: A standard Control Diagram.

We don’t want any of the external signals to translate into the output signal in an instable way.The System in Figure 4 shows the 3 common external signals: input (r), disturbance (w) andnoise (v). The Closed-Loop Transfer Function:

Y (s) = C (s)G(s)

1 + C (s)G(s)R(s)+

G(s)

1 + C (s)G(s)W (s)− C (s)G(s)

1 + C (s)G(s)V (s) = T (s)R(s)+S (s)G(s)W (s)−T (s)V (s

In which T (s) is the Complementary Sensitivity and S (s) the Sensitivity. From this wecan observe that T (s) and the product S (s)G(s) both should be stable. In this case the product

10



C (s)G(s) which is called the Open-Loop Transfer Function (L) may not turn −1: otherwisewe are dividing by 0. In other words, the poles have to lie in the LH-plane. So the roots of theequation:

1 + C (s)G(s) = 1 + L(s) = 0

have to lie in the LH-plane.

4.2 Stability Criteria of Routh

In the previous part we saw that the Stability of a system is determined by the position of thepoles in the pole-zero map. But sometimes a Systems Characteristic Equation is given in theform:

ansn + an−1sn−1 + . . . + a1s + a0 = 0

With the Stability Criteria of Routh we can state if the system is Stable without finding thepoles first. For this method first divide the entire equation by an:

sn + an−1sn−1 + . . . + a1s + a0 = 0

For the system to be Stable:

1. All coefficients have to be positive (> 0)

2. 3rd order: a1a2 − a0 > 0

3. 4th order: a1a2a3 − a0a23 − a2

1 > 0

4. 5th order: a3a4 − a2 > 0 and (a3a4 − a2)(a1a2 − a0a3)(a1a4 − a1)2 > 0

4.3 Stability Criteria of Nyquist

A Polar Figure or Nyquist Plot shows the Transfer Function of L(s) along the horizontal Real Axis and along the vertical Imaginary Axis. Normally only the curve L(s = jω) = L( jω = 0) isshown, because it can be determined from a Bode Diagram or from an Frequency Response-measurement. The points along the curve can be calculated using:

|L( jω)| =

Re[L( jω)]2 + Im[L( jω)]2 and argL( jω) = atan

Im[L( jω)]

Re[L( jω)]

A Closed Loop System is stable when the −1-point is at the left of the Polar Figure of theOpen Loop System L( jω), when this curve is observed in the direction of Increasing Frequen-cies. A Closed Loop System is Unstable when the −1-point is at the right of the Polar Figureof the Open Loop System L( jω).

Note: the Stability Criterium of Nyquist does not state whether the Open Loop Systemis stable. If the Transfer Function is measured through measurement it has to be stable, otherwisethe Stability Criteria of Routh has to be used.

11



4.4 Margins

In the previous sections stability was explained, but this was done with the assumption that wehave a very accurate description of the system. But some inaccuracies in this approach might becaused by:

1. Assumptions in high frequency behavior.

2. Linearization of Non-Linear behavior.

3. Inaccuracies in the system itself.

These inaccuracies can cause instability if the system does not have enough margin. In other words:the Polar curve has to lie far enough from the −1-point. The margin to this −1-point is describedby three margins:

1. Gain Margin: the factor with which the gain of the system can be increased before thesystem becomes instable.

(a) Common Values: GM > 2.

2. Phase Margin: the angle with which the phase-lag of the system can be increased beforethe system becomes instable.

(a) Common Values: 30 deg < P M < 60deg.

3. Modulus Margin: is the radius of the smallest circles with center −1, which touches thepolar figure.

(a) Common Values: M M > 0.5.

(b) The Modulus Margin can also be written as:

1M M

= 1min|(1 + L( jω)| = max| 1

(1 + L( jω)| = max|S ( jω)|

(c) This means that M M > 0.5 corresponds with a maximum value of < 6dB in theSensitivity Plot.

(d) Using the Modulus Margin both the Gain Margin and Phase Margin are alreadyadhered to. A Modulus Margin > 0.5 can only occur at a Stable system withGM > 2 and P M > 30 deg.

12



4.5 Loop Shaping

The design of a controller C (s) for a given system G(s) is focused mainly on Stability. Afterwardsthe focus is on decreasing the Sensitivity S (s) over an as large as possible area, this decreases theerror for any given reference but also more noise compression and more robust.

The following procedure describes Loop Shaping:

• Observe the Bode plot of the system which has to be controlled G( jω).

• From the requirements it should be clear what the Cross-Over Frequency should be. If an decrease in phase lag is required to meet the Phase Margin-requirement, use one of thetwo following tools:

Filter Transfer Function Use Breakpoint Other Instructions

PD-Controller

K (1 + τ ds) Decrease Phase

Lag

1

τ dAdvice: τ d =

1

2πωc

Lead-Filter K τ 1s + 1

τ 2s + 1Decrease Phase

Lag

1

τ 1,

1

τ 2

For Lead-Filteroperation τ 1 > τ 2

Advice: 1

τ 1=

ωc

3 and

1

τ 2= 3ωc

• Now that we have made sure that we do not have to much Phase Lag we increase the gainof the controller until: |C ( jω)G( jω)| = 1.

• If unwanted effects at high frequencies are present, the following filters can be used to suppressthese effects:


1st-orderLow-Pass

1

τ s + 1Passes

Low-Frequencies

1

τ

Notchτ 21 s2 + 2βτ 1s + 1

τ 22 s2 + 2βτ 2s + 1Remove

Frequencies

1

τ 1,

1

τ 1,

1

τ 2,

1

τ 2

τ 1 = τ 2β 1 < β 2 ≤ 1

Depth = β 1/β 2

InverseNotch

τ 21 s2 + 2βτ 1s + 1

τ 22 s2 + 2βτ 2s + 1Amplify

Frequencies

1

τ 1,

1

τ 1,

1

τ 2,

1

τ 2

τ 1 = τ 2β 2 < β 1 ≤ 1

Height = β 1/β 2

• Is less Sensitivity S ( jω) required at low frequencies, use some additional I -action:


PI-Controller

1

1 + 1τ is

Low-FrequencyGain

1

τ i

1

τ i≤ ωc

5

• Always make sure to check if |S ( jω)| < 6dB

13



4.6 Design for Performance

In Figure 4 a typical control system is displayed. When designing for performance we want theTracking Error and the Influence of Disturbances to be as low as possible. The Error of thesystem is described by:

E (s) = S (s)R(s)

−S (s)G(s)W (s)

In which the Reference R(s) and Disturbance W (s) are included. Because G(s) is a given plantwe can only adjust S (s). Many requirements can be translated to a specific Sensitivity at a certainfrequency. An example:

The Reference R(s) which is a sine-signal with amplitude 1[mm] and frequency 100[Hz] may not

influence the Error E (s) by more than 1[µm] In this case we can specify the Amplification of signals at a specific frequency:

1

1000 = −60dB at 100[Hz]

This is a direct requirement to the Sensitivity S (s), with this requirement we can make a guess

for an appropriate Cross-Over Frequency. At low frequencies |L( jω)| 1 so:

|S ( jω)| = 1

|1 + L( jω)| ≈ 1

|L( jω)|This means that if the system has a −2 slope at low frequencies in transfer function H (s) it willhave a slope of +2 for low frequencies in transfer function S (s). With one point on the S (s)-function and the slope of the S (s) we can calculate at which frequency this S (s)-function crossesthe 0dB-line. So at which frequency we have to place the Cross-Over Frequency in order tomeet the suppression of the Reference R(s) in the Error E (s).

4.7 Limits to Performance

To make sure the Reference R(s) and Disturbance W (s) do not influence the Error E (s),we want the Sensitivity S (s) to be as low as possible. The ideal case would be to have the entireSensitivity S (s) beneath 0dB this is however not possible due to the Bode Sensitivity Integral.This is however only valid for a System with at least 2 Poles more than Zero’s. This means thatthe Bode Sensitivity Integral applies on a system with n Zero’s and m Poles, if and only if:

n− m ≥ 2

The Bode Sensitivity Integral itself is described by:

∞

0

ln |S ( jω)|dω = 0

Ideally a system has a Relative Grade of 1 which means that that m = n − 1. In this case thesystem approaches the point (0, 0) with a phase lag of −90 deg.

14



Part II

Lagrangian Mechanics

This part of the document describes Lagrange’s Method for finding the Equations of Motion for aDynamical System. In this main part 3 sheets show the used work flow:

1. Equations of Motion

2. Linearization

3. Mechanical Vibrations of Undamped Systems

4. Mechanical Vibrations of Damped Systems

In the Appendix a detailed derivation of Lagrange’s equations of motion can be found.

15



Part III

Appendix

5 Detailed Lagrangian Mechanics

The drawback of Newton’s approach to Mechanical Analysis is that it considers the individualcomponents of a system separately, thus necessitating the calculation of interacting forces. TheLagrange approach to Mechanical Analysis is different in that it considers the system as a wholerather than its individual components.

Cartesian Coordinates The Motion of a Mechanical System and its individual parts aredescribed with respect to an Inertial Frame. The Inertial Frame is defined with axes x1, x2

and x3. Suppose a Mechanical System with n p discrete mass particles mi (i = 1, 2, . . . , n p. If wewould express the motion of this system by means of Cartesian Coordinates x1i, x2i and x3i foreach particle mi we would need 3n p coordinates to describe the total Mechanical System. Howeverthese Cartesian Coordinates are not independent due to connections between the mass particles.

Generalized Coordinates In general, the Mechanical System can be described by n in-dependent parameters q 1, q 2, . . . , q n which are called Generalized Coordinates. The numberDegrees of Freedom of this Mechanical System is n, it is equal to the minimum number of independent coordinates that are required to describe the systems position uniquely. Because of the connections between the masses n ≤ 3n p The Generalized Coordinates may not alwayshave a physical meaning nor will they be unique. This means that it is possible to describe theMechanical Systems motion with more sets of Generalized Coordinates, note however that thenumber of Degrees of Freedom is unique. A Coordinate Transformation is used to transformGeneralized Coordinates to Cartesian Coordinates, the latter ones will be a function of the

Generalized Coordinates:

x1i = x1i (q 1, q 2, . . . , q n)

x2i = x2i (q 1, q 2, . . . , q n)

x3i = x3i (q 1, q 2, . . . , q n)

i = 1, 2, . . . , n p

The Generalized Coordinates will be stored in a column matrix q , also shortly called column q .

q =

q 1q 2...

q 4

The position of a mass particle mi can be described by means of the Radius Vector −→ri , which isindependent of the Coordinate System (Cartesian, Cylindrical, Polar,. . . ). If we wish to quantifythe motion in a specific Coordinate System ri is used. So for the earlier introduced CartesianCoordinate System:

ri =

x1i

x2i

x3i

20



Both the Column ri and Radius Vector −→ri are functions of q :

ri = ri

q

and −→ri = −→ri

q

Prescribed Coordinates Sofar a Mechanical Systems position was completely expressed incolumn q . However some systems also depend on one or more geometrical parameters that are

known Explicit Functions of time. These Explicit Functions of time are called PrescribedCoordinates. Which are denoted:

s(t) =

s1(t)

s2(t)...

sm(t)

So the motion of a Mechanical System with Prescribed Coordinates (s(t)) depends on bothGeneralized Coordinates (q ) and Prescribed Coordinates. We can now write:

ri = ri

q, s(t)

and −→ri = −→ri

q, s(t)

Differentiation with respect to a Column Matrix In order to create a compact formu-

lation of the Equations of Motion, it appears to be useful to define an abbreviating notationfor the Derivatives of columns and vectors with respect to a Column Matrix. Consider theColumn Matrix Function f (q, t) of dimension m. This type of vectors have already been usedto describe the position of a mass with respect to a Cartesian Coordinate System. The PartialDerivative of this Column Matrix f to the column q and scalar t is defined as:

f ,g

=∂f

∂q =

f 1,q1f 1,q2

. . . f 1,qn

f 2,q1f 2,q2

. . . f 2,qn...

... . . .

...

f m,q1f m,q2

. . . f m,qn

with f i,qj=

∂f i∂q i

f ,t

=∂f

∂t =

f 1,t

f 2,t

...

f m,t

with f i,t =

∂f i∂t1

The Position of a mass in a Mechanical System with no Prescribed Coordinates can bedefined by r = r(q ). The derivative of position yields velocity so:

r = v = r ,q q

In Vector Form this can be written as:

−→r = −→v = −→r ,q q = q T −→r ,q

T

The Position of a mass in a Mechanical System with Prescribed Coordinates can be definedby r = r(q, t). The derivative of position yields velocity so:

r = v = r,t+ + r,q q

In Vector Form this can be written as:

−→r = −→v = −→r ,t + −→r ,q q = −→r ,t + q T −→r ,q

T

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Virtual Displacements Consider a Mechanical System and let −→r i be the position vector of mass particle mi of the system with respect to the Inertial Cartesian Coordinate System. Wethen freeze the system, if mi was moving it is not moving anymore. We then change −→r i by δ −→r i,this δ −→r i is called the Virtual Displacement, the Virtual Position −→r vis now defined as:

−→r v =

−→r + δ

−→r

The following requirements are imposed on Virtual Displacements:

• They are taken on a Fixed point in time.

• They must be consistent with Geometric or Kinematic Constraints of the system.

• They should be Arbitrary Small.

Virtual Displacements can also be applied to Generalized Coordinates:

−→q v = −→q + δ −→q

The Virtual Displacement of the Position Vector δ −→r can then be calculated by:

δ −→r = −→r ,q δq = q T −→r ,q

T

Virtual Work Consider a mass particle mi of a mechanical system and let −→r i be the position

vector of mi relative to the origin O of an inertial frame. If −→F ti is the total Resulting Force acting

upon mi then from Newton’s Second Law:

F = ma −→ F − mx = 0 −→ −→F ti − mi

−→r i = 0

If we introduce a Virtual Displacement δ −→r i(t) the system will still be in Equilibrium the above

equation still holds. The Virtual Work performed over the Virtual Displacement δ −→r i:

δW i =−→

F ti −mi −→r i

· δ −→r i = 0

This will still be equal to 0 because the system is still in Equilibrium. The Virtual Work forthe entire system is negligible:

δW =

npi=1

δW i =

npi=1

−→F ti −mi

−→r i

· δ −→r i = 0

The Total Resulting Force−→F ti is consists of Applied Forces

−→F appl

i and Constraint Forces−→F constr

i

, so that:

−→F ti =

−→F appl

i +−→F constr

i

Constraint Forces result from Geometric or Kinematic Constraints in the motion of anypart or particle of the system and are of a Reactive Nature. For example: forces that confinethe motion of a system to a given path or the Internal Forces in a rigid body.

Applied Forces are all forces except Constraint Forces. For example: gravitational forces,aerodynamic lift and drag, magnetic forces.

22



δW =

npi=1

−→F appl

i · δ −→r i +

npi=1

−→F constr

i · δ −→r i −np

i=1

−→F ti − mi

−→r i · δ −→r i = 0

For now we assume that the Work of the Constraint Forces through virtual displacementsare compatible with the kinematic system constrains is zero:

δW constr =

npi=1

−→F constr

i · δ −→r i = 0

If this is the case we can reduce the main expression to:

δW =

npi=1

δW i =

npi=1

−→F appl

i − mi −→r i

· δ −→r i = 0

Generalized Forces The Virtual Work of applied forces −→

F appli is given by:

δW appl =

npi=1

−→F appli · δ −→r i = 0

Our goal is to describe the motion of the system by means of the Generalized Coordinatesq , therefore we express the above equation in terms of q and δ q :

δW appl =

npi=1

−→F appl

i · −→r ,q

δq = δq T

npi=1

−→r ,q

T · −→F appli

= δq T Q = QT δq ; Q =

npi=1

−→r ,q

T ·−→F appli

Q denotes the Column Matrix of Generalized Forces, related to the virtual changes δq of the Generalized Coordinates q . In matrix notation:

Q =

npi=1

r,q

T · F appli

Applied Moments In practice it regularly happens that the resulting action of some of theapplied forces is (at least partially) equivalent to the action of so-called Applied Moments being

exerted at some or more distinct points of the system. Applied Moments are denoted by −→M appl

which is defined as, use a standard Cartesian Coordinate System:

−→M appl = M appl−→e 3 = Arm × Force

Note that with a standard Cartesian Coordinate System, Applied Moments rotatingCounter-Clock-Wise are defined as positive, this also follows from the cross-product. The VirtualWork of both the Applied Forces and Applied Moments is defined as:

δW appl =

nF i=1

δ −→r i · −→F appli +

nM j=1

δ −→θ j · −→M appl

j

As −→

θ j can be considered an explicit function of the Generalized Coordinates q and time t:

23



−→θ j =

−→θ j

q, t → −→

θ j,q δq = q T −→

θ j,q

T

Combining this knowledge in the formula for the Virtual Work of Applied Forces andApplied Moments:

δW appl = δq T

nF i=1

−→r i,q

T · −→F appli +

nM j=1

−→θ j,q

T · −→M appli

And the part between [ ] contains the definition for Generalized Forces which include a

combination of both Applied Forces and Applied Moments:

Q =

nF i=1

ri,q

T · F appli +

nM j=1

θ j,q

T ·M appli

Internal Energy Both the −→

F appli and

−→M appl

i can be split up into External- and Internal

Forces and Moments.

F appli = F ex

i + F ini

M appl j = M ex

j + M in j

Qappl

j = Qex

j + Qin

j ; with :

Qex =

nF i=1

ri,q

T · F exi +

nM j=1

θ j,q

T ·M exi and Qin =

nF i=1

ri,q

T · F ini +

nM j=1

θ j,q

T ·M ini

Generalized Internal Forces are defined to depend only on the Generalized Coordinates q and not on the Generalized Velocities q :

Qin = Qin(q )

Note that damping forces and friction forces depend on q these are External Forces. Otherexamples of External Forces are: gravitational forces, aerodynamic lift and drag, magnetic forces.The Generalized Internal Forces can always be related to a state quantity U in(q ) as follows:

Qin = −

∂U in

∂q

T

= −

U in,q

T

For the Virtual Work by Interal Forces we can write:

δW in = δq T Qin = −δq T

U in,q

T

= −U in,q δq = −δU in(q )

With this knowledge we can rewrite the Virtual Work of the Applied Forces:

δW appl = δW ex + δW in = δq T Qex − δU in

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Potential Energy Now we will split up the Generalized External Forces into Conser-vative Generalized External Forces and Non-conservative Generalized External Forces:

Qex = Qcons + Qnc

Note that all Generalized Internal Forces are Conservative. The Conservative Generalized

External Forces can be derived from a Potential Energy function V ex(q ), which depends onthe Generalized Coordinates only:

Qcons = −

∂V ex

∂q

T

= −

V ex,q

T

With this knowledge we can rewrite the Virtual Work of the Applied Forces:

δW appl = δq T Qnc − δV ex − δU in

We define the Total Potential Energy Function of the Mechanical System is defined:

V (q ) = U in(q ) + V ex(q )

So again we can rewrite the Virtual Work of the Applied Forces:

δW appl = δq T Qnc − V (q ) =

Qnc− V ,q

δq

Kinetic Energy In the next section we will show that in describing a systems motion theKinetic Energy of the system is required. The Kinetic Energy of our system mechanical systemby definition has the form:

T = 1

2

np

i=1

mi −→r i

· −→r i = T 0 + T 1 + T 2 = T 0 + mT q +

1

2

q T M q

T 0 is the term which represents the Transport Kinetic Energy of the system. It is the onlyterm remaining if q = 0.

T 0 = T 0(q, t) = 1

2

npi=1

mi−→r i,t · −→r i,t =

1

2

npi=1

miri,t · ri,t

T 1 is the term which represents the Mutual Kinetic Energy of the system. It occurs incombination with the Transport Kinetic Energy if Prescribed Motion is present.

T 1 = T 1(q, q, t) = mT q →

mT q, t =np

i=1

mi

−→r i,t

· ˙

−→r i,q =

np

i=1

mirT i,t

·ri,q

T 2 is the Square Kinetic Energy if no drivers are active in the system:

T 2 = T 2(q, q, t) = 1

2 q T M q → M

q

=

npi=1

mi

−→r i,q

T · −→r i,q =

npi=1

mi

ri,q

T · ri,q

M is called the Mass Matrix of the system. The Mass Matrix is Symmetric: M T = M .

25



Lagrange’s Equations of Motion Using some extra derivations we can find the followingexpressions:

δW inert =

− d

dt

T ,q

+ T ,q

δq

δW = δW appl+δW inert = Qnc− V ,q δq +−d

dtT ,q+ T ,q δq =

− d

dtT ,q− T ,q + V ,q

− QncT δq

This expression has to be equal to 0, so with some rearrangements we yield Lagrange’s Equa-tions of Motion:

d

dt

T ,q

− T ,q + V ,q =

QncT

Equilibrium Position An Equilibrium Position is used to linearize Lagranges Equa-tions of Motion. For this Equilibrium Position the following applies:

q (t) = q 0

= constant = time independent

q (t) = q (0) = 0; q (t) = q (0) = 0;

Because the above rules apply and the solution has to be time independent we can write LagrangesEquation of Motion as:

−T 0,q + V q

T

= 0

Because T 0 is the Transport Kinetic Energy this term will be zero when no drivers exist in thesystem.

V q

T

= 0

This means that when the derivative of the Potential Energy is zero a Equilibrium Positionhas been found. To find out if this is a Stable Position we need to find the Stiffness Matrix:

K 0 =

V ,q

T , q

q0

A Stable Position is found when K 0 is positive definite. This means that k11 and k22 are bothlarger than zero and the determinant is also larger then zero.

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