Ee2365 nol part 2

CONCEPT OF STABILITY

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Bode Plot

Frequency Response

We will now examine the response of a system to a sinusoidal input

We call this the frequency response of a system

Steady state response of a system to the sinusoidal input

We vary the frequency of the i/p signal over a certain range and study the resulting response

Frequency Response Consider a LTI system with T.F Y(s) /X(s) = G(s)

x(t) = X Sin wt Y(t)= Y sin (wt + Ø) where, Ø =tan -1 [imm G(jw)/ Real G(jw)] Y(jw) /X(jw) = G(jw)

G(S)X(S) Y(S)

Bode plot• Plots of the magnitude and phase characteristics are

used to fully describe the frequency response

• A Bode plot is a (semilog) plot of the transfer function

magnitude and phase angle as a function of

frequency

• The gain magnitude is many times expressed in

terms of decibels (dB)

dB = 20 log10 AMay 1, 2023 5BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING

There are 4 basic forms in an open-loop transfer function G(jω)H(jω)

Gain factor, K (jω)±p factor: pole and zero at origin (1+jωT)±q factor Quadratic factor

Bode plot procedure

The magnitude and phase plots for some typical factors

Factor Magnitude: Phase:

Gain, K

dBjG )( )( jG

0dB20logK

1 20 dB/decade

dBjG )( )( jG

-20 dB/decade

40logω

40 dB/decade

1/jω2

1+jωT90º

-40logω

-40 dB/decade

0.1/T 1/T 10/T

dBjG )( )( jG

1/(1+jωT

dBjG )(

Gain margin and Phase margin

Gain margin:The gain margin is the number of dB

that is below 0 dB at the phase crossover frequency (ø=-180º). It can also be increased before the closed-loop system becomes unstable

Phase margin:The phase margin is the number of

degrees the phase of that is above -180º at the gain crossover frequency

-180º

-270º

Gain margin

Phase marginPhase Crossover

with -180º line

Gain Crossover with 0dB line

Magnitude curve

Phase curve

Magnitude(dB)

Phase(Degree)

Log ω

Gain margin and Phase margin

Bode Plot - Example For the following T.F draw the Bode plot and obtain Gain cross over

frequency (wgc) , Phase cross over frequency , Gain Margin and Phase Margin.

G(s) = 20 / [s (1+3s) (1+4s)]

Solution: The sinusoidal T.F of G(s) is obtained by replacing s by jw in the given T.F

G(jw) = 20 / [jw (1+j3w) (1+j4w)] Corner frequencies: wc1= 1/4 = 0.25 rad /sec ; wc2 = 1/3 = 0.33 rad /sec Form a table

Term Corner Frequency Slope db/dB

Change in slope

20/jw --- -20

1/(1+j4w) wc1=1/4=0.25 -20 -20-20=-40

1/(1+j3w) wc2=1/3=0.33 -20 -40-20=-60

Choose a lower corner frequency and a higher Corner frequencywl= 0.025 rad/sec ; wh = 3.3 rad / sec

Calculation of Gain (A) (MAGNITUDE PLOT)

A @ wl ; A= 20 log [ 20 / 0.025 ] = 58 .06 dB

A @ wc1 ; A = [Slope from wl to wc1 x log (wc1 / wl ] + Gain (A)@wl

= - 20 log [ 0.25 / 0.025 ] + 58.06 = 38.06 dB

A @ wc2 ; A = [Slope from wc1 to wc2 x log (wc2 / wc1 ] + Gain (A)@ wc1

= - 40 log [ 0.33 / 0.25 ] + 38 = 33 dB

A @ wh ; A = [Slope from wc2 to wh x log (wh / wc2 ] + Gain (A) @ wc2

= - 60 log [ 3.3 / 0.33 ] + 33 = - 27 dB

Calculation of Phase angle for different values of frequencies [PHASE PLOT]

Ø = -90O- tan -1 3w – tan -1 4w

Frequency in rad / sec

Phase Angle in degrees

w=0 Ø= -90o

w=0.025 Ø= - 99o

w=0.25 Ø= -172o

w=0.33 Ø= -188o

w=3.3 Ø= - 259o

w=∞ Ø= - 270o

Wl =0.025W1 =0.25

W2 =0.33 Wh=3.3

Log w+[G

- 180o

Magnitude PlotY-Axis 1 unit =20 dB

Phase PlotY-Axis 1 unit = 50o

Wgc =1.1 rad/sec

Wpc =0.3 rad/sec

PMØgc=-240O

GM = - [20 log G(jwpc)] dB = - [32] dB = - 32 dB-

=180+gc = 180o+(-240o)=- 60o

Calculation of Gain cross over frequency The frequency at which the dB magnitude is Zero wgc = 1.1 rad / sec

Calculation of Phase cross over frequencyThe frequency at which the Phase of the system is - 180o

wpc = 0.3 rad / sec

Gain MarginThe gain margin in dB is given by the negative of dB magnitude of G(jw) at phase cross over frequency GM = - { 20 log [G( jwpc )] }

= - { 32 } = -32 dB

Phase Margin Ґ = 180o+ Øgc

= 180o + (- 240o) = -60o

Conclusion For this system GM and PM are negative in Values. Therefore the system is

unstable in nature.

The steady-state response of a linear system to sinusoidal excitation may be determined from the system transfer function T(s) by replacing s by j

The resulting frequency response function T(j) has the polar form :

)(je)(A)j(T

)s(X)s(Y

)s(X)s(Yarg

T(s)X(s) Y(s)

For any value of , T(j) reduces to a complex number whose amplitude, A(j) gives , and () gives

POLAR PLOT AND STABILITYPOLAR PLOT AND STABILITY

Open loop response : Polar plots

Polar plots are more useful in control system design work open-loop polar plot indicates whether the closed-loop system is

stable also gives a measure of how stable the system is

Represent T(j) on an Argand diagram :

Polar plot is traced out by the trajectory of T(j) as increases progressively from zero

A(j)(j)

T(s)-plane

BASICS OF POLAR PLOT

The polar plot of a sinusoidal transfer function G(jω) is a plot of the magnitude of G(jω) Vs the phase of G(jω) on polar co-ordinates as ω is varied from 0 to ∞.

(ie) |G(jω)| Vs G(jω) as ω → 0 to ∞. Polar graph sheet has concentric circles and radial lines.

Concentric circles represents the magnitude.

Radial lines represents the phase angles.In polar sheet

+ve phase angle is measured in ACW from 0

-ve phase angle is measured in CW from 0

PLOTTING PROCEDURE Express the given expression of OLTF in (1+sT) form.

Substitute s = jω in the expression for G(s)H(s) and get G(jω)H(jω).

Get the expressions for | G(jω)H(jω)| & G(jω)H(jω).

Tabulate various values of magnitude and phase angles for different values of ω ranging from 0 to ∞.

Usually the choice of frequencies will be the corner frequency and around corner frequencies.

Choose proper scale for the magnitude circles.

Fix all the points in the polar graph sheet and join the points by a smooth curve.

Write the frequency corresponding to each of the point of the plot.

MINIMUM PHASE SYSTEMS

Systems with all poles & zeros in the Left half of the s-plane – Minimum Phase Systems.

For Minimum Phase Systems with only poles Type No. determines at what quadrant the polar plot starts. Order determines at what quadrant the polar plot ends.

Type No. → No. of poles lying at the origin Order → Max power of ‘s’ in the denominator polynomial of the transfer function.

START & END OF POLAR PLOT

TYPICAL SKETCHES OF POLAR PLOT

GAIN MARGINGain Margin is defined as “the factor by which the system gain can be increased to drive the system to the verge of instability”.For stable systems,

ωgc ωpc

G(j)H(j) at ω=ωpc 1

GM = in positive dB

More positive the GM, more stable is the system.

For marginally stable systems,

ωgc = ωpc

G(j)H(j) at ω=ωpc = 1 GM = 0 dB

For Unstable systems,ωgc ωpc

G(j)H(j) at ω=ωpc > 1 GM = in negative dB Gain is to be reduced to make the system stable

Note: If the gain is high, the GM is low and the system’s step response shows high overshoots and long settling time. On the contrary, very low gains give high GM and PM, but also causes higher ess, higher values of rise time and settling time and in general give sluggish response.

Thus we should keep the gain as high as possible to reduce ess and obtain acceptable response speed and yet maintain adequate GM and PM. An adequate GM of 2 ( ie 6 dB ) and a PM of about 30 is generally considered good enough as a thumb rule.

Contd…

Contd… At = pc , G(j)H(j) = -180

Let G(j)H(j) at = pc be taken as B

If the gain of the system is increased by a factor 1/B, then the G(j)H(j) at = pc becomes B(1/B) = 1 and hence the G(j)H(j) locus pass through -1+j0 point driving the system to the verge of instability.

GM is defined as the reciprocal of the magnitude of the OLTF evaluated at the phase cross over frequency. GM = Kg = 1 / G(j)H(j) = pc

GM in dB = 20 log (1/B) = - 20 log BMay 1, 2023 29BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING

PHASE MARGINPhase Margin is defined as “ the additional phase lag that can be introduced before the system becomes unstable”.

Let ‘A’ be the point of intersection of G(j)H(j) plot and a unit circle centered at the origin.

Draw a line connecting the points ‘O’ & ‘A’ and measure the phase angle between the line OA and +ve real axis.

This angle is the phase angle of the system at the gain cross over frequency.

G(jgc)H(jgc) = gc

Contd… If an additional phase lag of PM is introduced at this frequency, then the phase angle G(jgc)H(jgc) will become 180 and the point ‘A’ coincides with (-1+j0) driving the system to the verge of instability.

This additional phase lag is known as the Phase Margin.

= 180 + G(jgc)H(jgc)

= 180 + gc

[Since gc is measured in CW direction, it is taken as negative]

Contd… For a stable system, the phase margin is positive.

A Phase margin close to zero corresponds to highly oscillatory system.

Polar plots contd.

A polar plot may be constructed from experimental data or from a system transfer function

If values of are marked along the contour, a polar plot has the same information as a Bode plot

Usually, the shape of a polar plot is of most interest

-j 0.50

j 0-0.5 0 0.5 1

ExampleProblem : Construct the polar plot for the (critically damped

system) defined by :

Solution :

Limiting conditions :(i)(ii)

0j1)j(T:0

e0)j(T.e.i

2j)j(T:1

Re-0.2 0 0.2 0.4 0.6 0.8 1

2)1s(1)s(T

2)j1(1)j(T

Example (contd.) For accurate work, results are usually calculated for a range of

values of :

0 0.25 0.5 0.75 1.0 1.5 2.0 3.0 5.0

T(j) 1+j0

0.83-j0.44

0.48-j0.64

0.18-j0.61

0-j0.5

-0.12-j0.28

-0.12-j0.16

-0.08-j0.06

-0.04-j0.01

-j 0.50

-0.5 0 0.5 1

2. G(s) = [(1+0.2s)(1+0.025s)] / [(s^3(1+0.005s)(1+0.001s)]

STABILITY OF CONTROL SYSTEM

Concepts of Stability

Stability - Definition

A system is stable if any bounded input produces a

bounded output for all bounded initial conditions

Basic concept of stability

Stability of the system and roots of characteristic equations

Characteristic EquationCharacteristic Equation

Consider an nth-order system whose the characteristic equation (which is also the denominator of the transfer function) is:

11 sasasasas)s(a n

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Routh Hurwitz Criterion

May 1, 2023

IntroductionIntroduction Goal: Determining whether the system is stable or

unstable from a characteristic equation in polynomial form without actually solving for the roots

Routh’s stability criterion is useful for determining the ranges of coefficients of polynomials for stability, especially when the coefficients are in symbolic (non numerical) form

To find Kmar & ωn

A A necessary necessary conditioncondition for Routh’s for Routh’s

StabilityStability A necessary condition for stability of the system is

that all of the roots of its characteristic equation have negative real parts, which in turn requires that all the coefficients be positive

A necessary (but not sufficient) condition for stability is that all the coefficients of the polynomial

characteristic equation be positive & none of the co-efficient vanishes

A A necessary and sufficient necessary and sufficient conditioncondition

for Stabilityfor Stability Routh’s formulation requires the computation of a

triangular array that is a function of the coefficients of the polynomial characteristic equation

A system is stable if and only if all the elements ofthe first column of the Routh array are positive

Characteristic EquationCharacteristic Equation Consider an nth-order system whose the

characteristic equation (which is also the denominator of the transfer function) is:

11 sasasasas)s(a n

Method for determining the Method for determining the Routh arrayRouth array

Consider the characteristic equation:

aaasaas

111 sasasasasas)s(a n

First arrange the coefficients of the

characteristic equation in two rows, beginning

with the first and second coefficients and followed by the even-numbered and odd-

numbered coefficients: May 1, 2023 48BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING

Routh array: method Routh array: method (cont’d)(cont’d) Then add

subsequent rows to

complete the Routh array:

cccsbbbsaaasaas

Routh array: method Routh array: method (cont’d)(cont’d)

Compute elements for the 3rd row:

cccsbbbsaaasaas

Routh array: method Routh array: method (cont’d)(cont’d)

Compute elements for the 4th row:

bbabac

cccsbbbsaaasaas

Example 1

Given the characteristic equation,

Is the system described by this characteristic equation stable?

Answer: One coefficient (-2) is negative

Therefore, the system does not satisfy the necessary condition for stability

4s4ss2s3s4s)s(a 23456

Example 2Given the characteristic equation,

Is the system described by this characteristic equation stable?

Answer:All the coefficients are positive and nonzeroTherefore, the system satisfies the necessary

condition for stabilityWe should determine whether any of the coefficients

of the first column of the Routh array are negative

44234)( 23456 sssssssa

Example 2 (cont’d): Routh array

?:??:??:

???:???:

0424:4131:

sssssss

44234)( 23456 sssssssa

Example 2 (cont’d): Routh array

?:??:??:

???:4025:

0424:4131:

sssssss

44234)( 23456 sssssssa

Example 2 (cont’d): Routh Example 2 (cont’d): Routh arrayarray

?:??:??:

05122:4025:

0424:4131:

sssssss

2)2(25

44234)( 23456 sssssssa

2525444

Example 2 (cont’d): Routh Example 2 (cont’d): Routh arrayarray

4:01576:43:

05122:4025:

0424:4131:

sssssss

44234)( 23456 sssssssa

The elements of the 1st column are not all positive:

the system is unstable

Special cases of Routh’s Special cases of Routh’s criteriacriteria Case 1: All the elements of a row in a RA are

zero Form Auxiliary equation by using the co-efficient of the row

which is just above the row of zeros Find derivative of the A.E. Replace the row of zeros by the co-efficient of dA(s)/ds complete the array in terms of these coefficients analyze for any sign change, if so, unstable no sign change, find the nature of roots of AE non-repeated imaginary roots - marginally stable repeated imaginary roots - unstable

Special cases of Routh’s Special cases of Routh’s criteriacriteria Case 2: First element of any of the rows

of RA is

Zero and the same remaining row contains atleast one non-zero element

Substitute a small positive no. ‘ε’ in place of zero and complete the array.

Examine the sign change by taking Lt ε 0

Example 3: Stability versus Parameter Range

Consider a feedback system such as:

The stability properties of this system are a function of the proportional feedback gain K. Determine the range of K over which the system is stable

Example 3 (cont’d)

The characteristic equation for the system is given by:

0)6)(1(

Expressing the characteristic equation in polynomial form, we obtain:

0)6(5 23 KsKss

0)6)(1(

The corresponding Routh array is:

Therefore, the system is stable if and only if

KsKsssa )6(5)( 23

:5)304(:

0Kand5.7K

0Kand05

Root Locus Technique

––44OO

INTRODUCTION TO ROOT LOCUS Introduced by W. R. Evans in 1948

Graphical method, in which movement of poles in the s-plane is sketched when some parameter is varied

The path taken by the roots of the characteristic equation when open loop gain K is varied from 0 to ∞ are called root loci

Direct Root Locus = 0 < k < ∞

Inverse Root Locus = - ∞ < k < 0

Root Locus Analysis The roots of the closed-loop characteristic equation

define the system characteristic responses

Their location in the complex s-plane lead to prediction of the characteristics of the time domain responses in terms of: damping ratio, natural frequency, wn

damping constant, first-order modes

Consider how these roots change as the loop gain is varied from 0 to

second-order modes

BASICS OF ROOT LOCUS Symmetrical about real axis

RL branch starts from OL poles and terminates at OL zeroes

No. of RL branches = No. of poles of OLTF

Centroid is common intersection point of all the asymptotes on the real axis

Asymptotes are straight lines which are parallel to RL going to ∞ and meet the RL at ∞

No. of asymptotes = No. of branches going to ∞

At Break Away point , the RL breaks from real axis to enter into the complex plane

At BI point, the RL enters the real axis from the complex plane

ANGLE AND MAGNITUDE CRITERIA Angle condition

G(s)H(s) = ±(2q+1)180º where q = 0, 1, 2 … = Odd multiple of 180º Use : Any point in s-plane which satisfies the angle

condition has to be on the root locus of the corresponding system

Magnitude condition |G(s)H(s)| = 1 Use: It gives the value of K corresponding to any point

which is on the root locusMay 1, 2023 68BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING

PROCEDURE FOR CONSTRUCTING ROOT LOCUS Locate the OL poles & zeros in the plot Find the branches on the real axis Find angle of asymptotes & centroid

Φa= ±180º(2q+1) / (n-m) σa = (Σpoles - Σzeroes) / (n-m) Find BA and BI points Find Angle Of departure (AOD) and Angle Of Arrival (AOA)

AOD = 180º- (sum of angles of vectors to the complex pole from all other poles) + (sum of angles of vectors to the complex pole from all zero)

AOA = 180º- (sum of angles of vectors to the complex zero from all other zeros) + (sum of angles of vectors to the complex zero from poles)

Find the point of intersection of RL with the imaginary axis

The Root Locus Procedure

0)(1 sF

Step 1: Write the characteristic equation as

Step 2: Rewrite preceding equation into the

form of poles and zeros as follows:

Step 3: Locate the poles and zeros with specific symbols, the root locus begins at the open-loop poles and ends at the open-loop zeros as K increases from 0 to infinity

If open-loop system has n-m zeros at infinity, there will be n-m branches of the root locus approaching the n-m zeros at infinity

Step 4: The root locus on the real axis lies in a section of the real axis to the left of an odd number of real poles and zeros

Step 5: The number of separate loci is equal to the number of open-loop poles

Step 6: The root loci must be continuous and symmetrical with respect to the horizontal real axis

Step 7: The loci proceed to zeros at infinity along asymptotes

centered at centroid and with angles

)1,2,1,0()12(

Step 8: The actual point at which the root locus crosses the imaginary axis is readily evaluated by using Routh’s criterion

Step 9: Determine the breakaway point d (usually on the real axis):

i ij pdzd1 1

Step 11: Plot the root locus that satisfy the phase criterion

Step 12: Determine the parameter value K1 at a specific root using the magnitude criterion

,2,1)12()( kksP

The Root Locus - example

Root Locus Plot - Example Loop Transfer function:Loop Transfer function:

Roots:Roots:s = 0s = 0, , s = –4s = –4, ,

s = –2 s = –2 4 4jj Real axisReal axis segments: segments:

between 0 and –4 .between 0 and –4 . Asymptotes:Asymptotes:

angles = angles =

))202044)()(44(())(( 22

ssssssssKKssGHGH

))00222244(( aa44

77,,44

55,,44

33,,440044

))1122(( kk

asymptotesasymptotes

––44XX

––22

––22jj

4545°°

––44jjXX

Root Locus Plot - Example Breakaway points:

Three points that breakaway at 90°

solving, jsb 45.22,2

ssss)ss(s

sssKssss

020186or

080368

)8072244(80368

––44XX

––22

––22jj

4545°°

––44jjXX

Root Locus Plot - Example The imaginary axis crossings:The imaginary axis crossings:

Characteristic equationCharacteristic equation

Routh tableRouth tabless44 1 36 1 36 KKss33 8 80 0 8 80 0ss22 26 26 KK 0 0s 80-8s 80-8KK/26 0 0/26 0 0ss00 KK 0 0 0 0

Condition for Condition for critical critical stabilitystability80-880-8KK/26 > 0 /26 > 0 or or KK<260<260The auxiliary equationThe auxiliary equation26 26 s s 22 + 260 = 0 + 260 = 0solvingsolving

008080363688 223344 KKssssssss

jjjjss 1616..331010

Root Locus Plot - Example The final root locus plot

is ---------

What is the value of the gain K corresponding to the breakaway point at sb = –2 ± 2.45j ?

––44XX

––22

––22jj

––44jjXX

XX3.163.16jj

Used to check the stability of CL system from OL response. OL stable system may become unstable in Closed loop and vice-versa.

Pole – Zero configuration from Nyquist point of view

Concept of Number of encirclements

Mapping theorem or Principle of Argument

Nyquist stability criteria

Single Valued Function

A single valued function is defined as for each point in the s-plane there is only one corresponding point in the q(s)-plane.

Let a function q(s) = (s-α1)(s-α2)…….. (s-αm) (s-β1)(s- β2)…….. (s- βn)

Let s = σ + jω, a complex variable in S-Plane, then q(s) is also complex and it may be defined as q(s) = u + jv on the complex plane.

A function q(s) is said to be Analytic in the S-plane if the function and all its derivatives exist.

The points in the S-plane where the function q(s) or its derivatives does not exist are called Singular points.

The poles of the function q(s) are singular points.

Encircled:If a point or region is found inside a closed path (closed contour) in a complex plane then it is said to be encircled

S1 is encircled by the closed path

S2 is not encircledMay 1, 2023 85BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING

Enclosed:Any point or region is said to be enclosed by a closed path, if it is found to lie to the right of the path when the path is traversed in the prescribed direction

The shaded region are the region enclosed by the closed path.

In Fig. 1, the point A is enclosed

In Fig. 2, the point B is enclosed

Fig. 1 Fig. 2

For a contour in s-plane, which does not go through any singular point, there corresponds a contour in the q(s) plane

A function G(s)H(s) = Open Loop Transfer Function

Poles of G(s)H(s) = Open Loop Poles

Zeros of G(s)H(s) =Open Loop Zeros

A function G(s) / [1+G(s)H(s)] = Closed Loop Transfer Function

Let a function F(s) = 1 + G(s)H(s)

Zeros of {F(s) = 1 + G(s)H(s)} = Closed Loop Poles of a systemPoles of {F(s) = 1 + G(s)H(s)} = Open Loop Poles of a system

For a stable system, the roots of the characteristic equation and hence the zeros of F(s) must lie in the Left Half of the S-Plane.

If q(s) is a single valued function, that has a finite number of poles & zeros in the s-plane and C is a contour in s-plane that does not go through any of the poles & zeros q(s); then the corresponding locus Cq mapped in the q(s)-plane will encircle the origin as many times as the difference between the number of zeros and poles of q(s) that are encclosed by the s-plane contour C.

N = Z – PIf N is positive, encirclement is CW direction.

If N is negative, encirclement is ACW direction.

1. N > 0 (Z > P)

If the contour in the s-plane encloses more no. of zeros than poles of q(s) in CW direction then the corresponding locus Cq mapped in the q(s)-plane will encircle the origin N times in the CW direction (same direction).

2. N < 0 (P > Z)

If the contour in the s-plane encloses more no. of poles than poles of q(s) in CW direction then the corresponding locus Cq mapped in the q(s)-plane will encircle the origin N times in the ACW direction (opposite direction).

3. N = 0 (P = Z)

If the contour in the s-plane encloses as many poles as zeros, or no poles and no zeros then the corresponding locus Cq mapped in the q(s)-plane will not encircle the origin.

s j R e jθ

Along C1, s varies from -j∞ to + j∞

Along C2, s = R e jθ ; where R→∞ and θ varies from + Π/2 to –Π/2

Let P = no. of poles of q(s)-plane lying on Right Half of s-plane and encircled by s-plane contour.

Let Z = no. of zeros of q(s)-plane lying on Right Half of s-plane and encircled by s-plane contour.

For the CL system to be stable, the no. of zeros of q(s) which are the CL poles, that lie in the right half of s-plane should be zero. That is Z = 0, which gives N = -P.

Therefore, for a stable system the no. of ACW encirclements of the origin in the q(s)-plane by the contour Cq must be equal to P.

We know that q(s) = 1+G(s)H(s)

Therefore G(s)H(s) = [1+G(s)H(s)] – 1

The contour Cq, which has obtained due to mapping of Nyquist contour from s-plane to q(s)-plane (ie)[1+G(s)H(s)] -plane, will encircle about the origin.

The contour CGH, which has obtained due to mapping of Nyquist contour from s-plane to G(s)H(s) -plane, will encircle about the point (-1+j0).

Therefore encircling the origin in the q(s)-plane is equivalent to encircling the point -1+j0 in the G(s)H(s)-plane.

If the contour CGH of the open loop transfer function G(s)H(s) in the G(s)H(s)–plane corresponding to Nyquist contour in the s-plane encircles the point -1+j0 in the counterclockwise direction as many times as the right half s-plane poles of G(s)H(s), then the closed loop system is stable

Nyquist Stability Criteria

Stable

Unstable

ExampleProblem : Construct the Nyquist stability plot for a feedback system with the following open-loop transfer characteristic :

Solution : For section ab, s = j , : 0

The polar plot may be constructed by calculating G(j )H(j ) for increasing values of :

)1s)(10s2s(20)s(H)s(G 2

)1j)(102j(20)j(H)j(G 2

(rad/s) 0 0.5 1.0 2.0 2.5 3.0 3.5 4.0 5.0

G(j )H(j ) 2.0+j0

1.54-j0.98

0.82-j1.29

-0.15-j1.23

-0.62-j1.02

-0.92-j0.49

-0.75+j0.02

-0.45+j0.19

-0.15+j0.15

0-270o

Example : contd.

On section bcd, s = RejR ; therefore( ie.), all of section bcd maps onto the origin

0R20)s(H)s(G 3

Re-1 0 1 2

G(s)H(s) - plane

Polar plot of G(s)H(s)

Example : solution contd.On section de, s = –j , : – 0

Therefore, the map of de is the image of the map of ab : G(–j)H(–j) = [G(j)H(j)]*

Re-1 0 1 2

-0.5 G(s)H(s) - plane

Note : the arrows indicate the direction along the contour on section de ; i.e decreasing frequency from – to zero

Example : solution contd.The complete Nyquist plot is presented as :

Clearly the point (-1,0) is outside the contour the system is stable

Note : the interior of the Nyquist contour is the region lying to the

right of the path (relative to the direction of the

arrows)G(s)H(s) - plane

Re-1 1 20

Poles lying on imaginary axis

If the open-loop transfer function has poles lying on the imaginary axis, and in particular at the origin, the Nyquist contour must be modified as shown :

» i.e. section efa bypasses the origin rather than passing through it

» On section efa : s = rejr0, : –90o to +90o

ExampleProblem : Sketch the Nyquist stability plot for a feedback system with the following open-loop transfer function :

Solution : For section ab, s = j, : 0

(i) 0 : G(j )H(j ) –1 – j(ii) = 1 : G(j )H(j ) –1+ j0(iii) : G(j )H(j ) 0–270o

)j1(j1)j(H)j(G 2

)1ss(s1)s(H)s(G 2

Example : contd.

On section bcd, s = RejR ; therefore

i.e. section bcd maps onto the origin of the G(s)H(s)-plane

Polar plot of G(s)H(s)

0R1)s(H)s(G 3

Re-1.5 -1 -0.5 0 0.5

Example : solution contd.Section de maps as the complex image of the polar plot as before :

Re-1.5 -1 -0.5 0 0.50

Example : solution contd.Finally on section efa :

, : –90o to +90o

This defines a semi-circle of infinite radius in the clockwise direction; i.e. from +90o to –90o

» Note : (-1,0) is neither inside nor outside the Nyquist plot The system is marginally stable

s = rejr0

1)s(H)s(G

Re-1.5 -1 -0.5 0.5

Polar plot : s = j, : 0

(i) 0 : G(j )H(j ) 0o

(ii) = 1 : G(j )H(j ) = 0.5 – j 0.5(iii) : G(j )H(j ) 0–90o

Example Consider a system with open-loop transfer function :

)1s(s1)s(H)s(G 4

)j1(1))H(jG(j 4

0 0.5 1 1.5

Contd… On section bcd, s = RejR ; therefore Section de maps as the complex image of the polar plot as

before However, on section efa :

s = rejr0, , : –90o to +90o

360to360

er1)s(H)s(G

01)()( 4 R

0 0.5 1 1.5

Since N=2; CL system is not stable

1. Nyquist plot for the system G(s)H(s) = K / [s(s+2)(s+10)]

SAMPLED DATA SYSTEMS

What is a compensator?

One that compensates any deficiencies in the system

One that compensates any inadequacies in the system

It is also a sub system placed inside the overall system

What do a compensator do?

Very simply, a compensator adds a zero or pole to a system

Study of the influence of a pole or zero addition inside a system

Improves STABILITY & PERFORMANCE

Compensator & Controller

What is the difference between a compensator and controller?

When two elements are doing mathematically the same function, what will be the difference?

PI, PD & PID Controllers

Lag, Lead & Lag Lead Compensator

Compensator - Definition

An additional sub system added in to the closed loop to improve the stability and performance of the overall system

Types of Compensators(Based on placement)

Cascade / Series Compensators

Feedback / Parallel Compensators

Input compensators

Output Compensators

Compensator

Process

Feedback

Compensator

Process

Feedback

Process

Feedback

Compensator

Process

FeedbackCompensator

Types of compensators

Feedback Compensator

Cascade Compensator

Output or Load Compensator

Input Compensator

Types of compensators (Based on working)

Lead Compensators

Lag Compensators

Lag – Lead Compensators

Lag compensator Have the nature of a lag network

For a sinusoidal input, output will have a phase lag (steady state)

Integrator or PI Controller or Low pass filter

Improves the steady state performance

What happens when we use a lag compensator in a system? Bandwidth reduced

Sluggish in nature that increases the rise time, settling time

System becomes more sensitive to parameter variations

Improves steady state accuracy

Decreases the stability of the system

Lead compensator Have the nature of a lead network

For a sinusoidal input, output will have a phase lead (steady state)

Differentiator or PD Controller or High pass filter

Improves the Transient state performance

What happens when we use a lead compensator in a

system? Addition of Zero and a pole to the system Zero added to the right of the pole Add more damping to the closed loop system Reduction of Rise time and Settling time –

Increase in Bandwidth Improves phase margin Improves stability Improves bandwidth No change in SS Error

Limitations of Lead Compensators

Single stage Lead Compensator fails if the phase angle required is above 60 degree. In that case use of multi stage compensators

As the bandwidth increases, the possibility of noise to enter in to the system also increases

Design methods Time Domain – Root Locus Frequency Domain – Bode Plot

Not like a PID tuning where sometimes, system TF not required

Essentially system TF is required for design of Compensators

Design Procedure1. Analysis of the existing uncompensated

system (Bode Plot / Root Locus)2. Altering the performance parameters

(Ex. Phase margin)3. Finding the compensated system function4. Analysis of the compensated system

(Bode / Root Locus)5. Design Verification

Lead Compensatoruses phase advance

at the zero-crossingincreases high

frequency gain and shifts wc to the right

increases bandwidthlimited amount of

compensation possible

Lag Compensator uses gain reduction to

shift the zero-crossing to the left

introduces phase lag at frequencies below wc

reduces bandwidth very large amount of

compensation possible

Characteristics of Lead andLag Compensators

Comparison Lead compensator, achieves

desired results through the merits of its property of phase lead contribution

Lag compensator achieves desired results through the merits of its property of attenuation at high frequency

Concluding Comment Selection of compensator

depends on the designer’s experience in the process

Another factor for consideration is the availability of components (Ex. Non Electrical systems)

E(s)R(s) C(s)

+ – cascade compensator

Design Criteria and Specifications

Design Criteria:accuracystability (absolute and

relative)speed of response

Time domain: steady-state error% overshoot rise time, Tr

time to 1st peak, Tp

settling time, Ts

Frequency domain:bandwidthphase margingain margin

s-plane:damping ratio, znatural frequency, wn

damping constant, s

Effects of Cascading a Controller

Increase Kc to reduce steady-state error

Increase Kc to increase the speed of response

What is the effect of increasing Kc on the closed-loop stability?

Examine the Nyquist plot and Bode plot

E(s)R(s) C(s)

+ – cascade compensator

Effects of Cascading a Controller Consider the following plant

characteristics.

1)( sH

)255)(2.0(50)( 2

If Kc = 1, then the position error constant is Kp = 10 and the steady-state error for a step input is

Increasing Kc to 3 will reduce the steady-state error to 3.2%

%1.9%1001

Effects of Increasing Kc For Kc = 3, the gain curve is

shifted up by about 10 dB, shifting the zero-crossing to the right and increasing the bandwidth

For Kc = 1, the phase margin is about 55, but with Kc = 3, the phase margin becomes -10, resulting in an unstable closed-loop systemFrequency (rad/sec)Frequency (rad/sec)

Bode Diagrams

Effects of Increasing Kc

On the Nyquist plot, increasing the gain Kc from 1 to 3 results in encirclements of the –1 point and the system becomes unstable. Note with Kc = 1, the gain margin is about 2.5 or about 8 dB

How can the loop characteristics be changed to allow a gain of Kc = 3 and keep the system stable?

Real AxisReal Axis

Nyquist DiagramsNyquist Diagrams

-1.5-1.5 -1-1 -0.5-0.5

-1.5-1.5

-0.5-0.5

0.50.5

1.51.5

––11

Kc = 3

Reshaping the Loop Frequency Characteristics

To achieve a stable system while keeping Kc = 3, phase advance may be added to the system in the frequency range where |GH(jw)| = 1

This may be done through cascade phase-lead compensation

Real AxisReal Axis

Nyquist DiagramsNyquist Diagrams

-1.5 -1 -0.5

Kc = 3

|GH|=1

Phase-Lead Compensation

Add phase-lead compensation to the controller.

This introduces phase advance of 70 and a gain increase of 15 dB at the new zero-crossing frequency, wc = 9.9 rad/s

sssGc 017.01

6.013)(

Frequency (rad/sec)Frequency (rad/sec)

Gm=8.1303 dB (at 15.324 rad/sec)Gm=8.1303 dB (at 15.324 rad/sec) Pm=15.772 deg. (at 9.9897 rad/sec)Pm=15.772 deg. (at 9.9897 rad/sec)

10-2 10-1 100 101 102 103

uncompensated

compensator

compensated

m = 16 at c = 9.9 rad/s

Phase-Lag Compensation

Add phase-lag compensation to the controller

This introduces gain reduction at and a small phase lag at the new zero-crossing frequency, wc = 3.8 rad/s

Gm=3.5624 dB (at 4.9746 rad/sec)Gm=3.5624 dB (at 4.9746 rad/sec) Pm=28.6 deg. (at 3.7882 rad/sec)Pm=28.6 deg. (at 3.7882 rad/sec)

compensator

uncompensated

compensated

sssGc 3.31

8.113)(

m = 29 at c = 3.8 rad/s

Phase-Lead Compensator Characteristics

Time constant form

where a > 1

Pole-zero form

Note: | z | < | p |

ssKsG cc

passivepassivenetworknetwork

amp.amp.gaingain

wherewhere1 , 1 pz

pszsKsG cc

Bode Plot Design of Phase-Lead Compensators

Three parameters to be chosen in the design

Kc : the compensator amplifier gain, chosen to meet the required low-frequency gain, or bandwidth requirements

t : the time constant of the compensator, chosen to place the phase advance at the desired frequency

a : the spread of the pole and zero of the compensator, defining the maximum phase advance

KsG cc

Bode Plot Design of Phase-Lead Compensators: Design Steps

1. Loop gain requirement: Determine the loop gain required to meet the

steady-state error specifications

2. Bode plot: Plot the loop transfer function including the

required gain increase from step 1

3.Required phase advance: (calculate a) Determine the required qm to meet the phase

margin specification at the predicted zero-crossing frequency

Bode Plot Design of Phase-Lead Compensators: Design Steps

Calculate from,

Check the new zero-crossing frequency based on the gain increase of ½| |dB to see if the phase margin will be met with the phase advance, m . Some iteration may be required

c , m m ½| |dB

specs.11sin

h.f. gainincreaseBode

Bode Plot Design of Phase-Lead Compensators: Design Steps4. Required t :

Select t such that, is at the new predicted zero-crossing frequency

5. Required Kc : Select Kc to give the required gain increase in the loop transfer

function. Remember, the compensator low-frequency gain is Kc/a

6. Implement:

KsG cc

Phase-Lead Design - Example 1

The plant

Specifications:

steady-state error <5% for a ramp input

Phase margin >45°

Steady-state error:

Draw the Bode plot of

05.02/

1)(lim0

sEsesss

1)()(1

Initial m = 18° , so choose m = 45° – 18° + 3° = 30° .

The gain increase is½| |dB =4.8 dB .

The new predicted c = 8.5 rad/s and m(uncomp) = 13°

50.0)30sin(

Phase-Lead Design ExamplePhase-Lead Design Example

10 0 10 1 10 2-180

Second iteration: choose qm = 45° – 13° + 3° = 35°

The gain increase is½| a |dB =5.7 dB

The new predicted wc = 8.7 rad/s and fm(uncomp) = 12°

Calculate t

Calculate the gain Kc

The final compensator

57.0)35sin(

0606..0077..3377..88

1481484040404011 cccc KKKK

amplifierssssssGGcc 0606..0011

2222..001177..33

11148148))((

Phase-Lead Design - Example 1Final Bode Plot

Phase-Lead Compensation DesignPhase-Lead Compensation Design

10 --1 10 0 10 1 10 2-180

8.7rad/s

compensateduncompensated

Time (sec.)Time (sec.)

Closed-Loop Step ResponseClosed-Loop Step Response

0 1 2 3 4 50

compensateduncompensated, K=40

uncompensated, K=1

Phase-Lead Design - Example 1Closed-Loop Step Response

Phase-Lag Compensator Characteristics

Time constant form

where ’ > 1

Pole-zero form

Note: | p | < | z |

passivenetwork

amp.gain where

KsG cc '11

ppsszzssKKssGG cc

11 ,,zz 11pp ''

Bode Plot Design of Phase-Lag Compensators

Three parameters to be chosen in the designKc : the compensator amplifier gain, chosen to

meet the required low-frequency gain requirements

t : the time constant of the compensator, chosen to place the zero of the compensator, one decade below the new zero-crossing frequency

a : the required high frequency gain reduction

KKssGG cccc ''1111

Bode Plot Design of Phase-Lag Compensators: Design Steps

1. Loop gain requirement:Determine the loop gain required to meet the steady-

state error specifications

2. Bode plot:Plot the loop transfer function including the required

gain increase from step 1

3. Required zero-crossing frequency: (calculate a’)Find the frequency, wc where the specified phase

margin + 5 would be met, if it were the zero-crossing frequency

Determine the gain reduction required to make wc the zero-crossing frequency

Calculate a from the required gain reduction = | a |dB

4. Required t :Select t such that the zero, 1/t is one decade

below the new zero-crossing frequency1010

''11 cc

5. Required Kc :Select Kc to give the required low-frequency

gain increase

6. Implement:

KKssGG cccc ''1111

Phase-Lag Design - Example 1

The plant

Specifications: steady-state error <5% for a

ramp input. Phase margin >45°

The first part of the design is the same as the phase-lead design

Steady-state error:

Draw the Bode plot of

4040 KK

0505..0022//

11))((limlim00

ssEEsseessssss

11))(())((11

11))((22

ssKKssss

ssRRssKGKG

ssEEpp

224040

ssGGpp

Initial fm = 18° at wc = 6.5 rad/s

Frequency at which fm = 45° + 5° = 50° (where the phase plot = –130°)wc = 1.7 rad/s

The required high frequency gain reduction is | a |dB = 20dB, a = 10

Phase-Lag Design ExamplePhase-Lag Design Example

10--2 10--1 10 0 10 1

1.7 6.5

Calculate t

Set the compensator zero at one decade below wc

The final compensator

17.010

7.1110

sssGc 8.581

88.5140)(

Phase-Lag Design - Example 1Final Bode Plot

Phase-Lag Design ExamplePhase-Lag Design Example

10--2 10--1 10 0 10 1

compensated

uncompensated

Phase-Lag Design - Example 1Closed-Loop Step Response

Time (sec.)Time (sec.)

Lead and Lag Design Step ResponseLead and Lag Design Step Response

0 1 2 3 4 50

uncompensated, K=40

uncompensated, K=1

lag compensatedlead compensated

Bode Plot Design of Lead-Lag Compensators

Five parameters to be chosen in the design are,Kc : amplifier gain, low-frequency gain requirements

a : the desired phase advance, qm

t : lead time constant, to place qm at the design zero- crossing frequency, wc´

a : the required high frequency gain reduction

t : lag time constant to place the lag zero, one decade below wc´

1. Loop gain requirementDetermine the loop gain required to meet the steady-

state error specifications

2. Bode plotPlot the loop transfer function including the required

gain increase from step 1

3. Desired Phase Advance (calculate a)Select a desirable amount of phase advance qm and

calculate the corresponding a

4. Required Gain Reduction (calculate a )Determine the frequency, wc where the design

phase margin would be achieved, accounting for the phase advance, qm and the 5° of the tail of the lag

fm(uncomp) = fm(spec) – qm + 5°

Determine the gain reduction required to make wc the zero-crossing frequency and calculate a

| gain reduction |dB = |1/ a |dB + ½| a |dB

5. Required t and t :Select t to place the qm at wc ; wc =

Select t such that the zero, 1/t is one decade below wc ;

6. Required Kc :Select Kc to give the required low-frequency gain increase.

7. Implement:

1010''11 ''

KKssGG cccc ))((ss

''ssss

BIBILIOGRAPHYTEXT BOOKS1. I.J. Nagrath and M. Gopal, ‘Control Systems Engineering’,

New Age International Publishers, 2003.2. Benjamin C. Kuo, Automatic Control systems, Pearson

Education, New Delhi, 2003.

REFERENCE BOOKS1. K. Ogata, ‘Modern Control Engineering’, 4th edition, PHI,

New Delhi, 2002.2. Norman S. Nise, Control Systems Engineering, 4th Edition,

John Wiley, New Delhi, 2007.3. Samarajit Ghosh, Control systems, Pearson Education, New

Delhi, 20044. M. Gopal, ‘Control Systems, Principles and Design’, Tata

McGraw Hill, New Delhi, 2002.

Thank You

Ee2365 nol part 2

Engineering

Transcript of Ee2365 nol part 2

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