10 elec3114

Dr Branislav HredzakControl Systems Engineering, Fourth Edition by Norman S. Nise

Copyright © 2004 by John Wiley & Sons. All rights reserved.

1

Frequency Response Techniques

• The definition of frequency response

• How to plot frequency response

• How to use frequency response to analyze stability

• How to use frequency response to analyze a system's transient and steady state error performance

• How to use frequency response to design the gain to meet stability specifications



2

Introduction

Frequency response technique has distinct advantages:

1. when modeling transfer functions from physical data

2. when designing compensators to meet a steady-state error requirement and a transient response requirement

3. when finding the stability of nonlinear systems

4. in settling ambiguities when sketching a root locus



3

The Concept of Frequency Response

The steady-state output sinusoid is:

… magnitude frequency response

… phasefrequency response

t



4

Analytical expression for the frequency response can be obtained from the transfer function G(s) of the system as

ωω jssGjG →= |)()(

Plotting of the frequency response )()( ωφω GGMjG ∠=

• As a function of frequency, with separate magnitude and phase plots



5

Problem Find the analytical expression for the magnitude frequency response and the phase frequency response for a system G(s) = 1 / ( s + 2). Also, plot the separate magnitude and phase diagrams.

Solution

ωjs =Substitute into2

1)(+

=s

sG

42

21)( 2 +

−=

+=

ωω

ωω j

jjG

41)()(2 +

==ω

ωω MjG

⎟⎠⎞

⎜⎝⎛−=

2tan)( ωωφ



6

4

1log20)(log202 +

=ω

ωM



7

Asymptotic Approximations: Bode Plots

• Bode plots or Bode diagrams: curves of the log-magnitude and phase frequency response as functions of log ω

Consider:

Then, the magnitude response is given by:

Converting the magnitude response into dB, we obtain:



8

Phase frequency response- is the sum of the phase frequency response curves of the zero terms minusthe sum of the phase frequency response curves of the pole terms.



9

Bode Plots for G(s) = (s + a)

The magnitude response in dB is:

At low frequencies when ω approaches zero:

- low-frequency asymptote



10

At high frequencies, where ω>>a :

- high-frequency asymptote



11

Phase response

From the above equation:

• at the break frequency, a, phase is 450

• at low frequencies the phase is 00

• at high frequencies the phase is 900



12

• It is often convenient to normalize the magnitude and scale the frequency so that the log-magnitude plot will be 0 dB at a break frequency of unity.

• Normalization and scaling makes it easier to add components to obtain the Bode plot of a function such as

• To normalize (s + a), factor out the quantity a and form a[(s/a)+ 1].

• The frequency is scaled by defining a new frequency variable, s1 = s/a.

• The magnitude is divided by the quantity a to yield 0 dB at the break frequency.

• Hence, the normalized and scaled function is (sl + 1).



13

… original magnitude response

… normalized and scaled magnitude response



14

… original phase response

… normalized and scaledphase response



15

Similarly, we can find the normalized and scaled Bode plots for:

G(s) = s G(s) = 1/s



16

G(s) = (s + a) G(s) = 1/(s + a)



17

Problem Draw the Bode plots for the system G(s) = K(s + 3)/[s(s + l)(s + 2)].

Solution

First we normalize G(s),

• the break frequencies are at 1, 2, and 3

To find value of G(s) at ω=0.1 we substitute s=0 for all (s/a+1) terms and

the actual value of s=j0.1 for term.s

K23

Hence KKjG 151.0/23)1.0( =≈

For K=1: dBjG 52.2315log20))1.0(log(20 ==



18



19

1



20

At low frequencies:

At high frequencies:

Break frequency: ωn

Bode Plots for G(s) = s2 + 2ζωns + ω2n

After scaling and normalization



21

At low frequencies: Phase = 00

At high frequencies: Phase = 1800

Phase at ωn :22)( nn jjG ζωω =

- hence phase at ωn is 900



22

Corrections to Second-Order Bode Plots G(s) = s2 + 2ζωns + ω2n

• there is an error between the actual response and the asymptotic approximation of the second-order polynomial

Actual response:

a magnitude correction of +20log2ζ can be made at the natural, or break, frequency



23

a magnitude correction of -20log2ζ can be made at the natural, or break, frequency

Bode Plots for G(s) = 1/ (s2 + 2ζωns + ω2n)



24

Problem Draw the Bode log-magnitude and phase plots of

)252)(2()3()( 2 +++

+=

sssssG

Solution



25

-the correction can be found by plotting a point-20log(2x0.2)=7.96 dB above the asymptotes at ωn=5

2.052)252( 222

=→=++↔++

ζωωζω

n

nnsss

dB4.24)50/3log(20 −=



26



27

lntroduction to the Nyquist Criterion

• relates the stability of a closed-loop system to open-loop poles location

• the Nyquist criterion can tell us how many closed-loop poles are in the right-half of the complex plane

• it is based on mapping of a contour through the function G(s)H(s)



28

Mapping

• Contour A can be mapped through F(s)=G(s)H(s) into contour B (Nyquistdiagram) by substituting each point of contour A into the function F(s)and plotting the resulting complex numbers.

Contour A

-1

Q



• Stability can be determined using equation Z = P – N

whereP… the number of open-loop poles of G(s)H(s) enclosed by the contour AN…the number of encirclements the Nyquist diagram (contour B) makes about (– 1)Z…the number of right-half-plane poles of the closed-loop system

Nyquist Criterion

Contour A

-1

Note: Counterclockwise encirclements are positive, and clockwise encirclements are negative



30

Applying the Nyquist Criterion to Determine Stability

P=0N=0Z=P-N=0 … system is stable

P=0N= - 2Z=P-N=2 … system is unstable

Note: Counterclockwise encirclements are positive, and clockwise encirclements are negative



31

Sketching the Nyquist Diagram

• The contour that encloses the right half-plane can be mapped through the function G(s)H(s) by substituting points along the contour into G(s)H(s)

• Approximations can be made to G(s)H(s) for points around the infinite semicircle by assuming that the vectors originate at the origin. Thus, their length is infinite, and their angles are easily evaluated.



32

Problem Sketch the Nyquist diagram for the system

G(s) = 500/[(s + 1)(s + 3)(s + 10)].

Solution

43=ω

14/30=ω



33

• If there are open-loop poles situated along the contour enclosing the right half-plane, then a detour around the poles on the contour is required; otherwise, a complete sketch of the Nyquist diagram cannot be made.

• each pole's vector rotates through + 1800 as we move around the contour near that pole

• detour must be only an infinitesimal distance into the right half-plane, or else some closed-loop, right-half plane poles will be excluded in the count



34

• each pole's vector rotates through - 1800

as we move around the contour near that pole

• detour must be only an infinitesimal distance into the left half-plane, or else some closed-loop, left-half plane poles will be excluded in the count



35

Problem Sketch the Nyquist diagram of the unity feedback system G(s) = (s + 2)/s2.

Solution



36

Stability via the Nyquist Diagram

• Stability can be determined using equation Z = P – N

whereP… the number of open-loop poles of G(s)H(s) enclosed by the contourN…the number of encirclements the Nyquist diagram makes about (– 1)Z…the number of right-half-plane poles of the closed-loop system

• If the closed-loop system has a variable gain K in the loop we can visualizethe Nyquist diagram as expanding (increased gain) or shrinking (decreased gain)

• OR

• we can visualize the Nyquist diagram as remains stationary and the (- 1) point is moved along the real axis. First, we set the gain to unity and position the critical point at (- 1/K) rather than (- 1).



37

Stability via the Nyquist Diagram

P = 2, hence in order to obtain a stable system Z = P – N = 0, therefore N = 2,which means there must be 2 counterclockwise crossings



38

• the encirclements of the critical point can be determined from the mapping of the positive jω-axis alone

• this system is stable for the range of loop gain, K, that ensures that the open-loop magnitude is less than unity at that frequency where the phase angle is 1800 (or equivalently, - 1800)

Stability via Mapping Only the Positive jω-Axis- this is a simplified approach to determine stability

- there must be noencirclements for this system to be stable



39

• this system is stable if the open-loop magnitude is greater than unity at that frequency where the phase angle is 1800 (or equivalently, -1800)

Stability via Mapping Only the Positive jω-Axis

- there must be twocounterclockwise encirclements for this system to be stable



40

Problem Find the range of gain for stability and instability, and the gain formarginal stability, for the unity feedback system G(s) = K / [ ( s2 + 2s + 2)(s+ 2)].For marginal stability find the radian frequency of oscillation. Use the Nyquistcriterion and the mapping of only the positive imaginary axis.

Solution First we draw the Nyquist diagram for K = 1

[ ] 6 when 0)()(Im == ωωω jHjG

06

180201

201)()( ∠=−=

=ωωω jHjG

Hence, stable for K < 20unstable K > 20marginally stable for K = 20, and oscillation is √6



41

Gain Margin and Phase Margin via the Nyquist Diagram

• Gain margin and phase margin can be used to determine how stable a system is.

• Systems with greater gain and phase margins can withstandgreater changes in system parameters before becoming unstable.



42

• Gain margin, GM , is the change in open-loop gain, expressed in decibels (dB), required at 1800 of phase shift to make the closed-loop system unstable.

• Phase margin, ΦM ,is the change in open-loop phase shift required at unity gain to make the closed-loop system unstable.



43

Problem Find the gain and phase margin for the systemG(s) = K / [ ( s2 + 2s + 2)(s+ 2)] if K = 6 .

Solution:

[ ] 6 when 0)()(Im == ωωω jHjG

3.0)()(6

−==ω

ωω jHjG

Hence dBGM 45.10)3.0/1log(20 ==

The phase angle at ω = 1.253 rad/s is -112.30

000 7.67)180(3.112 =−−−=ΦMHence

)()( ωω jHjG has a unity gain at ω =1.253 rad/sUsing computer it can be found that



44

Stability, Gain Margin, and Phase Margin via Bode Plots

• Bode plots are subsets of the complete Nyquist diagram but in another form

• They can be easily drawn

Evaluating Gain and Phase Margins using Bode Plots



45

Relation between Closed - Loop Transient and Closed - Loop Frequency Responses

Damping Ratio and Closed-Loop Frequency Response

• There will be no peak at frequencies above zero if ζ>0.707



46

Closed-loop frequency response peak vspercent overshoot for a two-pole system

Since ζ is related to percent overshoot, we can plot Mp vs. percent overshoot

)100/(%ln)100/ln(%

22 OSOS

+

−=

πζ



47

Response Speed and Closed-Loop Frequency Response

Bandwidth of the closed-loop frequency response is defined as the frequency, ωBW , at which the magnitude response curve is 3 dB down from its value at zero frequency

dB 3- toscorrespond 2/1=M



48



49



50Relation between Closed- and Open-Loop Frequency Responses

Damping Ratio from Phase Margin

• a phase margin of 65.520 (ζ = 0.707) or larger is required from the open-loop frequency response to ensure there is no peaking in the closed-loop frequency response

( )ωζωω

ωζωωω

ωω

n

n

n

n

jjjjG

2)2( 2

22

+−=

+=



51Response Speed from Open-Loop Frequency Response

• the closed-loop bandwidth can be estimated from the open-loop frequency response

• It can be shown that, approximately, the closed-loop bandwidth, ωBW,equals the frequency at which the open-loop magnitude response is between -6 and -7.5 dB if the open-loop phase response is between -1350

and -2250

ωBW=3.5rad/s



52

Steady-State Error Characteristics from Frequency Response

Type 0 system

Initial value:

For Type 0 system:

- which is the same as the initial value



53

Type 1 systemThe initial -20 dB/decade slope can be thought of as originating from a function,

which crosses the frequency axis (G’=0dB) at

For Type 1 system:

- which is the same as the frequency intercept



54

Type 2 system The initial -40 dB/decade slope can be thought of as originating from a function,

which crosses the frequency axis (G’=0dB) at

For type 2 system:

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