Frequency Response Analysis and Design

Control Systems
LEC 4 DESIGN VIA FREQUENCY RESPONSE
BEHZAD FARZANEGAN
3/1/2020 PROVIDED BY: BF(B.FARZANEGAN@AUT.AC.IR) 1

Frequency Response
Frequency response methods, developed by Nyquist and Bode in the 1930s, are
older than the root locus method, which was discovered by Evans in 1948
(Nyquist,1932; Bode, 1945). The older method, is not as intuitive as the root locus.
However, frequency response yields a new vantage point from which to view
feedback control systems.
This technique has distinct advantages in the following situations:
1. When modeling transfer functions from physical data.
2. When designing lead 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

Sinusoidal frequency response
The steady-state output sinusoid is
• M is called Magnitude Freq Response
• Ф is called Phase Freq Response
The system function is given by
( ) ( ) ( ) ( ) [ ( ) ( )]i iM M M          o o
( )
( )
( )
( ) ( ) ( )
i
i
M
M
M



  

    
o
o

Analytical Expression for
Frequency response
2 2
( ) ( )
( )
As B
C s G s
s





1 2
( )ss
K K
C s
s j s j 
 
 
2 2
( )
and ( )
i
G
where M R j A B
M G j


  

= ( )
( )( )
As B
G s
s j s j

 

 
2
( )
1
(= )
ss
C s
partial fraction terms from
j s j
G
K
s
s
K
 




1 4 4 2 4 4 3
( ) ( )
2 2
i G i Gj ji G i GM M M M
e e
s j s j 
    
 
 

Plotting Frequency Response
 𝐺(𝑗𝜔) = 𝑀 )𝐺(𝜔 < 𝛷 )𝐺(𝜔 can be plotted in several ways; two of them
are
• As a function of frequency, with separate magnitude and phase plots;
• As a polar plot, where the phasor length is the magnitude and the phasor
angle is the phase.
When plotting separate magnitude and phase plots, the magnitude
curve can be plotted in decibels (dB) vs. logω, where dB = 20 log M. The
phase curve is plotted as phase angle vs. log ω.

Frequency response plots Example
Problem: Find the analytical expression for the magnitude and phase
frequency response for a system G(s)=1/(s+2). Plot magnitude, phase and
polar diagrams.
Solution: substituting s=jw, we get
The magnitude frequency response is
The phase frequency response is
2
1 (2 )
( )
( 2) ( 4)
j
G j
j


 

 
 
2
( ) ( ) 1/ ( 4)G j M    
2
20log ( ) 20log(1/ 4) vs. logM    
1
( ) tan ( / 2) vs. log  
  

Frequency response plots Example
Problem: Find the analytical expression for the magnitude and phase
frequency response for a system G(s)=1/(s+2). Plot the polar diagrams.
Solution: substituting s=jw, we get
2 1
( ) ( ) 1/ ( 4) tan ( / 2)M    
    

))....()((
))....()((
)(
21
21
n
m
k
pspspss
zszszsK
sG



 js
pspspss
zszszsK
jG
n
m
k




)(....)()(
)(....)()(
)(
21
21


jspss
zszsKjG
m


.....)(log20log20.....
)(log20)(log20log20)(log20
1
21
Asymptotic Approximations:
Bode plots
Consider the transfer function
The magnitude frequency response is
Converting the magnitude response into dB, we obtain
We could make an approximation of each term that would consists only of
straight lines, then combine these approximations to yield the total
response in dB

Asymptotic Approximations:
Bode plots
 The straight-line approximations are called asymptotes.
We have low frequency asymptote and high frequency asymptote.
 a is called the break frequency.
Bode plots for G(s) = (s + a)
• At low frequency when ω approaches zero 𝐺(𝑗𝜔) ≈ 𝑎
• The magnitude response in dB is 20log M=20log from ω =0.01a to a
• At high frequencies where ω ˃˃ a
3/1/2020 PROVIDED BY: BF(B.FARZANEGAN@AUT.AC.IR)
𝐺(𝑗𝜔) ≈ 𝑎(
𝑗𝜔
𝑎
) = 𝑎(
𝜔
𝑎
)∠90 𝑜 = 𝜔∠90 𝑜
20log𝑀 = 20log𝑎 + 20log
𝜔
𝑎
= 20log𝜔
9

Normalized and scaled Bode
plots for different systems

HW!
Problem : Draw the Bode plot for the system defined as
2
( 3)
( )
[( 2)( 2 25)]
s
G s
s s s


  

Nyquist diagram showing gain
and phase margins
Gain margin, GM, The gain margin 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, 𝛷 𝑀 , The phase margin is the change in open-loop
phase shift required at unity gain to make the closed-loop system
unstable.

HW!
Problem: find the gain and phase margins of system
2
6
( ) ( )
( 23 2)( 2)
G j H j
s s
  
  

Relation between Closed-Loop Transient
and Closed-Loop Frequency Responses
Using the open loop transfer function
And closed-loop transfer function
We evaluate the magnitude of the closed-loop freq response as
𝐺(𝑠) =
𝜔 𝑛
2
)𝑠(𝑠 + 2𝜉𝜔 𝑛
𝑇(𝑠) =
)𝐶(𝑠
)𝑅(𝑠
=
𝜔 𝑛
2
𝑠2 + 2𝜉𝜔 𝑛 𝑠 + 𝜔 𝑛
2
𝑀 = | )𝑇(𝑗𝜔 | =
𝜔 𝑛
2
𝜔 𝑛
2
− 𝜔2 2 + 4𝜉2 𝜔 𝑛
2
𝜔2
14

Relation between Closed-Loop Transient
and Closed-Loop Frequency Responses
To relate the peak magnitude to the damping ratio we find
Since damping ratio is related to percent overshoot we can plot MP vs.
percent overshoot
𝑀 𝑃 =
1
2𝜉 1 − 𝜉2 𝜔 𝑃 = 𝜔 𝑛 1 − 2𝜉2
15

Response Speed and Closed-
Loop Frequency Response
We can relate the speed of the time response ( as measured in settling
time, rising time, and peak time) and the bandwidth , 𝜔 𝐵𝑊 of the
Frequency response.
The Bandwidth is defined as the frequency at which the magnitude
response curve is 3 dB down from its value at zero frequency.
𝜔 𝐵𝑊 = 𝜔 𝑛 1 − 2𝜉2) + 4𝜉4 − 4𝜉2 + 2
𝜔 𝐵𝑊 =
𝜋
𝑇𝑃 1 − 𝜉2
1 − 2𝜉2) + 4𝜉4 − 4𝜉2 + 2
𝜔 𝐵𝑊 =
4
𝑇𝑠 𝜉
1 − 2𝜉2) + 4𝜉4 − 4𝜉2 + 2
𝜔 𝑛 =
𝜋
𝑇𝑃 1 − 𝜉2 𝜔 𝑛 =
4
𝑇𝑠 𝜉
16

Relation between Closed-Loop and
Open-Loop Frequency Responses
These circles are called
constant M circles and are the
locus of the closed-loop
magnitude frequency
response for unity feed back
systems.
If the polar frequency
response of an open-loop
function, G(s), is plotted and
superimposed on top of the
constant M circles, the closed-
loop magnitude frequency
response is determined by
each intersection of this polar
plot with the M circles.

Relation between Closed-Loop and
Open-Loop Frequency Responses
These circles are called
constant N circles. If the
polar frequency response
of an open-loop function,
G(s), is plotted and
superimposed on top of
the constant N circles, the
closed-loop phase
response is determined by
each intersection of this
polar plot with the N
circles.

Example
Problem: Find the closed-loop frequency response of the unity feedback
system using G(s)=50/[s(s+3)(s+6)].
Solution
Evaluate the open-loop frequency function
The Polar plot of the open-loop freq response ( Nyquist diagram) is
shown superimposed over the M and N circles in the Figure.
The closed-loop magnitude frequency response is depicted as
𝐺(𝑗𝜔) =
50
)−9𝜔2 + 𝑗(18𝜔 − 𝜔3
19

Nichols Charts
Nichols Charts
displays the constant
M and N circles in dB,
so that changes in gain
are as simple to handle
as in the Bode plots.
The chart is a plot of
the open-loop
magnitude in dB vs.
open-loop phase angle
in degrees.

Example
Nichols chart with frequency response for G(s) = K/[s(s + 1)(s + 2)]
superimposed. Values for K = 1 and K = 3.16 are shown.

Time vs. Freq. Domain Analysis
Control system performance generally judged by time domain
response to certain test signals (step, etc.)
• Simple for < 3 OL poles or ~2nd order CL systems. systems.
• No unified methods for higher-order.
 Freq. response easy for higher order systems.
• Qualitatively related to time domain behavior.
• More natural for studying sensitivity and noise susceptibility.

Root locus vs. Frequency
Stability and transient response design via gain adjustment
◦ Frequency response design methods, unlike root locus methods, can be
implemented conveniently without a computer or other tool except for
testing the design.
Transient response design via cascade compensation
• Frequency response methods are not as intuitive as the root locus
Steady-state error design via cascade compensation
• With frequency response techniques, we build the steady-state error
requirement right into the design of the lead compensator. in using root
locus there are an infinite number of possible solutions to the design of a
lead compensator

Frequency Response Specifications
 Crossover frequency:
 Gain Margin:
 Phase Margin:
 Bandwidth (CL specification)

Performance Specifications
Three requirements enter into the design of a control system: transient response,
stability, and steady-state errors.
 The Nyquist criterion tells us how to determine if a system is stable.
• Typically, an open-loop stable system is stable in closed-loop if the open-loop magnitude
frequency response has a gain of less than 0 dB at the frequency where the phase frequency
response is 180°.
 Percent overshoot is reduced by increasing the phase margin.
 the speed of the response is increased by increasing the bandwidth.
 Steady-state error is improved by increasing the low-frequency magnitude
response, even if the high- frequency magnitude response is attenuated.
High frequency region indicates complexity.
e
Gp(s)
+ Gc(s)
r y

25

Performance Specifications
Requirements on open-loop frequency response
The gain at low frequency should be large enough to give a high value for
error constants.
 At medium frequencies the phase and gain margins should be large
enough.
 At high frequencies, the gain should be attenuated as rapidly as possible
to minimize noise effects.
Compensators
 lead: Improves the transient response.
 lag: improves the steady-state performance at the expense of slower
settling time.
 lead-lag: combines both

Transient Response via Gain
The root locus demonstrated that we can often design
controllers for a system via gain adjustment to meet a
particular transient response.
 We can effect a similar approach using the frequency
response by examining the relationship between phase
margin and damping.

Gain Adjustment and the
Frequency Response
20dB
Wc2
Wc1
𝝎 𝑩𝑾 𝟐
𝝎 𝑩𝑾 𝟏
𝑮(𝒔) =
𝝎 𝒏
𝟐
)𝒔(𝒔 + 𝟐𝝎 𝒏 𝜻
28

Design using Phase Margin
 Recall that the Phase Margin is closely related to the damping ratio of
the system.
 For a unity feedback system with open-loop function
We found that the relationship between PM and damping ratio is given
by
𝑮(𝒔) =
𝝎 𝒏
𝟐
)𝒔(𝒔 + 𝟐𝝎 𝒏 𝜻
𝜁 ≈
𝑃𝑀
100
, 𝑃𝑀 < 70∘Why?!
29

 Given a desired
overshoot, we can
convert this to a
required damping
ratio and hence PM
 Examining the
Bode plot we can
find the frequency
that gives the
desired PM

 The design procedure therefore consists of
• Draw the Bode Magnitude and phase plots.
• Determine the required phase margin from the percent
overshoot.
• Find the frequency on the Bode phase diagram that
yields the desired phase margin.
• Change the gain to force the magnitude curve to go
through 0dB.

Phase Margin Example
Problem: For the following position control system shown here, find the
preamplifier gain K to yield a 9.5% overshoot in the transient response
for a step input

Phase Margin Example
Solution:
 Draw the Bode plot
 For 9.5% overshoot, 𝜁 = 0.6 and PM must be 59.2∘
 Locate frequency with the required phase at 14.8 rad/s
 The magnitude must be raised by 55.3dB to yield the cross over
point at this frequency
 This yields a K = 583.9

HW!
PROBLEM: For a unity feedback system with a forward transfer function
use frequency response techniques to find the value of gain, K, to yield
a closed-loop step response with 20% overshoot.

Designing Compensation
 As we saw previously, not all specifications can be met via simple
gain adjustment.
 We examined a number of compensators that can bring the root locus
to a desired design point.
 A parallel design process exists in the frequency domain.
 In particular, we will look at the frequency characteristics for the.
• PI Controller
• Lag Controller
• PD Controller.
• Lead Controller
Understanding the frequency characteristics of these controllers allows
us to select the appropriate version for a given design.

PI Controller
 We saw that the integral compensator takes on the form
This results in infinite gain at low frequencies which reduces steady
state error.
 A decrease in phase at frequencies lower than the break will also
occur.
The Bode plot for a PI controller looks like this.
The break frequency is usually located at a frequency substantially
lower than the crossover frequency to minimize the effect on the
phase margin.
𝑮 𝒄(𝒔) = 𝑲
𝒔 + 𝒛 𝒄
𝒔
36

Steady-State Error Characteristics
from Frequency Response
For type 0 system (the initial slope is 0)
For un-normalized un-scaled Bode plot, the low frequency magnitude is
20 log KP
For type 1 system (the initial slope is -20dB/decade)
For un-normalized un-scaled Bode plot, the intersection of the initial -
20dB/decade slope with the frequency axis is KV
For type 2 system (the initial slope is -40dB/decade)
For un-normalized un-scaled Bode plot, the intersection of the initial -
40dB/decade slope with the frequency axis is √Ka

Lag Compensation
 Lag compensation approximates PI control
 This is often rewritten as
where α is the ratio between zero-pole break points
 The name Lag Compensation reflects the fact that this compensator
imparts a phase lag.
The Bode plot for a Lag compensator looks like this.
This compensator effectively raises the magnitude for low frequencies
without affecting the stability of the system.
 The effect of the phase lag can be minimized by careful selection of the
center frequency.
𝑮 𝒄(𝒔) = 𝑲
𝒔 + 𝒛 𝒄
𝒔 + 𝒑 𝒄
, 𝒛 𝒄 > 𝒑 𝒄
38

Lag Compensation
 The function of the lag compensator as seen on Bode diagrams is to
• improve the static error constant by increasing only the low-frequency gain
without any resulting instability,
• increase the phase margin of the system to yield the desired transient
response
Nise suggests setting the gain K for S.S error, then designing a lag
network to attain desired PM.
• Franklin suggests setting the gain K for PM, then designing a lag
network to raise the low freq. gain without affecting system stability.

Lag Compensation Design
The design procedure (Franklin et al.) consists of the following steps:
1. Find the open-loop gain K to satisfy the phase margin specification
without compensation
2. Draw the Bode plot and evaluate low frequency gain
3. Determine a to meet the low-frequency gain error requirement
4. Choose the corner frequency ω=1/T to be one octave to one decade
below the new crossover frequency
5. Evaluate the second corner frequency ω=1/αT
6. Simulate to evaluate the design and iterate as required

Lag Compensation Example
Problem: Returning again to the previous example, we will now design a
lag compensator to yield a ten fold improvement in steady-state error
over the gain compensated system while keeping the overshoot at 9.5%

Lag Compensation Example
 In the first example, we found the gain K=583.9 would yield our desired
9.5% overshoot with a PM of 59.2o at 14.8rad/s.
For this system we find that
We therefore require a Kv of 162.2 to meet our specification
We need to raise the low frequency magnitude by a factor of 10 (or 20dB)
without affecting the PM.
First we draw the Bode plot with K=583.9
Set the zero at one decade, 1.48rad/s, lower than the PM frequency
The pole will be at 1/α relative to this so
𝑮 𝒄(𝒔) =
𝒔 + 𝟏. 𝟒𝟖𝟑
𝒔 + 𝟎. 𝟏𝟒𝟖
42

Lag Compensation Design(Nise)

HW!
PROBLEM: For a unity feedback system with a forward transfer function
use frequency response techniques to design a lag controller to yield a
closed-loop step response with 20% overshoot, and improve the steady-
state error tenfold.

PD Controller
The ideal derivative compensator adds a pure differentiator, or zero, to the
forward path of the control system
 The root locus showed that this will tend to stabilize the system by
drawing the roots towards the zero location.
 We saw that the pole and zero locations give rise to the break points in
the Bode plot.
The Bode plot for a PD controller looks like this.
The stabilizing effect is seen by the increase in phase at frequencies above
the break frequency
 The stabilizing effect is seen by the increase in phase at frequencies above
the break frequency.
)𝑮 𝒄(𝒔) = 𝑲(𝒔 + 𝒛 𝒄
45

Lead Compensation
 Introducing a higher order pole yields the lead Compensator
 This is often rewritten as
where 1/B>1 is the ratio between pole-zero break Point
We want to
• increase the phase margin to reduce the percent overshoot
• increase the gain crossover to realize a faster transient response.
 The name Lead Compensation reflects the fact that this compensator
imparts a phase lead.
The Bode plot for a Lead compensator looks like this.
The high frequency magnitude is now limited
𝑮 𝒄(𝒔) =
)𝑲(𝒔 + 𝒛 𝒄
𝒔 + 𝒑 𝒄
𝒛 𝒄 < 𝒑 𝒄
46

Lead Compensation
The lead compensator can be used to change the damping ratio of the
system by manipulating the Phase Margin
 The phase contribution consists of
The peak occurs at with a phase shift and magnitude of
This compensator allows the designer to raise the phase of the system
in the vicinity of the crossover frequency
Why?!
47

Lead Compensation Design
The design procedure consists of the following steps:
1. Find the open-loop gain K to satisfy steady-state error or bandwidth
requirements.
2. Evaluate phase margin of the uncompensated system using the value
of gain chosen above.
3. Find the required phase lead to meet the damping requirements.
4. Determine the value of β to yield the required increase in phase

Lead Compensation Design
5. Determine the new crossover frequency
6. Determine the value of T such that ωmax lies at the new crossover
frequency
7. Draw the compensated frequency response and check the resulting
phase margin.
8. Check that the bandwidth requirements have been met.
9. Simulate to be sure that the system meets the specifications (recall
that the design criteria are based on a 2nd order system).

Lead Compensation Example
Problem: Returning to the previous example, we will now design a lead
compensator to yield a 20% overshoot and Kv=40, with a peak time of
0.1s.

From the specifications, we can determine the following requirements
• For a 20% overshoot we find ζ=0.456 and hence a Phase Margin of 48.1⁰
• For peak time of 0.1s with the given ζ, we can find the require closed loop
bandwidth to be 46.6rad/s.
• To meet the steady state error specification

From the Bode plot, we evaluate the PM to be 34⁰ for a gain of 1440
 We can’t simply increase the gain without violating the other design
constraints
 We use a Lead Compensator to raise the PM We require a phase margin of
48.1 ⁰ The lead compensator will also increase the phase margin frequency so
we add a correction factor to compensate for the lower uncompensated
system’s phase angle.
 The total phase contribution required is therefore
48.1⁰ – (34 ⁰ – 10 ⁰) = 24.1 ⁰

Lead Compensation
 Based on the phase requirement we find
 the lead compensator’s magnitude is 3.76 dB at ωmax.
 Examining the Bode magnitude, we find that the frequency at which
the magnitude is -3.77dB is ωmax=39rad/s
 The break frequencies can be found at 25.3 and 60.2.
The compensator is
The resulting system Bode plot shows the impact of the phase lead.

HW!
PROBLEM: Design a lead compensator for the previous system to meet
the following specifications: %OS = 20%, Ts =0.2s and Kv = 50.

Lag-Lead Compensation
 As with the Root Locus designs we considered previously, we often
require both lead and lag components to effect a particular design
 This provides simultaneous improvement in transient and steady-state
responses
One method is to design the lag compensation to lower the high-
frequency gain, stabilize the system, and improve the steady-state
error and then design a lead compensator.
 In this case we are trading off three primary design parameters
• Crossover frequency ωc which determines bandwidth, rise time and settling
time
• Phase margin which determines the damping coefficient and hence
overshoot
• Low frequency gain which determines steady state error characteristics

The transfer function of a single, passive lag-lead network is
Design Procedure
1. Using a second-order approximation, find the closed-loop
bandwidth required to meet the settling time, peak time, or rise
time requirement
2. Set the gain, K, to the value required by the steady-state error
specification.
3. Plot the Bode magnitude and phase diagrams for this value of gain.

4. Using a second-order approximation, calculate the phase margin to meet the
damping ratio or percent overshoot requirement.
5. Select a new phase-margin frequency near ωBW.
6. At the new phase-margin frequency, determine the additional amount of phase
lead required to meet the phase-margin requirement. Add a small contribution
that will be required after the addition of the lag compensator.
7. Design the lag compensator by selecting the higher break frequency one decade
below the new phase-margin frequency.
8. Design the lead compensator. Using the value of γ from the lag compensator
design and the value assumed for the new phase-margin frequency, find the
lower and upper break frequencies for the lead compensator and obtaining for
T.
9. Check the bandwidth to be sure the speed requirement in Step 1 has been met.
10. Redesign if phase-margin or transient specifications are not met, as shown by
analysis or simulation.

Lag-Lead Compensation Example

HW!
PROBLEM: Design a lag-lead compensator for a unity feedback system
with the forward-path transfer function
to meet the following specifications: %OS=10%; Tp = 0:6 s, and Kv=10.
Use frequency response techniques.

Design procedure via Nichols chart
1. Calculate the damping ratio from the percent overshoot
requirement .
2. Calculate the peak amplitude, Mp, of the closed-loop response
3. Calculate the minimum closed-loop bandwidth to meet the peak
time requirement by peak time and the damping ratio from (1).
4. Plot the open-loop response on the Nichols chart.
5. Raise the open-loop gain until the open-loop plot is tangent to the
required closed-loop magnitude curve, yielding the proper Mp.
6. Place the lead zero at this point of tangency and the lead pole at a
higher frequency. (Zeros and poles are added in SISOTOOL by
clicking either one on the tool bar and then clicking the position on
the open-loop frequency response curve where you desire to add
the zero or pole.)

Design procedure via Nichols chart
7. Adjust the positions of the lead zero and pole until the open-loop
frequency response plot is tangent to the same Mp curve, but at the
approximate frequency found in (3). This yields the proper closed-
loop peak and proper bandwidth to yield the desired percent
overshoot and peak time, respectively.
8. Evaluate the open-loop transfer function, which is the product of
the plant and the lead compensator, and determine the static error
constant.
9. If the static error constant is lower than required, a lag
compensator must now be designed. Determine how much
improvement in the static error constant is required.
10. Recalling that the lag pole is at a frequency below that of the lag
zero, place a lag pole and zero at frequencies below the lead
compensator and adjust to yield the desired improvement in static
error constant

Lag-Lead Design Using the
Nichols Chart Example
PROBLEM: Design a lag-lead compensator for the plant
to meet the following requirements:
• a maximum of 20% overshoot
• a peak time of no more than 0.5 seconds
• a static error constant of no less than 6.
SOLUTION: We follow the steps enumerated immediately above,
1. ζ . 0=456 for 20% overshoot.
2. Mp =1.23=1.81 dB for ζ = 0.456.
3. ωBW =9.3 r/s for ζ = 0.456 and Tp = 0.5.

4. Plot the open-loop frequency response curve on the Nichols chart for
K = 1.
5. Raise the open-loop frequency response curve until it is tangent to
the closed-loop peak of 1.81 dB curve
6. Place the lead zero at this point of tangency and the lead pole at a
higher frequency.
7. Adjust the positions of the lead zero and pole until the open-loop
frequency response plot is tangent to the same Mp curve, but at the
approximate frequency found in 3.
8. Checking Designs/Edit Compensator . . . Shows
which yields a Kv = 3.

9. We now add lag compensation to improve the static error constant
by at least 2. Now add a lag pole at 0.004 and a lag zero at 0.008.
Readjust the gain to yield the same tangency as after the insertion of
the lead.

HW!
PROBLEM: Design a lag-lead compensator using the Nichols chart for a
unity feedback system with the forward-path transfer function
to meet the following specifications: %OS=10%; Tp = 0:6 s, and Kv=10.
Use frequency response techniques.

Conclusions
 We have looked at techniques for designing controllers
using the frequency techniques
 There is once again a trade-off in the requirements of the
system
By selecting appropriate pole and zero locations we can
influence the system properties to meet particular design
requirements

Antenna Control: Gain Design
PROBLEM: Given the antenna azimuth position control system shown
below, Configuration 1, use frequency response techniques to do the
following:
a. Find the preamplifier gain required for a closed-loop response of 20%
overshoot for a step input.
b. Estimate the settling time.

Antenna Control: Cascade
Compensation Design
PROBLEM: Given the antenna azimuth position control system block
diagram shown before, Configuration 1, use frequency response
techniques and design cascade compensation for a closed-loop
response of 20% overshoot for a step input, a fivefold improvement in
steady-state error over the gain-compensated system operating at 20%
overshoot, and a settling time of 3.5 seconds.

Frequency Response Analysis and Design

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to Frequency Response Analysis and Design

Similar to Frequency Response Analysis and Design (20)

Recently uploaded

Recently uploaded (20)

Frequency Response Analysis and Design

Editor's Notes