DESIGN VIA FREQUENCY RESPONSE

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DESIGN VIA FREQUENCY RESPONSE

• OSMAN PARLAKTUNA
• OSMANGAZI UNIVERSITY
• ESKISEHIR,TURKEY
• www.ogu.edu.tr/oparlak

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GAIN AND PHASE MARGINS
M(dB)
0dB
log ?
GM
Phase (degrees)
?M
log ?
-1800
?GM
??M
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DAMPING RATIO AND CLOSED-LOOP FREQUENCY RESPONSE
R(s)
C(s)
Differentiating with respect to ?2, and setting
the derivative equal to zero yields the maximum
value M, Mp at ?p.
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A TYPICAL CLOSED-LOOP FREQUENCY RESPONSE
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RESPONSE SPEED AND CLOSED-LOOP FREQUENCY RESPONSE
Another relation between the frequency response
and time response is between the speed of the
time response (settling time, peak time) and the
bandwidth of the closed-loop frequency response
(?BW). This frequency can be found by finding
that frequency for which M1/?2 (-3dB).
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DAMPING RATIO FROM PHASE MARGIN
The relation between the phase margin and the
damping ratio will enable us to evaluate the
percent overshoot from the phase margin found
from the open-loop frequency response.
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TRANSIENT RESPONSE VIA GAIN ADJUSTMENT
We have derived the relation between damping
ratio (equivalently percent overshoot) and phase
margin. Thus, if we can vary the phase margin, we
can vary the percent overshoot.
M(dB)
If we desire a phase margin represented by CD, we
would have to raise the magnitude curve by AB.
Thus, a simple gain adjustment can be used to
design phase margin, and percent overshoot.
log ?
A
B
Phase (degrees)
C
?M
log ?
-1800
D
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EXAMPLE
R(s)
K
Find the value of K to yield a 9.5 overshoot in
the transient response for a step input.
9.5 overshoot implies ?0.6 for the closed-loop
dominant poles. Then
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BODE DIAGRAM USING MATLAB
Using K3.6, plot the Bode diagram num360 denco
nv(1 0,conv(1 100,1 36)) bode(num,den)
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Locate on the phase plot the frequency that
yields a 59.20 phase margin. This frequency is
found where the phase angle is -120.80. The value
of the frequency is 15 rad/s. At this frequency,
the gain is found to be -45 dB. The magnitude has
to be raised to 0 dB to yield the required phase
margin. K1alog10(45/20)177.8. Since the
log-magnitude plot was drawn for K3.6, with K1,
the required gain is (3.6)(177.8)640
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BODE DIAGRAM OF THE GAIN-COMPENSATED SYSTEM
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STEP RESPONSE OF THE COMPENSATED SYSTEM
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LAG COMPENSATION
M(dB)
Uncompensated
Compensated
Kv
log ?
Lag compensator
log ?
Phase
Desired phase
-1800
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PROPERTIES OF LAG COMPENSATION
The function of the lag compensator is to 1)
Improve the static error constant by increasing
only the low-frequency gain without resulting
instability, and 2) increase the phase margin of
the system to yield the desired transient
response. In the example Bode diagram, the
uncompensated system is unstable since the gain
at 1800 is greater than 0 dB. The lag
compensator, while not changing the low-frequency
gain, does reduce the high-frequency-gain. Thus,
the low-frequency gain of the system can be made
high to yield a large Kv without creating
instability. This stabilizing effect comes about
because the gain at 1800 of phase is reduced
below 0 dB.
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DESIGN PROCEDURE

• Set the gain K to that value that satisfies the
  steady-state specification and plot the magnitude
  and phase diagrams for this value of gain.
• Find the frequency where the phase margin is 50
  to 120 greater than the phase margin that yields
  the desired transient response. This step
  compensates for the fact that the phase of the
  lag compensator may contribute from -50 to -120
  of phase at the phase-margin frequency.

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• Select a lag compensator such that the composite
  magnitude diagram goes through 0 dB at the
  frequency found in step 2 Draw the compensators
  high-frequency asymptote to yield 0 dB at the
  frequency found in step 2 select the upper
  cutoff frequency to be 1 decade below the
  frequency found in step 2 select the
  low-frequency asymptote to be at 0 dB connect
  the compensators high- and low-frequency
  asymptotes with a -20 dB/decade line to locate
  the lower break frequency.

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• Reset the system gain K to compensate for any
  attenuation in the lag network in order to keep
  the static error constant to the same as that
  found in step 1.

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FREQUENCY RESPONSE OF A LAG COMPENSATOR
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EXAMPLE
R(s)
K
Use Bode diagrams to design a lag compensator to
yield a tenfold improvement in steady-state error
over the gain-compensated system while keeping
the percent overshoot at 9.5.
For the gain-compensated system, K640 yields a
9.5 overshoot. Thus, for this system Kv17.78.
For a tenfold improvement in steady-state error,
Kv must increase by a factor of 10. Therefore,
the value of K is 6400.
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The open-loop transfer function is
The Bode plots for K6400 are shown on the
following slide
The phase margin required for a 9.5 overshoot is
59.20. We increase this value of the phase margin
by 10.80 to 700 in order to compensate for the
phase angle contribution of the lag compensator.
From the phase plot, 700 phase margin occurs at
10 rad/s. At this frequency magnitude plot must
go through 0 dB
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The magnitude at 10 rad/s is 25 dB. Thus, the
lag compensator must provide -25 dB attenuation
at 10 rad/s. First draw the high frequency
asysmptote at -25 dB. Arbitrarily select the
higher break frequency to be about one decade
below the phase-margin frequency, or 1 rad/s.
Starting at the intersection of this frequency
with the lag compensators high frequency
asymptote, draw a -20 dB/dec line until 0 dB is
reached. The compensator must have a dc gain of
unity to retain the value of Kv that have already
designed by setting K6400.
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LAG COMPENSATOR DESIGN
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The lower cutoff frequency is found to be 0.06
rad/s. Hence the lag compensators transfer
function is
where the gain of the compensator is 0.06 to
yield a dc gain of unity. The compensated
systems forward transfer function is
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LEAD COMPENSATION
When we designed the lag network to improve the
steady-state error, we wanted a minimal effect on
the phase diagram in order not to change the
transient response. However, in designing lead
compensators we want to change the phase diagram,
increasing the phase margin to reduce the percent
overshoot, and increasing the gain crossover to
realize a faster transient response. Results of
a lead compensator In frequency domain larger
bandwidth, larger phase margin and larger phase
margin frequency. In time domain Lower percent
overshoots with smaller peak times.
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M(dB)
Compensator
A
C
log ?
Compensated system
Phase
log ?
00
D
B
-1800
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The uncompensated system has a small phase margin
(B) and a low phase-margin frequency (A). Using a
phase lead compensator, the phase angle plot
(compensated system) is raised for higher
frequencies. At the same time the gain crossover
frequency in the magnitude plot is increased from
A rad/s to C rad/s. These effects yield a larger
phase margin (D), a higher phase-margin frequency
(C), and a larger bandwidth. One advantage of
the frequency response technique over the root
locus is that we can implement a steady-state
requirement and then design a transient response.
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LEAD COMPENSATOR FREQUENCY RESPONSE
The transfer function of a lead compensator is
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DESIGN PROCEDURE

• Find the closed-loop bandwidth required to meet
  the settling time or peak time.
• Since the lead compensator has negligible effect
  at low frequencies, set the gain K of the
  uncompensated system to satisfy the steady-state
  error requirement
• Plot the Bode diagram for this value of K and
  determine the uncompensated systems phase
  margin.
• Find the phase margin to meet the damping ratio
  or percent overshoot requirement. Evaluate the
  additional phase contribution required from the
  compensator

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• Determine the value of ?
• Determine the compensators magnitude at the peak
  of the phase curve.
• Determine the new phase-margin frequency by
  finding where the uncompensated systems
  magnitude curve is the negative of the lead
  compensators magnitude at the peak of the
  compensators phase curve.
• Design the lead compensators break frequencies.
• Reset the system gain to compensate for the lead
  compensators gain

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EXAMPLE
R(s)
K
Design a lead compensator to yield a 20
overshoot and Kv40, with a peak time of 0.1 sec.
20 overshoot corresponds to ?0.456. Then using
Tp0.1 and the equation
?BW46.6 rad/s is required.
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In order to meet the specification of Kv40, K
must be set at 1440, yielding G(s)144000/s(s36)
(s100)
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Required phase margin for a 20 overshoot
(?0.456) can be determined using the following
equation as 480.
The uncompensated system with K1440 has a phase
margin of 350 at a phase-margin frequency of 30
rad/s. Since the lead compensator will also
increase the phase-margin frequency, we should
add a correction factor to compensate for the
lower uncompensated phase angle at this higher
phase-margin frequency.
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Since we do not know the higher phase-margin
frequency, we assume a correction factor of 100.
Then the maximum phase angle of the lead
compensator is 480-350100230.
equation yields ?0.44 for ?max230.
Equation yields lead compensators magnitude as
3.52 dB at ?max
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If we select ?max to be the new phase-margin
frequency, the uncompensated systems magnitude
at this frequency must be -3.5 dB to yield a 0 dB
crossover at ?max for the compensated system. The
uncompensated system passes through -3.52 dB at
?max 40 rad/s. This frequency is the new
phase-margin frequency.
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OPEN-LOOP FREQUENCY RESPONSE
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STEP RESPONSE
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CLOSED-LOOP FREQUENCY RESPONSE
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LAG-LEAD COMPENSATION
The transfer function of a single, passive
lag-lead network is
Where ?gt1. In this design, quantities for the lag
and lead compensators, ? and ? respectively, are
replaced by a single quantity ?. Here, ? and ?
must be reciprocals of each other. Following Bode
diagrams show the frequency response curves of
(s1)(s0.1)/(s ?)(s01./ ?) for some ?
values.
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DESIGN PROCEDURE

• Using a second-order approximation, find the
  closed-loop bandwidth to meet the settling time
  or peak time requirement.
• Set the gain K to meet the steady-state error
  specification.
• Plot the Bode diagrams for this value of gain.
• Using a second-order approximation, calculate the
  phase margin to meet the damping ratio or percent
  overshoot requirement.
• Select a new phase-margin frequency near ?BW

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• 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 lag compensator.
• Design the lag compensator by selecting the
  higher break frequency one decade below the new
  phase-margin frequency. Find the value of ? from
  the lead compensators requirements as ?1/?.
  This value, along with the previously found lags
  upper break frequency, allows us to find the
  lags lower break frequency.

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• Design the lead compensator. Using the value ?
  from the lag compensator design and the value
  assumed for the new phase-margin frequency, find
  the lower and upper break frequency for the lead
  compensator.
• Check the bandwidth to be sure that the speed
  requirement has been met.
• Redesign if phase-margin or transient
  specifications are not met.

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EXAMPLE
C(s)
R(s)
K
Design a passive lag-lead network using Bode
diagrams to yield a 13.25 overshoot, a peak time
of 2 seconds, and Kv12.
The bandwidth required for a 2-second peak time
is 2.29 rad/s. In order to have a Kv12, the
value of K should be 48. The Bode plots for the
uncompensated system with K48 are shown in the
following figure. The system is unstable.
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GAIN COMPENSATED SYSTEM
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The required phase margin to yield a 13.25
overshoot is 550 Let us select ?1.8 rad/s as the
new phase-margin frequency. At this frequency,
the uncompensated phase is -1750 and would
require a 550 contribution from the lead
compensator if we consider a -50 contribution
from the lag compensator. Now, let us design the
lag compensator. The lag compensator allows us to
keep the gain of 48 required for Kv12 and not
have to lower the gain to stabilize the system.
As long as the lag compensator stabilizes the
system, the design parameters are not critical
since the phase margin will be designed with the
lead compensator.
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Thus, choose the lag compensator so that its
phase response will have minimal effect at the
new phase-margin frequency. Choose the lag
compensators higher break frequency to be 1
decade below the new phase-margin frequency, at
0.18 rad/s. Since we need to add 550 of phase
shift with the lead compensator at 1.8rad/s,
?0.1, and ?10. Then, the transfer function of
the lag compensator is
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LAG COMPENSATED SYSTEM
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The lag-compensated system has a phase angle of
1800 at 1.8 rad/s. Using the values of ?max1.8
rad/s and ?0.1, we can calculate 1/T10.57 rad/s
and 1/(?T1)5.7 rad/s. The lead compensator is
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Lag-Lead Compensated System
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STEP RESPONSE
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