Department Of Industrial Engineering
Faculty Of Engineering And Information Technology
An-Najah National University
Control Systems
Dr. Mohammad Abuabiah
m.abuabiah@najah.edu
Office: 11-4-130
1
Chapter 11
Design Via
Frequency Response
2
Chapter Outcome
3
Dr. Mohammad Abuabiah
• Use frequency response techniques to adjust the gain to meet a
transient response specification.
• Use frequency response techniques to design
compensators to improve the steady-state error.
cascade
• Use frequency response techniques to design
compensators to improve the transient response.
cascade
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• Use frequency response techniques to design cascade
compensators to improve both the steady-state error and the
transient response.
11.1 Introduction
Dr. Mohammad Abuabiah
• When designing via frequency response methods, we use the
concepts of stability, transient response, and steady-state error
that we learned in Chapter 10.
• The emphasis in this chapter is on the design of gain, lag, lead,
and lag-lead compensation.
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• In the next extra chapter we will extrapolate the concepts and
designs problems involving PI, PD, and PID compensation.
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11.1 Tuning Vs Exact/Definitive Answer
Dr. Mohammad Abuabiah
• An Exact or definitive answer is a final one. Definitive
means authoritative, conclusive, final. A definitive
answer is usually a strong yes or no.
• Controller tuning is the process of determining the
controller parameters which produce the desired
output.
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• Controller tuning allows for optimization of a process
and minimizes the error between the variable of the
process and its set point.
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11.2 Transient Response via Gain Adjustment
Dr. Mohammad Abuabiah
• Let us begin our discussion of design via frequency response
methods by discussing the link between phase margin, transient
response, and gain.
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• In Section 10.10, the relationship between damping ratio
(equivalently percent overshoot) and phase margin is shown.
Thus, if we can vary the phase margin, we can vary the percent
overshoot.
• Thus, a simple gain adjustment can be used to design phase
margin and, hence, percent overshoot.
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11.2 Transient Response via Gain Adjustment
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❑Design Procedure:
Dr. Mohammad Abuabiah
1.
Draw the Bode magnitude and phase plots for a convenient value of gain.
2.
Using Eqs.
𝑶𝑺
𝜻 = − 𝐥𝐧(%
)/ 𝝅𝟐 + 𝐥𝐧𝟐 (%𝑶𝑺/𝟏𝟎𝟎)
𝟏𝟎𝟎
And
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𝝓𝑴 = (𝐭𝐚𝐧−𝟏 𝟐𝜻/ −𝟐𝜻𝟐 + 𝟏 + 𝟒𝜻𝟐 ) × 𝟏𝟖𝟎/𝝅
to, determine the required phase margin from the percent overshoot.
3.
Find the frequency, 𝝎𝝓𝑴 , on the Bode phase diagram that yields the desired phase
margin, CD, as shown on Figure.
4.
Change the gain by an amount AB to force the magnitude curve to go through 0 dB
at 𝝎𝝓𝑴 . The amount of gain adjustment is the additional gain needed to produce the
required phase margin.
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11.2 Transient Response via Gain Adjustment
Dr. Mohammad Abuabiah
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Example 11.1 Transient Response Design via Gain Adjustment
9
Dr. Mohammad Abuabiah
• PROBLEM: For the position control system shown in Figure, find the
value of preamplifier gain, K, to yield a 𝟗. 𝟓% overshoot in the
transient response for a step input. Use only frequency response
methods.
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Example 11.1 Transient Response Design via Gain Adjustment
10
Dr. Mohammad Abuabiah
• SOLUTION: We will now follow the previously described gain adjustment
design procedure.
• Write the simplified-subsystem transfer function:
clc
clear
close all
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%% Example 11.1
s = tf('s');
G1 = 100/(s+100);
G2 = 1/(s+36);
G3 = 1/s;
Ge = G1*G2*G3;
display(zpk(Ge))
1. Choose 𝑲 = 𝟑𝟔 (1/𝑑𝑐𝑔𝑎𝑖𝑛(𝑠 ∗ 𝐺𝑒 )) to start the magnitude plot at 𝟎 𝒅𝑩.
%% Step 1
K = 1/dcgain(s*Ge);
Example 11.1 Transient Response Design via Gain Adjustment
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2. Find the following:
Dr. Mohammad Abuabiah
a) 𝜻 = − 𝒍𝒏
%𝑶𝑺
𝟏𝟎𝟎
%𝑶𝑺
𝝅𝟐 +𝒍𝒏𝟐
𝟏𝟎𝟎
= 𝟎. 𝟔
𝟏𝟖𝟎
b) 𝝓𝑴 = 𝒕𝒂𝒏−𝟏 𝟐𝜻/ −𝟐𝜻𝟐 + 𝟏 + 𝟒𝜻𝟒 × 𝒑𝒊 = 𝟓𝟗. 𝟐𝒐
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%% Step 2
pos = 9.5;
z =(-log(pos/100))/(sqrt(pi^2+log(pos/100)^2));
Pm=atan(2*z/(sqrt(-2*z^2+sqrt(1+4*z^4))))*(180/pi);
display(Pm)
3. Locate on the phase plot the frequency that yields a 𝟓𝟗. 𝟐𝑜 phase
margin. This frequency is found where the phase angle is the
difference between −𝟏𝟖𝟎𝒐 and 𝟓𝟗. 𝟐𝒐 , or −𝟏𝟐𝟎. 𝟖𝒐 . The value of the
phase-margin frequency is 𝟏𝟒. 𝟖 𝒓𝒂𝒅/𝒔.
%% Step 3
Pmreq = Pm-180; % Answer -120.8
display(Pmreq)
figure, bode(K*Ge)
grid on
Example 11.1 Transient Response Design via Gain Adjustment
12
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An-Najah National University
Example 11.1 Transient Response Design via Gain Adjustment
13
Dr. Mohammad Abuabiah
4. At a frequency of 𝟏𝟒. 𝟖 𝒓𝒂𝒅/𝒔 on the magnitude plot, the gain is found to be
− 𝟐𝟒. 𝟐𝒅𝑩. This magnitude has to be raised to 𝟎 𝒅𝑩 to yield the required phase
margin. Since the log-magnitude plot was drawn for 𝑲 = 𝟑𝟔 , a
𝟐𝟒. 𝟐 𝒅𝑩 increase, or
𝑲 = 𝟑𝟔 × 𝟏𝟔. 𝟐 = 𝟓𝟖𝟑. 𝟐
• Thus, the controller transfer function is
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%% Step 4
M = 10^(24.2/20);
display(M); %16.2
K = K*M;
Gc = K;
display(Gc)
𝑮𝒄 𝒔 = 𝑲 = 𝟓𝟖𝟑. 𝟗
5. Finally, the results:
%% Result
T = feedback(Gc*Ge,1);
figure,step(T);
grid on
L = Gc*Ge;
[Gm,Pm,Wgm,Wpm] = margin(L);
display(Pm)
Example 11.1 Transient Response Design via Gain Adjustment
14
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Example 11.1 Transient Response Design via Gain Adjustment
15
Dr. Mohammad Abuabiah
• Final Result:
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Parameter
Phase Margin
Proposed Specification
𝟓𝟗. 𝟏𝟔𝟐𝒐
Actual Value
𝟓𝟗. 𝟏𝟗𝟕𝒐
Percent Overshoot
𝟗. 𝟓%
𝟖. 𝟔𝟒%
11.2 Gain Compensation: Analog
16
• Active-Circuit Realization:
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1𝐾𝞨
1𝐾𝞨
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𝐙
𝐙𝟏
𝑮𝒄 𝒔 = 𝑲 = 𝟓𝟖𝟑. 𝟗 = − 𝟐 =
𝐑𝟐
𝟕𝟓𝟎𝛀
≅−
𝐑𝟏
𝟏.𝟑𝛀
Example 11.1 Summary
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Dr. Mohammad Abuabiah
clc
clear
close all
%% Step 3
figure, bode(K*Ge)
grid on
%% Example 11.1
s = tf('s');
G1 = 100/(s+100);
G2 = 1/(s+36);
G3 = 1/s;
Ge = G1*G2*G3;
display(zpk(Ge))
%% Step 4
M = 10^(24.2/20);
display(M);
K = K*M;
Gc = K;
dsiplay(Gc)
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%% Step 1
K = 1/dcgain(s*Ge);
display(K)
%% Step 2
pos = 9.5;
z =(-log(pos/100))/(sqrt(pi^2+log(pos/100)^2));
dsiplay(z)
Pm=atan(2*z/(sqrt(-2*z^2+sqrt(1+4*z^4))))*(180/pi);
Pmreq = Pm-180;
display(Pmreq)
%% Result
T = feedback(Gc*Ge,1);
figure,step(T);
grid on
L = Gc*Ge;
[Gm,Pm,Wgm,Wpm] = margin(L);
display(Pm)
11.3 Lag Compensation
18
Dr. Mohammad Abuabiah
• In the next three sections, we will discuss the design of lag, lead,
and lag-lead compensation via Bode diagrams.
• The name lag compensator comes from the fact that the typical
phase angle response for the compensator is always negative, or
lagging in phase angle.
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• The function of the lag compensator as seen on Bode diagrams is
to improve the static error constant by increasing only the lowfrequency gain without any resulting instability.
• The transfer function of the lag compensator is
𝜶>𝟏
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11.3 Lag Compensation
Dr. Mohammad Abuabiah
An-Najah National University
20
11.3 Lag Compensation
Dr. Mohammad Abuabiah
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11.3 Lag Compensation
• Another form of The transfer function of the lag compensator is
Dr. Mohammad Abuabiah
𝑲𝒄 𝒔 + 𝝎𝒉
𝑮𝒄 𝒔 =
(𝒔 + 𝝎𝒍 )
Where,
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• 𝑲𝒄 =
𝝎𝒍
𝝎𝒉
• 𝝎𝒉 =
𝝎𝒇
• 𝝎𝒍 =
𝝎𝒉
𝑴
𝟏𝟎
• 𝝎𝒇 : At this frequency the magnitude plot must go through 0 dB.
• 𝑴: magnitude at the 𝝎𝒇 requency.
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11.3 Lag Compensation
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❑ Design Procedure:
Dr. Mohammad Abuabiah
1. Set the gain, 𝑲, to the value that satisfies the steady-state error specification and
plot the Bode magnitude and phase diagrams for this value of gain.
2. Find the frequency where the phase margin is 𝟓𝒐 to 𝟏𝟐𝒐 greater than the phase
margin that yields the desired transient response (Ogata, 1990). This step
compensates for the fact that the phase of the lag compensator may still
contribute anywhere from −5𝑜 to −12𝑜 of phase at the phase-margin frequency.
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3. Select a lag compensator whose magnitude response yields a composite Bode
magnitude diagram that goes through 0 dB at the frequency found in Step 2 as
follows: Draw the compensator’s high-frequency asymptote to yield 0 dB for the
compensated system at the frequency found in Step 2.
4. Reset the system gain, 𝑲, to compensate for any attenuation in the lag network in
order to keep the static error constant the same as that found in Step 1.
Example 11.2 Lag Compensation Design
Dr. Mohammad Abuabiah
• PROBLEM: Given the system of Figure, 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%.
23
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Example 11.2 Lag Compensation Design
Dr. Mohammad Abuabiah
• SOLUTION: We will follow the
compensation design procedure.
previously
described
1. From Example 11.1 a gain, K, of 583.9 yields a 9.5% overshoot.
clc
clear
close all
%% Example 11.2
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%% Step 1
s = tf('s');
G1 = 100/(s+100);
G2 = 1/(s+36);
G3 = 1/s;
K = 583.9;
Ge = G1*G2*G3*K;
display(zpk(Ge))
24
lag
Example 11.2 Lag Compensation Design
Dr. Mohammad Abuabiah
• SOLUTION: We will follow the
compensation design procedure.
previously
described
25
lag
58390
𝐾𝑣 = lim 𝑠𝐺(𝑠) = lim 𝑠
= 16.22
𝑠→0
𝑠→0 𝑠 𝑠 + 36 𝑠 + 100
For a tenfold improvement in steady-state error, 𝑲𝒗 must increase
by a factor of 𝟏𝟎, or 𝑲𝒗 = 𝟏𝟔𝟐. 𝟐. Therefore,
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%% Step 1
Kv = dcgain(s*Ge);
Kv = 10*Kv;
K = dcgain(Kv/(s*Ge));
Ge = Ge*K;
figure,bode(Ge)
Example 11.2 Lag Compensation Design
26
2. Step 2:
Dr. Mohammad Abuabiah
– The phase-margin required for a 𝟗. 𝟓% overshoot (𝜁 = 0.6) is found to be 𝟓𝟗. 𝟐𝒐 . We
increase this value of phase margin by 𝟏𝟎𝒐 to 𝟔𝟗. 𝟐𝒐 in order to compensate for the
phase angle contribution of the lag compensator.
– Now find the frequency where the phase margin is 𝟔𝟗. 𝟐𝒐 . This frequency occurs at a
phase angle of −𝟏𝟖𝟎𝒐 + 𝟔𝟗. 𝟐𝐨 = −𝟏𝟏𝟎. 𝟖 and is 𝟗. 𝟖 𝒓𝒂𝒅/𝒔.
– At this frequency, the magnitude plot must go through 0 dB. The magnitude at 9.8
rad/s is now +𝟐𝟒 𝒅𝑩 (exact, that is, non-asymptotic). Thus, the lag compensator must
provide −𝟐𝟒 𝒅𝑩 attenuation at 9.8 rad/s.
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%% Step 2
pos = 9.5;
z =(-log(pos/100))/(sqrt(pi^2+log(pos/100)^2));
Pm=atan(2*z/(sqrt(-2*z^2+sqrt(1+4*z^4))))*(180/pi);
Pm = Pm+10; %tuning parameter
Pmreq = Pm-180;
display(Pmreq)
wf = 9.8; %from Bode plot
Example 11.2 Lag Compensation Design
3. Step 3:
Dr. Mohammad Abuabiah
– Find the following:
o 𝟐𝟎𝒍𝒐𝒈𝑴 = 𝟐𝟒 → 𝑴 = 𝟏𝟎
𝟐𝟒
𝟐𝟎
= 𝟏𝟓. 𝟖𝟒𝟖𝟗
o 𝝎𝒇 = 𝟗. 𝟖 𝒓𝒂𝒅/𝒔
𝝎𝒇
o 𝝎𝒉 = 𝟏𝟎 = 𝟎. 𝟗𝟖 𝒓𝒂𝒅/𝒔
o 𝝎𝒍 =
𝝎𝒉
𝟎.𝟗𝟖
=
= 𝟎. 𝟎𝟔𝟏𝟖 𝒓𝒂𝒅/𝒔
𝑴
𝟏𝟓.𝟖𝟒𝟖𝟗
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𝝎
o 𝑲𝒄 = 𝝎 𝒍 =
𝒉
𝟎.𝟎𝟔𝟏𝟖
= 𝟎. 𝟎𝟔𝟑𝟎
𝟎.𝟗𝟖
%% Step 3
M = 10^(24/20); %The magnitude at 9.8 rad/s
wh = wf/10;
wl = wh/M;
Kc = wl/wh;
27
Example 11.2 Lag Compensation Design
•
Hence, the lag compensator’s transfer function is
Dr. Mohammad Abuabiah
𝑮𝒄 𝒔 =
•
𝑲𝒄 𝒔 + 𝝎𝒉
𝒔 + 𝟎. 𝟗𝟖
= 𝟎. 𝟎𝟔𝟑
(𝒔 + 𝝎𝒍 )
(𝒔 + 𝟎. 𝟎𝟔𝟐)
The compensated system’s forward transfer function is thus
𝑮 𝒔 𝑮𝒄 𝒔 =
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=
𝟓𝟖𝟑𝟗𝟎𝟎
𝒔 𝒔 + 𝟑𝟔 𝒔 + 𝟏𝟎𝟎
𝟎. 𝟎𝟔𝟑
𝒔 + 𝟎. 𝟗𝟖
(𝒔 + 𝟎. 𝟎𝟔𝟐)
𝟑𝟔𝟕𝟖𝟔 𝒔 + 𝟎. 𝟗𝟖
𝒔 𝒔 + 𝟑𝟔 𝒔 + 𝟏𝟎𝟎 𝒔 + 𝟎. 𝟎𝟔𝟐
%% Step 4
Gc = Kc*(s+wh)/(s+wl);
display(zpk(Gc))
%% Result
T = feedback(Gc*Ge,1);
figure,step(T);
[Gm,Pm,Wg,Wp]=margin(Gc*Ge)
Kv = dcgain(s*Gc*Ge);
28
29
Example 11.2 Lag Compensation Design
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30
Example 11.2 Lag Compensation Design
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Example 11.2 Lag Compensation Design
Dr. Mohammad Abuabiah
• Final Result:
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Parameter
𝑲𝒗
Proposed Specification
𝟏𝟔𝟐. 𝟐
Actual Value
𝟏𝟔𝟐. 𝟏𝟗𝟒
Phase Margin
Percent Overshoot
𝟓𝟗. 𝟐𝒐
𝟗. 𝟓%
𝟔𝟑. 𝟕𝟐𝒐
𝟗. 𝟐𝟏%
31
11.3 Lag Compensation: Analog
• Active-Circuit Realization:
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An-Najah National University
𝒔 + 𝟎. 𝟗𝟖
𝑮𝒄 𝒔 = 𝟎. 𝟎𝟔𝟑
(𝒔 + 𝟎. 𝟎𝟔𝟐)
𝒛𝟏
𝒛𝟐
32
Example 11.2 Summary
Dr. Mohammad Abuabiah
clc
clear
close all
%% Example 11.2
s = tf('s');
G1 = 100/(s+100);
G2 = 1/(s+36);
G3 = 1/s;
K = 583.9;
Ge = G1*G2*G3*K;
display(zpk(Ge))
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%% Step 1
Kv = dcgain(s*Ge);
Kv = 10*Kv;
K = dcgain(Kv/(s*Ge));
Ge = Ge*K;
figure,bode(Ge)
%% Step 2
pos = 9.5;
z =(-log(pos/100))/(sqrt(pi^2+log(pos/100)^2));
Pm=atan(2*z/(sqrt(-2*z^2+sqrt(1+4*z^4))))*(180/pi);
Pm = Pm+10;
Pmreq = 180-Pm;
display(Pmreq)
wf = 9.8; %from Bode plot
33
%% Step 3
M = 10^(24/20);
wh = wf/10;
wl = wh/M;
Kc = wl/wh;
%% Step 4
Gc = Kc*(s+wh)/(s+wl);
display(zpk(Gc))
%% Result
T = feedback(Gc*Ge,1);
figure,step(T);
stepinfo(T)
grid on
[Gm,Pm,Wg,Wp]=margin(Gc*Ge);
Kv = dcgain(s*Gc*Ge);
display(Kv)
11.4 Lead Compensation
34
Dr. Mohammad Abuabiah
• For second-order systems, we derived the relationship between phase margin and
percent overshoot as well as the relationship between closed-loop bandwidth and
other time-domain specifications, such as settling time, peak time, and rise time.
• When we designed the lag network to improve the steady-state error, we wanted a
minimal effect on the phase diagram in order to yield an imperceptible change in the
transient response.
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• However, in designing lead compensators via Bode plots, 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 (Improving Transient
-Response).
• The transfer function of the lead compensator is
𝜷<𝟏
11.4 Lead Compensation
•
Dr. Mohammad Abuabiah
In order to design a lead compensator and change both the phase margin and phasemargin frequency, it is helpful to have an analytical expression for the maximum value of
phase and the frequency at which the maximum value of phase occurs,
–
The frequency, 𝝎𝒎𝒂𝒙 , at which the maximum phase angle, 𝝓𝒎𝒂𝒙 , occurs is
–
the maximum phase shift of the compensator, 𝝓𝒎𝒂𝒙 , is
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𝟏 − 𝐬𝐢𝐧 𝝓𝒎𝒂𝒙
𝜷=
𝟏 + 𝐬𝐢𝐧 𝝓𝒎𝒂𝒙
–
the compensator’s magnitude at 𝝓𝒎𝒂𝒙 is
35
36
11.4 Lead Compensation
Dr. Mohammad Abuabiah
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11.4 Lead Compensation
• Another form of The transfer function of the lead compensator is
Dr. Mohammad Abuabiah
𝑲𝒄 𝒔 + 𝒛𝒄
𝑮𝒄 𝒔 =
(𝒔 + 𝒑𝒄 )
Where,
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• 𝑲𝒄 =
𝟏
𝜷
• 𝒛𝒄 = 𝝎𝒎𝒂𝒙 𝜷
•
𝒛𝒄
𝒑𝒄 =
𝜷
• 𝝎𝒎𝒂𝒙 : the frequency at which the maximum phase angle, 𝝓𝒎𝒂𝒙 .
37
11.4 Lead Compensation
38
❑ Design Procedure:
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1.
Find the closed-loop bandwidth required to meet the settling time, peak time, or rise time
requirement (using Equations discussed in Chapter 10).
2.
Since the lead compensator has negligible effect at low frequencies, set the gain, K, of the
uncompensated system to the value that satisfies the steady-state error requirement.
3.
Plot the Bode magnitude and phase diagrams for this value of gain and determine the
uncompensated system’s phase margin.
4.
Find the phase margin to meet the damping ratio or percent overshoot requirement. Then
evaluate the additional phase contribution required from the compensator.
5.
Determine the value of 𝜷 from the lead compensator’s required phase contribution.
6.
Determine the compensator’s magnitude at the peak of the phase curve.
7.
Determine the new phase-margin frequency by finding where the uncompensated system’s
magnitude curve is the negative of the lead compensator’s magnitude at the peak of the
compensator’s phase curve.
8.
Design the lead compensator’s break frequencies to find T and the break frequencies.
9.
Reset the system gain to compensate for the lead compensator’s gain.
10. Check the bandwidth to be sure the speed requirement in Step 1 has been met.
11. Simulate to be sure all requirements are met.
12. Redesign if necessary to meet requirements.
Example 11.3 Lead Compensation Design
Dr. Mohammad Abuabiah
• PROBLEM: Given the system of Figure, design a lead compensator
to yield 𝑶𝑺% ≤ 𝟐𝟎% overshoot and 𝑲𝒗 = 𝟒𝟎, with a peak time of ≤
0.1 second.
39
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Example 11.3 Lead Compensation Design
Dr. Mohammad Abuabiah
• SOLUTION: We will follow the previously described lead compensation design
procedure with the help of MATLAB.
• Step 1: write your system 𝑮𝒆 (𝒔):
clc
clear
close all
%% Example 11.3
%% Step 1
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s = tf('s');
G1 = 100/(s+100);
G2 = 1/(s+36);
G3 = 1/s;
Ge1 = G1*G2*G3;
display(zpk(Ge1))
• Step 2: find the value of 𝑲:
%% Step 2
Kv = 40;
K = dcgain(Kv/(s*Ge1));
display(K)
Ge2 = Ge1*K;
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Example 11.3 Lead Compensation Design
•
Step 3: Find the bandwidth frequency:
Dr. Mohammad Abuabiah
%% Step 3
pos = 20;
Tp = 0.1;
z=(-log(pos/100))/(sqrt(pi^2+log(pos/100)^2)); % Calculate required damping ration
wn=pi/ (Tp*sqrt(l-z^2)); %Calculate required natural frequency
wBW=wn*sqrt((1-2*z^2)+sqrt(4*z^4-4*z^2+2)); %determine the bandwidth
display(wBW)
•
Step 4: Find compensator beta:
%% Step 4
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[Gm,Pm,Wcg,Wcp]=margin(Ge2); %find the current phase margin
Pmreq =atan(2*z/(sqrt(-2*z^2+sqrt(l+4*z^4))))*(180/pi); % Calculate required phase margin.
Pmreq= Pmreq+15 %Add a correction factor (try & error)
Pmc = Pmreq-Pm % phase contribution required from lead compensator
beta=(1-sin(Pmc*pi/180))/(1+sin(Pmc*pi/180));
display(beta)
•
Step 5: Find the maximum frequency 𝒘𝒎𝒂𝒙 :
%% Step 5
magpc = 1/sqrt(beta); %Find compensator peak magnitude.
Mreq = 20*log10(1/magpc) % Mreq = -4.6127
figure, bode(Ge2);
wmax = 41.5; %This is the frequency at the peak magnitude (from bode plot).
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Example 11.3 Lead Compensation Design
Dr. Mohammad Abuabiah
• Step 6: Find the lead compensator controller:
%% Step 6
zc = wmax*sqrt(beta);
pc=zc/beta;
Kc=1/beta;
Gc = Kc*(s+zc)/(s+pc);
display(zpk(Gc));
• Step 7: Validate your result:
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%% Step 7
T = feedback(Gc*Ge2,1);
figure,step(T)
grid on
Kv = dcgain(s*Gc*Ge2);
display(Kv)
wBW = bandwidth(T);
display(wBW)
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Example 11.3 Lead Compensation Design
Dr. Mohammad Abuabiah
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Example 11.3 Lead Compensation Design
Dr. Mohammad Abuabiah
• Final Result:
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Parameter
𝑲𝒗
𝑻𝒑
Proposed Specification
𝟒𝟎
𝟎. 𝟏
Actual Value
𝟒𝟎
𝟎. 𝟎𝟔𝟖𝟓
Percent Overshoot
𝟐𝟎%
𝟏𝟗. 𝟗%
Bandwidth
𝟒𝟔. 𝟓𝟗𝟑𝟕
𝟕𝟑. 𝟑𝟖𝟓𝟑
𝟐. 𝟖𝟗𝟐𝟓 𝒔 + 𝟐𝟒. 𝟒
𝑮𝒄 =
𝒔 + 𝟕𝟎. 𝟓𝟖
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11.4 Lead Compensation: Analog
• Active-Circuit Realization:
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𝒛𝟏
𝒛𝟐
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11.5 Lag-Lead Compensation
Dr. Mohammad Abuabiah
• In this section, we improve both transient response and the
steady-state error by using a lead compensator and a lag
compensator. Our compensator is called a lag-lead
compensator.
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• We first 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 to meet the phase-margin
requirements.
• At the end of this section, we emphasize lag-lead design, using a
single, passive lag-lead network.
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11.5 Lag-Lead Compensation
47
Dr. Mohammad Abuabiah
• The transfer function of a single, passive lag-lead network is
𝑮𝒄 𝒔 = 𝑮𝑳𝒆𝒂𝒅 𝒔 𝑮𝑳𝒂𝒈 𝒔 =
𝟏
𝒔+𝑻
𝟏
𝒔+𝑻
𝜸
𝒔+
𝑻𝟏
𝒔+
𝟏
𝟐
𝟏
𝜸𝑻𝟐
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• where 𝜸 > 𝟏 . The first term in parentheses produces the lead
compensation, and the second term in parentheses produces the
lag compensation.
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11.5 Lag-Lead Compensation
Dr. Mohammad Abuabiah
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11.5 Lag-Lead Compensation
49
❑ Design Procedure:
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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. Find the value of 𝜸 from the lead compensator’s requirements (𝜸 = 𝟏/𝜷).
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 frequency for the
lead compensator.
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.
Example 11.4 lag-lead Compensation Design
Dr. Mohammad Abuabiah
• PROBLEM: Given a unity feedback system where
𝑲
𝑮 𝒔 =
𝒔 𝒔+𝟏 𝒔+𝟒
design a active lag-lead compensator using Bode diagrams to
yield a 13.25% overshoot, a peak time of 2 seconds, and 𝑲𝒗 = 𝟏𝟐.
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• SOLUTION: We will follow the steps previously mentioned in this
section for lag-lead design with the help of MATLAB.
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Example 11.4 lag-lead Compensation Design
Dr. Mohammad Abuabiah
• Step 1: write your system 𝑮(𝒔):
clc
clear
close all
%% Example 11.4
%% Step 1
s = tf('s’);
G = 1/(s*(s+1)*(s+4));
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• Step 2: Find the value of K:
%% Step 2
Kv = 12;
K = Kv*dcgain(1/(s*G));
Ge = K*G;
display(zpk(Ge));
display(K)
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Example 11.4 lag-lead Compensation Design
Dr. Mohammad Abuabiah
• Step 3: find the value of the Bandwidth 𝝎𝑩𝑾 :
%% Step 3
pos = 13.25;
Tp = 2;
z =(-log(pos/100))/(sqrt(pi^2+log(pos/100)^2));
Pmreq=atan(2*z/(sqrt(-2*z^2+sqrt(1+4*z^4))))*(180/pi);
wn = pi/(Tp*sqrt(1-z^2));
wBW = wn*sqrt((1-2*z^2)+sqrt(4*z^4-4*z^2+2));
display(wBW)
• Step 4: Find the value of 𝜷:
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%% Step 4
wpm = 0.8*wBW; %Choose new phase-margin frequency.
[M,P]=bode(Ge,wpm); % Get Bode data.
Pmreqc=Pmreq-(180+P)+5; % Find phase contribution required from lead compensator with additional 5 degrees.
beta=(1-sin(Pmreqc*pi/180))/(1+sin(Pmreqc*pi/180));
display(beta)
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Example 11.4 lag-lead Compensation Design
Dr. Mohammad Abuabiah
• Step 5: Compute the Lag-Compensator :
%% Step 5
zclag=wpm/10; % Calculate zero of lag compensator.
pclag=zclag*beta; % Calculate pole of lag compensator.
Kclag=beta; % Calculate gain of lag compensator.
Glag = Kclag*(s+zclag)/(s+pclag);
display(zpk(Glag));
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• Step 6: Compute the Lead-Compensator :
%% Step 6
zclead=wpm*sqrt(beta); % Calculate zero of lead compensator.
pclead=zclead/beta; %Calculate pole of lead compensator.
Kclead=1/beta; % Calculate gain of lead compensator.
Glead = Kclead*(s+zclead)/(s+pclead);
display(zpk(Glead));
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Example 11.4 lag-lead Compensation Design
Dr. Mohammad Abuabiah
• Step 7: Validate your result:
%% Step 7
Gc = Glag*Glead;
display(zpk(Gc))
Kv = dcgain(s*Gc*Ge);
display(Kv)
T = feedback(Gc*Ge,1);
figure,step(T)
grid on
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55
Example 11.4 lag-lead Compensation Design
Dr. Mohammad Abuabiah
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11.5 Active-Circuit Realization
Dr. Mohammad Abuabiah
• Lag-lead compensator can be formed by cascading the lag
compensator with the lead compensator, as shown in Figure
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𝑮𝒄 =
𝑪
𝑪𝟐
𝟏
𝑹 𝟏 𝑪𝟏
𝟏
𝒔+𝑹 𝑪
𝟐 𝟐
− 𝟏 𝒔+
×
𝑪
𝑪𝟒
𝟏
𝑹 𝟑 𝑪𝟑
𝟏
𝒔+𝑹 𝑪
𝟒 𝟒
− 𝟑 𝒔+
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