376
CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD
Therefore by applying the lead compensator with some gain adjustments
:
s
+1
D(s) = 0.12 × 4.5
s
+1
90
we get the compensated system with :
P M = 65◦ , ω c = 22 rad/sec, so that ω BW & 25 rad/sec.
The Bode plot with designed compensator is :
B ode D iagram s
60
G m =15.854 dB (at 72.55 rad/sec),P m =65.338 deg.(at 22.029 rad/sec)
40
20
P hase (deg);M agnitude (dB )
0
-20
-40
-60
-80
-50
-100
-150
-200
-250
-300
-1
10
10
0
10
1
10
2
10
3
F requency (rad/sec)
45. For the system shown in Fig. 6.104, suppose that
G(s) =
5
.
s(s + 1)(s/5 + 1)
Design a lead compensation D(s) with unity DC gain so that PM ≥
40◦ using Bode plot sketches, then verify and reÞne your design using
MATLAB. What is the approximate bandwidth of the system?
Solution :
Start with a lead compensator design with :
D(s) =
Ts + 1
αT s + 1
which has unity DC gain with α < 1.
377
Figure 6.104: Control system for Problem 45
The Bode plot of the given system is :
B ode D iagram s
100
G m =1.5836 dB (at 2.2361 rad/sec),P m =3.9431 deg.(at 2.0388 rad/sec)
P hase (deg);M agnitude (dB )
50
0
-50
-100
-50
-100
-150
-200
-250
-300
-2
10
10
-1
10
0
10
1
F requency (rad/sec)
Since P M = 3.9◦ , let’s add phase lead ≥ 60◦ . From Fig. 6.53,
1
' 20 =⇒ choose α = 0.05
α
To apply maximum phase lead at ω = 10 rad/sec,
1
1
1
ω=√
= 10 =⇒
= 2.2,
= 45
T
αT
αT
Therefore by applying the lead compensator :
s
+1
D(s) = 2.2
s
+1
45
10
2
378
CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD
we get the compensated system with :
P M = 40◦ , ω c = 2.5
The Bode plot with designed compensator is :
B ode D iagram s
100
G m =24.13 dB (at 12.756 rad/sec),P m =40.244 deg.(at 2.4901 rad/sec)
50
P hase (deg);M agnitude (dB )
0
-50
-100
-150
-50
-100
-150
-200
-250
-300
-2
10
10
-1
10
0
10
1
10
2
10
3
F requency (rad/sec)
From Fig. 6.50, we see that ω BW ' 2 × ω c ' 5 rad/sec.
46. Derive the transfer function from Td to θ for the system in Fig. 6.68. Then
apply the Final Value Theorem (assuming Td = constant) to determine
whether θ(∞) is nonzero for the following two cases:
(a) When D(s) has no integral term: lims→0 D(s) = constant;
(b) When D(s) has an integral term:
D(s) =
D0 (s)
,
s
where lims→0 D0 (s) = constant.
Solution :
The transfer function from Td to θ :
Θ(s)
=
Td (s)
1+
where Td (s) = |Td | /s.
0.9
s2
0.9 2
2
s s+2 D(s)
379
(a) Using the Þnal value theorem :
θ(∞) =
=
lim θ(t) = lim sΘ(t) = lim
t→∞
s→0
s→0
0.9
s2
s2 (s+2)+1.8D(s)
s2 (s+2)
|Td |
s
|Td |
|Td |
=
6= 0
lim D(s)
constant
s→0
Therefore, there will be a steady state error in θ for a constant Td
input if there is no integral term in D(s).
(b)
θ(∞) =
=
lim θ(t) = lim sΘ(t) = lim
t→∞
s→0
s→0
0.9
s2
s3 (s+2)+1.8D0 (s)
s3 (s+2)
|Td |
s
0
=0
1.8lim D0 (s)
s→0
So when D(s) contains an integral term, a constant Td input will
result in a zero steady state error in θ.
47. The inverted pendulum has a transfer function given by Eq. (2.35), which
is similiar to
1
G(s) = 2
.
s −1
(a) Design a lead compensator to achieve a PM of 30◦ using Bode plot
sketches, then verify and reÞne your design using MATLAB.
(b) Sketch a root locus and correlate it with the Bode plot of the system.
(c) Could you obtain the frequency response of this system experimentally?
Solution :
(a) Design the lead compensator :
D(s) = K
Ts + 1
αT s + 1
such that the compensated system has P M ' 30◦ & ω c ' 1 rad/sec.
(Actually, the bandwidth or speed of response was not speciÞed, so
any crossover frequency would satisfy the problem statement.)
α=
1 − sin(30◦ )
= 0.32
1 + sin(30◦ )
To apply maximum phase lead at ω = 1 rad/sec,
1
1
1
ω=√
= 1 =⇒
= 0.57,
= 1.77
T
αT
αT
CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD
Therefore by applying the lead compensator :
s
+1
0.57
D(s) = K s
+1
1.77
By adjusting the gain K so that the crossover frequency is around 1
rad/sec, K = 1.13 results in :
P M = 30.8◦
The Bode plot of compensated system is :
B ode D iagram s
5
G m =-1.0616 dB (at 0 rad/sec),P m =30.848 deg.(at 0.98625 rad/sec)
0
-5
-10
P hase (deg);M agnitude (dB )
380
-15
-20
-25
-30
-145
-150
-155
-160
-165
-170
-175
-180
-1
10
10
0
F requency (rad/sec)
(b) Root Locus of the compensated system is :
10
1
381
and conÞrms that the system yields all stable roots with reasonable
damping. However, it would be a better design if the gain was raised
some in order to increase the speed of response of the slow real root.
A small decrease in the damping of the complex roots will result.
(c) No, because the sinusoid input will cause the system to blow up because the open loop system is unstable. In fact, the system will
”blow up” even without the sinusoid applied. Or, a better description would be that the pendulum will fall over until it hits the table.
48. The open-loop transfer function of a unity feedback system is
G(s) =
K
.
s(s/5 + 1)(s/50 + 1)
(a) Design a lag compensator for G(s) using Bode plot sketches so that
the closed-loop system satisÞes the following speciÞcations:
i. The steady-state error to a unit ramp reference input is less than
0.01.
ii. PM ≥ = 40◦
(b) Verify and reÞne your design using MATLAB.
Solution :
Let’s design the lag compensator :
D(s) =
Ts + 1
, α>1
αT s + 1
From the Þrst speciÞcation,
¯
¯
¯ D(s)G(s) 1
1 ¯¯
¯
Steady-state error to unit ramp =
lims
− 2 ¯ < 0.01
s→0 ¯ 1 + D(s)G(s) s2
s
1
=⇒
< 0.01
K
=⇒ Choose K = 150
Uncompensated, the crossover frequency with K = 150 is too high for a
good P M . With some trial and error, we Þnd that the lag compensator,
s
+1
0.2
D(s) = s
+1
0.01
will lower the crossover frequency to ω c ' 4.46 rad/sec where the P M =
382
CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD
40.7◦ .
B ode D iagram s
150
G m =18.873 dB (at 15.477 rad/sec),P m =40.71 deg.(at 4.463 rad/sec)
100
P hase (deg);M agnitude (dB )
50
0
-50
-100
-150
-50
-100
-150
-200
-250
-300
-4
10
10
-3
10
-2
10
-1
10
0
10
1
10
2
10
3
F requency (rad/sec)
49. The open-loop transfer function of a unity feedback system is
G(s) =
K
.
s(s/5 + 1)(s/200 + 1)
(a) Design a lead compensator for G(s) using Bode plot sketches so that
the closed-loop system satisÞes the following speciÞcations:
i. The steady-state error to a unit ramp reference input is less than
0.01.
ii. For the dominant closed-loop poles the damping ratio ζ ≥ 0.4.
(b) Verify and reÞne your design using MATLAB including a direct computation of the damping of the dominant closed-loop poles.
Solution :
Let’s design the lead compensator :
D(s) =
From the Þrst speciÞcation,
Steady-state error to unit ramp
Ts + 1
, α<1
αT s + 1
¯
¯
¯ D(s)G(s) 1
1 ¯¯
< 0.01
−
lims ¯¯
s→0
1 + D(s)G(s) s2
s2 ¯
1
=⇒
< 0.01
K
=⇒ Choose K = 150
=
383
PM
From the approximation ζ '
, the second speciÞcation implies P M ≥
100
40. After trial and error, we Þnd that the compensator,
s
+1
10
D(s) = s
+1
100
results in a P M = 42.5◦ and a crossover frequency ω c ' 51.2 rad/sec as
shown by the margin output:
B ode D iagram s
80
G m =13.292 dB (at 136.03 rad/sec),P m =42.452 deg.(at 52.171 rad/sec)
60
40
P hase (deg);M agnitude (dB )
20
0
-20
-40
-60
-50
-100
-150
-200
-250
-300
-1
10
10
0
10
1
10
2
10
3
F requency (rad/sec)
and the use of damp veriÞes the damping to be ζ = 0.42 for the complex
closed-loop roots which exceeds the requirement.
50. A DC motor with negligible armature inductance is to be used in a position
control system. Its open-loop transfer function is given by
G(s) =
50
.
s(s/5 + 1)
(a) Design a compensator for the motor using Bode plot sketches so that
the closed-loop system satisÞes the following speciÞcations:
i. The steady-state error to a unit ramp input is less than 1/200.
ii. The unit step response has an overshoot of less than 20%.
iii. The bandwidth of the compensated system is no less than that
of the uncompensated system.