MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Mechanical Engineering
2.004 Dynamics and Control II
Fall 2007
Lecture 38
Overview of the frequency response and Bode plots
In this lecture, we’ll practice on the topic of frequency response and Bode plots. Se­
lect at least one partner, and work out the problems together. Call on the instructor(s)
for help.
1
1. Consider the following eight distinct systems whose pole–zero plots are provided.
Match them their bode plots given in the next page.
Pole−Zero Map 2 (double poles at zero)
Pole−Zero Map 1 (one pole at zero)
2
2
1.5
1.5
1
Imaginary Axis
Imaginary Axis
1
0.5
0
−0.5
0.5
0
−0.5
−1
−1
−1.5
−1.5
−2
−2
−1.5
−1
−0.5
1
0
Real Axis
0.5
1
1.5
−2
−2
2
−1.5
−1
2
1.5
1.5
Imaginary Axis
Imaginary Axis
0
−0.5
−1
−1.5
−1.5
−1
−0.5
0
Real Axis
0.5
1
1.5
−2
−2.5
2
−2
−1.5
−1
−0.5
4
0
0.5
Real Axis
1
1.5
2
2.5
1.5
2
1.5
2
Pole−Zero Map 6 (one pole at −1, one zero at 1.2)
4
2
3
1.5
1
Imaginary Axis
2
Imaginary Axis
2
0
Pole−Zero Map 5 (two poles at −1± j3)
1
0
−1
0.5
0
−0.5
−2
−1
−3
−1.5
−1.5
−1
−0.5
5
0
Real Axis
0.5
1
1.5
−2
−2
2
1.5
1.5
1
1
Imaginary Axis
2
0.5
0
−0.5
−1.5
0.5
1
0.5
1
−0.5
−1.5
0
Real Axis
0
Real Axis
0
−1
−0.5
−0.5
0.5
−1
−1
−1
Pole−Zero Map 7 (one pole at −1, one zero at zero)
2
−1.5
−1.5
6
Pole−Zero Map 7 (one pole at −1, one zero at −1.2)
Imaginary Axis
1.5
−0.5
−1
3
7
1
0.5
−1.5
−2
−2
0.5
1
1
0.5
−4
−2
0
Real Axis
Pole−Zero Map 4 (two poles at −1 and −2)
Pole−Zero Map 3 (one pole at −1)
2
−2
−2
−0.5
2
1.5
−2
−2
2
8
2
−1.5
−1
−0.5
0
Real Axis
0.5
1
Bode plots:
Bode Plot
Bode Plot
5
Magnitude (dB)
5
0
−5
0
−5
−10
−15
−10
180
−20
0
135
−45
Phase (deg)
Phase (deg)
Magnitude (dB)
10
90
45
0 −1
10
a
0
−180 0
10
1
10
Frequency (rad/sec) (rad/sec)
10
−90
−135
b
1
10
Frequency (rad/sec) (rad/sec)
Bode Plot
Bode Plot
0
0
Magnitude (dB)
Magnitude (dB)
10
−10
−20
−30
−90
Phase (deg)
Phase (deg)
−20
−30
−40
0
−40
0
−180
−270
−360 0
10
c
−10
−60
−90 −2
10
1
10
d
Frequency (rad/sec) (rad/sec)
−30
−1
10
0
10
Magnitude (dB)
Magnitude (dB)
−10
−20
−30
−40
Phase (deg)
Phase (deg)
5
0
−5
−10
0
−50
90
60
30
0 −2
10
−1
10
0
1
10
−30
−60
−90 −1
10
2
10
10
Frequency (rad/sec) (rad/sec)
f
0
10
Frequency (rad/sec) (rad/sec)
0
Magnitude (dB)
−20
−40
−60
−20
−40
−60
−80
−80
0
−100
0
−45
−45
Phase (deg)
Magnitude (dB)
Phase (deg)
g
1
10
Bode Plot
Bode Plot
0
−90
−135
−180 −1
10
2
10
Bode Plot
Bode Plot
0
e
1
10
10
Frequency (rad/sec) (rad/sec)
0
1
2
10
10
Frequency (rad/sec) (rad/sec)
10
3
h
−90
−135
−180 −1
10
0
1
10
10
Frequency (rad/sec) (rad/sec)
2
10
2. Nise chapter 10, problem 25 (page 679).
3. Nise chapter 10, problem 23 (page 679).
4
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Mechanical Engineering
2.004 Dynamics and Control II
Fall 2007
Supplement to Lecture 38
Solution to the practice problems on frequency response & Bode plots
1. Consider the following eight distinct systems whose pole–zero plots are provided.
Match them their bode plots given in the next page.
1.1 Answer:
Pole−Zero Map 1 (one pole at zero)
2
1.5
Imaginary Axis
1
0.5
0
−0.5
−1
−1.5
−2
−2
−1.5
−1
−0.5
0
Real Axis
0.5
1
1.5
2
This system has a pole at s = 0, its transfer function will be
Ga (s) =
K
,
s
where K is a gain. The Bode magnitude plot has -20dB/dec slope and the
Bode phase plot is always −90◦ .
Bode Plot
Magnitude (dB)
5
0
−5
−10
−15
Phase (deg)
−20
0
−45
−90
−135
−180 0
10
1
10
Frequency (rad/sec) (rad/sec)
1.2 Answer:
1
Pole−Zero Map 2 (double poles at zero)
2
1.5
Imaginary Axis
1
0.5
0
−0.5
−1
−1.5
−2
−2
−1.5
−1
−0.5
0
Real Axis
0.5
1
1.5
2
Because this system has double poles at s = 0, the transfer function will be
K
.
s2
The Bode magnitude plot has -40dB/dec slope and the Bode phase plot is
always −180◦ .
Gb (s) =
Bode Plot
Magnitude (dB)
10
0
−10
−20
−30
Phase (deg)
−40
0
−90
−180
−270
−360 0
10
1
10
Frequency (rad/sec) (rad/sec)
1.3 Answer:
Pole−Zero Map 3 (one pole at −1)
2
1.5
Imaginary Axis
1
0.5
0
−0.5
−1
−1.5
−2
−2
−1.5
−1
−0.5
0
Real Axis
0.5
1
1.5
2
This system has a pole at s = −1, the transfer function will be
G3 (s) =
K
.
s+1
The break frequency is ω = 1 (rad/s). Before the break frequency, the Bode
magnitude plot is flat and after the break frequency it decreases with −20
2
dB/dec slope. In the Bode phase plot, the phase at low frequencies is 0◦ and
decreases to −90◦ over 2 decade bandwidth around the break frequency.
Bode Plot
Magnitude (dB)
0
−10
−20
−30
Phase (deg)
−40
0
−30
−60
−90 −2
10
−1
10
0
1
2
10
10
Frequency (rad/sec) (rad/sec)
10
1.4 Answer:
Pole−Zero Map 4 (two poles at −1 and −2)
2
1.5
Imaginary Axis
1
0.5
0
−0.5
−1
−1.5
−2
−2.5
−2
−1.5
−1
−0.5
0
0.5
Real Axis
1
1.5
2
2.5
This system has two poles at s
= −1 and s = −2. The transfer function
will be
K
.
G4 (s) =
(s + 1)(s + 2)
The break frequencies are ω = 1 (rad/s) and ω = 2 (rad/s). The Bode
magnitude plots indicates −20dB/dec and the Bode phase plots indicates
−90◦ at each break frequency.
Bode Plot
Magnitude (dB)
0
−20
−40
−60
−80
Phase (deg)
−100
0
−45
−90
−135
−180 −1
10
0
1
10
10
Frequency (rad/sec) (rad/sec)
1.5 Answer:
3
2
10
Pole−Zero Map 5 (two poles at −1± j3)
4
3
Imaginary Axis
2
1
0
−1
−2
−3
−4
−2
−1.5
−1
−0.5
0
Real Axis
0.5
1
1.5
2
This system has a pair of complex poles at s = −1 ± j3. The transfer
function will be
K
K
,
= 2
(s − (−1 + j3))(s − (−1 − j3))
s + 2s + 10
√
√
√
where ωn = 10 and ζ = 1/ 10. Since ζ < 1/ �
2, the resonance peak
should be observable. The peak frequency is ωp = ωn 1 − 2ζ 2 = 2.82(rad/s)
and the correction is −20 log(2ζ) ≈ 4 dB. In Bode phase plot, at low fre­
quencies, the phase is 0◦ and it decreases to −180◦ over 2 decades bandwidth
around the peak frequency.
G5 (s) =
Bode Plot
Magnitude (dB)
0
−20
−40
−60
Phase (deg)
−80
0
−45
−90
−135
−180 −1
10
0
1
2
10
10
Frequency (rad/sec) (rad/sec)
10
1.6 Answer:
Pole−Zero Map 6 (one pole at −1, one zero at 1.2)
2
1.5
Imaginary Axis
1
0.5
0
−0.5
−1
−1.5
−2
−2
−1.5
−1
−0.5
0
Real Axis
0.5
1
1.5
2
This system has a pole at s =
−1 and a zero at s = 1.
2. The transfer
4
function will be
G6 (s) =
K(s − 1.2)
.
(s + 1)
The Bode magnitude plot gets +20 dB/dec at the zero and -20 dB/dec at
the pole. Therefore, at the low and high frequencies, the Bode magnitude
plot would be flat. Since the two break frequencies are very close, the
Bode magnitude plot looks almost flat. In the Bode phase plot, the zero
in the right–hand plane has the same effects as the pole in the left–hand
plane, the total phase change is −180◦ . When you plug in s = 0, then
g6 (s = 0) = −K/1.2, which is a negative value. Thus the Bode phase plot
starts at −180◦ or 180◦ and decreases by −180◦ .
Bode Plot
Magnitude (dB)
10
5
0
−5
Phase (deg)
−10
180
135
90
45
0 −1
10
0
1
10
Frequency (rad/sec) (rad/sec)
10
1.7 Answer:
Pole−Zero Map 7 (one pole at −1, one zero at −1.2)
2
1.5
Imaginary Axis
1
0.5
0
−0.5
−1
−1.5
−2
−2
−1.5
−1
−0.5
0
Real Axis
0.5
1
1.5
2
This system has a pole s = −1 and a zero at s = −1.2. The transfer function
will be
K
.
G7 (s) =
(s + 1)(s + 1.2)
The Bode magnitude plot is identical as the case of (g) since they have the
same break frequencies. Because the zero is on the left–hand plane, the total
phase is zero.
5
Bode Plot
Magnitude (dB)
10
5
0
−5
Phase (deg)
−10
0
−30
−60
−90 −1
10
0
1
10
Frequency (rad/sec) (rad/sec)
10
1.8 Answer:
Pole−Zero Map 7 (one pole at −1, one zero at zero)
2
1.5
Imaginary Axis
1
0.5
0
−0.5
−1
−1.5
−2
−2
−1.5
−1
−0.5
0
Real Axis
0.5
1
1.5
2
The system has a pole at s = −1 and a zero at s = 0. The transfer function
will be
Ks
.
G8 (s) =
s+1
Because of the zero at s = 0, the Bode magnitude plot starts with 20dB/dec
slope and it becomes flat after the break frequency at ω = 1 (rad/s). To find
the Bode phase plot, let’s compute the phase analytically.
G8 (jω) = K
jω(1 − jω)
ω 2 + jω
jω
=
=
.
1 + jω
(1 + jω)(1 − jω)
1 + ω2
Thus the phase is
−1
tan
� �
1
.
ω
If ω is very small, then 1/ω goes to the infinity, which the phase is 90◦ . After
the break frequency at ω = 1 (rad/s), the phase decreases by −90◦ due to
the pole.
6
Bode Plot
Magnitude (dB)
0
−10
−20
−30
−40
Phase (deg)
−50
90
60
30
0 −2
10
−1
10
0
1
10
10
Frequency (rad/sec) (rad/sec)
2
10
2. Nise chapter 10, problem 25 (page 679).
a. Find the gain margin, phase margin, zero dB frequency, 180◦ frequency, and
the closed–loop bandwidth.
Answer: From the given Bode plot, the zero dB frequency is ω ≈ 2 (rad/s)
and the phase at the zero dB frequency is −130◦ . Therefore the phase
margin is −130 − (−180) = 50◦ . The 180◦ frequency is about 6 (rad/s),
whose gain is about −12dB. Thus the gain margin is 12dB.
To find the closed–loop bandwidth, we find the frequency at which the
open–loop magnitude response is between −6 and −7.5 dB if the open–loop
phase response is between −135◦ and −225◦ . From the given bode plot, the
frequency is approximately 3 (rad/s), thus ωBW = 3 (rad/s). (Refer to page
653 of Nise)
b. Use your result in (a) to estimate the damping ratio, percent overshoot,
settling time, and peak time.
Answer: From the figure 10.48 in page 653 of Nise (or page 8 of Lecture note
32), ζ ≈ 0.45.
�
�
ζπ
= 20.5%.
%OS = exp − �
1 − ζ2
From the figure 10.41 in page 644 of Nise (or page 7 of Lecture note 32),
�
�
4
(1 − 2ζ 2 ) + 4ζ 4 − 4ζ 2 + 2 ≈ 3.92 (s)s.
Ts =
ωW B ζ
�
�
π
�
Tp =
(1 − 2ζ 2 ) + 4ζ 4 − 4ζ 2 + 2 ≈ 1.551 (s).
ωW B 1 − ζ 2
3. Nise chapter 10, problem 23 (page 679).
a. System 1 Answer: From the following Bode plot,
7
Phase (deg)
Magnitude (dB)
Bode Diagram
80
60
40
20
0
−20
−40
−90
System: sys1
Phase Margin (deg): 16.7
Delay Margin (sec): 0.03
At frequency (rad/sec): 9.69
Closed Loop Stable? Yes
−135
−180
−2
10
−1
0
10
1
10
2
10
10
Frequency (rad/sec)
the phase margin is 16.7◦ . The damping ratio ζ is approximately 0.15.
Using the same method in question (2), the closed–loop bandwidth ωW B =
14 (rad/s). Thus
�
�
ζπ
%OS = exp − �
≈ 62%.
1 − ζ2
Ts =
4
ωW B ζ
�
(1 − 2ζ 2 ) +
π
�
Tp =
ωW B 1 − ζ 2
�
4ζ 4 − 4ζ 2 + 2 ≈ 2.9 (s)s.
�
(1 − 2ζ 2 ) +
�
4ζ 4 − 4ζ 2 + 2 ≈ 0.35 (s).
Note that it approximately agrees with the exact step response.
System: Closed Loop r to y
I/O: r to y
Peak amplitude: 1.65
Overshoot (%): 64.6
At time (sec): 0.317
Step Response
1.6
1.4
Amplitude
1.2
1
System: Closed Loop r to y
I/O: r to y
Settling Time (sec): 2.62
0.8
0.6
0.4
0.2
0
0
0.5
1
1.5
Time (sec)
2
2.5
3
b. System 2 Answer: From the following Bode plot,
8
3.5
Bode Diagram
Magnitude (dB)
50
0
Phase (deg)
−50
−90
System: sys2
Phase Margin (deg): 29.3
Delay Margin (sec): 0.0794
At frequency (rad/sec): 6.44
Closed Loop Stable? Yes
−135
−180
−1
10
0
1
10
2
10
10
Frequency (rad/sec)
the phase margin is 29.3◦ . The damping ratio ζ is 0.25. Using the same
method in question (2), the closed–loop bandwidth ωW B = 10 (rad/s). Thus
�
�
ζπ
%OS = exp − �
≈ 44%.
1 − ζ2
Ts =
4
�
(1 − 2ζ 2 ) +
ωW B ζ
π
�
Tp =
ωW B 1 − ζ 2
�
4ζ 4 − 4ζ 2 + 2 ≈ 2.4 (s)s.
�
(1 − 2ζ 2 ) +
�
4ζ 4 − 4ζ 2 + 2 ≈ 0.48 (s).
Note that it approximately agrees with the exact step response.
I/O: r to y
Peak amplitude: 1.44
Overshoot (%): 44.1
At time (sec): 0.458
1.5
Step Response
1
Amplitude
System: Closed Loop r to y
I/O: r to y
Settling Time (sec): 2.03
0.5
0
0
0.5
1
1.5
2
Time (sec)
9
2.5
3
3.5
4