Hung-yi Lee

Announcement
• 第四次小考
• 時間 : 12/24
• 範圍 : Ch11.1, 11.2, 11.4

Reference
• Textbook: Chapter 10.4
• OnMyPhD: http://www.onmyphd.com/?p=bode.plot#h3_com plex
• Linear Physical Systems Analysis of at the Department of Engineering at Swarthmore
College: http://lpsa.swarthmore.edu/Bode/Bode.html

Bode Plot
• Draw magnitude and phase of transfer function g dB
Magnitude
Phase
10
 log
20 a log
  a
2
 
(refer to P500 of the textbook)

10 -1 10 0
Angular Frequency
(log scale)
10 1 http://en.wikipedia.org/wiki/File:Bode_plot_template.pdf

Drawing Bode Plot
• By Computer
• MATLAB
• http://web.mit.edu/6.302/www/pz/
• MIT 6.302 Feedback Systems
• http://www.wolframalpha.com
• Example Input: “Bode plot of -
(s+200)^2/(10s^2)”
• By Hand
• Drawing the asymptotic lines by some simple rules
• Drawing the correction terms

Magnitude
H

K
 s s

 z
1 p
1
 s s

 z p
2
2


  

H

K 
 j
 j
 
 z
1 p
1

 j
 j


 z p
2
2



 a (

)

| H |

K j j


 z
1
 p
1 dB

20 log a (

)

20 log K

20 log

20 log j
  p
1 j
  z
1

20 log j
 j

 z
2
 p
2



20 log j
  z
2
 j
  p
2

Draw each term individually, and then add them together.

Magnitude – Constant Term
20 log a (

)

20 log j


20 log K
 p
1

20 log

20 log j
  z
1 j
  p
2


20 log j
  z
2

20 log K

Magnitude – Real Pole
20 log a (

)

20 log j


20 log K
 p
1

20 log

20 log j
  z
1 j
  p
2


20 log j
  z
2

Suppose p
1 real number is a p
1


If ω >> |p
1
|

20 log j
  p
1
 
20 log
If ω = 10Hz
If ω = 100Hz j
  
20 log

Magnitude
Magnitude




20
40 dB dB
Decrease 20dB per decade
| p
1
|
If ω << |p
1
|

20 log j
  p
1
 
20 log p
1
Constant

Magnitude – Real Pole
20 log a (

)

20 log j


20 log K
 p
1

20 log

20 log j
  z
1 j
  p
2


20 log j
  z
2

Constant
|p
1
|
Decrease
20dB per decade
If ω >> |p
1
|

20 log j
  p
1
 
20 log
If ω = 10Hz
If ω = 100Hz j
  
20 log

Magnitude
Magnitude




20
40 dB dB
Decrease 20dB per decade
Asymptotic Bode Plot
If ω << |p
1
|

20 log j
  p
1
 
20 log p
1
Constant

Magnitude – Real Pole
Cut-off Frequency
(-3dB)
If ω << |p
1
|

20 log p
1
If ω = |p
1
|

20 log j
  p
1



20

20 log log
 jp
1
2 |
 p
1 p
1
|

 
20 log |
 
20 log | p
1
| p
1

20
|

3 log
3dB lower
2

Magnitude – Real Zero
20 log a (

)

20 log j


20 log K
 p
1

20 log

20 log j
  z
1 j
  p
2


20 log j
  z
2

Suppose z
1 real number
 is a
| z
1
| z
1

If ω >> |z
1
|
20 log j
  z
1

20 log
If ω = 10Hz
If ω = 100Hz j
 
20 log

Magnitude

20
Magnitude

40 dB dB
Increase 20dB per decade
If ω << |z
1
|
20 log j
  z
1

20 log z
1
Constant

Magnitude – Real Zero
20 log a (

)

20 log j


20 log K
 p
1

20 log

20 log j
  z
1 j
  p
2


20 log j
  z
2

Constant
Asymptotic
Bode Plot
Increase
20dB per decade
If ω >> |z
1
|
20 log j
  z
1

20 log
If ω = 10Hz
If ω = 100Hz j
 
20 log

Magnitude

20
Magnitude

40 dB dB
Increase 20dB per decade
If ω << |z
1
|
|z
1
| 20 log j
  z
1

20 log z
1
Constant

Magnitude – Real Zero
• Problem: What if |z
1
| is 0?
Asymptotic
Bode Plot
 z
1

|z
1
|
If |z
1
|=0, we cannot find the point on the Bode plot

Magnitude – Real Zero
• Problem: What if |z
1
| is 0?
If |z
1
|=0
20 log j
  z
1

20 log j


20 log

If ω = 0.1Hz
Magnitude=-20dB
If ω = 1Hz Magnitude=0dB
If ω = 10Hz Magnitude=20dB
0 .
1
 
1 rad / s

10

Simple Examples
 p
1 p
2

| p
1
|
-20dB
+
| p
2
|
-20dB
 z
1 p
1

+20dB
+
| z
1
| | p
1
|
-20dB
-20dB
| p
2
| | p
1
|
-40dB
-20dB
| p
1
| | z
1
|

Simple Examples
 p
1 z
1 p
1

| z
1
|
+20dB +
 z
1

+20dB
+
-20dB
| p
1
|
+20dB
| z
1
|
| p
1
|
| p
1
|
-20dB
+20dB
| p
1
|

H
Magnitude – Complex Poles
H
 s
2 

0
1 s
Q
 
0
2

 
2 
1

0
Q j
  
0
2

1


0
2  
2

 j

0
Q

Q

0 .
5
The transfer function has complex poles
If
If
 
H
20

0
  log |

H
1 j

0

2
 
|
 
40 log

0
 
H j

 
0
20 log |

H
1 j
 
 
|
 
40 constant log

-40dB per decade

H
Magnitude – Complex Poles
H
 s
2 

0
1 s
Q
 
0
2

 
2 
1

0
Q j
  
0
2
H

1


0
2  
2

 j

0
Q

 
0 j
1

0
2
Q
The asymptotic line for conjugate complex pole pair.
If
If
 
20 log

0
| H constant
 
|
 
40 log

0
  
0
-40dB per decade
20 log | H
 
|
 
40 log

The approximation is not good enough peak at ω=ω
0
20 log

| H

40 log

|

20 log
Q

0
2

20 log Q
0

Magnitude – Complex Poles
H
 s
2 

0
1 s
Q
 
0
2
Height of peak:
Q
 dB
20 log Q constant
Only draw the peak when Q>1
-40dB per decade

Magnitude – Complex Poles
• Draw a peak with height 20logQ at ω
0 approximation is only an
• Actually,
The peak is at

0
1

1
2 Q
2
The height is
20 log

 Q 1

1
4 Q
2

 
0
Q

1
Q

1 .
67
Q

2 .
5
Q

5
Q

10

Magnitude – Complex Zeros constant
+40dB per decade
Q
 dB
20 log Q

Phase
H

K
 s s

 z
1 p
1
 s s

 z p
2
2


  

H

K 
 j
 j
 
 z
1 p
1

 j
 j


 z p
2
2





(

)
 
K





 j
 j
 
 z
1 p
1 j
 j
 
 z
2 p
2




Again, draw each term individually, and then add them together.

Phase - Constant

(

)
 
K
 
 j
  z
1 j
  z
2


 
 j
  p
1

K
K

0
Two answers

K
K

0 j
  p
2



Phase – Real Poles

(

)
 
K
 
 j
  z
1 j
  z
2


 
 j
  p
1 p
1 is a real number
 p
1
 j
  p
2


If ω << |p
1
|
 
 j
  p
1


?
0

If ω = |p
1
|
 
 j
  p
1


?

45

If ω >> |p
1
|
 
 j
  p
1


?

90


Phase – Real Poles

(

)
 
K
 
 j
  z
1 j
  z
2


 
 j
  p
1 p
1 is a real number
 
 j
  p
1

0.1|p
1
|
0
 |p
1
|
10|p
1
| j
  p
2


If ω << |p
1
|
 
 j
  p
1


?
0

If ω = |p
1
|
 
 j
  p
1


?

45

If ω >> |p
1
|
 
 j
  p
1


?

90


Phase – Real Poles
0 .
1 | z
1
|
| z
1
|
Exact
Bode Plot
Asymptotic
Bode Plot
10 | z
1
|

Phase – Real Zeros

(

)
 
K
 
 j
  z
1 z
1 is a real number
 z
1
 j
  z
2


 
 j
  p
1 j
  p
2


If z
1
< 0
If ω << |z
1
|
If ω = |z
1
|
If ω >> |z
1
|
 
 j
  z
1

90






 j
 j
 j
 

 z
1 z
1 z
1


?


?


?
45

0

|z
1
|
0

45

90


Phase – Pole at the Origin
• Problem: What if |z
1
| is 0?
 z
1


90


Phase – Complex Poles
If
  
0
0

H
 s
2 

0
1 s
Q

 
0
2
If
  
0

180
 p
1

0
If
  
0


90
 p
2

Phase – Complex Poles

Phase – Complex Poles
The red line is a very bad approximation.
(The phase for complex zeros are trivial.)

Magnitude – Real poles and zeros
0.1|P|
Given a pole p
0.5|P| |P| 2|P| 10|P|

Magnitude
– Complex poles and Zeros
H p
1
 s
2 
1

0
Q
 s
 
0
2

0 p
2

0
2 Q

Computing the correction terms at 0.5ω
0 and 2ω
0

Phase – Real poles and zeros
Given a pole p
0.1|P| 0.5|P| |P| 2|P| 10|P|
0 。
(We are not going to discuss the correction terms for the phase of complex poles and zeros.)

Exercise 11.58
K
 z
1 p
1
,
, z
2
100

0 ,

50 p
2
, p
3
 
100 ,

100 ,

400
• Draw the asymptotic Bode plot of the gain for H(s)
= 100s(s+50)/(s+100) 2 (s+400)
K 20log | K |

40dB
If ω << |p|

20 log p
If ω >> |p| Decrease 20dB per decade p
1
40dB p
2
40dB p
3
52dB
100 100 400

Exercise 11.58
K
 z
1 p
1
,
, z
2
100

0 ,

50 p
2
, p
3
 
100 ,

100 ,

400
K
40dB p
1
40dB
100 z
1

10

1 Hz , 0

100 Hz , 40
Hz , 20 dB dB

 dB
 p
2
40dB p
3
52dB
100 400
If ω << |z
1
| 20 log z
1
If ω >> |z
1
| Increase 20dB per decade z
2
34dB
50

Exercise 11.58
z
1
40dB p
1
40dB
K
Compute the gain at ω=100
100

40dB/decad e

20dB/decad e
50 100 p
2
40dB
100
?
400 z
2
34dB
50 p
3
52dB
400
20dB/decad e

Exercise 11.58
Compute the gain at ω=100
K
40dB p
1
40dB p
2
40dB
100 z
1

100 Hz , 40 dB


1 Hz , 0 dB

100 p
3
52dB
400 z
2
34dB
50
40dB

6dB
100
40dB

40dB

40dB

52dB

40dB

40dB
 
12 dB

Exercise 11.58
z
1
K
40dB p
1
40dB
100

40dB/decad e

20dB/decad e
50 100 p
2
40dB
-12dB
100
400 z
2
34dB
50 p
3
52dB
400
20dB/decad e

Exercise 11.58
• MATLAB

K
Exercise 11.52
K
 z

1
8000
0 p
1
, p
2
, p
3
 
10 ,

40 ,

80
• Draw the asymptotic Bode plot of the gain for H(s)
= 8000s/(s+10) (s+40)(s+80). Add the dB correction to find the maximum value of a(ω)
78dB p
1
20dB p
2
32dB p
3
38dB z
1
10
8dB
40 80
10 40
80

Exercise 11.52
• Draw the asymptotic Bode plot of the gain for H(s)
= 8000s/(s+10) (s+40)(s+80). Add the dB correction to find the maximum value of a(ω)
8dB
10 40
80
Is 8dB the maximum value?

Exercise 11.52
• Draw the asymptotic Bode plot of the gain for H(s)
= 8000s/(s+10) (s+40)(s+80). Add the dB correction to find the maximum value of a(ω)
K
78dB p
1
20dB p
2
32dB p
3
38dB z
1
Correction
P
1 p
2 p
3
Total
10 40 80
5 10 20
-1dB -3dB -1dB
40 80 160
-1dB -3dB -1dB
-1dB -3dB -1dB
-1dB -3dB -2dB -4dB -4dB -1dB

Exercise 11.52
• Draw the asymptotic Bode plot of the gain for H(s)
= 8000s/(s+10) (s+40)(s+80). Add the dB correction to find the maximum value of a(ω)
8dB
Maximum gain is about 6dB
20loga
  a


6dB
 

10
6
20

2 10 40
80 Correction
P
1 p
2 p
3
Total
5 10 20 40 80
-1dB -3dB -1dB
-1dB -3dB -1dB
160
-1dB -3dB -1dB
-1dB -3dB -2dB -4dB -4dB -1dB

Homework
• 11.59, 11.60, 11.63

Answer
• 11.59

Answer
• 11.60

Answer
• 11.63

• http://lpsa.swarthmore.edu/Bode/underdamped/u nderdampedApprox.html

Examples
• http://lpsa.swarthmore.edu/Bode/BodeExamples.h
tml