Transcript Lecture 24 Bode Plot

Lecture 24
Bode Plot
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
  
Magnitude
dB
Degree
g  
10 log a  
 20 log a  
(refer to P500
of the textbook)
Phase
10-1
100
Angular Frequency
(log scale)
2
101
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
Asymptotic Lines:
Magnitude
Magnitude
H s   K
 s  z1  s  z 2 
 s  p1  s  p 2 
a ( )  | H  j   | K
H  j   K
 j   z1  j 
 j   p1  j 
 z 2 
 p 2 
j   z1 j   z 2 
j   p1 j   p 2 
dB  20 log a ( ) 
20 log K  20 log j   z 1  20 log j   z 2 
 20 log j   p 1  20 log j   p 2 
Draw each term individually, and then add them together.
Magnitude – Constant Term
20 log a ( )  20 log K  20 log j   z 1  20 log j   z 2 
 20 log j   p 1  20 log j   p 2 
20 log K
Magnitude – Real Pole
20 log a ( )  20 log K  20 log j   z 1  20 log j   z 2 
 20 log j   p 1  20 log j   p 2 
If ω >> |p1|
Suppose p1 is a
real number
p1
 20 log j   p 1

  20 log j    20 log 

| p1 |
If ω = 10Hz Magnitude   20 dB
If ω = 100Hz Magnitude   40 dB
Decrease 20dB per decade
If ω << |p1|
 20 log j   p 1   20 log p 1
Constant
Magnitude – Real Pole
20 log a ( )  20 log K  20 log j   z 1  20 log j   z 2 
 20 log j   p 1  20 log j   p 2 
Magnitude
Constant
If ω >> |p1|
 20 log j   p 1
|p1|
Decrease
20dB per
decade
  20 log j    20 log 
If ω = 10Hz Magnitude   20 dB
If ω = 100Hz Magnitude   40 dB
Decrease 20dB per decade
If ω << |p1|
Asymptotic Bode Plot
 20 log j   p 1   20 log p 1
Constant
Magnitude – Real Pole
If ω << |p1|
 20 log p 1
If ω = |p1|
 20 log j   p 1
  20 log jp 1  p 1
Cut-off Frequency
(-3dB)
  20 log

2 | p1 |

  20 log | p 1 |  20 log
  20 log | p 1 |  3
3dB lower
2
Magnitude – Real Zero
20 log a ( )  20 log K  20 log j   z 1  20 log j   z 2 
 20 log j   p 1  20 log j   p 2 
Suppose z1 is a
real number
20 log j   z 1
 20 log j   20 log 

z1 
| z1 |
If ω >> |z1|
If ω = 10Hz Magnitude  20 dB
If ω = 100Hz Magnitude  40 dB
Increase 20dB per decade
If ω << |z1|
20 log j   z 1  20 log z 1
Constant
Magnitude – Real Zero
20 log a ( )  20 log K  20 log j   z 1  20 log j   z 2 
Magnitude
 20 log j   p 1  20 log j   p 2 
Constant
Asymptotic
Bode Plot
If ω >> |z1|
20 log j   z 1
Increase
20dB per
decade
|z1|
 20 log j   20 log 
If ω = 10Hz Magnitude  20 dB
If ω = 100Hz Magnitude  40 dB
Increase 20dB per decade
If ω << |z1|
20 log j   z 1  20 log z 1
Constant
Magnitude – Real Zero
Magnitude
• Problem: What if |z1| is 0?

z1
Asymptotic
Bode Plot

|z1|
If |z1|=0, we cannot find
the point on the Bode plot
Magnitude – Real Zero
If |z1|=0
20 log j   z 1
 20 log j 
 20 log 
If ω = 0.1Hz
If ω = 1Hz
If ω = 10Hz
Magnitude=-20dB
Magnitude=0dB
Magnitude=20dB
Magnitude (dB)
• Problem: What if |z1| is 0?
0 .1
1
  rad / s 
10
Simple Examples

-20dB
p1
+

p2
| p1 |
-20dB
-20dB
| p 2 | | p1 |
| p2 |
-40dB

+20dB
z1
p1

| z1 |
-20dB
-20dB
+
| p1 |
| p1 |
| z1 |
Simple Examples

-20dB
p1
+20dB +

z1
| z1 |
| p1 |
+20dB
| z1 |
| p1 |

+20dB
p1
z1

-20dB
+
| p1 |
+20dB
| p1 |
Magnitude – Complex Poles
1
H s  
s 
2
0
s
Q
If    0
1
H  j  
 j 

Q  0 .5
2
0

2
0
2
0


j  
Q
1
2

The transfer function
has complex poles
j
0
Q
H  j   1  0
2
2
0
20 log | H  j   |  40 log  0
If    0

constant

H  j   1  
2

20 log | H  j   |  40 log 
-40dB per decade
Magnitude – Complex Poles
H s  
s 
2
0
Q
s  0
2
If    0
 j 

2
0
2
0

j  
Q
1

2

1
H  j 0  
0
2
j
Q
constant
20 log | H  j   |  40 log  0
1
H  j  

The asymptotic line for
conjugate complex pole pair.
1
j
0
Q

2
0
If    0 -40dB per decade
20 log | H  j   |  40 log 
The approximation is not good
peak at ω=ω0
enough
20 log | H  j   | 20 log
Q
0
2
  40 log  0  20 log Q
Magnitude – Complex Poles
1
H s  
s 
2
0
Q
s  0
2
Height of peak:
Q dB  20 log Q
Only draw the
peak when Q>1
constant
-40dB per decade
Magnitude – Complex Poles
• Draw a peak with height 20logQ at ω0 is only an
approximation
• Actually,
Q  1 . 67
Q  2 .5
The peak is at
0 1
Q 5
1
2Q
Q  10
2
The height is

20 log  Q


Q 1
1
1
4Q
2




0
Magnitude – Complex Zeros
+40dB per decade
constant
Q dB  20 log Q
Asymptotic Lines:
Phase
Phase
H s   K
 s  z1  s  z 2 
 s  p1  s  p 2 
H  j   K
 j   z1  j 
 j   p1  j 
 ( )   K
   j   z 1     j   z 2 
   j   p 1     j   p 2 
Again, draw each term individually,
and then add them together.
 z 2 
 p 2 
Phase - Constant
 ( )   K    j   z 1     j   z 2     j   p 1     j   p 2 
Two answers
K
K 0
K
K 0
Phase – Real Poles
 ( )   K    j   z 1     j   z 2     j   p 1     j   p 2 
p1 is a real number
If ω << |p1|

   j   p1   ?
0

If ω = |p1|
p1

   j   p 1   ?  45 
If ω >> |p1|
   j   p 1   ?  90 
Phase – Real Poles
 ( )   K    j   z 1     j   z 2     j   p 1     j   p 2 
p1 is a real number
If ω << |p1|
   j   p1 
   j   p1   ?
0.1|p1|
0

|p1|
10|p1|
0

If ω = |p1|
   j   p 1   ?  45 
If ω >> |p1|
   j   p 1   ?  90 
Phase – Real Poles
Exact
Bode Plot
0 .1 | z1 |
Asymptotic
Bode Plot
| z1 |
10 | z 1 |
Phase – Real Zeros
 ( )   K    j   z 1     j   z 2     j   p 1     j   p 2 
z1 is a real number

If z1 < 0
If ω << |z1|   j   z 1   ?
If ω = |z1|   j   z 1   ?
If ω >> |z1|   j   z 1   ?
   j   z1 
z1

90
45
0
0

45
90



|z1|


Phase – Pole at the Origin
• Problem: What if |z1| is 0?

z1

 90

Phase – Complex Poles
1
H s  
If    0
s 
2
0
Q
0

s  0
2

If    0
p1
 180
0


If    0
 90

p2
Phase – Complex Poles
Phase – Complex Poles
The red line is a very bad approximation.
(The phase for complex zeros are trivial.)
Correction Terms
Magnitude – Real poles and zeros
Given a pole p
0.1|P|
0.5|P| |P| 2|P|
10|P|
Magnitude
– Complex poles and Zeros
1
H s  
s 
2
0
Q
s  0
2

p1
0
0
2Q
p2

Computing the correction
terms at 0.5ω0 and 2ω0
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.)
Examples
Exercise 11.58
K  100
z 1 , z 2  0 ,  50
p 1 , 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
100
p 2 - 40dB
100
p 3- 52dB
400
Exercise 11.58
K
40dB
p 1- 40dB
K  100
z 1 , z 2  0 ,  50
p 1 , p 2 , p 3   100 ,  100 ,  400
p 2 - 40dB
100
z1
100 Hz , 40 dB 
10 Hz , 20 dB 
1 Hz , 0 dB 
p 3- 52dB
100
400
If ω << |z1| 20 log z 1
If ω >> |z1| Increase 20dB
per decade
z2
34dB
50
Exercise 11.58
z1
z2
34dB
50
K
40dB
p 1- 40dB
Compute the
gain at ω=100
p 2 - 40dB
100
100
400
?
- 20dB/decad e
 40dB/decad e
 20dB/decad e
50
p 3- 52dB
100
400
Compute the
gain at ω=100
Exercise 11.58
K
40dB
p 1- 40dB
p 2 - 40dB
100
z1
100 Hz , 40 dB 
1 Hz , 0 dB 
p 3- 52dB
100
400
40dB
z2
 6dB
34dB
50
40dB  40dB  40dB  52dB  40dB  40dB
100
  12 dB
Exercise 11.58
z1
z2
34dB
50
K
40dB
p 1- 40dB
p 2 - 40dB
100
100
p 3- 52dB
400
-12dB
- 20dB/decad e
 40dB/decad e
 20dB/decad e
50
100
400
Exercise 11.58
• MATLAB
K  8000
z1  0
Exercise 11.52
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)
K
78dB
z1
p 1- 20dB
p 2 - 32dB
10
40
10
p 3- 38dB
80
8dB
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
10
40
z1
Correction
5
10
20
P1
-1dB
-3dB
-1dB
p2
-1dB
p3
Total
-1dB
-3dB
-2dB
p 3- 38dB
80
40
80
160
-3dB
-1dB
-1dB
-3dB
-1dB
-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    6dB
6
a    10 20  2
Correction
5
10
20
P1
-1dB
-3dB
-1dB
p2
-1dB
p3
Total
-1dB
-3dB
-2dB
40
10
40
80
160
-3dB
-1dB
-1dB
-3dB
-1dB
-4dB
-4dB
-1dB
80
Homework
• 11.59, 11.60, 11.63
Thank you!
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