### frequency response

```Lecture 22
Frequency Response
Hung-yi Lee
Outline (Chapter 11)
Frequency Response
Amplitude
Ratio
Phase
Shift
Ch 11.1
Filter
Lowpass
Filter
Highpass
Filter
Bandpss
Filter
Notch
Filter
Ch 11.2
Bode
Plot
Ch 11.4
Draw frequency
response
Review
• Network Function
X
Y
Hs  
X
Network
Function
Y  Hs X
Hs 
Output depends on input and
complex frequency s.
Frequency Response
In this lecture, we only focus on s=jω.
That is, only
consider H(jω)
Y  Hs X
| Y || Hs  || X |
Y  Hs   X
Only focus on the
change of frequency
H(jω) is the frequency response.
Frequency Response
• H(jω) is a complex number, so frequency response is usually
represented by two curves
|Y|
| H j  |
|X|
a  
Magnitude
Ratio
H j 
 Y  X
Phase
Shift
  
Example 11.1
R  8 L  0.2H
vin (t )  10 cos 20t  10 cos 300t
vout (t )  ?
Network Function:
Vout
R
H (s) 

Vin sL  R
Frequency Response:
R
8
H ( j ) 

j  L  R j   0.2  8
0
40
H ( j ) 
j  40
tan 1
a ( )  H ( j ) 
40

40
1600   2
 ( )  H ( j )   tan
1

40
Example 11.1
R  8 L  0.2H
vin (t )  10 cos 20t  10 cos 300t
vout (t )  ?
a ( ) 
40
1600  
 ( )   tan 1
2

40
if   20 a (20)  0.894
 (20)  26.6
if   300 a (300)  0.132
 (300)  82.4
Based on Superposition:
vout  8.94 cos(20t  26.6 )  1.32 cos(300t  82.4 )
Example 11.1
Input
vin (t )  10 cos 20t  10 cos 300t
Output
vout  8.94 cos(20t  26.6 )
 1.32 cos(300t  82.4 )
Example 11.2
Find Network Function
1
sC
Vout  VC  Vx
1
sC
1
Vin  Vin
1
2
R
sC
2 
1 
R 

sC 
sC 

Vin
1


2 R 

sC 

1
R
sC V

1  in

2 R 

sC



H (s) 
Vout
Vin
1
1
s
sC   1 
RC

1
1  2

s

2 R 

RC
sC


R
Assume C=1/aR
a=1/RC
1 sa
H (s)   
2 sa
1
K
, z  a, p   a
2
Example 11.2
1 sa
1
H (s)   
a
2 sa
RC
1
K
, z  a, p   a
2
Set s=jω
 1 j  a
H ( j ) 

2 j  a
a  j

2a  j 
  a 
a 2   2  tan 1  a 
a    tan
2

2
2
1
1
sC
a ( )  H ( j ) 
a2   2
2 a2   2

1
2
The amplitude ratio is constant.
 ( )  H ( j )
  tan 1   a    tan 1  a 
 2 tan 1  a 
Example 11.2
 1 j  a
H ( j ) 

2 j  a
Distance of jω to a
Magnitude Ratio
1 | j  a | 1
| H ( j ) ||  | 

2 | j  a | 2
Distance of jω to -a
Phase Shift
    2 tan 1  a 
Example 11.2
• Phase shift network: Waveform distortion
1
a ( ) 
2
    2 tan  a 
vin (t )  10 cos10t  10 cos 20t
1
1
a
 10
RC
vout  5 cos(10t  90 )
 5 cos(20t  127  )
Another Example
for Phase Shift Network
1
sC
vout
vin
1
sC
V2 
1
sC
1
R
sC
Vin
V1 
R
R
1
sC
Vin
Vout  V2  V1
1

 R

 sC
R 1 R 1

sC
sC

1

R
sC

 R 1

sC



Vin



 sCR  1 

Vin
 sCR  1 


Vin



Another Example
for Phase Shift Network
vin
vout
Vout
 sCR  1 

Vin
 sCR  1 
sCR  1
Hs  

Vin sCR  1
Vout
jCR  1
H  j  
jCR  1
| jCR  1 |
a  
1
| jCR  1 |
     jCR  1   jCR  1
 tan 1  CR   tan 1 CR 
 2 tan 1 CR 
Computing
Frequency Response
by Poles and Zeros
Frequency Response

s  z1 s  z 2 
Hs   K
s  p1 s  p2 
K: gain factor
p1, p2, …: poles
z1, z2, …: zeros

p1
z1
z2

j  z1  j  z 2 
H  j   K
 j  p1  j  p2 
j
p2

Frequency Response
- Magnitude Ratio

j  z1  j  z 2 
H  j   K
 j  p1  j  p2 
Distance of jω to z1
a ( ) | H j  | K
Distance of jω to z2
j  z1 j  z 2 
j  p1 j  p2 
Distance of jω to p1
Distance of jω to p2
Frequency Response
- Magnitude Ratio
Magnitude ratio at ω

p1
d4
j
=| Gain factor|
X
d2
all distance to zeros
all distance to poles

z1
z2
Gain factor is K
d1d 2
a ( )  K
d 3d 4
d1
p2
d3
Frequency Response
- Phase Shift

j  z1  j  z 2 
H  j   K
 j  p1  j  p2 
 ( )  K
0 or  180
 1   2    3   4 
3
j
1
z1
z2
  j  p1    j  p2 
 ( )  K
p1
2
  j  z1    j  z 2 



p2
4

Example 11.3
H ( j ) 
20 j  25
  2  j 20  500
H (s) 
20s  25
s 2  20 s  500
y  tan 1 x
2025  j 

500   2  j 20

a ( ) 

20 625   2
500   
2 2
 400 2
20
 1 
1
2
tan

tan

 500
2

25
500  
 ( )  
tan 1   tan 1 20  180  2  500

25
500   2
Example 11.3
20s  25
H (s)  2
s  20 s  500
K  20 z1  25 p1 , p2  10  j 20
Example 11.3
20s  25
H (s)  2
s  20 s  500
K  20 z1  25 p1 , p2  10  j 20
Homework
• 11.2, 11.4
Application: Hearing
Reference: http://hyperphysics.phy-astr.gsu.edu/hbase/sound/hearcon.html
dB
Range of Hearing
Range of hearing: 20Hz - 20KHz
Range of Audio: 300Hz – 3.4KHz
Frequency
Equal Loudness Contour
Blue Line:
Same Energy
dB
Red Curve:
Equal loudness
phon (unit):
60 phon means
as loud as the
60dB 1kHz sound.
Hearing v.s. Age
Application: 柯南 68 集 (黑暗中消失的麒麟之角)
Phase and Sound
Pass a Phase
shift network
Structure of Ear
Cochlea (耳蝸)
Pass 20kHz
Each neuron only passes
a specific frequency.
Pass 20Hz
to
brain
Pass high
frequency
Pass low
frequency
Cochlea (耳蝸)
vin (t )  10 cos 20t
 10 cos 300t
Neuron for
ω=300
vout (t )  10 cos 300t
Neuron for
ω=100
vout (t )  0
ω=300, Amplitude=10
ω=20, Amplitude=10
to
brain
Neuron for
ω=20
vout (t )  10 cos 20t
Machine Hearing
 ( )
A set of filters (filter bank)
Ear of
machine
Feature
Send to computer
How?
Each triangle is
the amplitude
ratio of a filter
Energy of the
signal of the
output of the filter
How to write a program a. Observe data, and write down
for speech recognition? some rules
b. Simulate human brain, and let
the computer learn by itself.
Speech
Signal
Signal
Processing
(Filter Bank)
Features
Acoustic
Model
Machine Learning
ㄨㄣˊ ㄧˋ …

Text
Language
Model
Deep Neural
Network (Brain)
Speech Recognition
Thank you!
• 11.2:











1.075 cos 2t  3.9  1.346 cos cos 10t  21.8  0.482 cos 50t  71.9
• 11.4:

0.186 cos 2t  71  0.485 cos cos 10t  39.1  0.837 cos 50t  22.6
Acknowledgement
• 感謝 陳俞兆(b02)
• 上課時指出投影片的錯誤
• 感謝 陳均彥(b02)
• 上課時指出投影片的錯誤
• 感謝 林楷恩(b02)
• 指出投影片方程式的錯誤
```