### Lecture 21

```Lecture 21
Network Function
and s-domain Analysis
Hung-yi Lee
Outline
• Chapter 10 (Out of the scope)
• Frequency (chapter 6) → Complex Frequency (sdomain)
• Impedance (chapter 6) → Generalized
Impedance
• Network function
What are we considering?
Final target
Chapter 6
and 7
Complete
Response
Chapter 5 and 9
What really
observed
Zero State
Response
Zero Input
Response
Forced
Response
Natural
Response
Response
Transient
Response
x ( t )  A cos  t   
What are we considering?
Complete
Response
Final target
Chapter 5 and 9
What really
observed
Zero State
Response
Zero Input
Response
Forced
Response
Natural
Response
Response
Transient
Response
x ( t )  Ae
t
This lecture
cos  t   
Complex Frequency
Complex Frequency
In Chapter 6
Current or Voltage Sources
x ( t )  A cos  t   
Currents or Voltages in the circuit
x ( t )  A  cos  t    
 The same frequency
 Different magnitude and phase
In Chapter 10
Current or Voltage Sources
Currents or Voltages in the circuit
x ( t )  Ae
t
cos  t   
x ( t )  A e
t
cos  t    
 The same frequency and exponential term
 Different magnitude and phase
You can observe the results from differential equation.
Complex Frequency
s plane
x ( t )  Ae
t
cos  t   
Phasor:
X  A
Complex
Frequency:
s    j
Generalized Impedance
Complex Frequency - Inductor
For AC Analysis
(Chapter 6)
I L  I m  i
i L ( t )  I m cos  t   i 
v L ( t )    L I m sin  t   i 
iL (t )
v L (t )

  L I m cos  t   i  90



V L   LI m   i  90
Impedance of inductor
v L (t )  L
di L ( t )
dt
ZL 
VL
IL

 LI m   i  90
Im  i

   L  90

 j L


Complex Frequency - Inductor
x (t ) 
Ae
t
iL (t )  I m e
t
cos  t   i 
cos  t   
v L (t )  L I m e
iL (t )
v L (t )
 LIme
t
 LIme
t

v L (t )  L
di L ( t )
dt
t
cos  t   i    L I m e

2

 LIme

t
sin  t   i 
cos  t   i    sin  t   i 


t
2

2

 



2

2

2
cos  t   i 

sin  t   i  


2

2

cos   t   i  tan

1
 

 
Complex Frequency - Inductor
iL (t )  I m e
t
v L t   L I m e
cos  t   i 
I L  I m  i
t
V L  LI m


cos   t   i  tan

2
1

2

2
 

2
     i  tan

1
 

 

 
Generalized Impedance of inductor
ZL 
VL
LI m


     i  tan

Im  i
2
2

IL
 L 
2

 



2

2

 j
2
1
 

  L 


2

2
2

2
    tan


    j  L  s L


s    j
1
 

 
Generalized Impedance
• Generalized Impedance (Table 10.1)
Element
Impedance
Generalized
Impedance
Resistor
R
R
Inductor
j L
sL
Capacitor
1 / j C
1 / sC
Special case: s  j 
The circuit analysis for DC circuits can be used.
Example 10.2
L  1H, R  5  , C 
1
s domain
diagram
V  20  90 V

F
10
v ( t )  20 e
2t

1 
R

Z ( s )  sL   R //

sL


sC
sCR  1


 2 . 5  143
s  2  j4
cos( 4 t  90 ) V

I
V

Z (s)
i (t )  8 e
20  90

2 . 5  143
2 t

 8   53
cos( 4 t  53 ) A


Network Function
Network Function / Transfer
Function
• Given the phasors of two branch variables, the ratio of the
two phasors is the network function/transfer function
H s  
Y
phasors of current or voltage
X
Complex number
The ratio depends on complex frequency
Y  H s  X
X : input
Y : output
Network Function / Transfer
Function
• Network Function/Transfer Function is not new
idea
I
Impedance: Z  s  
V
Admittance: Y  s  
H s  
Y
X
V
I
1
Z s 

I
V
special cases for network function
Network Function / Transfer
Function
H s  
Y
Current or voltage
X
Current or voltage
In general, network function can have four meaning
H s  
V2
H s  
I2
Voltage Gain
H s  
V
H s  
I
V1
I1
Current Gain
I
V
“Impedance”
Example 10.5
Vs
IL
VL
VR
sL
R
IC
1
VC
sC
H 1(s) 
IL
1 

V s  I L  sL  R 

sC 

Vs
Polynomial of s
H 1 s  
IL
Vs

1
sL  R 
1
sC

sC
s LC  sRC  1
2
Polynomial of s
Example 10.5
Vs
IL
VL
VR
sL
R
IC
1
VC
sC
H 2 (s) 
VL
Vs
sL

1
sL  R 
sC
Polynomial of s
2

s LC
s LC  sRC  1
2
Polynomial of s
Network Function / Transfer
Function
| Y | | H  s  || X |
Y  H s  X
|H(s)| is complex
frequency
dependent
 Y     H  s     X 
 4  j7
s    j
s  0 represents
dc
H ( 0 ) is the dc gain
Output will be
very large when
x ( t )  Ae
4 t
cos 7 t   
Example 10.5 –
Check your results by DC Gain
Vs
IL
VL
VR
sL
R
IC
1
VC
sC
For DC
H 1(s) 
H 2 (s) 
IL
Vs
VL
Vs

sC
s LC  sRC  1
2
2

s LC
s LC  sRC  1
2
H 1 0   0
Capacitor =
open circuit
H 2 0   0
Inductor =
short circuit
Example 10.5 –
IL
Vs
VL
VR
sL
R
IC
VC
1
sC
A
C:F  t
L:H  t
V
A
V
R : 
A
1
H 1(s) 
V
t
Vs

A
V
sC
H 2 (s) 
s LC  sRC  1
2
2
 1   V  A 
   t  t 
 t   A  V 
s:?
A
t V
IL
V
1V
t
A
t A V
V
VL
Vs
1
t
2
 1   V  A 
   t  t 
 t   A  V 
2

2
s LC
s LC  sRC  1
2
 1   V  A 
   t  t 
 t   A  V 
1V
t
A
t A V
Zeros/Poles
Poles/Zeros
• General form of network function
H (s) 
bm s
m
 b m 1 s
a n s  a n 1 s
n
m 1
n 1
   b1 s  b 0
   a1 s  a 0

The “zeros” is the root of N(s).
If z is a zero, H(z) is zero.
The “poles” is the root of D(s).
If p is a pole, H(s) is infinite.
N s 
D s 
Poles/Zeros
• General form of network function
b m s  b m 1 s
m 1
a n s  a n 1 s
n 1
m
H (s) 
s

m

b m 1
s
m 1
n
 
b1
s
bm
bm
bm
an
a n 1
a1
a0
n
s
n 1
 
an
an
K 
bm
an
s
   a1 s  a 0
b0
bm
s 
   b1 s  b 0
a1
K
 s  z1  s  z 2   s  z m 
 s  p1  s  p 2   s  p n 
m zeros: z1, z2, … ,zm
n poles: p1, p2, … ,pn
Example 10.8
s  16 s  164 s
4
H (s)  5
 s  32 s
3
2

s  16 s  164 s
3

Find its zeros and poles
Denominator:
s  32  p 1   32
2
 s  s  s  16 s  164
2

 36 s  40 s  400
2
Numerator:
4
2

  s  0  s  0  s  z 3  s  z 4 
z3 and z4 are the two
roots of s2+16s+164
Zeros:
z1=0, z2=0,
z3=-8+j10, z4=-8-j10
s  36  p 2 , p 3   j 6
2
s  40 s  400  p 4  p 5   20
2
Poles:
p1=-32, p2=j6, p3=-6j,
p4=-20, p5=-20
Example 10.8
s  16 s  164 s
4
H (s)  5
 s  32 s
3
2

2
 36 s  40 s  400
2

Find its zeros and poles
Zeros:
Poles:
z1=0, z2=0,
p1=-32, p2=j6, p3=-6j,
z3=-8+j10, z4=-8-j10 p4=-20, p5=-20
Pole and Zero Diagram
Zero (O), pole (X)
We can read the characteristics of the
network function from this diagram
For example, stability of network
Stability
• A network is stable when all of its poles fall within
the left half of the s plane
Y  H s  X
If p = σp + jωp is a pole
If the output is
X 
Y
Hp

Y

y t   Ae
0
H(p)=∞

p
cos  p  

Complex
frequency is p
No input ……
The waveforms corresponding to the complex
frequencies of the poles can appear without input.
Stability
• A network is stable when all of its poles fall within
the left half of the s plane
The poles are at
the right plane.
Appear
automatically
Unstable
The poles are at
the left plane.
Appear
automatically
Stable
Stability
• A network is stable when all of its poles fall within
the left half of the s plane
σp = 0
The poles are on
the jω axis.
Appear
automatically
Marginally stable
oscillator
Thank you!
Acknowledgement
• 感謝 趙祐毅(b02)
• 在上課時指出投影片中的錯誤
Appendix
What is Network/Transfer
Function considered?
Input
Output
Natural
Response
Forced
Response
Network
Function H(s)
Natural Response
• It is also possible to observe natural response from
network function.
Y  H s  X
Differential Equation
Fix ω0, decrease α
2
2
1 ,  2        0
The position of the two roots λ1 and λ2.
α=0
Undamped
Time-Domain Response of a
System Versus Position of Poles
(unstable)
(constant magnitude
Oscillation)
(exponential decay)
The location of the
poles of a closed
Loop system is shown.
Cancellation
Example 10.3 - Miller Effect
Z in 
V in

I1
V in
V in  V out
1
sC



V in

sC V in  V out

V in

sC V in    A V in

1
sC 1  A 
Capacitor with
capacitance C(1+A)
Example 10.3 - Miller Effect
Z out 
V out

I2
V out
V out  V in
1




V out
sC
sC V out  V in

  A V in
sC   A V in  V in 
A
sC 1  A 

Capacitor with
capacitance
1
1 

sC  1  
A

1 

C 1  
A

Example 10.3 - Miller Effect
```