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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
Steady State
Response
Transient
Response
x(t )  A cost   
What are we considering?
Final target
Complete
Response
Chapter 5 and 9
What really
observed
Zero State
Response
Zero Input
Response
Forced
Response
Natural
Response
Steady State
Response
Transient
Response
This lecture
x(t )  Aet 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
t
t

x
(
t
)

A
e
cost   
x(t )  Ae 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)
iL (t )  I m cost  i 
iL (t )
vL (t )  LI m sin t  i 
vL (t )

 LI m cos t  i  90
I L  I m i

Impedance of inductor
vL (t )  L
diL (t )
dt


VL  LI m  i  90
LI m i  90 
ZL 

 L90  jL
IL
I m i
VL

Complex Frequency - Inductor
x(t ) 
Aet cost   
iL (t )  I m et cost  i 
vL (t )  LI met cost  i   LI m et sin t  i 
 LI m et  cost  i    sin t  i 
iL (t )
vL (t )
vL (t )  L
diL (t )
dt


 LI m e    
cost  i 
2
2
  



sin t  i 
2
2
 

t
2
2


 LI m et  2   2 cos t  i  tan 1 


Complex Frequency - Inductor
iL (t )  I m et cost  i 
I L  I m i


VL  LI m  2   2  i  tan 1 


vL t   LI m et  2   2

1  
cos t  i  tan



Generalized Impedance of inductor
ZL 
VL
IL

LI m

1  
    i  tan

   L  2   2  tan 1  



I m i
2
2



 L    
j
2
2
 2 2
  
2
2

    j L  sL


s    j
Generalized Impedance
• Generalized Impedance (Table 10.1)
Element
Impedance
Generalized
Impedance
Resistor
Inductor
R
j L
R
sL
Capacitor
1 / j C
1 / sC
Special case:
s  j
The circuit analysis for DC circuits can be used.
Example 10.2
1
F
10
v(t )  20e  2t cos(4t  90 )V
L  1H, R  5, C 
1 
R

Z ( s )  sL   R //

sL


sC
sCR  1


 2.5143
s domain
diagram
V  2090 V
s  2  j 4
V
2090

I


8


53
Z ( s ) 2.5143
i (t )  8e 2t cos(4t  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
Y
Hs  
X
phasors of current or voltage
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
V
Impedance: Zs  
I
1
I
Admittance: Ys  

Z s  V
V
Y
Hs  
X
Impedance and admittance are
special cases for network function
Network Function / Transfer
Function
Y
Hs  
X
Current or voltage
Current or voltage
In general, network function can have four meaning
Hs  
V2
Hs  
I2
V1
I1
Voltage Gain
V
Hs  
I
“Impedance”
Current Gain
I
Hs  
V
“Admittance”
Example 10.5
VL
Vs
I L sL
VR
R IC
1
sC
H1 ( s ) 
H1 s  
1 

Vs  I L  sL  R 

sC 

IL
Vs
IL
Vs
VC
Polynomial of s

1
sL  R 
1
sC

sC
s 2 LC  sRC  1
Polynomial of s
Example 10.5
VL
Vs
I L sL
VR
R IC
1
sC
H 2 (s) 
VL
Vs

VC
sL
1
sL  R 
sC
Polynomial of s
s 2 LC
 2
s LC  sRC  1
Polynomial of s
Network Function / Transfer
Function
Y  Hs X
| Y || Hs  || X | Y   Hs   X 
|H(s)| is complex
frequency
dependent
 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7t   
Example 10.5 –
Check your results by DC Gain
VL
Vs
I L sL
VR
R IC
1
sC
VC
For DC
IL
sC
s 2 LC  sRC  1
H1 0   0
Capacitor =
open circuit
2
s
LC
H 2 (s) 
 2
Vs
s LC  sRC  1
H 2 0   0
Inductor =
short circuit
H1 ( s ) 
Vs
VL

Example 10.5 –
Check your results by Units
VL
VR
R IC
I L sL
Vs
1
sC
A
C:F  t
V
A
V
L:H  t
A
V
R : 
A
1 A
t
t V
V
IL
sC
H1 ( s ) 
 2
Vs
s LC  sRC  1
V
2
 1   V  A 
   t  t 
 t   A  V 
1V A
t
t A V
1
s:?
t
VC
2
 1   V  A 
   t  t 
 t   A  V 
VL
2
s
LC
H 2 (s) 
 2
Vs
s LC  sRC  1
V
1
 
t
2
 V  A 
 t  t 
 A  V 
1V A
t
t A V
Zeros/Poles
Poles/Zeros
• General form of network function
bm s m  bm 1s m 1    b1s  b0 Ns 
H (s) 

n
n 1
an s  an 1s    a1s  a0 Ds 
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.
Poles/Zeros
• General form of network function
bm s m  bm 1s m 1    b1s  b0
H (s) 
an s n  an 1s n 1    a1s  a0
bm 1 m 1
b0
b1
s 
s  s 
bm
bm
bm
bm

s  z1 s  z 2  s  z m 

K
a
a
a
an s n  n 1 s n 1    1 s  0
s  p1 s  p2  s  pn 
an
an
a1
m
bm
K
an
m zeros: z1, z2, … ,zm
n poles: p1, p2, … ,pn
Example 10.8
s 4  16s 3  164s 2
H (s)  5
s  32 s 2  36 s 2  40s  400


Denominator:
Numerator:
s 4  16 s 3  164 s 2

 s  s  s  16s  164
2

Find its zeros and poles

 s  0s  0s  z3 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  32  p1  32
s 2  36  p2 , p3   j 6
s 2  40 s  400  p4  p5  20
Poles:
p1=-32, p2=j6, p3=-6j,
p4=-20, p5=-20
Example 10.8
s 4  16s 3  164s 2
Find its zeros and poles
H (s)  5
2
2
s  32 s  36 s  40s  400
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
H(p)=∞
If the output is y t   Ae cos p   
p
Y
Y
X
 0
H 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
1 , 2     2  02
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
Vin
Vin
Zin 

Vin  Vout
I1
1
sC
Vin

sC Vin  Vout


Vin

sC Vin   AVin

1
sC 1  A


Capacitor with
capacitance C(1+A)
Example 10.3 - Miller Effect
Vout
Vout
Zout 

Vout  Vin
I2
1
sC
Vout

sC Vout  Vin


 AVin

sC  AVin  Vin 

A

sC 1  A
1
 1
sC 1  
 A
Capacitor with C 1  1 


capacitance
 A
Example 10.3 - Miller Effect