Chapter 2

Network Transforms &

Analysis
Outline
2.1. Network transform equations and analysis
Reference: F.F. Kuo, "Network Analysis and Synthesis“: Chapter 7

2.2. Network functions for 1- and 2-port networks

2.3. Poles and Zeros of Network

2.4. Frequency response and Bode plot

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The transformed circuit
• When analyzing a network in time domain we will be dealing with
• Derivation and Integration
• However, when transformed to complex frequency domain these become
• Derivation  multiplication by ‘ ’
• Integration  division by ‘ ’
• Hence, it is easier to do network analysis in complex frequency domain.
• A Linear Time-invariant (LTI) system can be described by a set of ordinary
linear differential equations (OLDEs)
• Initial Conditions (ICs) : Capacitor voltage and Inductor current
• Zero Input Response (ZIR) and Zero State Response (ZSR)
• Complementary function and Particular Integral
• Transient and steady state responses

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The transformed circuit…

N
d ( n ) r (t ) M
d ( m ) e(t )
OLDE :  an n
 a0   bm m
 b0
n 1 dt m 1 dt
N M
Laplace Transform :  an s R ( s )   bm s m E ( s )
n

n0 m0
M

R( s) m
b
m0
s m

Transfer function : H ( s )   N
E ( s)
 n
a
n0
s n

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The transformed circuit…
• If the network is lumped LTI:
R(s)
H (s)  is a rational function in s with real cofficents.
E (s)
N ( s)
H (s)  , D ( s )  0 : where N ( s ) and D ( s ) are polynomial s
D( s )
M bM 1 M 1 bM  2 M  2 b1 b0
s  s  s  ...  s
N ( s) bM bM bM bM
H (s)  
D ( s ) s N  a N 1 s M 1  a N  2 s N  2  ...  a1 s  a0
M aN aN aN aN

m 1
( s  zm ) = : Scale Factor
H (s)  k N
: Transmission zeros
 ( s  pn )
n 1
: Poles
= max( , ): Order
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Network Elements
• The voltage current relationships of network elements in time domain
and complex frequency domain are given as:
• Ideal energy sources
=ℒ and = ℒ[ ]
• Resistor
v(t )  Ri(t ) V ( s)  RI ( s )
• Inductor
di(t)
v(t)  L V (s)  sLI(s)  Li(0 )
dt
1 t V (s) i(0 )
i(t)   v( )d  i(0) I (s)  
L  sL s
0

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Network Elements…
• An inductor is represented in frequency domain as
• An impedance sL in series with a voltage source : used in mesh analysis.
• An admittance 1/sL in parallel with a current source: used in nodal analysis.

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Network Elements…
• Capacitor t
1
v(t )   i ( )d  v(0  ) I ( s ) v (0  )
C 0
V ( s)  
sC s
dv(t ) I ( s )  sCV ( s )  Cv(0  )
i (t )  C
dt
• A capacitor is represented in frequency domain as
• An impedance 1/sC in series with a voltage source :used in mesh analysis
• An admittance sC in parallel with a current source: used in nodal analysis

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Example 1
• In the figure below, the switch is switched from position 1 to 2 at = 0.
Draw its transformed circuit and write the transformed equations using
mesh analysis.

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Example 1…Solution
• The transformed circuit is

• Equation:

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Example 2
• In the figure below the switch is thrown to position 2 at = 0. Find i(t).
iL (0  )  2amp
vC (0  )  2V

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Example 2…Solution
• The transformed circuit is

• Transformed equation :
5 2  2
 2    3  s   I ( s)
s s  s

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Example 2…Solution…
• Solving for ( )
2s  3
I ( s) 
( s  2)( s  1)
1 1
I ( s)  
s  2 s 1
• Inverse transforming

i (t )  e 2t  e t

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Example 3
• Consider the following circuit. At = 0, the switch is opened. Find the node
voltages 1( ) and 2( )?
1
L h C  1f
2
G  1 mho V  1v

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Example 3…Solution
• The transformed circuit becomes

• Transformed equations :

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Example 3…Solution…
• Solving the previous equations

• Inverse transforming

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Outline
2.1. Network transform equations and analysis

2.2. Network functions for 1- and 2-port networks

Reference: F.F. Kuo, "Network Analysis and Synthesis“: Chapter 7

V.K. Aatre, "Network Theory and Filter Design“: Chapter 6

2.3. Poles and Zeros of Network

2.4. Frequency response and Bode plot

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Network Function
• The excitation , ( ), and response, ( ), of a linear system are related by a
linear differential equation.
• When transformed to complex frequency domain the relationship between
excitation and response is algebraic one.
• When the system is initially inert, the excitation and response are related by
the system function ( ) given by
R( s)  H ( s) E ( s)
• A network function is defined as the ratio of the zero-state response to the
input, both the response and the input expressed in Laplace domain.
R( s)
H (s) 
E (s)
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Network Function…
• The system function may have many different forms and may have special
names. Such as:
• Driving point admittance
• Transfer impedance
• Voltage or current ratio transfer function
• This is because
• excitation and response may be taken from the same port or different ports and
• excitation and response can be either voltage or current

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Driving Point Function
• If the excitation and the response are measured at the same set of terminals,
then the network function is called the driving point (DP) function.
Impedance
• When the excitation is a current source and the response is a voltage, then
the system function is an impedance.
V1 ( s )
z dp ( s )  (DP impedance)
Admittance I1 ( s )
• When the excitation is a voltage source and the response is a current, ( ) is
an admittance. I ( s)
ydp ( s )  1 (DP admittance)
V1 ( s )
1
y
• Note: dp ( s ) 
z dp ( s )
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Driving Point Function …

V0 ( s ) I 0 (s)
H ( s)  H ( s) 
I g ( s) Vg ( s )
 1 
  sL 1
sC
H ( s )  z dp ( s )  R   
H ( s )  ydp ( s ) 
1 1
 sL sL  R
sC sC
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Driving Point Function …
• Ladder Network

• DP function: continued fraction expansion

1
z11 ( s )  z1 ( s ) 
1
y2 ( s) 
1
z3 ( s ) 
1
y4 ( s) 
...
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Transfer Functions
• If the excitation and response are measured at different sets of terminals,
then the corresponding network function is called a transfer function.
Transfer Impedance
V2 ( s )
z 21 ( s )  (Transfer impedance)
I1 ( s )

Transfer Admittance

I 2 (s)
y21 ( s )  (Transfer admittance)
V1 ( s )

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Transfer Functions…
Voltage-ratio transfer function
• When the excitation is a voltage source and the response is also a voltage,
then ( ) is a voltage-ratio transfer function.
V2 ( s )
G21 ( s )  (Transfer voltage ratio)
V1 ( s )

V0 ( s ) V0 ( s )  Z 2 ( s ) I ( s )
H (s) 
Vg ( s )
Vg ( s ) Z 2 (s)
I (s)  H (s) 
Z1 ( s )  Z 2 ( s ) Z1 ( s )  Z 2 ( s )

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Transfer Functions…
Current-ratio transfer function
• When the excitation is a current source and the response is another current
in the network, then ( ) is called a current-ratio transfer function.
I 2 ( s)
 21 ( s)  (Transfer current ratio)
I1 ( s )

I 0 ( s)
H (s)   R  sL 
I g (s) I g ( s )  I 0 ( s )1  
 1 sC 
I g ( s )  I1 ( s )  I 0 ( s )
I 0 (s) 1 sC
1 
I1 ( s )  I 0 ( s )( R  sL) I g ( s ) R  sL  1 sC
sC

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Transfer Functions…
• Note that, the network function is a function of the system elements
only.
• It is obtained from the network by using the standard circuit laws.
Such as:
• Kirchhoff’s laws
• Nodal analysis
• Mesh analysis

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Example 4
• Obtain the driving point impedance of the network. Then using the
following excitations determine the response.
1. ig (t )  Sino t u (t )
2. ig (t ) is a square pulse in fig.b b

3. ig (t ) has waveform in fig.c

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Example 4…Solution
• First lets find the driving point impedance
• Note that it is the equivalent impedance of the 3 elements
1
z dp ( s ) 
ydp ( s )
1
ydp ( s )  sC  G
sL

1 s
H (s)  
1
sC   G C  s 2   G  s  1 
sL   C  CL 

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Example 4…Solution…
1. ig (t )  Sinot u (t )

Its transform is 0
I g (S )  2
s  o2

Hence, the response is

o s
Vo ( s )  I g ( s ) H ( s )  2 2

s  o  2 G 1 
C s  s 
 C LC 

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Example 4…Solution…
2. The excitation is given as
i (t )  u (t )  u (t  a )
1 1  as
I (s)   e
s s

Hence, the response is

1  e  as s
Vo ( s )  I g ( s ) H ( s )  
s  2 G 1 
C s  s 
 C LC 

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Example 4…Solution…
3. The excitation is given as
t ta
ig (t )  u (t )  u (t )  u (t  a )
a a
1 1 e  as
I (s)   2  2
s as as

Hence, the response is

1  e  as s
Vo ( s )  I g ( s ) H ( s )  
s  2 G 1 
C s  s 
 C LC 

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Natural and Forced Response
• Consider the partial fraction expansion of ( )
Ai Bj
R( s)   
i s  si i s  sj

where are the poles of ( ) and are the poles of ( ).

• Taking the inverse Laplace transform of ( )
s jt
r (t )   Ai e   B j e
si t

i i
si t
• The terms Ai e are associated with the system ( ) and are called the free
response terms.
• The terms B j e s j t are due to the excitation ( ) and are called the forced
response terms.
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Natural and Forced frequencies
• The frequencies are the natural frequency of the system, while the
frequencies are the frequencies of the excitation (forced frequencies).
• Example: Find the natural frequency of the following network.

di
L  Ri  0
dt
( sL  R ) I ( S )  0
sL  R  0
R
s
L

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Exercise
• Find the natural frequencies of the following networks?

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Exercise
• Find the free response and the forced response for the circuit below. The
system is inert before applying the source.
1
v g (t )  (cos t )u (t )
2

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Exercise
• Find the voltage transfer ratio of the active network shown below.

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Outline
2.1. Network transform equations and analysis

2.2. Network functions for 1- and 2-port networks

2.3. Poles and Zeros of Network

Reference: F.F. Kuo, "Network Analysis and Synthesis“: Chapter 6

V.K. Aatre, "Network Theory and Filter Design“: Chapter 6

2.4. Frequency response and Bode plot

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Poles and Zeros

N
d ( n ) r (t ) M
d ( m ) e(t )

n 1
an
dt n
 a0   bm
m 1 dt m
 b0
Zeros: roots of ( )
=0
M bM 1 M 1 bM  2 M  2 b1 b0 M
s  s  s  ...  s  (s  z m )
N ( s) bM bM bM bM m 1
H (s)   k N
D ( s ) s N  a N 1 s M 1  a N  2 s N  2  ...  a1 s  a0
aN aN aN aN
 (s  p )
n 1
n

Poles: roots of ( )
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Poles and Zeros…
• For lumped LTI network
• numerator polynomial ( ) & denominator polynomial ( ) have real
coefficients
• zeros and poles must be real or occur in complex conjugate pairs
• In the complex plane, a pole is denoted by a small cross, and a zero by a
small circle.
• For example if,

• Zeros: Poles:

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Pole-Zero Diagram

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Pole-Zero Diagram…

a. f (t )  u (t )
b. f (t )  e  t
0

c. f (t )  cos 0t
d . f (t )  e  t cos 0t
0

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Exercise
• Consider the network shown in figure below. Let ( ) be the input and ( )
(mesh current of the mesh II) be the output. Find ( ) and draw the pole-
zero diagram?

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Pole Location
• Effect of pole location upon exponential decay: > >0

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Pole Location…
• Pole locations corresponding to sin and sin : > >0

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Pole Location…
• Pole locations corresponding damped sinusoidal: >

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Pole Location…
• Effect of right-half plane poles upon time response

• Unstable!

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Pole Location…
• Effect of double poles on the axis

• Unstable!

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OCNF and SCNF
• For dp-impedance function let ( ) = 0.
N ( s ) V1 ( s)
H ( s )  z dp ( s )  
D( s ) I1 ( s )
D( s )V1 ( s )  0
• Thus the roots of ( ) are the natural frequencies of the network under open
circuit conditions & are called the open-circuit natural frequencies (OCNF).
• On the other hand for a dp-admittance network function if we make the
voltage ( ) go to zero, then the zero input current response is governed by
the equation 1 D ( s ) I1 ( s )
ydp ( s )   
z dp ( s ) N ( s ) V1 ( s )
N ( s ) I1 ( s )  0
• The network is being tested under short circuit conditions and the roots of
( ) are called the short-circuit natural frequencies (SCNF).
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Example 5
• Find the OCNF, SCNF and the input impedance of the network shown
in figure below.

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Example 5…Solution
• OCNF  Poles
1
sL  0
sC
s 2 LC  1  0
1
sj
LC
j
p1 , p2  
8

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Example 5…Solution…
• SCNF  Zeros
• From the inductor circuit 4 ||1 = → z1  0
• The network reduces to
1
sj
LC
j
s
4
2
5
5
z 2 , z3   j
8
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Example 5…Solution…
• To find
( s  z1 )( s  z 2 )( s  z3 )
Z in  k
( s  p1 )( s  p2 )
 2 5
s s  
8
z dp ( s )  k 
 2 1
s  
 8
• As → 0, the capacitor acts like an open circuit and the impedance of
network is 5 shown in figure. 2
s s  5 8
• Thus, =1 and Z in  2
s  1 8
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Exercise
• Find the OCNF, SCNF and the input admittance of the network shown
in figure below.

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Outline
2.1. Network transform equations and analysis

2.2. Network functions for 1- and 2-port networks

2.3. Poles and Zeros of Network

2.4. Frequency response and Bode plot

Reference: F.F. Kuo, "Network Analysis and Synthesis“: Chapter 8

V.K. Aatre, "Network Theory and Filter Design“: Chapter 6

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Frequency Response
• Let ( ) be any network function. Then
H ( s ) s  j   H ( j  )  H ( j  ) e j (  )
is known as the frequency response of the network.
• where H ( j ) : amplitude or magnitude response
 ( ) : phase response
• The frequency response of the network is the value of the network function
evaluated on the imaginary axis (real frequency axis) of the complex -plane.
• The amplitude and phase response of a system provide valuable information
in the analysis and design of transmission circuits.
• Most often the frequency response curves are plotted with frequency ( or )
represented on logarithmic scale. H ( j ) dB  20 log H ( j )
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Low-pass filter
• The cutoff frequency of the filter is
indicated on the amplitude response
curves as .
H ( jC )  0.707 H ( jmax )
20 log H ( jC )  20 log H ( jmax )  3dB
• The system will not “pass” frequencies
that are greater than .
• The phase response is almost linear till .
• Hence, if all the significant harmonic
terms are less than , then the system
will produce minimum phase distortion.

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Example 6
• Consider RC low-pass filter shown in figure below. Plot the amplitude and
phase characteristics of the network.

V2 ( s ) 1 / sC 1 RC
H ( s)   
V1 ( s ) R  1 / sC s  1 RC
1 RC
H ( j ) 
j  1 RC
1 RC
H ( j ) 
 2  1 RC 
2

 ( j )   tan 1 (RC )

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Example 6…
• The amplitude is unity and the phase
is zero degrees at = 0.
• The amplitude and phase decrease 1
monotonically as we increase . =

• When = 1/ , the amplitude is

0.707 and phase is 45°.
• Half power point

• As increases to infinity ( )
goes to zero and the phase
approaches −90°.

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Amplitude and phase from pole-zero diagram…
• For the system function
A0 ( s  z0 )( s  z1 )
H ( s) 
( s  p0 )( s  p1 )( s  p2 )
• ( ) can be written as
A0 ( j  z0 )( j  z1 )
H ( j ) 
( j  p0 )( j  p1 )( j  p2 )

• Each one of the − or − represent a vector from the zero or pole to

any on the imaginary axis.
j j
• If we express j  zi  N i e j , i
j  p j  M j e
• Then ( ) can be given as
A0 N 0 N1 j   1  2  0 1  2 
H ( j )  e 0

M 0 M 1M 2
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Amplitude and phase from pole-zero diagram…

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Amplitude and phase from pole-zero diagram…
• In general

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Example 7
• For
4s
F (s) 
s2  2s  2
find the magnitude and phase for = 2.
Solution:
• First let us find the zeros and poles
4 j
F ( j ) 
( j  1  j )( j  1  j )
• Zero j  0
• Poles: j  1  j and j   1  j

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Example 7…
• Magnitude
2 4
F ( j 2)  4 * 
2 * 10 5

• Phase

 ( j 2)  900  71.80  450  26.80

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Exercise (Reading Assignment)
• Examine the property of ( ) around the poles and zeroes.
• Reference: F.F. Kuo, "Network Analysis and Synthesis“: Chapter 8

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Bode Plot
• Bode plots are semi-logarithmic plots of amplitude and phase versus
frequency.
• In these plots we take the logarithm of the amplitude/phase and plot it on
linear frequency scale.
• For amplitude ( ) , if we express in terms of decibel it becomes
H ( j ) dB  20 log H ( j )
N (s)
• For system function H (s) 
D( s)
| N ( j ) |
| H ( j ) |
| D ( j ) |
• Expressing the amplitude in terms of decibels we have
20 log | H ( j ) | 20 log | N ( j ) | 20 log | D ( j ) |
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Bode Plot…
• In factored from both N(s) and D(s) are made up of 4 kinds of terms
• Constant
• A root at origin,
• A simple real root, s  a
• Complex set of roots, s 2  2s   2   2
• To understand the nature of log-amplitude plots, we only need to discuss the
amplitude response of these 4 terms.
• If the term is on the numerator it carries positive sign, if on denominator
negative sign.

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Constant
• The dB gain or loss is
20 log K  K 2
• is either positive | | > 1 or negative | | < 1
• The phase is either 0° for > 0, or 180° for < 0.

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Single root at origin,
• The loss or gain of a single root at origin is
 20 log | j | 20 log 
• Thus the plot of magnitude in dB vs frequency is a straight line with slope of
± 20 B/ .
• 20 when is in the numerator.
• -20 when is in the denominator.
• The phase is ±90°:
• 90° when s is in the numerator.
• −90° when s is in the denominator

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Single root at origin, …

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The factor +
• For convenience lets set =1. Then the magnitude is
1
 20 log | j  1 | 20 log  1
2 2

• The phase is
arg( j  1)  tan 1 

• A straight line approximation can be obtained by examining the asymptotic

behavior of the factor +1
• For << 1, the low frequency asymptote is
1
 20 log  1  20 log 1  0dB
2 2

• For >> 1, the high frequency asymptote is

1
 20 log  1  20 log 
2 2

• These 2 asymptotic approximations meet at = 1: break frequency or cutoff frequency

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The factor + …

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The factor + …
• Note that the maximum error is for = 1 (for non normalized = .)
• For the general case different from 1, we normalize the term by dividing
by .
• The low frequency asymptote is
1
2
  2
20 log 2  1  20 log 1  0dB
 

• The high frequency asymptote is

1
2
  2
20 log 2  1  20 log   20 log 
 

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Complex conjugates
• For complex conjugates it is convenient to adopt a standard symbol.
• We describe the pole (zero) in terms of magnitude 0 and angle measured
from the negative real axis.
• These parameters that describe the pole (zero) are
• 0, the undamped frequency of oscillation, and
• , the damping factor.
• If the pole (zero) pair is given as
p1, 2    j
• and are related to 0 and with
  0 cos   0
  0 sin   0 1   2

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Complex conjugates…
• Substituting these terms in the conjugate equation: ( s  p1 )( s  p2 )
 
( j    j )( j    j )  j  0  j0 1   2 j  0  j0 1   2 

   2  2 j 0  0
2

• For = 1 (for convenience), the magnitude of conjugate pairs can be
expressed as
1

 20 log 1    j 2  20 log 1  

2
 
2 2 2
 4  
2 2

• The phase is
2
 ( )  tan 1
1 2

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Complex conjugates…
• The asymptotic behavior is
• For low frequency, << 1
1


 20 log 1   
2 2
 4 2
  20 log1  0dB
2 2

• For high frequency, >> 1

1


 20 log 1   
2 2 2
 4    40 log 
2 2

which is a straight line with slope of 40 / .

• These 2 asymptotes meet at =1

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Complex conjugates…
• =1

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Magnitude: second-order pole

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Phase: second-order pole

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Example 8
• Using Bode plot asymptotes, draw the magnitude vs. frequency for the
following system function. Plot the actual response using Matlab.
0.1s
G(s)  2
 s  s s 
  1 
 4
 3
 1 
 50  16 *10 10 

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Example 8 …

  0 and   50 : first - order break

frequencies
  400 : second - order break frequency
  0.2

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Example 8… Matlab Plot

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Assignment
• V.K. Aatre, "Network Theory and Filter Design“:
Chapter 5: 5.3, 5.6, 5.9
Chapter 6: 6.1, 6.4, 6.3, 6.5, 6.10
• F.F. Kuo, "Network Analysis and Synthesis“:
Chapter 7: 7.3, 7.4, 7.6
Chapter 8: 8.1, 8.3
 Matlab  8.6, 8.7, 8.8

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