This document discusses transfer functions, frequency response, and bode plots in the context of linear time-invariant systems within electronic engineering. It covers fundamental concepts such as the properties of linear systems, application of transfer functions, and the significance of poles and zeros in determining system stability and response. Additionally, it includes examples to illustrate how to calculate transfer functions and their implications in circuit design.

1. Network Analysis
Chapter 5
The Transfer Functions,
Frequency Response, and Bode Plot
Chien-Jung Li
Department of Electronic Engineering
National Taipei University of Technology

2. Linear Systems
• Linear System
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Linear Time-invariant
Lumped (LTIL)
System
( )x t ( )y t
Input
(excitation)
Output
(response)
A system is said to be linear if the following two properties are hold:
Amplitude linearity: then( ) ( )x t y t→ ( ) ( )Kx t Ky t→
Superposition principle: and( ) ( )1 1x t y t→
( ) ( ) ( ) ( )1 2 1 2x t x t y t y t   + → +   
( ) ( )2 2x t y t→
then
• Time-invariant System
If the characteristics or properties of a system do not change with time,
then the system is said to be time invariant. (Note that time invariance is
defined for systems, not for signals.)

3. Transfer Function
( ) ( )X s x t =  L
( ) ( )Y s y t =  L
( )
( )
( )X
Y s
G s
s
=
( ) ( ) ( )Y s G s X s=
• Assume an initially relaxed linear system excited at t=0 by an input
x(t), and assume that y(t) is the corresponding output. Let
• For a linear system
where is called the transfer function of the circuit or
system, and it provides a direct mathematical relationship
between the input and the output for any arbitrary input.
Linear
system
( )x t ( )y t
( )X s ( )Y s( )G s
Transform into s-domain
(frequency domain)
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4. Application of Transfer Function
• Once the transfer function of the circuit or system is known, the
output may be determined for any arbitrary input.
• The transfer function is fixed by the nature of the system or circuit
and is not dependent on the type of excitation.
• The transfer function describes the input/output relationship which
can include frequency dependent parameters, i.e., the gain of an
amplifier.
• The transfer concept has been developed with the assumption that
the circuit has been initially relaxed. Unless state otherwise, initially
relaxed circuit conditions will be assumed in all further
developments relative to the transfer function.
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5. Example
( ) ( )
( ) ( )1 1
2 1
40 40 10
40 4 40 104
V s V ssV s V s
s s
s
= = =
+ ++
( )
( )
( )
2
1
10
10
V s
G s
V s s
= =
+
( )1V s ( )2V s( )
10
10
G s
s
=
+
+
−
( )1v t
+
−
( )2v t
1
F
40
4 Ω
+
−
( )1V s
+
−
( )2V s
40
s
4
• Determine the transfer function of the circuit:
5/64 Department of Electronic Engineering, NTUT

6. Example
( )
( )
( )
2
1
20 10
2 20 10
V s
G s
V s s s
= = =
+ +
+
−
( )1v t
+
−
( )2v t20 Ω
2 H
+
−
( )1V s
+
−
( )2V s20
2s
• Determine the transfer function of the circuit:
( ) ( ) ( ) ( )2 1 1
10
10
V s G s V s V s
s
= =
+
1( ) 20 for 0v t t= >
( )1
20
V s
s
=
( )
( )2
10 20 200
10 10
V s
s s s s
= =
+ +
10
2( ) 20 20 t
v t e−
= −
• If the input 1( ) 20sin10 for 0v t t t= >
( )
( )
( )
1 2 22
20 10 200
10010
V s
ss
= =
++
( )
( )( )2 2
2000
10 100
V s
s s
=
+ +
( )10
2( ) 10 10 2 sin 10 45t
v t e t−
= + −
• If the input
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7. Natural and Forced Response
( ) ( ) ( )Y s G s X s=
Total response ( )y t
Poles Forced response
Poles Natural response
• Consider the input/output relationship
• In general both G(s) and X(s) will have poles (as well as zeros). Thus
the poles of Y(s) will consist of the poles of G(s) and the poles of X(s).
Recall that each pole (or pair of complex poles) may be considered
as producing one of the terms in the associated output function y(t).
• The poles of G(s) are a function of the system parameters, whereas
poles of X(s) are a function of excitation or input.
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8. Example
( )
( )
( )
( )( )2 2
( )
5 6 2 17
Y s N s
G s
X s s s s s
= =
+ + + +
( )
( )
( )
( )( )( )2
( )
2 3 2 17
Y s N s
G s
X s s s s s
= =
+ + + +
( )
( )
( )( )( )2
( )
2 3 2 17
N s X s
Y s
s s s s
=
+ + + +
( ) 10cos6x t t=
( ) 2
10
36
s
X s
s
=
+
( )
( )
( )( )( )( )2 2
10
36 2 3 2 17
sN s
Y s
s s s s s
=
+ + + + +
( ) ( ) ( )2 3
1 1 2 3 4 2sin 6 sin 4t t t
y t B t B e B e B e tφ φ− − −
= + + + + +
• The transfer function of a certain system is given by
write the general form of the response y(t) for
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9. Poles and Zeros of Transfer Functions
( )
( )
( )
N s
G s
D s
=
( ) 1
1 1 0
n n
n nN s a s a s a s a−
−= + + …+ +
( ) 1
1 1 0
m m
m mD s b s b s b s b−
−= + + …+ +
• In general, the transfer function of a LTI system is a ratio of
polynomials in the variable s and can be expressed as
• Poles (finite): The m roots of D(s) are called the finite poles of the G(s)
• Zeros (finite): The n roots of N(s) are called the finite zeros of the G(s)
• The order of a transfer function is the value of the larger of the two
integers m and n. Thus, if m > n, the transfer function is of order n.
Most transfer function occurring in circuits are characterized by the
condition m ≥ n.
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10. Zeros and Poles at Infinite
in m n= −
( ) for 1n
nN s a s s≈ >>
( ) for 1m
mD s b s s≈ >>
( ) for 1i
n n
n
n m m
nm m n
m
a a
a s b b
G s s
b s s s−
≈ = = >>
( ) for 1i
n
mn mn n n
m
m m m
a s a a
G s s s s
b s b b
−
≈ = = >>
• If m > n, there are one or more zeros at infinite. In this case, the transfer
function is said to have ni zeros at infinite, where
Consider that s approaches infinity, N(s) and D(s) may each closely
approximately by their highest term:
im n m= −
• If m < n, there are one or more poles at infinite. In this case, the
transfer function is said to have mi poles at infinite, where
• We can conclude that the number of zeros of a transfer function is
equal to poles if zeros and poles at are included in the total.s = ∞
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11. Factored Form of Transfer Function
Since is a complex variable
(also called the complex frequency), the
poles and zeros can be plotted on the
plane (except the ones at infinite).
s jσ ω= +
• Let z1, z2, z3, …, zn represent the n finite zeros of G(s), and let p1, p2, …,
pm represent the m finite poles of G(s). G(s) may be expressed in
factored form:
( ) 1 2
1 2
( )( ) ( )
( )( ) ( )
n
m
A s z s z s z
G s
s p s p s p
− − −
=
− − −
⋯
⋯
• s-plane s-plane
: pole
: zero
1z
2z
4z
2p
1p
3p
Im
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3z
Re

12. Example
( )
( )[ ]
[ ][ ][ ]
0 ( 2)
( 6) ( 4 3) ( 4 3)
A s s
G s
s s j s j
− − −
=
− − − − − − − +
( )
( )( )( )
( )
( )( )2
2 2
6 4 3 4 3 6 8 25
As s As s
s s s j s j s s s
+ +
= =
+ + + + − + + +
• Write the transfer function corresponds to the s-plane
s-plane
6−
4 3j− +
4 3j− −
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−2 0
Re
Im

13. Example
• Construct an s-plane of finite poles and zeros with the transfer
function:
( )
( )2
3 2
2 6 25
7 10
s s
G s
s s s
+ +
=
+ +
s-plane
5−
3 4j− +
3 4j− −
2− 0
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Re
Im

14. Stability (I)
• The concept of stability is very important in the design and analysis
of active electronic circuits and closed-loop feedback control system.
For example, an amplifier may turn out to oscillate or to move into
saturation if circuit condition are not favorable.
• Relationship to Natural Response:
When a system is excited by an arbitrary input signal, the natural response
turns appear in the output and if such a response vanishes after a sufficient
period of time, the circuit settles into a of operation in which the forced
response assume a steady-state.
The question of stability can be related to whether or not the natural
response terms banish, remain at a fixed level, or possibly even grow without
bound.
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15. Stability (II)
• In the next few slides, we will see that stability can be related to the
locations of poles in the complex s-plane.
• Definitions of Stability:
Stable system: all natural response terms vanish or approach zero after a
sufficiently long time.
Unstable system: at least one term in the natural response grows without
bound (i.e., approaches infinity) as time increases.
Marginally stable system: there are no unstable terms and if at least one
term approaches a constant nonzero value or a constant amplitude
oscillation as time increases.
( )ny t ( )ny t ( )ny t
t t
t
Stable Unstable Marginally stable
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16. Practical Perspective – The S -Plane
s-plane
LHHP RHHP
Left-hand half-plane (LHHP):
All points to the left of the -axis, but not
including the -axis.
jω
jω
Right-hand half-plane (RHHP):
All points to the right of the -axis, but not
including the -axis.
jω
jω
jω-axis:
The -axis will be considered as a separated
area of the s-plane for reasons that will be
clear shortly. The -axis includes the point
s = 0.
jω
jω
• The effects of different pole locations will be investigated as they
relate to the natural response. In each case, the form of a natural
response term yn(t) will be shown based on the particular pole or pair
of poles assumed.
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Re
Im

17. Poles on the Negative Real Axis
LHHP
1 1s α= −
• The natural response corresponds to a pole
on the negative real axis is of the form:
( ) 1t
ny t Ae α−
=
where A is some arbitrary constant. This term
approaches zero as t increases, so it is a
stable response.
( )ny t
t
Stable
• Time constant . So, if is very close
to the origin, is very large, and the time
that it takes for the natural response to
become negligible can be rather long, and
vise versa.
s-plane
Time response
(natural, from circuit)
( )11τ α= 1α
τ
• Multiple-order pole:
( ) k t
ny t At e α−
=
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Re
Im

18. Complex Poles in the LHHP
LHHP
1 1 1s jα ω= − +
( )ny t
t
Stable
s-plane
Time response
(natural, from circuit)
1 1 1s jα ω∗
= − −
• The natural response corresponds to a pair of
complex poles in the LHHP is of the form:
( ) ( )1
1sint
ny t Ae tα
ω θ−
= +
where A and are arbitrary constants. This term
approaches zero as t increases, so it is a stable
response.
θ
• Time constant . The closer is to
the jω-axis, the longer will be the duration o
the natural response, and vice versa.
( )11τ α= α
• If the poles are very close to the negative real
axis, the oscillation frequency ω1 is small,
and the period of the oscillation will be long,
and vice versa.
( ) ( )1
1sintk
ny t At e tα
ω θ−
= +
• Multiple-order complex-pole pairs:
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Re
Im

19. Poles on Positive Real Axis and RHHP
RHHP
1 1s α=
( )ny t
t
Unstable
s-plane
Time response
(natural, from circuit)
• A pole on the positive real axis: The natural
response is of the form
( ) 1t
ny t Aeα
=
where A is some arbitrary constant. This term
grows without bound, so it is a unstable response.
2 2 2s jα ω= +
2 2 2s jα ω∗
= −
• Complex poles in the RHHP The natural
response is of the form:
( ) ( )2
2sint
ny t Ae tα
ω θ= +
where A is some arbitrary constant. This term
grows without bound, so it is a unstable response.
19/64 Department of Electronic Engineering, NTUT
Re
Im

20. Poles at the Origin and on jω-axis
• A pole at the origin: The natural response is
of the form
( )ny t A=
The response remains constant as time increases,
so it is marginally stable. 1 0s =
s-plane
2 2s jω=
2 2s jω∗
= −
( )ny t
t
Marginally stable
Time response
(natural, from circuit)
• Poles on jω-axis: The natural response is of
the form
( ) ( )2sinny t A tω θ= +
This function oscillates with a constant amplitude
for all time, so it is a marginally stable response.
In many systems, jω-axis poles are undesirable.
• Multiple Pole-pairs on jω-axis:
( ) ( )3sink
ny t At tω θ= +
which is an unstable response.
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Re
Im

21. Summary of Stability
s-plane
LHHP RHHP
Unstable
Stable
Poles at originPoles at originPoles at originPoles at origin
1st order: marginally stable
≥ 2nd order: unstable
Poles onPoles onPoles onPoles on jω-axis
1st order pole-pair: marginally stable
≥ 2nd order pole-pair: unstable
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Re
Im

22. Transfer Function Algebra
• Complete linear circuit blocks are connected to achieve a composite
system function and this complete transfer function may be
expressed as a combination of the individual transfer functions.
• Each transfer function given is either assumed to be unaffected by
the interconnection used, or the transfer function is defined under
the loaded conditions given.
• There may be loading effects between blocks, and if the loading
effect occurs, a modified transfer function could be defined under
such conditions. The point is that one cannot simply “throw together”
blocks and assume that transfer functions remain unchanged. The
loading effect will be discussed later.
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23. Cascade Connection
( )1G s ( )2G s ( )2G s( )X s ( )Y s
( )1Y s ( )2Y s
( ) ( ) ( )1 1Y s G s X s=
( ) ( ) ( )2 2 1Y s G s Y s= ( ) ( ) ( )1n n nY s G s Y s−=
( ) ( ) ( ) ( ) ( )1 2 3 nG s G s G s G s G s= ⋯( )X s ( )Y s
( ) ( ) 1 2( ) ( ) ( ) ( ) ( )nY s G s X s G s G s G s X s= = ⋯
• Here, we stress that the preceding individual transfer functions are
either unaffected by the connections, or the transfer function are
defined under loaded conditions.
• The composite transfer function of a cascade connection is the product of all
the individual transfer functions.
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24. Parallel Connection
( )1G s
( )2G s
( )nG s
∑( )X s ( )Y s 1 2( ) ( ) ( ) ( )nG s G s G s G s= + + +⋯( )X s ( )Y s
1 1( ) ( ) ( )Y s G s X s=
2 2( ) ( ) ( )Y s G s X s=
( ) ( ) ( )n nY s G s X s=
1 2( ) ( ) ( ) ( )nY s Y s Y s Y s= + + +⋯
[ ]1 2( ) ( ) ( ) ( ) ( )nY s G s G s G s X s= + + +⋯
1 2
( )
( ) ( ) ( ) ( )
( )
n
Y s
G s G s G s G s
X s
= = + + +⋯
• The composite transfer function of a parallel connection is the sum of all the
individual transfer functions.
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25. Feedback Loop
( )G s
( )H s
∑( )X s ( )Y s
+
−
( )F s
( )D s
( ) ( ) ( )D s X s F s= −
( ) ( ) ( )Y s G s D s=
( ) ( ) ( )F s H s Y s=
( )
( ) ( )
1 ( ) ( )
G s
Y s X s
G s H s
 
=  + 
( ) ( )
( )
( ) 1 ( ) ( )
Y s G s
T s
X s G s H s
= =
+
( )
( )
1 ( ) ( )
G s
T s
G s H s
=
+
( )X s ( )Y s
• A feedback loop consisting of a forward transfer function G (s) and a
feedback transfer function H(s). The subtraction of the input X(s) and
the feedback signal F(s) yield a difference variable D(s).
• The result is one of the most important
relationships of linear system theory, and it
servers as the basis for much of the design
work of stable linear circuit and closed-loop
feedback control system.
25/64 Department of Electronic Engineering, NTUT

26. Example
• A simple resistive network of voltage divider is used to illustrate how
loading effects can affect the transfer results.
+
−
( )1V s
1000 1
( )
1000 1000 2
AG s = =
+
1 kΩ
1 kΩ
+
−
( )V s
+
−
( )1V s
1 kΩ
1 kΩ
1 kΩ
1 kΩ
+
−
( )2V s
2
1
V ( ) 1000 1 1
( )
V ( ) 1000 1500 2 5
s
G s
s
= = =
+
cascade
Why not ?
1 1 1
( ) ( ) ( )
2 2 4
A AG s G s G s= ⋅ = ⋅ =
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27. Example (I)
( )1G s
( )1H s
∑( )X s
( )Y s
+
−
( )2G s
∑
( )3G s
+
+
• Determine a single transfer function equivalent to the system shown.
( )1G s
( )1H s
∑( )X s
( )Y s
+
−
( )2G s
∑
( )3G s
+
+
Feedback loop
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28. Example (II)
( )X s
( )Y s∑
( )3G s
( ) ( )
( ) ( ) ( )
1 2
1 2 11
G s G s
G s G s H s+
+
+
( )X s ( )Y s
( ) ( )
( ) ( ) ( )
( )1 2
3
1 2 11
G s G s
G s
G s G s H s
+
+
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29. Generalized 2nd-order Transfer Function
• Let P(s) represent a quadratic denominator factor with unity
coefficient for the s2 term:
α ω= + +2 2
( ) 2 nP s s s
ζω ω= + +2 2
( ) 2 n nP s s s
α
ζ
ω
=
n
where
or
is the damping ratio and ωn is the natural frequency.
• No Damping Circuit: α = 0 and also ζ = 0
ω= +2 2
( ) nP s s
The response will then contain an undamped sinusoidal function of
radian frequency .ωn
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30. Roots of the 2nd-oredr Denominator
α ω= + +2 2
( ) 2 nP s s s ζω ω= + +2 2
( ) 2 n nP s s s• The Quadratic Denominator: or
roots
α α ω ζω ω ζ= − ± − = − ± −1 2 2 2
2
1n n n
s
s
• Overdamped Circuit: α ω> n or when ζ > 1
In this case, is not too meaningful because no oscillations actually
occur. The two roots s1 and s2 are real that represent two damping
factors in the exponential terms of the natural response, respectively.
ωn
• Critically Damped Circuit: α ω= n or when ζ = 1
• Underdamped Circuit: α ω< n or when ζ < 1
α ω α α ω ζω ω ζ= − ± − = − ± = ± −1 2 2 2
2
1n d n n
s
j j j
s
roots
represents the damped oscillation frequency,
which is the actual oscillation frequency in the natural response for the
underdamped case.
ω ω α ω ζ= − = −2 2 2
1d n n
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31. Common Form of a 2nd-order Transfer Function
• The common form of a second-order transfer function that arises
frequently in practice, which is a low-pass second-order response, is
given by
ω
ζω ω
= =
+ +
2
2 2
( )
( )
( ) 2
n
n n
Y s
G s
X s s s
• Unit Step Response:
Let input x(t) = u(t), such that X(s) = 1/s
( )
ω
ζω ω
= =
+ +
2
2 2
( ) ( ) ( )
2
n
n n
Y s G s X s
s s s
• When the system is critically damped: ω
ω ζ−
= − + =( ) 1 ( 1) for 1nt
ny t t e
• When the system is underdamped:
ζω
ζ ω ζ
ζ
−
−
= − − +
−
2 1
2
( ) 1 sin( 1 cos )
1
nt
n
e
y t t
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32. Step Response of a 2nd-order System
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ωnt
v(t)

33. What is Frequency Response
• Frequency response is the quantitative measure of a system or device
in frequency domain to show how a system affects the frequency
components of the input signal.
• A Bode plot is a graph of the transfer function (LTI system) versus
frequency, plotted with a log-frequency axis, to show the frequency
response of the system.
• We may want to know what is the relationship between the frequency
response and s-plane.
ω
( )ωX
ω
( )ωY
Transfer function?
s-plane? Frequency Response?
(System)
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Linear Network
(Signal) (Signal)

34. Laplace Transform of a Signal
( ) ( ) t j t
X s x t e e dtσ ω
∞
− −
−∞
 = ⋅ ∫
0σ = 1σ = 2σ =1σ = −2σ = −
( )x t
t
F F F F F
( )σRe-axis r
σ = 0r
σ = 1r σ = 2rσ = −2r σ = −1r
( )Im-axis jω
Signal
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35. Fourier Transform is the Special Case
0σ =
Signal Spectrum
( )Re-axis σ ( )Im-axis jω
Amplitude
Department of Electronic Engineering, NTUT35/64
s-domain

36. Laplace Transform of a System – s-plane
Pole
Zero
System
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Linear Network
Pole-zero plot
Frequency response
s-domain

37. Frequency Response is the Special Case
Department of Electronic Engineering, NTUT37/64
Pole-zero plot
Frequency responsePhysical system
R C
L
Laplace
transform
Evaluate
at σ=0
Phasor transform
Real
Image
Frequency
Amplitude

38. Steady-state Impedance and Admittance
( ) ( )s j
Z Z s Z jω
ω=
= =
( ) ( ) ( )Z j R jXω ω ω= +
( ) ( )s j
Y Y s Y jω
ω=
= =
• Impedance Z
• Admittance Y
( ) ( ) ( )Y j G jBω ω ω= +
resistance
reactance
conductance
susceptance
Department of Electronic Engineering, NTUT
represents a complex value( )ωZ j
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39. Example
20sin2t
+
−
0t =
4 Ω
( )i t
1
F
6
2
40
4s +
+
−
4
( )I s
6
s
20 0∠
+
−
4
( )I s
6
3j
jω
= −
• Use (a) the s-domain model to find
the complete time-domain current i(t)
and (b) phasor-domain model to
determined the steady-state current
iss(t).
( )
( )( )
2 2
2
40 40
104 4
6 4 6 1.5 44
ss sI s
s s s
s s
+ += = =
+ + ++
(a)
( ) ( )1.5
2.4 4sin 2 36.87t
i t e t−
= − + +
(b) 20 0 20 0
4 36.87
4 3 5 36.87
I
j
∠ ∠
= = = ∠ −
− ∠ −
( ) ( )4sin 2 36.87ssi t t= +
Transient response Steady-state response
Department of Electronic Engineering, NTUT39/64

40. Steady-State Transfer Function
( )
( )
( )
Y s
G s
X s
=
( ) ( )
( )
( )
( ) ( )ω
ω
ω ω β ω
ω=
= = = ∠s j
Y j
G j G s A
X j
Department of Electronic Engineering, NTUT
• Steady-state Transfer function (Frequency Response): ,thusσ = 0 ω=s j
Amplitude response Phase response
• Input and output relationship
( ) ( )sin xx t X tω θ= +
( ) ( )sin yy t Y tω θ= +
Input:
Output:
phasor
phasor
xX X θ= ∠
yY Y θ= ∠
( )Y G j Xω= ( ) ( ) ( )y xY Y A Xθ ω β ω θ = ∠ = ∠ ∠  ( ) ( )( )ω θ β ω= ∠ +xX A
( )ω=Y X A ( )y xθ θ β ω= +Output amplitude: Output phase:
( ) ( )
( )
( )
ω
ω ω
ω
= =
Y j
A G j
X j
40/64

41. Decibel Response
• Whereas the linear amplitude response is useful for many purpose, the
decibel amplitude response are widely used in many applications.
( )A ω
( )
( )
( )
( )
ω
ω ω
ω
= =dB 10 1020log 20log
Y j
A A
X j
( )dBA ω
• For example: 0.01 FC µ=10 kR = Ω
R
+
−
( )1v t
+
−
( )2v tC
R
+
−
1V
+
−
2V1
j Cω
( )
( )
ω
ωω
= =
+ ⋅+
8 8
2 1 18 44 8
10 10
10 1010 10
j
V V V
jj
( )
8 4
2
8 4 4
1
10 10
10 10 10
V
G j
V j j
ω
ω ω
= = =
+ ⋅ +
( ) ( )
4
8 2
10
10
A G jω ω
ω
= =
+
( ) ( )
4
dB 10 10 8 2
10
20log 20log ( )
10
A Aω ω
ω
= =
+
( ) 1
4
tan
10
ω
β ω −
= −
Department of Electronic Engineering, NTUT41/64

42. Development of Bode Plot Approach
( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
1 2 3
0
1 2 3
N N N Nn
D D D Dm
G j G j G j G j
G j A
G j G j G j G j
ω ω ω ω
ω
ω ω ω ω
⋅ ⋅
= ⋅
⋅ ⋅
⋯
⋯
( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
1 2 3
0
1 2 3
N N N Nn
D D D Dm
G j G j G j G j
G j A
G j G j G j G j
ω ω ω ω
ω
ω ω ω ω
⋅ ⋅
= ⋅
⋅ ⋅
⋯
⋯
( )
     
+ + +     
+ + +      = =
+ + +      
+ + +     
     
⋯
⋯
⋯
⋯
1 2
1 21 2
1 2
1 2
1 2
1 1 1
( )( ) ( )
( )( ) ( )
1 1 1
n
nn
m
n
n
s s s
N N N
N N Ns N s N s N
G s A A
s D s D s D s s s
D D D
D D D
• Normalized Factored Form
• Steady-state Transfer Function
• Amplitude Response
( ) ( ) for any
( ) ( ) for any
Nk Nk
Dk Dk
A G j k
A G j k
ω ω
ω ω
=
=
The net amplitude can be represented as( ) ( )A G jω ω=
( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
1 2 3
0
1 2 3
N N N Nn
D D D Dm
A A A A
A A
A A A A
ω ω ω ω
ω
ω ω ω ω
⋅ ⋅
= ⋅
⋅ ⋅
⋯
⋯
Let
Department of Electronic Engineering, NTUT
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
⋅ ⋅
= ⋅
⋅ ⋅
⋯
⋯
1 2 3
0
1 2 3
N N N Nn
D D D Dm
G s G s G s G s
A
G s G s G s G s
Force the constant
terms to be unity.
42/64

43. Example
• Write the normalized factored form of ( )
( )+
=
+ +2
2000 4
116 1600
s
G s
s s
( )
( )
( )( )
+
=
+ +
2000 4
16 100
s
G s
s s
 
× + 
 =
   
+ × +   
   
4
2000 4
4 4
16 100
16 100
16 16 100 100
s
s s
 
+ 
 =
  
+ +  
  
5 1
4
1 1
16 100
s
s s
Normalized factored form ready for
Bode plot.
Department of Electronic Engineering, NTUT43/64

44. Linear Scale to Log Scale
10 10 10log log logxy x y= +
10 10 10log log log
x
x y
y
= −
( ) ( ) ( ) ( )
( ) ( ) ( )
10 10 0 10 1 10 2 10
10 1 10 2 10
20log 20log 20log 20log 20log
20log 20log 20log
N N Nn
D D Dm
A A A A A
A A A
ω ω ω ω
ω ω ω
= + + + +
− − − −
⋯
⋯
( ) ( )
( ) ( )
(dB)
10
(dB)
10
20log for any
20log for any
Nk Nk
Dk Dk
A A k
A A k
ω ω
ω ω
=
=
(dB)
0 10 020logA A=
( ) ( ) ( ) ( ) ( ) ( ) ( )(dB) (dB) (dB) (dB) (dB) (dB) (dB)
(dB) 0 1 2 1 2N N Nn D D DmA A A A A A A Aω ω ω ω ω ω ω= + + + + − − − −⋯ ⋯
( ) ( ) ( ) ( ) ( ) ( ) ( )1 2 1 1N N Nn D D Dmβ ω β ω β ω β ω β ω β ω β ω= + + + − − − −⋯ ⋯
• Recall that
• Represent the amplitude response in dB
Let
• Phase Response
KeyKeyKeyKey 1111:::: The decibel response makes it easy to
express the amplitude response as the sum
and difference of simpler functions. (This
means the amplitude response can be drawn
with dB on the y-axis)
Department of Electronic Engineering, NTUT44/64

45. 1 2 3 4 5 6 7 8 910 2 3 4 5 6 7 8 910 2 3 4 5 6 7 8 910 2 3 4 5 6 7 8 910
0.01 0.1 1 10 100
0
10
20
30
40
50
Semi-Log Plot
KeyKeyKeyKey 2222:::: Normalized factored
form is easy to show that
how ω affects AdB when ω is
2 times of α (1 octave) or
when ω is 10 times of α (1
decade). This is ready on x-
axis.
( ) 1 with 0
s
G s α
α
= + > ( ) 1
j
G j
ω
ω
α
= +
( ) ( )
ω
ω ω
α
 
= = +  
 
2
1A G j ( )
2
dB 1020log 1A
ω
ω
α
 
= +  
 
• Normalized factored form
frequency response
σ = 0
decibel response
AdB(ω)(dB)
Department of Electronic Engineering, NTUT45/64

46. Bode Plot Forms - Negative Real Zero (I)
ω α
1 2 3 4 5 6 7 8 910 2 3 4 5 6 7 8 910 2 3 4 5 6 7 8 910 2 3 4 5 6 7 8 910
0.01 0.1 1 10 100
0
10
20
30
40
50
E BP
• Zero on Negative Real Axis – Amplitude Response
( ) 1 with 0
s
G s α
α
= + > ( ) 1
j
G j
ω
ω
α
= + ( )
2
1A
ω
ω
α
 
= +  
 
( )
2
dB 1020log 1A
ω
ω
α
 
= +  
 
( )dB 1020log 1 0 dB forA ω ω α= <<≃ ( )
2
dB 10 1020log 20log forA
ω ω
ω ω α
α α
 
= >> 
 
≃
ω = α
Department of Electronic Engineering, NTUT
+6 dB/oct. (or +20 dB/dec.)
AdB(ω)(dB)
46/64

47. Bode Plot Forms - Negative Real Zero (II)
ω
α
1 2 3 4 5 6 7 8 910 2 3 4 5 6 7 8 910 2 3 4 5 6 7 8 910 2 3 4 5 6 7 8 910
0.01 0.1 1 10 100
0
20
40
60
80
100
E
BP
45
90
( ) 1
tan
ω
β ω
α
−
=
• Zero on Negative Real Axis – Phase Response
( ) 1 with 0
s
G s α
α
= + > ( ) 1
j
G j
ω
ω
α
= +
ω = α
Department of Electronic Engineering, NTUT
β(ω)(deg)
47/64

48. Bode Plot Forms - Negative Real Pole (I)
• Pole on Negative Real Axis – Amplitude Response
( )
1
with 0
1
G s
s
α
α
= >
+
( )
1
1
G j
j
ω
ω
α
=
+
( ) 2
1
1
A ω
ω
α
=
 
+  
 
( )
ω
ω
α
 
= − +  
 
2
dB 1020log 1A
ω α
1 2 3 4 5 6 7 8 910 2 3 4 5 6 7 8 910 2 3 4 5 6 7 8 910 2 3 4 5 6 7 8 910
0.01 0.1 1 10 100
−50
−40
−30
−20
−10
0
E
BP
ω = α
− 6 dB/oct. (or − 20 dB/dec.)
Department of Electronic Engineering, NTUT
AdB(ω)(dB)
48/64

49. Bode Plot Forms - Negative Real Pole (II)
ω
α
1 2 3 4 5 6 7 8 910 2 3 4 5 6 7 8 910 2 3 4 5 6 7 8 910 2 3 4 5 6 7 8 910
0.01 0.1 1 10 100
−90
−80
−60
−40
−20
100
E
BP
−45
0
( ) 1
tan
ω
β ω
α
−
= −
• Pole on Negative Real Axis – Phase Response
( )
1
with 0
1
G s
s
α
α
= >
+
( )
1
1
G j
j
ω
ω
α
=
+
ω = α
Department of Electronic Engineering, NTUT
β(ω)(deg)
49/64

50. Bode Plot Forms - Zero at Origin (I)
• Zero at Origin – Amplitude Response
( )G s s= ( )G j jω ω= ( )A ω ω= ( )dB 1020logA ω ω=
ω
1 2 3 4 5 6 7 8 910 2 3 4 5 6 7 8 910 2 3 4 5 6 7 8 910 2 3 4 5 6 7 8 910
0.01 0.1 1 10 100
20−
10−
0
10
20
( )ω ω= = =dB 1020log 1 0 dB for 1 rad/sA
+6 dB/oct. (or +20 dB/dec.)
Department of Electronic Engineering, NTUT
AdB(ω)(dB)
50/64

51. Bode Plot Forms - Zero at Origin (II)
ω
1 2 3 4 5 6 7 8 910 2 3 4 5 6 7 8 910 2 3 4 5 6 7 8 910 2 3 4 5 6 7 8 910
0.01 0.1 1 10 100
0
20
40
60
80
100
45
90
( ) 90β ω =
• Zero at Origin – Phase Response
( )G s s= ( )G j jω ω=
Department of Electronic Engineering, NTUT
β(ω)(deg)
51/64

52. Bode Plot Forms - Pole at Origin (I)
( )
1
G s
s
= ( )
1
G j
j
ω
ω
= ( )
1
A ω
ω
= ( )dB 10 10
1
20log 20logA ω ω
ω
= = −
• Pole at Origin – Amplitude Response
ω
1 2 3 4 5 6 7 8 910 2 3 4 5 6 7 8 910 2 3 4 5 6 7 8 910 2 3 4 5 6 7 8 910
0.01 0.1 1 10 100
20−
10−
0
10
20
( )ω ω= − = =dB 1020log 1 0 dB for 1 rad/sA
− 6 dB/oct. (or − 20 dB/dec.)
Department of Electronic Engineering, NTUT
AdB(ω)(dB)
52/64

53. Bode Plot Forms - Pole at Origin (II)
ω
1 2 3 4 5 6 7 8 910 2 3 4 5 6 7 8 910 2 3 4 5 6 7 8 910 2 3 4 5 6 7 8 910
0.01 0.1 1 10 100
80−
60−
40−
20−
0
90−
( ) 90β ω = −( )
1
G s
s
= ( )
1
G j
j
ω
ω
=
• Pole at Origin – Phase Response
Department of Electronic Engineering, NTUT
β(ω)(deg)
53/64

54. Example (I)
( )
 
+ 
 =
  
+ +  
  
5 1
4
1 1
16 100
s
G s
s s
( )
ω
ω
ω ω
 
+ 
 =
  
+ +  
  
5 1
4
1 1
16 100
j
G j
j j
( )
ω
ω
ω ω
 
+  
 
=
   
+ +   
   
2
2 2
5 1
4
1 1
16 100
A
( ) ( )
ω ω ω
ω ω
     
= = + + − + − +     
     
2 2 2
10 10 1020log 14 20log 1 20log 1 20log 1
4 16 100
dBA A
( )
ω ω ω
β ω − − −
= −1 1 1
tan tan tan
4 16 100
• Consider , plot the amplitude frequency response.( )
( )+
=
+ +2
2000 4
116 1600
s
G s
s s
Department of Electronic Engineering, NTUT54/64

55. Example (II)
( )rad/sω
1 2 3 4 5 6 7 8 910 2 3 4 5 6 7 8 910 2 3 4 5 6 7 8 910 2 3 4 5 6 7 8 910
1 10 100 1 k 10 k
0
10
20
30
14 dB
14
+6 dB/oct.
26 dB
− 6 dB/oct.
Department of Electronic Engineering, NTUT
AdB(ω)(dB)
55/64

56. RC Low-pass Filter
ω
45−
0
90−
0.1ωb
1
ω =b
RC 10ωb
0.1 bf 10 bf
1
2π
=bf
RC
f
R
+
−
1v
+
−
2vC
( )
( )
( )
= =
+
2
1
1
1
V s
G s
V s sRC
ω =
1
b
RC
ω
=
+
1
( )
1
b
G s
s
Department of Electronic Engineering, NTUT56/64

57. RC High-pass Filter
( )
( )
( )
= =
+
2
1 1
V s sRC
G s
V s sRC
ω =
1
b
RC
( )
ω
ω
=
+1
b
b
s
G s
s
π
=
1
2
bf
RC
ω
45+
90+
0
0.1ωb
1
ω =b
RC 10ωb
0.1 bf 10 bf
1
2π
=bf
RC
f
R
+
−
1v
+
−
2vC
Department of Electronic Engineering, NTUT57/64

58. Roots of the 2nd-oredr Denominator
α ω= + +2 2
( ) 2 nP s s s ζω ω= + +2 2
( ) 2 n nP s s s• The Quadratic Denominator: or
roots
α α ω ζω ω ζ= − ± − = − ± −1 2 2 2
2
1n n n
s
s
• Overdamped Circuit: α ω> n or when ζ > 1
• Critically Damped Circuit: α ω= n or when ζ = 1
• Underdamped Circuit: α ω< n or when ζ < 1
α ω α α ω ζω ω ζ= − ± − = − ± = − ± −1 2 2 2
2
1n d n n
s
j j j
s
roots
58/64 Department of Electronic Engineering, NTUT
Two real roots
Double real roots (重根)
Complex conjugate roots
• The Normalized Form for Bode Plot:
( )
ω
ζω ω
=
+ +
2
2 2
2
n
n n
G s
s s
( )
ζ
ω ω
=
 
+ +  
 
2
1
1 2
n n
G s
s s
( )ω
ω ω
ζ
ω ω
=
 
− + 
 
2
1
1 2
n n
G j
j

59. Second-order Lowpass Function
( )ω
ω ω
ζ
ω ω
=
 
− + 
 
2
1
1 2
n n
G j
j
( )
( )
ω
ω ωω ω ζζ ω ωω ω
= =
         + − +    − +   
        
2 2 42 2
22
1 1
1 4 21 4
n nn n
A
( ) ( ) ω ω
ω ζ
ω ω
     = − + − +       
 
2 4
2
dB 1020log 1 4 2
n n
A
( )
ω
ζ
ω
β ω
ω
ω
−
= −
 
−  
 
1
2
2
tan
1
n
n
• Second-order Amplitude and Phase Responses:
• It seems not easy to draw the Bode plot, how do you think?
Let’s go back to the roots of the quadratic denominator.
Department of Electronic Engineering, NTUT59/64

60. Root Locations and Frequency Response (I)
α ω= + +2 2
( ) 2 nP s s s ζω ω= + +2 2
( ) 2 n nP s s s• The Quadratic Denominator: or
roots
α α ω ζω ω ζ= − ± − = − ± −1 2 2 2
2
1n n n
s
s
• For damping ratio ζ > 1 c
v
roots ζω ω ζ α
ζω ω ζ α
− + − = −
=
− − − = −
2
11
2
2
2
1
1
n n
n n
s
s
σ
s-plane
jω
1s2s
σ
s-plane
jω
1s2s
ζ increases ω
( )ωdBA
α1 α2
−6 dB/oct.
c
v
−12 dB/oct.
ω
( )ωdBA
α1 α2
−6 dB/oct.
c
v
−12 dB/oct.
Department of Electronic Engineering, NTUT60/64

61. Root Locations and Frequency Response (II)
• For damping ratio ζ = 1 c
v
roots ζω ω ζ ζω α
ζω ω ζ ζω α
− + − = − = −
=
− − − = − = −
2
11
2
2
1
1
1
n n n
n n n
s
s
σ
s-plane
jω
1s
ω
( )ωdBA
α1
c
v
−12 dB/oct.
• So, when you get double roots while you get two different
real roots.
ζ = 1 ζ > 1
• For , the same principle as first-order poles (or zeros) for Bode
plot can be applied. (two 1st-order roots)
ζ ≥ 1
Department of Electronic Engineering, NTUT
• For , a higher means that two real roots depart farther.ζ > 1 ζ
61/64

62. Root Locations and Frequency Response (III)
Chien-Jung, Li, Dept. E.E. & Grad. Inst.
Computer and Comm. Engineering, NTUT
• For damping ratio ζ < 1 c
v
roots ζω ω ζ α ω
ζω ω ζ α ω
− + − = − +
=
− − − = − −
2
1
2
2
1
1
n n d
n n d
j js
s j j
σ
jω
1s
2s
α−
ωdj
ω− dj
α ζω= n
α
ζ
ω
=
n
ω
( )ωdBA
ωn
ζ = 1
c
v
ωnj
ω− nj
ζ < 1
ζ decreases
σ
jω
1s
2s
α−
ωdj
ω− dj
ω
( )ωdBA
ωn
ζ = 1
c
v
ωnj
ω− nj
ζ < 1
Department of Electronic Engineering, NTUT62/64

63. Frequency Response of a Second-order System
Department of Electronic Engineering, NTUT63/64
ζ = 0.1
ζ = 0.2
ζ = 0.5
ζ = 0.707
ζ = 1
0
10
5
15
−10AdB(ω)(dB)
−5
−15
−20
−25
−30
−35
−40
0.1 0.2 0.5 1 2 5 10
−12 dB/octave
0.1 0.2 0.5 1 2 5 10
ω ωn
ζ = 0.1
ζ = 0.2
ζ = 0.5
ζ = 0.707
ζ = 1
β(ω)(dB)
0
−45
−90
−135
−180

64. Example
• Determine (a) The transfer function G(s) = V2(s)/V1(s) (b) , , andζ ωn nf
(c) Frequency response of the given RLC circuit.
+
−
+
−
0.1 H
µ0.1 F( )1v t ( )2v tΩ5 k
+
−
+
−
0.1s
7
10
s
( )1V s ( )2V s5000
×
×
= = =
+ +
+
7
7 7
7 7
10
5000
5000 10 10
10 5000 10 2000
5000
p
sZ
s s
s
( )
( )
( )
= = =
+ + +
7
2
2 7
1
10
0.1 0.1 200 10
p
p
ZV s
G s
V s Z s s s
ω
ζω ω
= =
+ + + +
28
2 8 2 2
10
2000 10 2
n
n ns s s s
ω = =8 4
10 10 rad/sn
( )
π
= =
4
10 rad/s
1.592 kHz
2
nf
ζ = =
⋅ 4
2000
0.1
2 10
Department of Electronic Engineering, NTUT64/64