Circuit Network Analysis - [Chapter5] Transfer function, frequency response, and Bode plot

Network Analysis

Chapter 5

The Transfer Functions,

Frequency Response, and Bode PlotChien-Jung Li

Department of Electronic Engineering

National Taipei University of Technology

Linear Systems

• Linear System

2/64 Department of Electronic Engineering, NTUT

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.)

Transfer Function

( ) ( )X s x t = L

( ) ( )Y s y t = L

( ) ( )( )X

Y sG 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)


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.


Example

( ) ( ) ( ) ( )1 12 1

40 40 1040 4 40 104

V s V ssV s V ss s

s= = =

+ ++

( ) ( )( )

2

1

1010

V sG s

V s s= =

+( )1V s ( )2V s( ) 10

10G s

s=

+

+

−

( )1v t

+

−

( )2v t1 F

40

4 Ω

+

−

( )1V s

+

−

( )2V s40s

4

• Determine the transfer function of the circuit:


Example

( ) ( )( )

2

1

20 102 20 10

V sG 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

1010

V s G s V s V ss

= =+

1( ) 20 for 0v t t= >

( )1

20V s

s=

( ) ( )2

10 20 20010 10

V ss s s s

= =+ +

102( ) 20 20 tv t e−= −

• If the input 1( ) 20sin10 for 0v t t t= >

( ) ( )( )1 2 22

20 10 20010010

V sss

= =++

( ) ( ) ( )2 2

2000

10 100V s

s s=

+ +

( )102( ) 10 10 2 sin 10 45tv t e t−= + −

• If the input


Natural and Forced Response

( ) ( ) ( )Y s G s X s=

Total response ( )y tPoles 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.


Example

( )( )

( )( ) ( )2 2

( )5 6 2 17

Y s N sG s

X s s s s s= =

+ + + +

( )( )

( )( ) ( ) ( )2

( )2 3 2 17

Y s N sG s

X s s s s s= =

+ + + +

( ) ( )( )( ) ( )2

( )

2 3 2 17

N s X sY s

s s s s=

+ + + +

( ) 10cos6x t t=

( ) 2

1036s

X ss

=+

( ) ( )( ) ( ) ( ) ( )2 2

10

36 2 3 2 17

sN sY s

s s s s s=

+ + + + +

( ) ( ) ( )2 31 1 2 3 4 2sin 6 sin 4t t ty 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


Poles and Zeros of Transfer Functions

( ) ( )( )

N sG s

D s=

( ) 11 1 0

n nn nN s a s a s a s a−

−= + + … + +

( ) 11 1 0

m mm 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.


Zeros and Poles at Infinite

in m n= −

( ) for 1nnN s a s s≈ >>

( ) for 1mmD s b s s≈ >>

( ) for 1i

n nn

n m mnm m n

m

a aa s b b

G s sb s s s−≈ = = >>

( ) for 1i

nmn mn n n

mm m m

a s a aG 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 = ∞


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 zG s

s p s p s p− − −=

− − −⋯

⋯

• s-planes-plane

: pole

: zero

1z

2z

4z

2p

1p

3p

Im


3z

Re

Example

( ) ( )[ ][ ][ ][ ]

0 ( 2)

( 6) ( 4 3) ( 4 3)

A s sG 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− −


−2 0Re

Im

Example

• Construct an s-plane of finite poles and zeros with the transfer

function:

( ) ( )2

3 2

2 6 25

7 10

s sG s

s s s

+ +=

+ +

s-plane

5−

3 4j− +

3 4j− −

2− 0


Re

Im

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.


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


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.

jωjω

Right-hand half-plane (RHHP):

All points to the right of the -axis, but not

including the -axis.

jωjω

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.


Re

Im

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:

( ) 1tny 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= 1ατ

• Multiple-order pole:

( ) k tny t At e α−=


Re

Im

Complex Poles in the LHHP

LHHP

1 1 1s jα ω= − +

( )ny t

t

Stable

s-plane

Time response


1 1 1s jα ω∗ = − −

• The natural response corresponds to a pair of

complex poles in the LHHP is of the form:

( ) ( )11sint

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.

( ) ( )11sintk

ny t At e tα ω θ−= +

• Multiple-order complex-pole pairs:


Re

Im

Poles on Positive Real Axis and RHHP

RHHP

1 1s α=

( )ny t

t

Unstable

s-plane

Time response


• A pole on the positive real axis: The natural

response is of the form

( ) 1tny t Aeα=


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:

( ) ( )22sint

ny t Ae tα ω θ= +


grows without bound, so it is a unstable response.


Re

Im

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


• 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:

( ) ( )3sinkny t At tω θ= +

which is an unstable response.


Re

Im

Summary of Stability

s-plane

LHHP RHHP

UnstableStable

Poles at originPoles at originPoles at originPoles at origin

1st order: marginally stable

≥ 2nd order: unstable

Poles on Poles on Poles on Poles on jω-axis

1st order pole-pair: marginally stable

≥ 2nd order pole-pair: unstable


Re

Im

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.


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.


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 sG 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.


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 sG s H s

= +

( ) ( )( )

( ) 1 ( ) ( )Y s G s

T sX s G s H s

= =+

( )( )

1 ( ) ( )G s

T sG 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.


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 2AG 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 5s

G ss

= = =+

cascade

Why not ?1 1 1

( ) ( ) ( )2 2 4A AG s G s G s= ⋅ = ⋅ =


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


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 23

1 2 11

G s G sG s

G s G s H s+

+


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

αζω

=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


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

rootsα 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

sj 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 21d n n


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

ωζω ω

= =+ +

2

2 2

( )( )

( ) 2n

n n

Y sG s

X s s s

• Unit Step Response:

Let input x(t) = u(t), such that X(s) = 1/s

( )ωζω ω

= =+ +

2

2 2( ) ( ) ( )

2n

n n

Y s G s X ss s s

• When the system is critically damped: ωω ζ−= − + =( ) 1 ( 1) for 1ntny t t e

• When the system is underdamped:ζω

ζ ω ζζ

−−= − − +

−2 1

2( ) 1 sin( 1 cos )

1

nt

n

ey t t


Step Response of a 2nd-order System


ωnt

v(t)

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)


Linear Network

(Signal) (Signal)

Laplace Transform of a Signal

( ) ( ) t j tX 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

Department of Electronic Engineering, NTUT34/64

Fourier Transform is the Special Case

0σ =

Signal Spectrum

( )Re-axis σ ( )Im-axis jω

Am

plitu

de


s-domain

Laplace Transform of a System – s-plane

Pole

Zero

System


Linear Network

Pole-zero plot

Frequency response

s-domain

Frequency Response is the Special Case


Pole-zero plot

Frequency responsePhysical system

R C

L

Laplace

transform

Evaluate

at σ=0

Phasor transform

Real

Ima

ge

Frequency

Am

plitu

de

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

38/64

Example

20sin2t +−

0t =

4 Ω

( )i t1

F6

2

404s +

+−

4

( )I s6s

20 0∠ +−

4

( )I s

63j

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 40104 4

6 4 6 1.5 44

ss sI ss s s

s s

+ += = =+ + ++(a)

( ) ( )1.52.4 4sin 2 36.87ti t e t−= − + +

(b) 20 0 20 04 36.87

4 3 5 36.87I

j∠ ∠= = = ∠ −− ∠ −

( ) ( )4sin 2 36.87ssi t t= +

Transient response Steady-state response


Steady-State Transfer Function

( )( ) ( )Y s

G sX s

=

( ) ( ) ( )( ) ( ) ( )

ω

ωω ω β ω

ω== = = ∠

s j

Y jG j G s A

X j


• 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:

( ) ( ) ( )( )

ωω ω

ω= =

Y jA G j

X j

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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 ωω

= =dB 10 1020log 20logY j

A AX j

( )dBA ω

• For example: 0.01 FC µ=10 kR = Ω

R+

−( )1v t

+

−( )2v tC

R+

−1V

+

−2V1

j Cω

( )( )

ωωω

= =+ ⋅+

8 8

2 1 18 44 8

10 1010 1010 10

jV V V

jj

( )8 4

28 4 4

1

10 1010 10 10

VG j

V j jω

ω ω= = =

+ ⋅ +

( ) ( )4

8 2

10

10A G jω ω

ω= =

+

( ) ( )4

dB 10 10 8 2

1020log 20log ( )

10A Aω ω

ω= =

+

( ) 14tan

10ωβ ω −= −


Development of Bode Plot Approach

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )

1 2 30

1 2 3

N N N Nn

D D D Dm

G j G j G j G jG j A

G j G j G j G j

ω ω ω ωω

ω ω ω ω⋅ ⋅

= ⋅⋅ ⋅

⋯

⋯

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )

1 2 30

1 2 3

N N N Nn

D D D Dm

G j G j G j G jG j A

G j G j G j G j

ω ω ω ωω

ω ω ω ω⋅ ⋅

= ⋅⋅ ⋅

⋯

⋯

( )

+ + + + + + = =

+ + + + + +

⋯

⋯

⋯

⋯

1 21 21 2

1 21 2

1 2

1 1 1( )( ) ( )( )( ) ( )

1 1 1

nnn

mn

n

s s sN N N

N N Ns N s N s NG s A A

s D s D s D s s sD 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 30

1 2 3

N N N Nn

D D D Dm

A A A AA A

A A A A

ω ω ω ωω

ω ω ω ω⋅ ⋅

= ⋅⋅ ⋅

⋯

⋯

Let


( ) ( ) ( ) ( )( ) ( ) ( ) ( )

⋅ ⋅= ⋅

⋅ ⋅⋯

⋯

1 2 30

1 2 3

N N N Nn

D D D Dm

G s G s G s G sA

G s G s G s G s

Force the constant

terms to be unity.

42/64

Example

• Write the normalized factored form of ( ) ( )+=

+ +2

2000 4

116 1600

sG s

s s

( ) ( )( )( )

+=

+ +2000 4

16 100

sG s

s s

× + =

+ × +

42000 4

4 416 100

16 10016 16 100 100

s

s s

+ =

+ +

5 14

1 116 100

s

s s

Normalized factored form ready for

Bode plot.


Linear Scale to Log Scale

10 10 10log log logxy x y= +

10 10 10log log logx

x yy

= −

( ) ( ) ( ) ( )( ) ( ) ( )

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)


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 0s

G s αα

= + > ( ) 1j

G jωωα

= +

( ) ( ) ωω ωα = = +

2

1A G j ( )2

dB 1020log 1Aωωα = +

• Normalized factored form

frequency response

σ = 0

decibel response

Ad

B(ω

) (

dB

)


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 0s

G s αα

= + > ( ) 1j

G jωωα

= + ( )2

1Aωωα = +

( )2

dB 1020log 1Aωωα = +

( )dB 1020log 1 0 dB forA ω ω α= <<≃ ( )2

dB 10 1020log 20log forAω ωω ω αα α = >>

≃

ω = α


+6 dB/oct. (or +20 dB/dec.)

Ad

B(ω

) (

dB

)

46/64

Bode Plot Forms - Negative Real Zero (II)

ωα

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

BP45

90

( ) 1tanωβ ωα

−=

• Zero on Negative Real Axis – Phase Response

( ) 1 with 0s

G s αα

= + > ( ) 1j

G jωωα

= +

ω = α


β(ω

) (

de

g)

47/64

Bode Plot Forms - Negative Real Pole (I)

• Pole on Negative Real Axis – Amplitude Response

( ) 1with 0

1G s

sα

α

= >+

( ) 1

1G j

jω ω

α

=+

( )2

1

1

A ωωα

= +

( ) ωωα = − +

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.)


Ad

B(ω

) (

dB

)

48/64

Bode Plot Forms - Negative Real Pole (II)

ωα

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

( ) 1tanωβ ωα

−= −

• Pole on Negative Real Axis – Phase Response

( ) 1with 0

1G s

sα

α

= >+

( ) 1

1G j

jω ω

α

=+

ω = α


β(ω

) (

de

g)

49/64

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.)


Ad

B(ω

) (

dB

)

50/64

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=


β(ω

) (

de

g)

51/64

Bode Plot Forms - Pole at Origin (I)

( ) 1G s

s= ( ) 1

G jj

ωω

= ( ) 1A ω

ω= ( )dB 10 10

120log 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.)


Ad

B(ω

) (

dB

)

52/64

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 = − ( ) 1G s

s= ( ) 1

G jj

ωω

=

• Pole at Origin – Phase Response


β(ω

) (

de

g)

53/64

Example (I)

( ) + =

+ +

5 14

1 116 100

s

G ss s

( )ω

ωω ω

+ =

+ +

5 14

1 116 100

j

G jj j

( )ω

ωω ω

+ =

+ +

2

2 2

5 14

1 116 100

A

( ) ( ) ω ω ωω ω = = + + − + − +

2 2 2

10 10 1020log 14 20log 1 20log 1 20log 14 16 100dBA A

( ) ω ω ωβ ω − − −= −1 1 1tan tan tan4 16 100

• Consider , plot the amplitude frequency response.( ) ( )+=

+ +2

2000 4

116 1600

sG s

s s


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.


Ad

B(ω

) (

dB

)

55/64

RC Low-pass Filter

ω

45−

0

90−

0.1ωb

1ω =b RC 10ωb

0.1 bf 10 bf1

2π=bf RC f

R

+

−

1v

+

−

2vC

( ) ( )( )= =

+2

1

11

V sG s

V s sRC

ω = 1b RC

ω

=+

1( )

1b

G ss


RC High-pass Filter

( ) ( )( )= =

+2

1 1

V s sRCG s

V s sRC

ω = 1b RC

( ) ω

ω

=+1

b

b

s

G ss

π= 1

2bf RC

ω

45+

90+

0

0.1ωb

1ω =b RC 10ωb

0.1 bf 10 bf1

2π=bf RC f

R

+

−

1v

+

−

2vC


Roots of the 2nd-oredr Denominator


rootsα 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

sj j j

s

roots


Two real roots

Double real roots (重根)

Complex conjugate roots

• The Normalized Form for Bode Plot:

( ) ωζω ω

=+ +

2

2 22n

n n

G ss s

( )ζ

ω ω

=

+ +

2

1

1 2n n

G ss s

( )ωω ωζω ω

=

− +

2

1

1 2n n

G j

j

Second-order Lowpass Function

( )ωω ωζω ω

=

− +

2

1

1 2n n

G j

j

( )( )

ωω ωω ω ζζ ω ωω ω

= = + − + − +

2 2 42 222

1 1

1 4 21 4n nn n

A

( ) ( ) ω ωω ζω ω

= − + − +

2 4

2dB 1020log 1 4 2

n n

A

( )ωζωβ ωωω

−= −

−

12

2tan

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.


Root Locations and Frequency Response (I)


rootsα 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

− + − = −=

− − − = −

211

22 2

1

1

n n

n n

s

s

σ

s-planejω

1s2s

σ

s-planejω

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.


Root Locations and Frequency Response (II)

• For damping ratio ζ = 1 c

v

roots ζω ω ζ ζω α

ζω ω ζ ζω α

− + − = − = −=

− − − = − = −

211

22 1

1

1

n n n

n n n

s

s

σ

s-planejω

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


• For , a higher means that two real roots depart farther.ζ > 1 ζ

61/64

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

− + − = − +=

− − − = − −

21

22

1

1

n n d

n n d

j js

s j j

σ

jω

1s

2sα−

ωdj

ω− dj

α ζω= n

αζω

=n

ω

( )ωdBA

ωn

ζ = 1

c

v

ωnj

ω− nj

ζ < 1

ζ decreases

σ

jω

1s

2s

α−

ωdj

ω− djω

( )ωdBA

ωn

ζ = 1

c

v

ωnj

ω− nj

ζ < 1


Frequency Response of a Second-order System


ζ = 0.1

ζ = 0.2

ζ = 0.5

ζ = 0.707

ζ = 1

0

10

5

15

−10A

dB(ω

) (

dB

)

−5

−15

−20

−25

−30

−35

−400.1 0.2 0.5 1 2 5 10

−12 dB/octave

0.1 0.2 0.5 1 2 5 10ω ωn

ζ = 0.1

ζ = 0.2

ζ = 0.5ζ = 0.707

ζ = 1

β(ω

) (

dB

)

0

− 45

− 90

− 135

− 180

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

710s

( )1V s ( )2V s5000

× ×= = =+ ++

7

7 7

7 7

105000 5000 10 10

10 5000 10 20005000

psZ

s ss

( ) ( )( )= = =

+ + +

72

2 71

100.1 0.1 200 10p

p

ZV sG s

V s Z s s s

ωζω ω

= =+ + + +

28

2 8 2 2

102000 10 2

n

n ns s s s

ω = =8 410 10 rad/sn

( )π

= =410 rad/s

1.592 kHz2nf

ζ = =⋅ 4

20000.1

2 10
