Frequency Response

Instructor: Chia-Ming TsaiElectronics Engineering

National Chiao Tung UniversityHsinchu, Taiwan, R.O.C.

Contents• Introduction

• Transfer Function

• The Decibel Scale

• Bode Plots

• Series Resonance

• Parallel Rosonance

• Passive Filters

• Active Filters

• Applications

Introduction• Analysis with a constant frequency is already

learned.

• To obtain the frequency response

– Keep the amplitude and the phase of the sinusoidal source constant (amplitude=1, phase=0)

– Sweep the frequency from a starting frequency to a stop frequency

– Plot the amplitude and the phase of the desired voltage or current versus frequency

Transfer Function

)(

)(AdmittanceTransfer )(

)(

)(ImpedanceTransfer )(

)(

)(gainCurrent )(

)(

)(gain Voltage)(

)()(

)()(

i

o

i

o

i

o

i

o

H

V

IH

I

VH

I

IH

V

VH

X

YH

• The transfer function H()

can be expressed as

• Zeros: the roots of N()=0

• Poles: the roots of D()=0

)(

)()(

D

NH

.... , , 21 zzj

.... , , 21 ppj

Example 1

RC

H

RCjCjR

Cj

s

o

1 where

tan1

1

1

1

1

1)(

0

0

1

20

V

VH

Phasor domainTime domain

90)(

45)(

0)0(

0

0)(

707.0)(

1)0(

0

H

H

H

Example 2

)( Lj

Example 3

1 012 :Poles

2 ,0 0)2( :Zeros

, 12

)2(

1)24(5.0

)24(5.0

)(

)(

)(5.0124

24)(

:Sol

zeros. and poles its

and )()( Find

2

2

sss

sss

jsss

ss

jj

jj

jj

j

i

o

io

io

I

I

II

II

The Thought of Bode Plots• It is quite difficult to handle the plotting of the

transfer function in a linear scale.

• If the transfer function is transformed to a logarithmic scale, then the plotting becomes much more easy.

)log()log(

)log()log(log

))((

))((log)(log

))((

))((

)(

)()( If

21

21

21

2110

21

21

pjpj

zjzj

pjpj

zjzj

pjpj

zjzj

A

AH

AD

NH

The Decibel (dB) Scale

dB 35.0log10 ,5.0 If

dB 32log10 ,2 If

(dB) log10

dB) (10 log(bel) log

bel 10

1dB 1 decibel 1

logbels ofNumber

is, that ,gain

power themeasure toused is bel The

dB12

dB12

dB1

2

1

2

1

2

1

2

GPP

GPP

GP

P

P

P

P

PG

P

PG

G

1

2dB

1

2dB

21

1

2

1

2

12

1

22

2dB

22

log 20

levels,current

compare toused be alsocan It

log 20

levels, voltage

comparing when Assume

log10log 20

log10

I

IG

V

VG

RR

R

R

V

V

RV

RVG

RIRVP

Bode Plots

2

1

22

11

1

221

211

1

10dB

)(21

zeroor )(211 pole quadraticA 4.

)1( zeroor )1(1 pole simpleA 3.

origin at the )( zeroor )( pole simpleA 2.

gain A 1.

are These function. transfer ain appear can factors basicSeven

)(21)1(

)(21)1()()(

asgiven is )( oftion representa form standard The

log20

definesplot magnitude Bode The

lnlnlnln

kk

nn

nn

kk

j

j

jj

jj

zjpj

jj

K

jjpj

jjzjjK

HH

jHeH

HeH

H

H

H

H

Bode Plots• Steps to construct a Bode plot:

– Plot each factor separately

– Additively combine all of them graphically because of the logarithms involved

• The mathematical convenience of the logarithm makes the Bode plots a powerful tool

• Straight-line plots used instead of actual plots

Bode Plots: A Gain K

0 if 180

0 if 0

log20

)(

10dB

K

K

KH

K

H

Bode Plots: Zero/Pole at Origin

90

log20

90)(

10dB

H

jH

90

log20

90)(

10dB

11

H

jH

Bode Plots: Simple Zero

, 90

, 45

0 , 0

tan

, log20

0 , 01log20

tan

1log20

1)(

11

1

110

10

dB

1

1

110dB

1

zz

zH

z

z

jH

zjH 20 dB/decade

7.5)1.0(tan 1 3.84)10(tan 1

Bode Plots: Simple Pole

, 90

, 45

0 , 0

tan

, log20

0 , 01log20

tan

1log20

11)(

11

1

110

10

dB

1

1

110dB

1

zp

pH

p

p

jH

pjH

-20 dB/decade

-45/decade

Bode Plots: Quadratic Pole

,180

,90

0 , 0

1

2tan

, log40

0 , 0

)1for polescomplex (

21log20

)(211)(

2221

10dB

2

2

210dB

22

n

n

n

n

nn

nn

H

jjH

jjH

Bode Plots: Quadratic Zero

,180

,90

0 , 0

1

2tan

, log40

0 , 0

)1for zeros(complex

21log20

)(21)(

2211

10dB

1

2

110dB

21

k

k

k

n

kk

kk

H

jjH

jjH

Summary

Summary

Summary

Example 1

)101(

1

)21(

1)(10

)101)(21(

10

)10)(2(

200)(

jjj

jj

j

jj

j

H

Example 1 (Cont’d)

)101(

1

)21(

1)(10

)101)(21(

10

)10)(2(

200)(

jjj

jj

j

jj

j

H

Example 2

2

22

)51(

1)101(

14.0

)51(

)101(4.0

)5(

10)(

jj

j

jj

j

jj

j

H

Example 2 (Cont’d)

2

22

)51(

1)101(

14.0

)51(

)101(4.0

)5(

10)(

jj

j

jj

j

jj

j

H

Example 3

2

22

)10(1061

1)1(

100

1

)10(1061

)1(1001

10060

1)(

sss

ss

s

ss

s

H

Example 3 (Cont’d)

2

22

)10(1061

1)1(

100

1

)10(1061

)1(1001

10060

1)(

sss

ss

s

ss

s

H

Series Resonance

Hz 2

1

2

)( rad/s 1

01

)Im(

iscondition resonance The

1

1)(

00

0

00

LCf

requencyresonant fLC

CL

CLjR

CjLjRs

Z

I

VHZ

When Resonance Occurs1. The impedance is purely resistive. The LC

series combination acts like a short circuit.

2. The voltage and the current are in phase, so the power factor is unity.

3. The impedance Z() is minimum.

4. The voltage across L and C can be much larger than the source voltage.

CR

VL

R

VLI m

Cm

L0

00

1

VV

Half-Power Frequencies

is, that ),( at

maximum thehalf ispower dissipated The2

1

2

1)(

is, that highest, theis

power dissipated theresonance,At 2

1)(

iscircuit the

by dissipatedpower average The

1

1

2 1,

2

2

2

0

2

22

es frequencihalf-power

R

VR

R

VP

RIP

RLC

CLR

VI

CLjR

mm

m

I

Z

LCL

R

L

R

RC

LR

R

R

VPP m

1

22

21

2)()(

4

1)()(

2

2,1

2

2,12,1

2

21

2

21

ZZ

L

RB

12

210

Quality Factor: Q

) ( 1

2

121

21

2

1

2

1

2

1

is period onein dissipatedenergy The2

1 is storedenergy peak The

resonanceat period onein

circuit by the dissipatedEnergy

circuit in the storedenergy Peak 2

0

0

0

0

0

2

2

0

22

2

L

RB

BCRR

LQ

R

Lf

fRI

LIQ

fRITRI

LI

Q

Summary

Voltage across L and C QVm

Parallel Resonance

L

RRC

BQ

RCB

LCRCRC

LC

LC

LCj

R

LjCj

R

00

0

22

2

2,1

0

00

1

1

2

1

2

1

rad/s 1

01

)Im(

iscondition resonance The

11

11)(

Y

V

IHY

When Resonance Occurs1. The impedance is purely resistive. The LC

parallel combination acts like an open circuit.

2. The voltage and the current are in phase, so the power factor is unity.

3. The admittance Y() is minimum.

4. The current flowing through L and C can be much larger than the source current.

mmCm

L QIRCIL

RI

L

V 0

00

II

ComparisonsSeries circuit Parallel circuit

Passive FiltersLowpass

Highpass

Bandpass

Bandstop

Lowpass Filter

, 09

, 54

0 , 0

, 0

, 2

1 0 , 1

)(

1 where

1

1

1

1

1

1)(

c

c

c

c

i

o

RC

jRCj

CjR

Cj

Η

V

VH

Highpass Filter

, 0

, 54

0 , 09

, 1

, 2

1 0 , 0

)(

1 where

1

1

1

1)(

c

c

c

cc

i

o

RC

j

j

RCj

RCj

CjR

R

H

V

VH

Bandpass Filter

, 90

, 0

0 , 90

, 0

, 1

0 , 0

)(

1 where

)(1

)(1

1)(

0

0

210

20

2

H

V

VH

LC

jRCj

jRC

LCjRCj

RCj

CjLjR

R

i

o

B

Bandstop Filter

, 0

, 0

0 , 0

, 1

: , 0

0 , 1

)(

1 where

)(1

)(1

)(1

)(1

1

1)(

0

0

210

20

20

2

2

H

V

VH

LC

jRCj

j

LCjRCj

LCj

CjLjR

CjLj

i

o

rejectionfrequency

B

Active Filters

i

f

i

o

Z

Z

V

VH )(

ffc

ci

f

ffi

f

i

f

ff

f

fff

ii

CRjR

R

RCjR

R

RCj

R

CjR

R

1 ,

1

1

1

1)(

1

1||

Z

ZH

Z

Z

A general 1st-order active filter

Active 1st-order lowpass filter

Active 1st-Order Highpass Filter

1

ff

iii

RCj

R

Z

Z

iic

cfi

ii

fi

ii

f

i

f

CRj

jRC

RCj

RCj

CjR

R

1 ,

1

1

1)(

Z

ZH

Active Bandpass Filter

Active Bandpass Filter

21

2210

12

21

1

1

2

)()(

1 &

1 ,

11

1)(

i

f

i

f

i

o

R

RKGainPassband

RCRCR

R

j

j

j

HH

V

VH

Bandreject (or Notch) Filter

Bandreject (or Notch) Filter

i

f

i

f

i

o

R

RKGainPassband

RCRCR

R

j

j

j

)()0(

1 &

1 ,

11

1)(

12

21

1

1

2

HH

V

VH

Applications: Radio Receiver

2055kHzrejected

2055kHz

Touch-Tone Telephone (1/2)

Touch-Tone Telephone (2/2)

Crossover Network

(Lowpass)

(Highpass)