EMLAB

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12. Variable frequency network performance

EMLAB

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Variable-Frequency Response AnalysisNetwork performance as function of frequency.Transfer function

Sinusoidal Frequency AnalysisBode plots to display frequency response data

Resonant CircuitsThe resonance phenomenon and its characterization

ScalingImpedance and frequency scaling

Filter NetworksNetworks with frequency selective characteristics:low-pass, high-pass, band-pass

Learning Goals

EMLAB

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0RRZRResistor

Variable frequency-response analysis

In AC steady state analysis the frequency is assumed constant (e.g., 60Hz).Here we consider the frequency as a variable and examine how the performancevaries with the frequency.

Variation in impedance of basic components

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90LLjZL

Inductor

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Capacitor

9011

CCjZc

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Series RLC network

Cj

RCjLCj

CjLjRZeq

1)(1 2

C

LCjRC

j

j

)1( 2

C

LCRCZeq

222 )1()(||

RC

LCZeq

1tan

21

sC

sRCLCssZeq

1)(

2

Driving Point functions

Simplification in notation sj

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7

Pulse represented by sum of sinusoidal signal

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General form

011

1

011

1

...

...

)(

)()(

bsbsbsb

asasasa

sI

sVsZ

nn

nn

mm

mm

sCZsLsZRsZ CLR

1,)(,)(

Network functions for basic components

Network functions

When voltages and currents are defined at different terminal pairs we define the ratios as Transfer Functions

If voltage and current are defined at the same terminals we defineDriving Point functions (Impedance/Admittance)

Nomenclature

INPUT OUTPUT TRANSFER FUNCTION SYMBOLVoltage Voltage Voltage Gain Gv(s)Current Voltage Transimpedance Z(s)

Current Current Current Gain Gi(s)

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Nomenclature

CircuitCircuit

terminals

Port #1

terminals

Port #2

} ,{

},{)(

11

22

IV

IVsH

V2 or I2

Transfer function

)2,1()(,)( iV

IsY

I

VsZ

i

i

i

iDriving point function

1I2I

2V

1V

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Example 12.1

sL

sC

1

R

So VsCsLR

RsV

/1)(

SV

sRCLCs

sRC

12

So VRCjLCj

RCjV

js

1)( 2

0101)1053.215()1053.21.0()(

)1053.215(332

3

jj

jVo

MATLAB can be effectively used to compute frequency response characteristics

0 200 400 600 800 10000

0.2

0.4

0.6

0.8

1

0 200 400 600 800 1000-100

-50

0

50

100

|| 0V 0V

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Some Matlab command

754

132A A= [2,3,1; 4, 5, 7]

97531A A= [1:2:9]

TAA A= A’

01

11

1 axaxaxay nn

nn

);x,a(polyvaly

];a,a,,a,a[a

1.e3];:2:[1x

011nn

01

11

1

01

11

1

bxbxbxb

axaxaxaH

nn

nn

mm

mm

./den;numH

x);polyval(b,den

x);polyval(a,num

];b,b,,b,[bb

];a,a,,a,[aa

1e3];:2:1[x

011nn

011mm

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0 200 400 600 800 10000

0.2

0.4

0.6

0.8

1

|| 0V

clear; % clear all variables.

clf; % clear current figure.

w = [1:1:1.e3];

b=[15*2.53*1e-3 0];

a=[0.1*2.53*1e-3 15*2.53*1e-3 1];

num = polyval(b,i*w);

den = polyval(a,i*w);

H = num./den;

mag = abs(H);

plot(w,mag);

Matlab example

EMLAB

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Example

)104(

000,20,50 :poles

0 :zero

70

21

1

K

HzpHzp

z

)(

)()(

sV

sVsG

S

o

ooinin

inin

S

O

RsCRsC

RsC

V

VsH

1

1

1)(

000,40

000,40

100 ss

s

For this case the gain was shown to be

))...()((

))...()(()(

21

210

n

m

pspsps

zszszsKsH

Zeros = roots of numeratorPoles = roots of denominator

Find the pole and zero locations and the value of K0 for the voltage gain

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0 2 4 6 8 10

x 105

-100

-80

-60

-40

-20

0

20

40

60

80

100

0 2 4 6 8 10

x 105

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

clear;

clf;

w = logspace(1,6,200);

b = [1.e0 * 40e3*pi, 0];

a = [1, 40.1e3*pi, 4e6*pi*pi];

num = polyval(b,i*w);

den = polyval(a,i*w);

H = num./den;

%arg = angle(H)*180/pi;

mag = abs(H);

plot(w,mag);

%semilogx(w,mag);

%freqs(num, den,w);

clear;

clf;

w = logspace(1,6,200);

b = [1.e0 * 40e3*pi, 0];

a = [1, 40.1e3*pi, 4e6*pi*pi];

num = polyval(b,i*w);

den = polyval(a,i*w);

H = num./den;

%arg = angle(H)*180/pi;

mag = abs(H);

plot(w,mag);

%semilogx(w,mag);

%freqs(num, den,w);

|| H

H

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Poles and Zeros

(More nomenclature)

011

1

011

1

...

...)(

bsbsbsb

asasasasH n

nn

n

mm

mm

; Arbitrary network function

Using the roots, every (monic) polynomial can be expressed as aproduct of first order terms

))...()((

))...()(()(

21

210

n

m

pspsps

zszszsKsH

functionnetwork theof poles,...,,

functionnetwork theof zeros,...,,

21

21

n

m

ppp

zzz

The network function is uniquely determined by its poles and zerosand its value at some other value of s (to compute the gain)

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011

1

011

1

...

...)(

bsbsbsb

asasasasH n

nn

n

mm

mm

Positive real function

1. The coefficients of the polynomial are related to the values of R, L, C and they are all real and positive.

2. At the same time, the real part of the poles are all negative if the circuit contains only passive components.

))...()((

))...()(()(

21

210

n

m

pspsps

zszszsKsH

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Sinusoidal frequency analysis

)(sH

Circuit represented bynetwork function

)cos(0

)(0

tB

eA tj

)(cos|)(|

)(

0

)(0

jHtjHB

ejHA tj

)()()(

)()(

|)(|)(

Notation

jeMjH

jH

jHM

stics.characteri phase and magnitude

calledgenerally are offunction as ),(),( of Plots M

)(log)(

))(log2010

10

PLOTS BODE vs

(M

. of functiona as )( functionnetwork the

analyze wefrequency theof functiona asnetwork a ofbehavior study the To

jH

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Transfer function

H(s)H(s)+

Vin-

+Vout

-

0 2 4 6 8 10

x 105

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

)(sH

)cos(0

)(0

tB

eA tj

)(cos|)(|

)(

0

)(0

jHtjHB

ejHA tj

)()()( sVsHsV inout

|| H

0 2 4 6 8 10

x 105

-100

-80

-60

-40

-20

0

20

40

60

80

100

H

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Drawing transfer function is difficult!

))...()((

))...()(()(

21

210

n

m

pspsps

zszszsKsH

||log...||log||log

||log...||log||log||log

))...()((

))...()((log|)(|log

10210110

10210110010

21

2101010

n

m

n

m

pspsps

zszszsK

pspsps

zszszsKsH

|)(|log|)(|log|)(||)(|log

|)(||)(|)()(

|)(|)(

2121

)]()([2121

)(

21

sHsHsHsH

esHsHsHsH

esHsHsHsHj

sHj

)(...)()(

)(...)()(

))...()((

))...()(()(

21

210

21

210

n

m

n

m

pspsps

zszszsK

pspsps

zszszsKsH

Magnitude :

Phase :

Bode plot

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History of the Decibel

Originated as a measure of relative (radio) power

1

2)2 log10(|P

PP dB 1Pover

21

22

21

22)2

22 log10log10(|

I

I

V

VP

R

VRIP dB 1Pover

By extension

||log20|

||log20|

||log20|

10

10

10

GG

II

VV

dB

dB

dB

Using log scales the frequency characteristics of network functionshave simple asymptotic behavior.The asymptotes can be used as reasonable and efficient approximations

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Standard form of a transfer function

n

m

ps

ps

ps

zs

zs

zs

KsH

1...11

1...11

)(

21

210

m

m

p

s

p

s

p

s

z

s

z

s

z

sKsH

1log20...1log201log20

1log20...1log201log20||log20|)(|log20

102

101

10

102

101

1001010

On applying to log function, terms containing poles and zeroes can be separated. Then, it becomes easier to draw the transfer function.

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2

110

110 1log101log20

zz

sy

1

1

11

tan11zz

jz

s

Magnitude Phase

0

90

110log z

45dB0

110log z

y

dB3 dB/dec20

How to draw Bode plot of simple zero or polesBode plots can be drawn simply following the procedures below.

1. Draw first two extreme regions where and .

2. Connect those two regions with a smooth curve.

3. Mark the exact point where .

1/ 1 z 1/ 1 z

1/ 1 z

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Bode plot of quadratic polesDrawing a transfer function containing quadratic polynomial deserves a special care. Depending on the value of ζ, the shape of the transfer function curves changes significantly.

1. p1 and p2 are real numbers and have different values.2. p1 and p2 are real numbers and have the same value. 3. p1 and p2 are complex number and have different values.

21

2

00

11

1

21

1)(

ps

psss

sH

)1( )1( )1(

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0

2

0

2

21

2

00

11

1

11

1

21

1)(

ssps

psss

sH

(1) p1, p2 same

dB/dec40

|)(|log20 10 jH

dB/dec40

|)(|log20 10 jH

110log p110log p

210log p

dB/dec20

(2) p1, p2 distinct

When the poles are real valued )1(

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When the poles are complex valued )1(

0

2

0

2

21

2

00

11

1

11

1

21

1)(

sj

sj

ps

psss

sH

dB0

110log pdB3

dB/dec40

dB0

110log pdB3

dB/dec40

With smaller

202,1 1 jp

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Standard form of quadratic functions

11

1)(

0

2

0

s

Qs

sHQ

12

1. p1 and p2 are real numbers and have different values.2. p1 and p2 are real numbers and have the same value. 3. p1 and p2 are complex number and have different values.

)2/1( Q

)2/1( Q

)2/1( Q

Instead of ζ, Q is more widely used and is called ‘Quality factor’. The Q factor is related with the power loss of a passive component.

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]...)()(21)[1(

]...)()(21)[1()()( 2

233310

bbba

N

jjj

jjjjKjH

General form of a network function

Frequency independentPoles/zeros at the origin

First order terms Quadratic terms for complex conjugate poles/zeros

..|)()(21|log20|1|log20

...|)()(21|log20|1|log20

||log20log20

21010

233310110

10010

bbba jjj

jjj

jNK

DND

N

BAAB

loglog)log(

loglog)log(

|)(|log20|)(| 10 jHjH dB

212

1

2121

zzz

z

zzzz

...)(1

2tantan

...)(1

2tantan

900)(

211

23

3311

1

b

bba

NjH

Display each basic Display each basic term separately and term separately and add the results to add the results to obtain final answerobtain final answer

Let’s examine each basic termLet’s examine each basic term

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a. Constant Term

b. Poles/Zeros at the origin

90)(

)(log20|)(|)( 10

Nj

Njj

NdB

NN

linestraight a is this

log is axis- xthe 10

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c. Simple pole or zeroj1

1

210

tan)1(

)(1log20|1|

j

j dB

asymptotefrequency low 0|1| dBj

(20dB/dec) asymptotefrequency high 10log20|1| dBj

frequency) akcorner/bre1 whenmeet asymptotes two The (

Behavior in the neighborhood of the corner

FrequencyAsymptoteCurvedistance to asymptote Argument

corner 0dB 3dB 3 45

octave above 6dB 7db 1 63.4

octave below 0dB 1dB 1 26.6

125.0

0)1( j

90)1( j

1

1

Asymptote for phase

High freq. asymptoteLow freq. Asym.

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Simple zero

Simple pole

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d. Quadratic pole or zero ])()(21[ 22 jjt ])()(21[ 2 j

222102 2)(1log20|| dBt 2

12

)(1

2tan

t

1 asymptotefrequency low 0|| 2 dBt 02t

1 asymptote freq. high 2102 )(log20|| dBt 1802t

1 )2(log20|| 102 dBt 902tCorner/break frequency221 2

102 12log20|| dBt

2

12

21tan

t

2

2Resonance frequency

Magnitude for quadratic pole Phase for quadratic pole

dB/dec40

These graphs are inverted for a zero

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Example Generate magnitude and phase plots

)102.0)(1(

)11.0(10)(

jj

jjGv

Draw asymptotesfor each term 1,10,50 :nersBreaks/cor

Draw composites

40

20

0

20

dB

90

90

1.0 1 10 100 1000

dB|10

decdB /20

dec/45

decdB /20

dec/45

EMLAB

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EMLAB

34Generate magnitude and phase plots

)11.0()(

)1(25)( 2

jj

jjGv 10 1, :(corners) Breaks

40

20

0

20

dB

90

270

90

1.0 1 10 100

Draw asymptotes for each

dB28

decdB /40

180

dec/45

45

Form composites

Example

EMLAB

35A function with complex conjugate poles

1004)()5.0(

25)( 2

jjj

jjG

Put in standard form

125/)10/()15.0/(

5.0)( 2

jjj

jjG

40

20

0

20

dB

90

90

01.0 1.0 1 10 100 270

1 )2(log20|| 102 dBt

2.01.0

25/12

])()(21[ 2

2 jjt

dB8

Draw composite asymptote

Behavior close to corner of conjugate pole/zerois too dependent on damping ratio.Computer evaluation is better

Example

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Using MATLAB to compute Magnitude & Phase information

011

1

011

1

...

...)(

bsbsbsb

asasasasV n

nn

n

mm

mm

o

),(

];,,...,,[

];,,...,,[

011

011

dennumfreqs

bbbbden

aaaanum

nn

mm

MATLAB commands required to display magnitudeand phase as function of frequency

NOTE: Instead of comma (,) one can use space toseparate numbers in the array

1)1053.215()1053.21.0()(

)1053.215(332

3

jj

jVo

EXAMPLE

» num=[15*2.53*1e-3,0];» den=[0.1*2.53*1e-3,15*2.53*1e-3,1];» freqs(num,den)

1a

2b1b

0b

Missing coefficients mustbe entered as zeros

» num=[15*2.53*1e-3 0];» den=[0.1*2.53*1e-3 15*2.53*1e-3 1];» freqs(num,den)

This sequence will alsowork. Must be careful notto insert blanks elsewhere

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EMLAB

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clear;

clf;

w = logspace(1,7,200);

num = [1.e0 * 40e3*pi, 0];

den = [1, 40.1e3*pi, 4e6*pi*pi];

freqs(num, den,w);

EMLAB

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Evaluation of frequency response using MATLAB

1004)()5.0(

25)( 2

jjj

jjG

» num=[25,0]; %define numerator polynomial» den=conv([1,0.5],[1,4,100]) %use CONV for polynomial multiplicationden = 1.0000 4.5000 102.0000 50.0000» freqs(num,den)

Using default options

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Determining Transfer function from Bode plotThis is the inverse problem of determining frequency characteristics. We will use only the composite asymptotes plot of the magnitude to postulate a transfer function. The slopes will provide information on the order

A

A. different from 0dB.There is a constant Ko

B

B. Simple pole at 0.11)11.0/( j

C

C. Simple zero at 0.5

)15.0/( j

D

D. Simple pole at 3

1)13/( j

E

E. Simple pole at 20

1)120/( j

)120/)(13/)(11.0/(

)15.0/(10)(

jjj

jjG

20

|

00

0

1020|dBK

dB KK

If the slope is -40dB we assume double real pole. Unless we are given more data

EMLAB

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Example Determine a transfer function from the composite magnitude asymptotes plot

A

A. Pole at the origin. Crosses 0dB line at 5

j5

B

B. Zero at 5

C

C. Pole at 20

D

D. Zero at 50

E

E. Pole at 100

)1100/)(120/(

)150/)(15/(5)(

jjj

jjjG

EMLAB

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12.3 Resonant circuits

H(s)H(s)+

Vout-

)()()( sVsHsV inout

A circuit that selects one frequency is called a resonant circuit.

SR

LR+

Vin-

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Series resonance

H(s)H(s)

L C

V_ACSRC1

RR2R=50 Ohm

CC1

LL1

RR1

ss

QssQ

Qs

CsRLCs

CsR

sCsLR

R

V

VsH

in

out

0

0

2

00

0

22

2

2

2

1

1

11

11)(

0

0

1

1)(

jQ

sH

+Vout

-

CRQ

LC 200

1,

1

200

220

02 2:form Standard sss

Qs

EMLAB

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Spectrum analyzer

Transfer function of a resonator

101

102

103

104

-100

-50

0

50

100

Frequency (radians)

Pha

se (

de

gre

es)

101

102

103

104

10-3

10-2

10-1

100

Frequency (radians)

Ma

gni

tud

e

clear;clf;w0=10^2;Q=10;H0=1;w = logspace(1,4,200);num = [1/(Q*w0), 0];den = [1/(w0*w0), 1/(Q*w0),1];freqs(num, den,w);

2|)(| H

)(H

EMLAB

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80 90 100 110 1200

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

2

0

0

21

1|)(|

Q

sH

BWQ

Qx

xQQ

x

QQxx

Qx

Q

0

0

2

22

0

0

1

)0(14

1

2

1

14

1

2

1,01

1

1

Half Power Band-width

80 90 100 110 120-80

-60

-40

-20

0

20

40

60

80

2|)(| H

100

Q

Half power band-width

Power 가 1/2 이 되는 지점 .

)(H

EMLAB

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101

102

103

104

-100

-50

0

50

100

Frequency (radians)

Pha

se (

de

gre

es)

101

102

103

104

10-3

10-2

10-1

100

Frequency (radians)

Ma

gni

tud

e

101

102

103

104

-100

-50

0

50

100

Frequency (radians)

Pha

se (

de

gre

es)

101

102

103

104

10-4

10-2

100

Frequency (radians)

Ma

gni

tud

e

101

102

103

104

-100

-50

0

50

100

Frequency (radians)

Pha

se (

de

gre

es)

101

102

103

104

10-3

10-2

10-1

100

Frequency (radians)

Ma

gni

tud

e

2Q 10Q 100Q

On a log scale, a point where |H| is lowered by 3dB from the peak value is called a half power point.

HPBW vs. Q

With large Q, BW gets narrow.

3dB-BW

Q : Quality factor

2|)(| H

)(H

EMLAB

47

RR1

I_ACSRC1

RR2L

L1CC1

Parallel resonance

H(s)H(s) LL1

CC1

ss

QssQ

Qs

sCsL

LCR

sCsLR

sCsL

I

IsH

in

out

0

0

2

00

0

22

1

1

11

11

11

||

1||

)(

outIinI

EMLAB

48

Resonant circuits

These are circuits with very special frequency characteristics. And resonance is a very important physical phenomenon

CjLjRjZ

1

)(

circuit RLCSeries

LjCjGjY

1

)(

circuit RLCParallel

LCCL

11

whenzero iscircuit each of reactance The

0

The frequency at which the circuit becomes purely resistive is calledthe resonance frequency

EMLAB

49

Properties of resonant circuitsAt resonance the impedance/admittance is minimal

Current through the serial circuit/voltage across the parallel circuit canbecome very large (if resistance is small)

CRR

LQ

0

0 1

:FactorQuality

222 )1

(||

1)(

CLRZ

CjLjRjZ

222 )1

(||

1)(

LCGY

CjLj

GjY

Given the similarities between series and parallel resonant circuits, we will focus on serial circuits

EMLAB

50

Properties of resonant circuitsAt resonance the power factor is unity

CIRCUIT BELOW RESONANCE ABOVE RESONANCESERIES CAPACITIVE INDUCTIVEPARALLEL INDUCTIVE CAPACITIVE

Phasor diagram for series circuit Phasor diagram for parallel circuit

RV

C

IjVC

Lj

1GV1CVj

L

Vj

1

EMLAB

51

Example 12.7 Determine the resonant frequency, the voltage across eachelement at resonance and the value of the quality factor

LC

10 sec/2000

)1010)(1025(

163

radFH

I

AZ

VI S 5

2

010

2Z resonanceAt

50)1025)(102( 330L

)(902505500 VjLIjVL

902505501

501

0

00

jICj

V

LC

C

R

LQ 0 25

2

50

||||

|||| 0

SC

SS

L

VQV

VQR

VLV

resonanceAt

111

111

0

2

0

2

s

Q

s

sC

sRCLCssC

RsC

sLVS

EMLAB

52

Given L = 0.02H with a Q factor of 200, determine the capacitornecessary to form a circuit resonant at 1000Hz

R

L0200200Q withL

LC

10

C02.0

110002 FC 27.1

What is the rating for the capacitor if the circuit is tested with a 10V supply?

VVC 2000|| ||||

|||| 0

SC

SS

L

VQV

VQR

VLV

resonanceAt

59.1200

02.010002R

AI 28.659.1

10

The reactive power on the capacitorexceeds 12kVA

Example 12.8

V010

2

00

2

11

111

s

Q

s

sC

LCssRCsCsC

sLRZ

EMLAB

53

Normalizationfactor

( ) cos [ ]m

O

Vi t t A

R

Energy transfer in resonant circuits

EMLAB

54

2

2

1LiWL

TRiWD 2

2

1

D

L

D

L

W

WQ

Q

R

L

TR

L

TRi

Li

W

W

2

22

1

21

21

0

2

2

Energy stored in L

Energy dissipated in one cycle

Quality factor in terms of Energy

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55

Example 12.10 A series RLC circuit as the following properties:

sec/100sec,/4000,4 0 radBWradR

Determine the values of L,C.

CRR

LQ

0

0 1

LC

10

QBW 0

1. Given resonant frequency and bandwidth determine Q.2. Given R, resonant frequency and Q determine L, C.

40100

40000 BW

Q

HQR

L 040.04000

440

0

FRQL

C 662

020

1056.11016104

111

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56

The Tacoma Narrows Bridge Opened: July 1, 1940Collapsed: Nov 7, 1940

Likely cause: windvarying at frequencysimilar to bridgenatural frequency

2.020

Example 12.12

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57

x

tbkmamF cos0Fxxx

02 Fxxx bskms

0.44’

1.07’

'77.3

Displacement vs. voltage

b

kQ

m

k

ks

k

bs

k

m

00

02

,

F

xxx

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58

Increasing selectivity by cascading low Q circuits

Single stage tuned amplifier

MHzsradFHLC

9.99/10275.61054.210

11 8

1260

398.010

1054.2250 6

12

L

CR

L

C

GBWQ 10

Example 12.15

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59

12.5 Filter networksNetworks designed to have frequency selective behavior

COMMON FILTERS

Low-pass filterLow-pass filterHigh-pass filterHigh-pass filter

Band-pass filterBand-pass filter

Band-reject filterBand-reject filter

We focus first onPASSIVE filters

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60

Examples : Examples : voice signal spectrum

Time domain signal Frequency domain spectrum

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61

Example : ADSL signal spectrum

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62

Filter

H(s)

Filter

H(s)

+Vout

-

SR

LR+

Vin-

CC1

V_ACSRC1

RR1

RR2

Low-pass filterLow-pass filter

1-st order1-st order

LL1

RR2

V_ACSRC1

RR1

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63

Prototype low pass filter

RR1

V_ACSRC1

LL1

RR2

CC1

V_ACSRC1

RR1

RR2

12

2

2

1

1

1

1)(

ps

RsLsLR

R

V

VsH

in

out

1

22 1

1

1

11

1

)(

psCsR

sCR

sCI

IsH

in

out

CRp

21

1

L

Rp 2

1

dB0110log p

y

dB3

dB/dec20

Type #1

Type #2

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64

1

1

1)(

ps

sH

LPF-to-HPF transformation

dB0

y

dB3

dB/dec20

sp

sp

sH11 1

1

1

1)'(

dB0

y

dB3

dB/dec20

onsubstituti'1

1 s

p

p

s

10log110log p110log p

LPF→HPF

11

p

s1

1

p

s

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65

RR1

V_ACSRC1

LL1

RR2

CC1

LPF→HPF

'replace

21

s

ps

Lpss

LpsL

21

121

1 1'

1

' L

L11

21

1

1

LpC

LsCp

s

sC'

'121

CC1

LL1

CC1

RR2

V_ACSRC1

RR1

CC1

RR2

V_ACSRC1

RR1

LL1

R=L=1.0 nH

RR2

V_ACSRC1

RR1

121

1

1

CpL

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66

1

1

1)(

ps

sH

dB0

y

dB3

dB/dec20

''

1

1)'(

0

0

0

ss

BW

sH

dB0

y

dB3

dB/dec20

'

' replace 0

0

0

1 s

s

BWp

s

10log

010log

110log p

LPF→BPF

1'

' 0

0

0

1

s

s

BWp

s

jBW

js

s

BW

js

jBW

js

s

BW

js

0

1200

0

0

2

0

1100

0

0

1

'

'

'

'

'

'

dB/dec20

2' 1'

LPF-to-BPF transformation

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67

1201

110

0

0111

'

1'

'

'

LpBW

s

sLBW

p

s

s

BW

pLsL

LL1

'1

'

1

''

11120110

0

01

sBW

Cps

BWCp

ss

BW

pCsC

'

' Replace 0

0

0

1 s

s

BWp

s

LL1

CC1

CC1

RR1

V_ACSRC1

LL1

RR2 V_AC

SRC1

RR2R=50 Ohm

CC1

LL1

RR1

CC1

RR2

V_ACSRC1

RR1

CC1

LL1 R

R2V_ACSRC1

RR1

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68

1

1

1)(

ps

sH

dB0

y

dB3

dB/dec20

''

11

1)'(

0

0

0

ss

BW

sH

dB0

y

dB3

dB/dec20

''

1 Replace

0

0

01

ss

BW

p

s

10log

010log

110log p

LPF→BPF

1'

' 0

0

0

1

s

s

BWp

s

jBW

js

s

BW

js

0

2100

0

0

1

'

'

'

dB/dec20

LPF-to-BRF transformation

2' 1'

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69

'1

'1

1

''

11

20

11

0

0

0

111

sBWpLs

BWpLss

BW

pLsL

LL1

'

1'

1

'

'11

1

20

1

0

01

0

sBWCps

BWCps

s

BWpCsC

''

1 replace

0

0

01

ss

BW

p

s

LL1

CC1

CC1

RR1

V_ACSRC1

LL1

RR2

CC1

RR2

V_ACSRC1

RR1

CC1

LL1

RR2

V_ACSRC1

RR1

RR2

RR1

CC1

LL1

V_ACSRC1

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Example 12.19 A simple notch filter to eliminate 60Hz interference

)1

(1

1

CLj

CL

CjLj

CjLj

ZR

inReq

eq VZR

RV

0

LCZR

1 01

0

LCV

tttvin 10002sin2.0602sin)( FCmHL 100,3.70

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71

Active filter

Passive filters have several limitations

1. Cannot generate gains greater than one

2. Loading effect makes them difficult to interconnect

3. Use of inductance makes them difficult to handle

Using operational amplifiers one can design all basic filters, and more, with only resistors and capacitors

The linear models developed for operational amplifiers circuits are valid, in amore general framework, if one replaces the resistors by impedances

Ideal Op-Amp

These currents are zero

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Basic Inverting Amplifier

0V

VV gain Infinite

0V

0 II -impedanceinput Infinite

02

2

1

1 Z

V

Z

V

11

22 V

Z

ZV

Linear circuit equivalent

0I

1

2

Z

ZG

1

11 Z

VI

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73

Basic Non-inverting amplifier1V

1V

0I

1

1

2

10

Z

V

Z

VV

11

120 V

Z

ZZV

1

21Z

ZG

01 I

Basic Non-inverting Amplifier

Due to the internal op-amp circuitry, it haslimitations, e.g., for high frequency and/orlow voltage situations. The OperationalTransductance Amplifier (OTA) performswell in those situations

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(non-inverting op-amp)Example 12.40 “BASS-BOOST” AMPLIFIER

DESIRED BODE PLOT

OPEN SWITCH

(6dB)

500

2Pf

gain voltagefor the

of value theand locations zero and pole theFind oK

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75

Example 12.41 TREBLE BOOST

Original player responseDesired boost

Proposed boost circuit

Non-inverting amplifier