Variable Frequency Response Analysis 8 Ed

7/27/2019 Variable Frequency Response Analysis 8 Ed

1/105

Variable-Frequency Response Analysis

Network performance as function of frequency.

Transfer function

Sinusoidal Frequency AnalysisBode plots to display frequency response data

Resonant Circuits

The resonance phenomenon and its characterization

Scaling Impedance and frequency scaling

Filter Networks

Networks with frequency selective characteristics:

low-pass, high-pass, band-pass

VARIABLE-FREQUENCY NETWORK

PERFORMANCE

LEARNING GOALS


2/105

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 performance

varies with the frequency.

Variation in impedance of basic components


3/105

90LLjZL Inductor


4/105

Capacitor 9011

CCjZc


5/105

Frequency dependent behavior of series RLC network

Cj

RCjLCj

CjLjRZeq

1)(1 2

C

LCjRC

j

j

)1( 2

C

LCRCZeq

222 )1()(||

RC

LCZeq

1tan

21

sC

sRCLCssZ

sj

eq

1)(

2

notation"intionSimplifica"


6/105

For all cases seen, and all cases to be studied, the impedance is of the form

011

1

01

1

1

......)(

bsbsbsbasasasasZ n

n

n

n

m

m

m

m

sCZsLsZRsZ CLR

1,)(,)(

Simplified notation for basic components

Moreover, if the circuit elements (L,R,C, dependent sources) are real then the

expression for any voltage or current will also be a rational function in s

LEARNING EXAMPLE

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


7/105

USING MATLAB TO COMPUTE MAGNITUDE AND PHASE INFORMATION

011

1

011

1

...

...)(

bsbsbsb

asasasasV

n

n

n

n

m

m

m

mo

),(

];,,...,,[

];,,...,,[

011

011

dennumfreqs

bbbbden

aaaanum

nn

mm

MATLAB commands required to display magnitude

and phase as function of frequency

NOTE: Instead of comma (,) one can use space to

separate 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 must

be 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 also

work. Must be careful not

to insert blanks elsewhere


8/105

GRAPHIC OUTPUT PRODUCED BY MATLAB

Log-log

plot

Semi-log

plot


9/105

LEARNING EXAMPLE A possible stereo amplifier

Desired frequency characteristic

(flat between 50Hz and 15KHz)

Postulated amplifier

Log frequency scale


10/105

Frequency domain equivalent circuit

Frequency Analysis of Amplifier

)()(

)()(

sVsV

sVsV

i n

o

S

i n

)(/1

)( sVsCR

RsV S

i ni n

i ni n

]1000[/1

/1)( i n

oo

oo V

RsC

sCsV

ooi ni n

i ni n

RsCRsCRsCsG

11]1000[

1)(

000,40

000,40]1000[100 sss

000,401001058.79

100101018.3

191

1691

oo

i ni n

RC

RC

required

actual

000,40000,40]1000[)(000,40||100

sssGs

Frequency dependent behavior is

caused by reactive elements

)()()(

sVsVsG

S

o

Voltage Gain

)50( H z

)20( kH z


11/105

NETWORK FUNCTIONS

INPUT OUTPUT TRANSFER FUNCTION SYMBOL

Voltage Voltage Voltage Gain Gv(s)Current Voltage Transimpedance Z(s)

Current Current Current Gain Gi(s)

Voltage Current Transadmittance Y(s)

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 define

Driving Point Impedance/Admittance

Some nomenclature

EXAMPLE

admittanceTransfer

tanceTransadmit

)(

)()(

1

2

sV

sIsYT

gainVoltage)(

)(

)(1

2

sV

sV

sGv

To compute the transfer functions one must solve

the circuit. Any valid technique is acceptable


12/105

LEARNING EXAMPLE

admittanceTransfer

tanceTransadmit

)(

)()(

1

2

sV

sIsYT

gainVoltage

)(

)()(

1

2

sV

sVsGv

The textbook uses mesh analysis. We will

use Thevenins theorem

sLRsC

sZT H ||1)( 1

1

11

RsL

sLR

sC

)()(

1

112

RsLsC

RsLLCRssZT H

)()( 11

sVRsL

sLsVOC

)(sVOC

)(sZTH

)(2 sV2R

)(2 sI

)(

)()(

2

2sZR

sVsI

T H

OC

)(

)(

1

112

2

1

1

RsLsC

RsLLCRsR

sVRsL

sL

121212

2

)()()( RCRRLsLCRRs

LCssYT

)()(

)(

)(

)()( 2

1

22

1

sYRsV

sIR

sV

sVsG T

sv

)(

)(

1

1

RsLsC

RsLsC


13/105

POLES AND ZEROS (More nomenclature)

011

1

011

1

...

...)(

bsbsbsb

asasasasH

n

n

n

n

m

m

m

m

Arbitrary network function

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

))...()((

))...()(()(

21

210

n

m

pspsps

zszszsKsH

functionnetworktheofpolesfunctionnetworktheofzeros

n

m

ppp

zzz

,...,,

,...,,

21

21

The network function is uniquely determined by its poles and zeros

and its value at some other value of s (to compute the gain)

EXAMPLE

1)0(

22,22

,1

21

1

H

jpjp

z

:poles

:zeros

)22)(22()1()( 0

jsjs

sKsH84

120

ss

sK

18

1)0( 0KH

84

18)(

2

ss

ssH


14/105

LEARNING EXTENSION Find the driving point impedance at )(sVS

)(

)()(

sI

sVsZ S

)(sI

)(1

)()(: sIsC

sIRsVi n

i nS KVL

i n

i nsC

RsZ1

)(

Replace numerical values

M

s

1001

flhdldlhd


15/105

LEARNING EXTENSION

)104(

000,20,50

0

7

0

21

1

K

H zpH zp

z

:poles

:zero

)(

)()(

sV

sVsG

K

S

o

o

gainvoltagethefor

ofvaluetheandlocationszeroandpoletheFind

ooi ni n

i ni n

RsCRsCRsCsG

11]1000[

1)(

000,40

000,40]1000[100 sss

For this case the gain was shown to be

))...()((

))...()(()(

21

210

n

m

pspsps

zszszsKsH

Zeros = roots of numerator

Poles = roots of denominator

VariableFrequencyResponse

SINUSOIDAL FREQUENCY ANALYSIS


16/105

SINUSOIDAL FREQUENCY ANALYSIS

)(sH

Circuit represented by

network function

)cos(0

)(0

tB

eAtj

)(cos|)(|

)(

0

)(0

jHtjHB

ejHAtj

)()()(

)()(

|)(|)(

jeMjH

jH

jHM

Notation

stics.characteriphaseandmagnitudecalledgenerallyareoffunctionasofPlots ),(),(M

)(log)(

))(log2010

10

PLOTSBODE vs

(M

.offunctionaasfunctionnetworkthe

analyzewefrequencytheoffunctionaasnetworkaofbehaviorthestudyTo

)(jH


17/105

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 functions

have simple asymptotic behavior.The asymptotes can be used as reasonable and efficient approximations


18/105

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

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

)( 2

233310

bbba

N

jjj

jjjjK

jH

General form of a network function showing basic terms

Frequency independent

Poles/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)(

2

11

23

331

11

b

bba

NjH

Display each basic term

separately and add theresults to obtain final

answer

Lets examine each basic term


19/105

Constant Term

Poles/Zeros at the origin

90)(

)(log20|)(|

)(

10

Nj

Nj

j NdB

NN

linestraightaisthis

logisaxis-xthe 10

2


20/105

Simple pole or zero j1

1

210

tan)1(

)(1log20|1|

j

j dB

asymptotefrequencylow0|1| dBj

(20dB/dec)asymptotefrequencyhigh 10log20|1| dBj

frequency)akcorner/bre1whenmeetasymptotestwoThe (

Behavior in the neighborhood of the corner

Frequenc Asymptot Curve

distance to

asymptote Argument

corner 0dB 3dB 3 45

octave above 6dB 7db 1 63.4

octave below 0dB 1dB 1 26.6

12

5.0

0)1( j

90)1( j

1

1

Asymptote for phase

High freq. asymptoteLow freq. Asym.


21/105

Simple zero

Simple pole

Q d ti l ])()(21[2

jjt ])()(21[2

j


22/105

Quadratic pole or zero ])()(21[2

2 jjt ])()(21[2

j

222102 2)(1log20|| dBt 21

2)(1

2tan

t

1 asymptotefrequencylow0|| 2 dBt 02t

1 asymptotefreq.high2102 )(log20|| dBt 1802

t

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

221 2102 12log20|| dBt

2

12

21tan

t

2

2

Resonance frequency

Magnitude for quadratic pole Phase for quadratic pole

dB/dec40

These graphs are inverted for a zero

G G t it d d h l t


23/105

LEARNING EXAMPLE Generate magnitude and phase plots

)102.0)(1(

)11.0(10)(

jj

jjGvDraw asymptotes

for each term1,10,50:nersBreaks/cor

40

20

0

20

dB

90

90

1.0 1 10 100 1000

dB|10

decdB /20

dec/45

decdB/20

dec/45

Draw composites


24/105

asymptotes

LEARNING EXAMPLE G t it d d h l t


25/105

LEARNING EXAMPLE Generate magnitude and phase plots

)11.0()(

)1(25)(

2

jj

jjGv 101,:(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


26/105

21

020 0)(

KjK

dB

Final results . . . And an extra hint on poles at the origin

dec

dB40

dec

dB20

dec

dB40

LEARNING EXTENSION Sketch the magnitude characteristic


27/105


)100)(10(

)2(10)(

4

jj

jjG

formstandardinNOTisfunctiontheBut

10010,2,:breaks

Put in standard form

)1100/)(110/(

)12/(20)(

jj

jjG

We need to show about

4 decades

40

20

0

20

dB

90

90

1 10 100 1000

dB|25



28/105


2)(

102.0(100)(

j

jjG

origintheatpoleDouble

50atbreak

formstandardinisIt

40

20

0

20

dB

90

270

90

1 10 100 1000

Once each term is drawn we form the composites



29/105


)110/)(1()(

jj

jjG


)10)(1(

10)(

jj

jjG

101,:breaks

origintheatzero

formstandardinnot

40

20

0

20

dB

90

270

90

1.0 110

100Once each term is drawn we form the composites

decdB/20

decdB/20

LEARNING EXAMPLE A function with complex conjugate poles


30/105

LEARNING EXAMPLE A function with complex conjugate poles

1004)()5.0(25

)(2

jjj

jjG


125/)10/()15.0/(

5.0)(

2

jjj

jjG

40

20

020

dB

90

90

01.0 1.0 1 10 100270

1 )2(log20|| 102 dBt

2.01.0

25/12

])()(21[ 22 jjt

dB8

Draw composite asymptote

Behavior close to corner of conjugate pole/zero

is too dependent on damping ratio.

Computer evaluation is better

Evaluation of frequency response using MATLAB


31/105


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 multiplication

den =1.0000 4.5000 102.0000 50.0000

freqs(num,den)

Using default options



32/105

Evaluation of frequency response using MATLAB User controlled

>> clear all; close all %clear workspace and close any open figure>> figure(1) %open one figure window (not STRICTLY necessary)>> w=logspace(-1,3,200);%define x-axis, [10^{-1} - 10^3], 200pts total

1004)()5.0(25

)(2

jjj

jjG

>> G=25*j*w./((j*w+0.5).*((j*w).^2+4*j*w+100)); %compute transfer function>> subplot(211) %divide figure in two. This is top part>> semilogx(w,20*log10(abs(G))); %put magnitude here

>> grid %put a grid and give proper title and labels>> ylabel('|G(j\omega)|(dB)'), title('Bode Plot: Magnitude response')



33/105

>> semilogx(w,unwrap(angle(G)*180/pi)) %unwrap avoids jumps from +180 to -180>> grid, ylabel('Angle H(j\omega)(\circ)'), xlabel('\omega (rad/s)')>> title('Bode Plot: Phase Response')

Evaluation of frequency response using MATLAB User controlled ContinuedRepeat for phase

No xlabel here to avoid clutter

USE TO ZOOM IN A SPECIFIC REGION OF INTEREST

Compare with default!



34/105


]136/)12/[(

)1(2.0)(

2

jjj

jjG 6/136/12

12/1

])()(21[ 22 jjt

40

20

0

20

dB

90

270

90

1.0 110

100

1 )2(log20|| 102 dBt

decdB/20

decdB/40

decdB/0

12

dB5.9

)1(2.0)(

jjG


35/105

]136/)12/[(

)(.0)(

2

jjj

jjG

num=0.2*[1,1];

den=conv([1,0],[1/144,1/36,1]);

freqs(num,den)

DETERMINING THE TRANSFER FUNCTION FROM THE BODE PLOT


36/105

DETERMINING THE TRANSFER FUNCTION FROM THE BODE PLOT

This 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 0.1

1

)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

Determine a transfer function from the composite


37/105

LEARNING EXTENSIONDetermine a transfer function from the composite

magnitude asymptotes plot

A

A. Pole at the origin.

Crosses 0dB line at 5

j

5

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

Sinusoidal RESONANT CIRCUITS - SERIES RESONANCE


38/105

Im{ } 0Z

RESONANT FREQUENCY

PHASOR DIAGRAMQUALITY FACTOR

RESONANT CIRCUITS


39/105

RESONANT CIRCUITS

These are circuits with very special frequency characteristics

And resonance is a very important physical phenomenon

CjLjRjZ

1)(

circuitRLCSeries

LjCjGjY

1)(

circuitRLCParallel

L CCL

110

whenzeroiscircuiteachofreactanceThe

The frequency at which the circuit becomes purely resistive is called

the resonance frequency

Properties of resonant circuits


40/105


At resonance the impedance/admittance is minimal

Current through the serial circuit/

voltage across the parallel circuit can

become very large (if resistance is small)

CRR

LQ

0

0 1

:FactorQuality

222)

1(||

1)(

CLRZ

Cj

LjRjZ

222 )1

(||

1)(

LCGY

Cj

Lj

GjY

Given the similarities between series and parallel resonant circuits,

we will focus on serial circuits



41/105


At resonance the power factor is unity

CIRCUIT BELOW RESONANCE ABOVE RESONANCE

SERIES CAPACITIVE INDUCTIVE

PARALLEL INDUCTIVE CAPACITIVE

Phasor diagram for series circuit Phasor diagram for parallel circuit

RV

C

IjVC

Lj

1GV1CVj

L

Vj

1

LEARNING EXAMPLE Determine the resonant frequency, the voltage across each


42/105

q y g

element at resonance and the value of the quality factor

LC

10 sec/2000

)1010)(1025(

163

radFH

I

A

Z

VI S 5

2

010

2ZresonanceAt

50)1025)(102( 330L

)(902505500 VjLIjVL

902505501

501

0

0

0

jICj

V

LC

C

R

LQ 0

25

2

50

||||

|||| 0

SC

SS

L

VQV

VQR

VLV

resonanceAt

LEARNING EXAMPLE Given L = 0.02H with a Q factor of 200, determine the capacitor


43/105

necessary to form a circuit resonant at 1000Hz

R

L0200

200QwithL

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

S

S

L

VQV

VQR

VLV

resonanceAt

59.1200

02.010002R

AI 28.659.1

10

The reactive power on the capacitor

exceeds 12kVA

LEARNING EXTENSION Find the value of C that will place the circuit in resonance


44/105

Find the value of C that will place the circuit in resonance

at 1800rad/sec

LC

10 218001.0

1

)(1.0

11800

C

CH

FC 86.3

Find the Q for the network and the magnitude of the voltage across the

capacitor

R

LQ 0

60

3

1.01800

Q

||||

|||| 0

SC

SS

L

VQV

VQR

VLV

resonanceAt

VVC 600||

Resonance for the series circuit

2/1

1)( M


45/105

222)

1(||

1)(

C

LRZ

CjLjRjZ

QR

CQRL1

, 00

:resonanceAt

)(1

)(

0

0

0

0

j QR

QRjQRjRjZ

)(1

1

0

0

1

jQV

VG Rv

isgainvoltageThe:Claim

)(1

jZ

R

CjLjR

RGv

vv GGM |)(|,|)(

2/1

20

0

2 )(1

)(

Q

(tan)( 0

0

1

Q

QBW 0

12

1

2

12

0 QQL O

sfrequenciepowerHalf

Z

RGv

The Q factor LQ 0

1


46/105

CRRQ

0

RLowQHigh:circuitseriesFor

G)(lowRHighQHigh:circuitparallelFor

MBWSmallQHigh

dissipates

Stores as Efield

Stores as M

field

Capacitor and inductor exchange storedenergy. When one is at maximum the

other is at zero

D

S

W

WQ 2

cyclebydissipatedenergy

storedenergymaximum2

Q can also be interpreted from anenergy point of view

221

20202 m xeffD RIRIW

22

2

1

2

1m xm xS CVL IW

22

0 Q

R

L

W

W

D

s

ENERGY TRANSFER IN RESONANT CIRCUITS


47/105

Normalizationfactor

( ) cos [ ]mO

Vi t t A

R

LEARNING EXAMPLE Determine the resonant frequency, quality factor and


48/105

2

mH2

F5

Determine the resonant frequency, quality factor and

bandwidth when R=2 and when R=0.2

CRR

LQ

0

0 1

LC

10

QBW 0

sec/10)105)(102(

1 4630

rad

R Q

2 10

0.2 100

R Q BW(rad/sec)

2 10 1000

0.2 100 100

Evaluated with EXCEL

RQ 002.010000 QBW /10000

LEARNING EXTENSION A series RLC circuit as the following properties:


49/105

sec/100sec,/4000,4 0 radBWr adR

Determine the values of L,C.

CRR

LQ

0

0 1

LC

1

0

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

C6

620

20

1056.11016104

111

LEARNING EXAMPLE Find R, L, C so that the circuit operates as a band-pass filter


50/105

with center frequency of 1000rad/s and bandwidth of 100rad/s

)(1

jZ

R

CjLjR

RGv

CRR

LQ

0

0 1

LC

10

QBW 0

dependent

Strategy:

1. Determine Q2. Use value of resonant frequency and Q to set up two equations in the three

unknowns

3. Assign a value to one of the unknowns

10

100

10000

BW

Q

R

L

R

LQ

1000100

LCLC

1)10(

1 230

For example FFC6

101

HL 1

100R

PROPERTIES OF RESONANT CIRCUITS: VOLTAGE ACROSS CAPACITOR


51/105

|||| 0 RVQV

resonanceAt

But this is NOT the maximum value for the

voltage across the capacitor

CRjL C

CjLjR

Cj

V

V

S

20

1

1

1

1

2

221

1)(

Q

uu

ug

2

0

0

;SV

Vgu

22

22

2

1

)/1)(/(2)2)(1(20

Q

u

u

QQuuu

du

dg

CRR

LQ

0

0 1

LC

10

2

2 1)1(2Q

u

20

maxmax

2

11

Qu

2

2

424

max

4

11

2

11

4

1

1

Q

Q

QQQ

g

2

0

4

11

||||

Q

VQV S

LEARNING EXAMPLE 150, RR andwhenDetermine max0


52/105

mH50

F5

Natural frequency depends only on L, C.

Resonant frequency depends on Q.

sr adLC

/2000)105)(105(

11620

CRR

LQ

0

0 1

LC

1

0

20

maxmax

2

11

Qu

RQ

050.02000 2max 2112000

Q

R Q Wmax

50 2 1871

1 100 2000

Evaluated with EXCEL and rounded to zero decimals

Using MATLAB one can display the frequency response


53/105

R=50Low Q

Poor selectivity

R=1

High Q

Good selectivity

LEARNING EXAMPLE The Tacoma Narrows Bridge Opened: July 1, 1940


54/105

Collapsed: Nov 7, 1940

Likely cause: wind

varying at frequency

similar to bridgenatural frequency

2.020

Tacoma Narrows Bridge Simulator Assume a low Q=2.39


55/105

)11( f tV

42

40

BA

B

i n RR

R

v

v

resonanceatmodeltheFor

.deflection4'causedwind42mphafailureAt1

5.9H20

F66.31

42mxVi n

0.44

1.07

'77.3

PARALLEL RLC RESONANT CIRCUITS


56/105

222 )1(||

1)(

CLRZ

CjLjRjZ

222 )1(||

1)(

LCGY

CjLj

GjY

Impedance of series RLC Admittance of parallel RLC

IVYZLCCLGR

,,,

esequivalencNotice

SS YVI

SSL

SSC

SSG

IY

LjV

LjI

IY

CjCVjI

IY

GGVI

11

||1

||

||||

1

0

0

SL

SC

LC

SG

I

LG

I

IG

CI

II

II

GYL

C

00

resonanceAt

CRR

LQ

0

0 1

LC

10

Series RLC

Parallel RLC

LGG

CQ

0

0 1

LC

10

|| SIQ

Series RLCQ

BW 0

Parallel RLCQ

BW 0

VARIATION OF IMPEDANCE AND PHASOR DIAGRAM PARALLEL CIRCUIT


57/105

LEARNING EXAMPLE If the source operates at the resonant frequency of the

network compute all the branch currents


58/105

mHLFC

SGVS

120,600

01.0,0120

network, compute all the branch currents

SSL

SSC

SSG

IY

LjV

LjI

IY

CjCVjI

IY

GGVI

11

||1

||

||||

1

0

0

SL

SC

LC

SG

ILG

I

I

G

CI

II

II

GYL

C

0

0

resonanceAt

|| SIQ

SG IAI )(02.1012001.0

sr ad

LC

/85.117

)106(120.0

1140

)(9049.80120)10600()85.117()901( 6 AIC

)(9049.8 AIL

_______xI

LEARNING EXAMPLE Derive expressions for the resonant frequency, half power

frequencies bandwidth and quality factor for the transfer


59/105

frequencies, bandwidth and quality factor for the transfer

characteristic

i n

ou t

I

VH

LjCjGYT

1

Ti n

ou t

T

i nou t

YI

VH

Y

IV

1

22 1

1

1

1||

LCGLj

CjG

H

LC

1

0 :frequencyResonant22

max||5.0|)(| HjH h sfrequenciepowerHalf

2

2

22

1G

LCG

h

h

RGH

1

|| max

GL

Ch

h

1

LCC

G

C

Gh

1

22

2

C

GBW L OH I

L

CR

L

C

GBW

Q 10

L GG

CQ

0

0 1

12

121

2

0QQ

L O

Replace and show

LEARNING EXAMPLE Increasing selectivity by cascading low Q circuits


60/105

Single stage tuned amplifier

M Hsr ad

FHLC

9.99/10275.6

1054.210

11 81260

398.010

1054.2250

6

12

L

CR

L

C

GBWQ

10

LEARNING EXTENSION Determine the resonant frequency, Q factor and bandwidth


61/105

FCmHLkR 150,20,2

Parallel RLC

L GG

CQ

0

0 1

LC

10

QBW 0

srad/577)10150)(1020(

1630

173

2000/1

10150577 6

Q

sradBW /33.3173

577

LEARNING EXTENSION 0C,L,Determine


62/105

120,/1000,6 QsradBWkR

Parallel RLC

L GG

CQ

0

0 1

LC

10

Q

BW 0

sr adBWQ /102.11000120 50

FR

QC

167.0

102.16000

1205

0

HQ

RL

417102.1120

60005

0

Can be used to verify computations

PRACTICAL RESONANT CIRCUIT The resistance of the inductor coils cannot be

neglected


63/105

neglected

LjRCjjY

1)( LjR

LjR

22 )()(

LR

LjRCjjY

2222

)()(

)(

LR

LCj

LR

RjY

2

22

10

)(

L

R

LCLR

LCY R

real

R

LQ

LC

000 ,

1 2

0

0

11

QR

maximaareimpedanceandvoltagetheresonanceAt.Y

IZIV

2

0

2

0

222

11 R

L

RR

L

RR

LR

ZRRR

M AX

20RQZM AX

How do you define a quality factor forthis circuit?

LEARNING EXAMPLE 5,50, RR forbothDetermine 0


64/105

R

LQ

LC

000 ,

1 2

0

0

11

QR

sr adFH

/2000)105)(1050(

1630

20

0112000,050.02000

QRQ R

R Q0 Wr(rad/s) f(Hz)

50 2 1732 275.7

5 20 1997 317.8

RESONANCE IN A MORE GENERAL VIEW


65/105

222)

1(||

1)(

C

LRZ

CjLjRjZ

222 )1

(||

1)(

L

CGY

CjLj

GjY

For series connection the impedance reaches maximum at resonance. For parallel

connection the impedance reaches maximum

1)(1)( 22

LGjLCj

LjZ

CRjLCj

CjY ps

12)(2 jj

aswrittenwastermquadratictheplotsBodeIn

0

1

L C

QCRCR

122 0

series

QLGLG

122 0

parallel

2

1QA high Q circuit is highly

under damped

Resonance

SCALING


66/105

Scaling techniques are used to change an idealized network into a more

realistic one or to adjust the values of the components

M

M

M

K

CC

LKL

RKR

'

'

'

scalingimpedanceorMagnitude

''11'' 0

CLLCCLLC

'

'00

R

L

R

LQ

Magnitude scaling does not change the

frequency characteristics nor the qualityof the network.

CCLL

K' F

1

''

1,''

unchangediscomponenteachofImpedance

scalingtimeorFrequency

F

F

K

CC

K

LL

RR

'

'

'

0

'

0 FK

)('

''0 BWK

QBW F

QR

LQ

'

''

'0 Constant Q

networks

LEARNING EXAMPLE Determine the value of the elements and the characterisitcs

of the network if the circuit is magnitude scaled by 100 and


67/105

2

H1

F2

1

of the network if the circuit is magnitude scaled by 100 and

frequency scaled by 1,000,000

2,

2

2,/20 BWQsr ad

FC

HL

R

200

1'

100'

200'

M

M

M

K

C

C

LKL

RKR

'

'

'

scalingimpedanceorMagnitude F

F

KCC

K

LL

RR

'

'

'

FC

mHL

R

200

1

''

100''

200''

0'0 FK

)('

''0 BWK

QBW

F

sr ad/10414.1 6''0

unchangedare0,Q

LEARNING EXTENSIONDetermine10,000.byscaledfrequencyand100byscaledmagnitudeis2FC1H,L,10RwithnetworkRLCAn


68/105

elementscircuitresultingthe

Determine10,000.byscaledfrequencyand100byscaled

M

M

M

K

CC

LKL

RKR

'

'

'

scalingimpedanceorMagnitude

FC

HL

R

02.0'

100'

1000'

F

F

K

CC

K

LL

RR

'

'

'

scalingFrequency

FC

HL

kR

2''

01.0''

1''

Scaling

FILTER NETWORKS


69/105

Networks designed to have frequency selective behavior

COMMON FILTERS

Low-pass filterHigh-pass filter

Band-pass filter

Band-reject filter

We focus first on

PASSIVE filters

Simple low-pass filter


70/105

RCj

CjR

Cj

V

VGv

1

1

1

1

1

0

RCj

Gv

;1

1

1

2

tan)(

1

1||)(

v

v

G

GM

2

11,1max

MM

frequencypowerhalf 1

1BW

Simple high-pass filter


71/105

CRj

CRj

CjR

R

V

VGv

111

0

RCj

jGv

;1

1

2

tan2)(

1||)(

v

v

G

GM

2

11,1max

MM

frequencypowerhalf

1

1

L O

Simple band-pass filter


72/105

Band-pass

CLjR

R

V

VGv

11

0

222 1)(

LCRC

RCM

11

LCM 0)()0( MM

2

4/)/( 202

LRLRL O

LC

10

2

4/)/(20

2

LRLRH I

L

RBW L OH I

)(2

1)( H IL O MM

Simple band-reject filter


73/105

0110

00

C

LjLC

10 VV circuitopenasactscapacitorthe0at

10 VV circuitopenasactsinductortheat

filterpass-band

theinasdeterminedareH IL O ,

LEARNING EXAMPLE Depending on where the output is taken, this circuit

can produce low-pass, high-pass or band-pass or band-


74/105

reject filters

Band-pass

Band-reject filter

CLjR

Lj

V

V

S

L

1 1)(,00 S

L

S

L

V

V

V

VHigh-pass

CLjR

Cj

V

V

S

C

1

1

0)(,10 S

C

S

C

V

V

V

VLow-pass

FCHLR 159,159,10 forplotBode

LEARNING EXAMPLE A simple notch filter to eliminate 60Hz interference


75/105

)1(1

1

CLj

CL

CjLj

CjLj

ZR

i n

Req

eqV

ZR

RV

0

LCZR 1 010

LCV

tttvi n 10002sin2.0602sin)(

FCmHL 100,3.70

LEARNING EXTENSION )( jGvforplotBodetheofsticcharacterimagnitudetheSketch


76/105

RCj

Cj

R

CjjGv

1

1

1

1

)(

sr adFRC /2.0)1020)(1010(63 20dB/dec-ofasymptotefrequencyHigh

0dB/decofasymptotefrequencylow

5rad/s:frequencyerBreak/corn



77/105

RCj

RCj

CjR

R

jGv

11)(

sr adFRC /5.0)1020)(1025(63

srad/21

at0dBCrosses20dB/dec.

20dB/dec-ofasymptotefrequencyHigh


2rad/s:frequencyerBreak/corn


Band pass 1


78/105

Band-pass

LCjRCj

RCj

LjCj

R

RjGv 2

)(11

)(

5.0

102

1010102

,1010

3

363

362

RC

L C

sradRC

/10001

at0dBCrosses20dB/dec.

40dB/dec-ofasymptotefrequencyHigh


rad/s1000:frequencyerBreak/corn

2

4/)/( 202

LRLRL O

2

4/)/( 202

LRLRH I

1000

1

0

LC

sr ad/618

sr ad/1618

decdB/40

ACTIVE FILTERS


79/105

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 a

more general framework, if one replaces the resistors by impedances

Ideal Op-Amp

These currents are

zero

Basic Inverting Amplifier


80/105

0V

VVgainInfinite

0V

0 II-impedanceinputInfinite

02

2

1

1 Z

V

Z

V

1

1

22 V

Z

ZV

Linear circuit equivalent

0I

1

2

Z

ZG

1

11

Z

VI

Basic Non-inverting amplifier

V0I


81/105

1V

1V

0I

1

1

2

10

Z

V

Z

VV

1

1

120 V

Z

ZZV

1

2

1 Z

Z

G

01 I

Basic Non-inverting Amplifier

Due to the internal op-amp circuitry, it has

limitations, e.g., for high frequency and/or

low voltage situations. The Operational

Transd uctance Am pl i f ier(OTA) performs

well in those situations

Operational Transductance Amplifier (OTA)


82/105

0RRin:OTAIdeal

COMPARISON BETWEEN OP-AMPS AND OTAs PHYSICAL CONSTRUCTION

Comparison of Op-Amp and OTA - Parameters

Amplifier Type Ideal Rin Ideal Ro Ideal Gain Input Current input Voltage

Op-Amp 0 0 0

OTA gm 0 nonzero


83/105

Basic Op-Amp Circuit Basic OTA Circuit

i nS

i nv

L

L

S

Si nS

i n

i n

i nv

L

L

RR

RA

RR

R

V

vA

VRR

R

v

vARR

Rv

0

0

0

0

i nS

i nm

Li n

m

Si nS

i n

i n

i nm

L

RR

Rg

RR

R

v

iG

VRR

Rv

vgRR

Ri

0

00

0

00

vAAAmp-OpIdeal mm gG

OTAIdeal

Basic OTA Circuits


84/105

)0()(1

0

0

00

10

vdxxiC

v

vgi

t

m

)0()()( 0

0

10 vdxxvC

gtv

t

m

Integrator

In the frequency domain

10 VCj

gV m

00

0

ii

vgi

in

i nm polarity)(notice

m

eq

i n

i n

gR

i

v 1

ResistorSimulated

OTA APPLICATION


85/105

Basic OTA Adder

11vgm

22vgm

21 21vgvg mm

ResistorSimulated

Equivalent representation

)(1 213

0 21vgvg

gv mm

m

mgofilityProgrammab


86/105

)10

10.,.(

10

7

mSgge

g

mSg

m

m

m

decades7-3:range

valuesTypical

AB Cm IA

Sg

201

Controlling transconductance

LEARNING EXAMPLE resistoraProduce k25


87/105

AB Cm

m

m

Ig

SmS

g

mSg

20

10410

4

4

7

4

SSgg

m

m

753 1041041

1025

m

eq

i n

i n

gR

i

v 1

ResistorSimulated

)(201045

AIA

SS AB C

AAI AB C 21026

LEARNING EXAMPLE Floating simulated resistor


88/105

1101 vgi m 1202 vgi m

i nmvgi 0

One grounded terminal

011ii

102ii

21 mm gg

operationproperFor

AB Cm

m

m

Ig

SmS

g

mSg

20

10410

4

4

7

4

resistor10MaProduce

Sgm 76 101010

1 S

7104

The resistor cannot be produced

with this OTA!

LEARNING EXAMPLE

210

321

210

,,,

vvv

ggg mmm

a)producetoSelect

m

SmS

g

mSg

1044

4

7


89/105

210 210 vvv b)

AB Cm

m

Ig

Sg

20

10410

7

4

)(1

21

3

0 21vgvg

gv mm

m

Case a

2;103

2

3

1 m

m

m

m

g

g

g

g

Two equations in three unknowns.

Select one transductance

)(1020

11.0 433 AImSg ABCm

A5

AImSg AB Cm 102.0 22

AImSg AB Cm 501 11

Case b

Reverse polarity of v2!

ANALOG MULTIPLIERBased on modulating the control current


90/105

ASSUMES VG IS ZERO

AUTOMATIC GAIN CONTROL


91/105

For simplicity of analysiswe drop the absolute valueI N O I N

I N O

v small v Av

Av big v

B

OTA-C CIRCUITS

Circuits created using capacitors, simulated resistors, adders and integrators


92/105

Circuits created using capacitors, simulated resistors, adders and integrators

integrator

resistor

Frequency domain analysis assuming

ideal OTAs

1101 im VgI 0202 VgI m

0201 IIIC CICj

V

10

021101

VgVgCj

V mim

1

2

2

1

01

i

m

m

m

V

g

Cj

gg

V

Magnitude Bode plot

1

0

i

vV

VG

2

1

m

mdc

g

gA

Cgf

C

g

mC

mC

2

2

2

LEARNING EXAMPLE

1

4

51

0

jV

VG

i

v

:Desired

m

m

SmS

g

mSg

101

1

6

3

OK


93/105

)10(2 5

ABCm

m

Ig

g

20

103

4dcA kH zfCC 100)10(25

2

1

m

mdc

g

gA

C

gf

C

g

mC

mC

2

2

2

Two equations in three unknowns.

Select the capacitor value

pFC 25 Sgm6125

2 107.15)1025)(10(2

OK

AISg AB Cm 14.38.62 11

AI AB C 785.020

7.152

biasesandncestransductatheFind

TOW-THOMAS OTA-C BIQUAD FILTER biquad ~ biquadratic

20 )()(

CjBjAV


94/105

20

02 )()(

jQ

jVi

Cj

VVgV im

021101

)( 201202 im VVgI )( 23303 oim VVgI

)(1

0302

2

02 IICj

V

03i

)(unknownsfourandequationsFour 02010201 ,,, IIVV

1)()(12

132

21

21

3

2

321

2

3

2

2

01

jgg

Cgj

gg

CC

Vg

gVV

g

g

g

Cj

V

mm

m

mm

i

m

mii

m

m

m

1)()(12

132

21

21

3

21

312

1

11

02

jgg

Cgj

gg

CC

Vgg

gCjV

g

CjV

V

mm

m

mm

i

mm

mi

m

i

1

22

3

21

2

30

21

210 ,,

C

C

g

ggQ

C

g

QCC

gg

m

mmmmm

Filter Type A B C

Low-pass 0 0 nonzero

Band-pass 0 nonzero 0

High-pass nonzero 0 0

C

gBW

g

gQ

Cg

CC

gg

m

m

m

m

mm

3

3

0

21

21

LEARNING EXAMPLE

5.-gainfrequencycenterand75kHz,ofbandwidth500kHz,offrequencycenterwithfilterpass-bandaDesign

capacitorspF50andionconfiguratThomasTowtheUse


95/105

AB Cm

m

m

Ig

S

mS

g

mSg

20

10410

4

4

7

4

1)()(12

132

21

21

3

21

312

1

11

02

jgg

Cgj

gg

CC

Vgg

gCjV

g

CjV

V

mm

m

mm

i

mm

mi

m

i

1

22

3

21

2

30

21

210 ,,

C

C

g

ggQ

C

g

QCC

gg

m

mmmmm

capacitorspF-50andionconfiguratThomas-TowtheUse

BW

03 iV

3

20 |)(|

m

mv

g

gjG

0)(1 221

210

jgg

CCmm

21 CC

Sg

BW m 56.23107521050

3

123

Sgm 8.1175 2

213

61

2520

)105(

108.1171052

m

g Sgm 5.2091

AI

AI

AI

AB C

AB C

AB C

18.1

89.5

47.10

3

2

1

Bode plots for resulting amplifier


96/105

Bode plots for resulting amplifier

LEARNING BY APPLICATION Using a low-pass filter to reduce 60Hz ripple


97/105

Thevenin equivalent for AC/DC

converter

Using a capacitor to create a low-

pass filter

Design criterion: place the corner frequency

at least a decade lower

T HT H

OF VCRjV 1

1

21

||||

CR

VV

T H

T HOF

CRT HC

1

||1.0|| T HOF VV

FCC

05.5362

1500


98/105

Filtered output

LEARNING EXAMPLE Single stage tuned transistor amplifier

Select the capacitor for maximum

gain at 91 1MHz


99/105

gain at 91.1MHz

Antenna

VoltageTransistor Parallel resonant circuit

CjLjR

V

V

A

1||||1000

40

LCRC

jj

Cj

V

V

Cj

Cj

CjLjR

A1

)(

/

1000

4

//

111

10004

2

0

LC/1frequencycenterwithpass-Band

C6

6

10

1101.912 pFC 05.3

100

1000

410

R

LCV

V

A

AVV0forplotBodeMagnitude

LEARNING BY DESIGN Anti-aliasing filter

Nyquist Criterion


100/105

When digitizing an analog signal, such as music, any frequency components

greater than half the sampling rate will be distorted

In fact they may appear as spurious components. The phenomenon is known as

aliasing.

SOLUTION: Filter the signal before digitizing, and remove all components higher

than half the sampling rate. Such a filter is an anti-aliasing filter

For CD recording the industry standard is to sample at 44.1kHz.

An anti-aliasing filter will be a low-pass with cutoff frequency of 22.05kHz

Single-pole low-pass filter

RCjV

V

i n

1

101

050,2221

RC

C

kRnFC 18.721

Resulting magnitude Bode plot

Attenuationin audio range

Improved anti-aliasing filter Two-stage buffered filter


101/105

01v

RCjV

V

1

1

01

02

RCjV

V

i n

1

101

One-stage

Two-stage

Four-stage

ni nn

RCjV

V

1

10

stage-n

LEARNING BY DESIGN Notch filter to eliminate 60Hz hum

Notch filter characteristic


102/105

mHL

FC

704.0

10

Magnitude Bode plot

sCsLRR

R

V

V

tapeam p

am p

tape

am p

/1||

1

1

2

2

tapeam p

tapeam p

am p

tape

am p

RR

LsLCs

LCs

RR

R

V

V

LC

1frequencynotch To design, pick one, e.g., C and determine the other


103/105

(non-inverting op-amp)DESIGN EXAMPLE BASS-BOOST AMPLIFIER DESIRED BODE PLOT


104/105

OPEN SWITCH(6dB)

500

2P

f

Switch closed??

DESIGN EXAMPLE TREBLE BOOSTOriginal player response Desired boost


105/105

Proposed boost circuit

Design equations

Documents

Variable Frequency Response Analysis 8 Ed