variable-frequency response analysis network performance as function of frequency

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

VARIABLE-FREQUENCY NETWORKPERFORMANCE

LEARNING GOALS

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

90LLjZL Inductor

Capacitor 9011

CCjZc

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

eq1

)(2

notation" in tionSimplifica"

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

011

1

011

1

...

...)(

bsbsbsb

asasasasZ n

nn

n

mm

mm

sCZsLsZRsZ CLR

1,)(,)(

Simplified notation for basic components

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

USING MATLAB TO COMPUTE MAGNITUDE AND 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

GRAPHIC OUTPUT PRODUCED BY MATLAB

Log-logplot

Semi-logplot

LEARNING EXAMPLE A possible stereo amplifier

Desired frequency characteristic(flat between 50Hz and 15KHz)

Postulated amplifier

Log frequency scale

Frequency domain equivalent circuit

Frequency Analysis of Amplifier

)(

)(

)(

)(

sV

sV

sV

sV

in

o

S

in

)(/1

)( sVsCR

RsV S

inin

inin

]1000[/1

/1)( in

oo

oo V

RsC

sCsV

ooinin

inin

RsCRsC

RsCsG

1

1]1000[

1)(

000,40

000,40]1000[

100 ss

s

000,401001058.79

100101018.3191

1691

oo

inin

RC

RC

required

actual

000,40

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

s

ssGs

Frequency dependent behavior iscaused by reactive elements

)(

)()(

sV

sVsG

S

o

Voltage Gain

)50( Hz

)20( kHz

NETWORK FUNCTIONS

INPUT OUTPUT TRANSFER FUNCTION SYMBOLVoltage 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 defineDriving Point Impedance/Admittance

Some nomenclature

EXAMPLE

admittanceTransfer

tanceTransadmit

)(

)()(

1

2

sV

sIsYT

gain Voltage)(

)()(

1

2

sV

sVsGv

To compute the transfer functions one must solvethe circuit. Any valid technique is acceptable

LEARNING EXAMPLE

admittanceTransfer

tanceTransadmit

)(

)()(

1

2

sV

sIsYT

gain Voltage)(

)()(

1

2

sV

sVsGv

The textbook uses mesh analysis. We willuse Thevenin’s theorem

sLRsC

sZTH ||1

)( 11

11

RsL

sLR

sC

)()(

1

112

RsLsC

RsLLCRssZTH

)()( 11

sVRsL

sLsVOC

)(sVOC

)(sZTH

)(2 sV

2R)(2 sI

)(

)()(

22 sZR

sVsI

TH

OC

)(

)(

1

112

2

11

RsLsCRsLLCRs

R

sVRsL

sL

121212

2

)()()(

RCRRLsLCRRs

LCssYT

)()(

)(

)(

)()( 2

1

22

1

sYRsV

sIR

sV

sVsG T

sv

)(

)(

1

1

RsLsC

RsLsC

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

function network the of poles

function network the of zeros

n

m

ppp

zzz

,...,,

,...,,

21

21

The network function is uniquely determined by its poles and zerosand 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

sKsH

84

120

ss

sK

18

1)0( 0KH

84

18)( 2

ss

ssH

LEARNING EXTENSIONFind the driving point impedance at )(sVS

)(

)()(

sI

sVsZ S

)(sI

)(1

)()(: sIsC

sIRsVin

inS KVL

in

in sCRsZ

1)(

Replace numerical values

M

s

1001

LEARNING EXTENSION

)104(

000,20,50

0

70

21

1

K

HzpHzp

z

:poles

:zero

)(

)()(

sV

sVsG

K

S

o

o

gain voltagethefor

of valuethe and locations zero and pole the Find

ooinin

inin

RsCRsC

RsCsG

1

1]1000[

1)(

000,40

000,40]1000[

100 ss

sFor this case the gain was shown to be

))...()((

))...()(()(

21

210

n

m

pspsps

zszszsKsH

Zeros = roots of numeratorPoles = roots of denominator

VariableFrequencyResponse

SINUSOIDAL FREQUENCY ANALYSIS

)(sH

Circuit represented bynetwork function

)cos(0

)(0

tB

eA tj

)(cos|)(|

)(

0

)(0

jHtjHB

ejHA tj

)()()(

)()(

|)(|)(

jeMjH

jH

jHM

Notation

stics.characteri phase and magnitude

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

)(log)(

))(log2010

10

PLOTS BODE vs

(M

. of function a as function network the

analyze wefrequency the of function a as network a ofbehavior thestudy To

)( jH

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

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

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

233310

bbba

N

jjj

jjjjKjH

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)(

211

23

3311

1

b

bba

NjH

Display each basic termseparately and add theresults to obtain final answer

Let’s examine each basic term

Constant Term

Poles/Zeros at the origin

90)(

)(log20|)(|)( 10

Nj

Njj

NdB

NN

linestraight a is this

log is axis- xthe 10

Simple pole or zero j1

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.

Simple zero

Simple pole

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 frequency

221 2102 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

LEARNING EXAMPLEGenerate 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

asymptotes

LEARNING EXAMPLEGenerate 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

21

020 0)(

Kj

K

dB

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

dec

dB40

dec

dB20

dec

dB40

LEARNING EXTENSION Sketch the magnitude characteristic

)100)(10(

)2(10)(

4

jj

jjG

form standard in NOT is function theBut

100 10, 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


2)(

102.0(100)(

j

jjG

origin theat pole Double

50at break

form standard in isIt

40

20

0

20

dB

90

270

90

1 10 100 1000

Once each term is drawn we form the composites

Put in standard form

)110/)(1()(

jj

jjG


)10)(1(

10)(

jj

jjG

10 1, :breaks

origin theat zero

form standard innot

40

20

0

20

dB

90

270

90

1.0 110

100Once each term is drawn we form the composites

decdB /20decdB /20

LEARNING EXAMPLEA 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[ 22 jjt

dB8

Draw composite asymptote

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

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)

LEARNING EXTENSIONSketch the magnitude characteristic

]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

]136/)12/[(

)1(2.0)( 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

This is the inverse problem of determining frequency characteristics. We will use only the composite asymptotes plot of the magnitude to postulatea 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

LEARNING EXTENSIONDetermine 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

Sinusoidal

RESONANT CIRCUITS

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

CjLjRjZ

1

)(

circuit RLC Series

LjCjGjY

1

)(

circuit RLC Parallel

LCCL

110

when zero iscircuit each of reactance The

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

Properties of resonant circuits

At 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

Properties of resonant circuits

At 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

LEARNING EXAMPLEDetermine 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

LEARNING EXAMPLEGiven 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

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

R

LQ 0 60

3

1.01800

Q

||||

|||| 0

SC

SS

L

VQV

VQR

VLV

resonanceAt

VVC 600||

Resonance for the series circuit

222 )1

(||

1)(

CLRZ

CjLjRjZ

QRCQRL

1, 00

:resonanceAt

)(1

)(

0

0

0

0

jQR

QRjQRjRjZ

)(1

1

0

0

1

jQV

VG Rv

is gain voltageThe :Claim

)(1

jZ

R

CjLjR

RGv

vv GGM |)(|,|)(

2/120

0

2 )(1

1)(

Q

M

)(tan)( 0

0

1

Q

QBW 0

12

1

2

12

0 QQLO

sfrequenciepower Half

Z

RGv

The Q factorCRR

LQ

0

0 1

R LowQ High :circuit seriesFor G) (low R HighQ High :circuit parallelFor

M

BW SmallQ High

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

cycleby dissipatedenergy storedenergy maximum2

Q can also be interpreted from anenergy point of view

22

1

20202 mxeffD RIRIW

22

2

1

2

1mxmxS CVLIW

220 Q

R

L

W

W

D

s

LEARNING EXAMPLE

2

mH2

F5

Determine the resonant frequency, quality factor andbandwidth when R=2 and when R=0.2

CRR

LQ

0

0 1

LC

10

QBW 0

sec/10)105)(102(

1 4

630 rad

R Q2 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 EXTENSIONA 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

LEARNING EXAMPLEFind R, L, C so that the circuit operates as a band-pass filterwith 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 unknowns3. Assign a value to one of the unknowns

10100

10000 BW

Q

R

L

R

LQ

1000100

LCLC

1)10(

1 230

For example FFC 6101

HL 1

100R

PROPERTIES OF RESONANT CIRCUITS: VOLTAGE ACROSS CAPACITOR

|||| 0 RVQV resonanceAt

But this is NOT the maximum value for thevoltage across the capacitor

CRjLCCj

LjR

CjV

V

S

20

1

11

1

2

221

1)(

Qu

u

ug

2

0

0

;SV

Vgu

22

22

2

1

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

Qu

u

QQuuu

du

dg

CRR

LQ

0

0 1

LC

10

22 1)1(2

Qu

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

mH50

F5

150, RR and when Determine max0

Natural frequency depends only on L, C.Resonant frequency depends on Q.

sradLC

/2000)105)(105(

11620

CRR

LQ

0

0 1

LC

10

20

maxmax

2

11

Qu

RQ

050.02000 2max 2

112000Q

R Q Wmax50 2 18711 100 2000

Evaluated with EXCEL and rounded to zero decimals

Using MATLAB one can display the frequency response

R=50Low QPoor selectivity

R=1High QGood selectivity

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

Likely cause: windvarying at frequencysimilar to bridgenatural frequency

2.020

Tacoma Narrows Bridge Simulator

)11( ftV

42

40

BA

B

in RR

R

v

v

resonanceat model theFor

.deflection 4' caused wind 42mph a failureAt

1

5.9H20

F66.31

42mxVin

Assume a low Q=2.39

0.44’

1.07’

'77.3

PARALLEL RLC RESONANT CIRCUITS

222 )1

(||

1)(

CLRZ

CjLjRjZ

222 )1

(||

1)(

LCGY

CjLj

GjY

Impedance of series RLC Admittance of parallel RLC

IVYZ

LCCLGR

,

,,

esequivalenc Notice

SS YVI

SSL

SSC

SSG

IYLj

VLj

I

IY

CjCVjI

IY

GGVI

11

||1

||

||||

1

0

0

SL

SC

LC

SG

ILG

I

IG

CI

II

II

GYL

C

0

0

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

LEARNING EXAMPLE

mHLFC

SGVS120,600

01.0,0120

If the source operates at the resonant frequency of the network, compute all the branch currents

SSL

SSC

SSG

IYLj

VLj

I

IY

CjCVjI

IY

GGVI

11

||1

||

||||

1

0

0

SL

SC

LC

SG

ILG

I

IG

CI

II

II

GYL

C

0

0

resonanceAt

|| SIQ

SG IAI )(02.1012001.0

sradLC

/85.117)106(120.0

1140

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

)(9049.8 AIL

_______xI

LEARNING EXAMPLEDerive expressions for the resonant frequency, half powerfrequencies, bandwidth and quality factor for the transfercharacteristic

in

out

I

VH

LjCjGYT

1

Tin

out

T

inout YI

VH

Y

IV

1

22 1

11

1||

LCGLj

CjGH

LC

10 :frequencyResonant

22max

||5.0|)(| HjH h sfrequenciepower Half

22

2 21

GL

CGh

h

RG

H 1

|| max

GL

Ch

h

1

LCC

G

C

Gh

1

22

2

C

GBW LOHI

L

CR

L

C

GBWQ

10

LGG

CQ

0

0 1

12

1

2

12

0 QQLO

Replace and show

LEARNING EXAMPLE 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

LEARNING EXTENSIONDetermine the resonant frequency, Q factor and bandwidth

FCmHLkR 150,20,2

Parallel RLC

LGG

CQ

0

0 1

LC

10

QBW 0

srad /577)10150)(1020(

1630

1732000/1

10150577 6

Q

sradBW /33.3173

577

LEARNING EXTENSION 0 C, L, Determine

120,/1000,6 QsradBWkR

Parallel RLC

LGG

CQ

0

0 1

LC

10

QBW 0

sradBWQ /102.11000120 50

FR

QC

167.0

102.16000

1205

0

HQ

RL

417

102.1120

60005

0

Can be used to verify computations

PRACTICAL RESONANT CIRCUITThe resistance of the inductor coils cannot beneglected

LjRCjjY

1)(

LjR

LjR

22 )()(

LR

LjRCjjY

2222 )()()(

LR

LCj

LR

RjY

2

22

10

)(

L

R

LCLR

LCY R

real

R

LQ

LC0

00 ,1 2

00

11

QR

maxima are impedance and voltagethe resonanceAt .Y

IZIV

20

2

0

222

11R

LR

R

LR

R

LRZ RRRMAX

20RQZMAX

How do you define a quality factor for this circuit?

LEARNING EXAMPLE 5,50, RR for both Determine 0

R

LQ

LC0

00 ,1 2

00

11

QR

sradFH

/2000)105)(1050(

1630

20

01

12000,050.02000

QRQ R

R Q0 Wr(rad/s) f(Hz)50 2 1732 275.75 20 1997 317.8

RESONANCE IN A MORE GENERAL VIEW

222 )1

(||

1)(

CLRZ

CjLjRjZ

222 )1

(||

1)(

LCGY

CjLj

GjY

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

1)(1)( 22

LGjLCj

LjZ

CRjLCj

CjY ps

12)( 2 jj

as written was term quadratic the plots Bode In

0

1

LC

QCRCR

122 0

series

QLGLG

122 0

parallel

2

1QA high Q circuit is highly

under damped

Resonance

SCALING

Scaling techniques are used to change an idealized network into a morerealistic one or to adjust the values of the components

M

M

M

K

CC

LKL

RKR

'

'

'

scaling impedanceor Magnitude

''

11'' 0 CLLC

CLLC

'

'00

R

L

R

LQ

Magnitude scaling does not change thefrequency characteristics nor the qualityof the network.

CCLL

ωKω' F

1

''

1,''

unchanged iscomponent each of Impedance

scaling timeor Frequency

F

F

K

CC

K

LL

RR

'

'

'

0'0 FK

)('

''0 BWKQ

BW F

QR

LQ

'

''

'0 Constant Q

networks

LEARNING EXAMPLE

2

H1

F2

1

Determine the value of the elements and the characterisitcsof the network if the circuit is magnitude scaled by 100 andfrequency scaled by 1,000,000

2,2

2,/20 BWQsrad

FC

HL

R

200

1'

100'

200'

M

M

M

K

CC

LKL

RKR

'

'

'

scaling impedanceor Magnitude F

F

K

CC

K

LL

RR

'

'

'

FC

mHL

R

200

1''

100''

200''

0'0 FK

)('

''0 BWKQ

BW F

srad /10414.1 6''0

unchanged are 0,Q

LEARNING EXTENSION

elementscircuit resulting the

Determine 10,000.by scaledfrequency and 100by scaled

magnitude is 2FC 1H,L ,10R with network RLC An

M

M

M

K

CC

LKL

RKR

'

'

'

scaling impedanceor Magnitude

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

Networks designed to have frequency selective behavior

COMMON FILTERS

Low-pass filterHigh-pass filter

Band-pass filter

Band-reject filter

We focus first onPASSIVE filters

Simple low-pass filter

RCjCj

R

CjV

VGv

1

11

1

1

0

RCj

Gv

;1

1

1

2

tan)(

1

1||)(

v

v

G

GM

2

11,1max

MM

frequencypower half

1

1

BW

Simple high-pass filter

CRj

CRj

CjR

R

V

VGv

11

1

0

RCj

jGv

;1

1

2

tan2

)(

1||)(

v

v

G

GM

2

11,1max

MM

frequencypower half

1

1

LO

Simple band-pass filter

Band-pass

CLjR

R

V

VGv

11

0

222 1)(

LCRC

RCM

11

LCM 0)()0( MM

2

4/)/( 20

2

LRLRLO

LC

10

2

4/)/( 20

2

LRLRHI

L

RBW LOHI

)(2

1)( HILO MM

Simple band-reject filter

011

000

CLj

LC

10 VV circuit open as actscapacitor the 0at

10 VV circuit open as actsinductor the at

filter pass-band

the in as determined are HILO ,

LEARNING EXAMPLEDepending on where the output is taken, this circuitcan produce low-pass, high-pass or band-pass or band-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

CjV

V

S

C

1

1

0)(,10 S

C

S

C

V

V

V

VLow-pass

FCHLR 159,159,10 for plot Bode

LEARNING EXAMPLEA 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

LEARNING EXTENSION )( jGvfor plot Bode the of sticcharacteri magnitude the Sketch

RCjCj

R

CjjGv

1

11

1

)(

sradFRC /2.0)1020)(1010( 63 20dB/dec- of asymptotefrequency High

0dB/dec of asymptotefrequency low

5rad/s :frequencyer Break/corn


RCj

RCj

CjR

RjGv

11

)(

sradFRC /5.0)1020)(1025( 63

srad /21

at 0dB Crosses 20dB/dec.

20dB/dec- of asymptotefrequency High


2rad/s :frequencyer Break/corn


Band-pass

LCjRCj

RCj

LjCj

R

RjGv 2)(11

)(

5.0102

1010102

,1010

3

363

362

RC

LC

sradRC

/10001

at 0dB Crosses 20dB/dec.

40dB/dec- of asymptotefrequency High


rad/s 1000 :frequencyer Break/corn

2

4/)/( 20

2

LRLRLO

2

4/)/( 20

2

LRLRHI

10001

0 LC

srad /618

srad /1618

decdB /40

ACTIVE FILTERS

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 arezero

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

Basic Non-inverting amplifier

1V

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

Operational Transductance Amplifier

0RRin :OTA Ideal

Comparison of Op-Amp and OTA

Amplifier Type Ideal Rin Ideal Ro Ideal Gain Input Current input VoltageOp-Amp 0 0 0

OTA gm 0 nonzero

Basic Op-Amp Circuit Basic OTA Circuit

inS

inv

L

L

S

SinS

inin

invL

L

RR

RA

RR

R

V

vA

VRR

Rv

vARR

Rv

0

0

00

inS

inm

Linm

SinS

inin

inmL

RR

Rg

RR

R

v

iG

VRR

Rv

vgRR

Ri

0

00

0

00

vAA

Amp-Op Idealmm gG

OTA Ideal

Basic OTA Circuits

)0()(1

00

00

10

vdxxiC

v

vgit

m

)0()()( 00

10 vdxxvC

gtv

tm

Integrator

In the frequency domain

10 VCj

gV m

00

0

ii

vgi

in

inm polarity) (notice

meq

in

in

gR

i

v 1

Resistor Simulated

Basic OTA Adder

11vgm

22vgm

21 21vgvg mm

Resistor Simulated)(1

213

0 21vgvg

gv mm

m

Equivalent representation

mg ofility Programmab

)10

10.,.(

10

7

mSgge

g

mSg

m

m

m

decades 7 - 3 :range

valuesTypical

ABCm IA

Sg

201

LEARNING EXAMPLE

ABCm

m

m

Ig

SmS

g

mSg

20

10410

4

4

74

resistor a Produce k25

SSgg mm

753 1041041

1025

meq

in

in

gR

i

v 1

Resistor Simulated

)(20104 5 AIA

SS ABC

AAIABC 2102 6

LEARNING EXAMPLE Floating simulated resistor

1101 vgi m 1202 vgi m

inmvgi 0

One grounded terminal

011 ii 102 ii

21 mm gg

operationproper For

ABCm

m

m

Ig

SmS

g

mSg

20

10410

4

4

74

resistor 10M a Produce

Sgm7

6 101010

1

S7104

The resistor cannot be producedwith this OTA!

LEARNING EXAMPLE

210

210

321

210

210

,,,

vvv

vvv

ggg mmm

b)

a)

produce to Select

ABCm

m

m

Ig

SmS

g

mSg

20

10410

4

4

74

)(1

213

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 4

33 AImSg ABCm A5

AImSg ABCm 102.0 22

AImSg ABCm 501 11

Case b

Reverse polarity of v2!

OTA-C CIRCUITS

Circuits created using capacitors, simulated resistors, adders and integrators

integrator

resistor

Frequency domain analysis assumingideal OTAs

1101 im VgI 0202 VgI m

0201 IIIC CICjV

1

0

021101

VgVgCj

V mim

1

2

2

1

01

i

m

m

m

V

gCj

gg

V

Magnitude Bode plot

1

0

iv V

VG

2

1

m

mdc g

gA

C

gf

C

g

mC

mC

2

2

2

LEARNING EXAMPLE

)10(21

4

51

0

jV

VG

iv

:Desired

ABCm

m

m

Ig

SmS

g

mSg

20

1010

1

1

63

4dcA kHzfCC 100)10(2 5

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 ABCm 14.38.62 11

AIABC 785.020

7.152

biases and ncestransducta the Find

TOW-THOMAS OTA-C BIQUAD FILTER biquad ~ biquadratic

20

02

20

)()(

)()(

jQ

j

CjBjA

V

V

i

Cj

VVgV im

021101

)( 201202 im VVgI )( 23303 oim VVgI

)(1

03022

02 IICj

V

03i

)( unknownsfour and equationsFour 02010201 ,,, IIVV

1)()(12

132

21

21

32

321

2

3

2

2

01

jggCg

jggCC

Vgg

VVgg

gCj

V

mm

m

mm

im

mii

m

m

m

1)()(12

132

21

21

321

312

1

11

02

jggCg

jggCC

VgggCj

VgCj

V

V

mm

m

mm

imm

mi

mi

1

22

3

21

2

30

21

210 ,,

C

C

g

ggQ

C

g

QCC

gg

m

mmmmm

Filter Type A B CLow-pass 0 0 nonzeroBand-pass 0 nonzero 0High-pass nonzero 0 0

C

gBW

g

gQ

Cg

CC

gg

m

m

m

m

mm

3

3

0

21

21

LEARNING EXAMPLE

ABCm

m

m

Ig

SmS

g

mSg

20

10410

4

4

74

5.- gainfrequency center and 75kHz, of bandwidth

500kHz, offrequency center withfilter pass-band a Design

1)()(12

132

21

21

321

312

1

11

02

jggCg

jggCC

VgggCj

VgCj

V

V

mm

m

mm

imm

mi

mi

1

22

3

21

2

30

21

210 ,,

C

C

g

ggQ

C

g

QCC

gg

m

mmmmm

capacitors pF-50 and ionconfigurat Thomas-Tow the Use

BW

03 iV

3

20 |)(|

m

mv g

gjG

0)(1 2

21

210

j

gg

CC

mm

21 CC

Sg

BW m 56.23107521050

312

3

Sgm 8.1175 2

213

61252

0)105(

108.1171052

mg Sgm 5.2091 AI

AI

AI

ABC

ABC

ABC

18.1

89.5

47.10

3

2

1

Bode plots for resulting amplifier

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

Thevenin equivalent for AC/DCconverter

Using a capacitor to create a low-pass filter

Design criterion: place the corner frequencyat least a decade lower

THTH

OF VCRj

V

1

1

21

||||

CR

VV

TH

THOF

CRTHC

1

||1.0|| THOF VV

FCC

05.5362

1500

Filtered output

LEARNING EXAMPLESingle stage tuned transistor amplifier

Select the capacitor for maximumgain at 91.1MHz

AntennaVoltage

Transistor Parallel resonant circuit

CjLjR

V

V

A 1||||

1000

40

LCRCj

j

Cj

V

V

Cj

Cj

CjLjR

A1

)(

/

1000

4

/

/11

1

1000

4

2

0

LC/1frequency center with pass-Band

C6

6

10

1101.912 pFC 05.3

1001000

410

R

LCV

V

A

AVV0for plot Bode Magnitude

LEARNING BY DESIGN Anti-aliasing filter

Nyquist CriterionWhen 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 higherthan 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

in

1

101

050,2221

RCC

kRnFC 18.721

Resulting magnitude Bode plot

Attenuationin audio range

Improved anti-aliasing filter Two-stage buffered filter

01v

RCjV

V

1

1

01

02

RCjV

V

in

1

101

One-stage

Two-stage

Four-stage

nin

n

RCjV

V

1

10

stage-n

mHL

FC

704.0

10

Magnitude Bode plot

sCsLRR

R

V

V

tapeamp

amp

tape

amp

/1||

1

1

2

2

tapeamp

tapeamp

amp

tape

amp

RRL

sLCs

LCs

RR

R

V

V

LC

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

LEARNING BY DESIGNNotch filter to eliminate 60Hz hum

Notch filter characteristic

Filters

variable-frequency response analysis network performance as function of frequency

Documents

function of frequency

basic componentsmoreover

basic termsdisplay

rational function

bynetwork function history

dependent sources

mesh analysis

transfer functionsif