Variable frequency networks - basic engineering circuit analysis Irwin

8/10/2019 Variable frequency networks - basic engineering circuit analysis Irwin

1/51

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

Filter NetworksNetworks with frequency selective characteristics:

low-pass, high-pass, band-pass

VARIABLE-FREQUENCY NETWORK

PERFORMANCE

LEARNING GOALS


2/51

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

0RRZRResistor


3/51

90LLjZL Inductor


4/51

Capacitor 9011

CCjZc


5/51

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"


6/51

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


7/51

LEARNING EXAMPLE A possible stereo amplifier

Desired frequency characteristic

(flat between 50Hz and 15KHz)

Postulated amplifier

Log frequency scale


8/51

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


9/51

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

admittanceTransfer

tanceTransadmit

)(

)()(

1

2

sV

sIsYT

gain Voltage)(

)(

)(1

2

sV

sV

sGv

To compute the transfer functions one must solve

the circuit. Any valid technique is acceptable

EXAMPLE


10/51

LEARNING EXAMPLE

admittanceTransfer

tanceTransadmit

)(

)()(

1

2

sV

sIsYT

gain Voltage

)(

)()(

1

2

sV

sVsGv

The textbook uses mesh analysis. We will

use Thevenins theorem

sLRsC

sZT H ||1

)( 11

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 2 22

1 1

( ) ( )( ) ( )

( ) ( )v T

V s R I s G s R Y s

V s V s

)(

)(

1

1

RsLsC

RsLsC


11/51

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 a

product 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


12/51

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

Variable

Frequency

Response


13/51

SINUSOIDAL FREQUENCY ANALYSIS

)(sH

Circuit represented by

network function

)cos(0

)(0

tB

eA tj

)(cos|)(|

)(

0

)(0

jHtjHB

ejHA tj

)()()(

)()(

|)(|)(

jeMjH

jH

jHM

Notation

stics.characteriphaseandmagnitudecalledgenerallyareoffunctionasofPlots ),(),(M

)(log)(

))(log2010

10

PLOTSBODE vs

(M

.offunctionaasfunctionnetworkthe

analyzewefrequencytheoffunctionaasnetworkaofbehaviorthestudyTo

)(jH


14/51

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


15/51

]...)()(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


16/51

Constant Term

Poles/Zeros at the origin

90)(

)(log20|)(|

)(

10

Nj

Nj

j NdB

NN

linestraightaisthis

logisaxis-xthe 10


17/51

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.


18/51

Simple zero

Simple pole

2 2


19/51

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

Magnitude for quadratic pole Phase for quadratic pole

dB/dec40

These graphs are inverted for a zero


20/51

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


21/51

asymptotes

G t it d d h l t


22/51

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


23/51

21

020 0)(

KjK

dB

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

dec

dB40

dec

dB20

dec

dB40

LEARNING EXTENSION Sk t h th it d h t i ti


24/51

LEARNING EXTENSION Sketch the magnitude characteristic

)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

26 |dB

LEARNING EXTENSION Sk t h th it d h t i ti


25/51


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



26/51

Put in standard form

)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

DETERMINING THE TRANSFER FUNCTION FROM THE BODE PLOT


27/51

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


28/51

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


29/51

RESONANT CIRCUITS

These are circuits with very special frequency characteristics

And resonance is a very important physical phenomenon

L CCL

110

whenzeroiscircuiteachofreactanceThe

The frequency at which the circuit becomes purely resistive is called

the resonance frequency

CjLjRjZ

1)(

circuitRLCSeries

LjCjGjY

1)(

circuitRLCParallel

P i f i i


30/51

Properties of resonant circuits

At resonance the impedance/admittance is minimal

Current through the serial circuit/

voltage across the parallel circuit can

become very large

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

P ti f t i it


31/51

Properties of resonant circuits

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

1GV 1CVj

L

Vj

1

LEARNING EXAMPLE Determine the resonant frequency the voltage across each


32/51

LEARNING EXAMPLE Determine the resonant frequency, the voltage across each

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( 330 L

)(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 EXTENSION Fi d th l f C th t ill l th i it i


33/51

LEARNING EXTENSION 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

1)( M


34/51

Resonance for the series circuit

222)

1(||

1)(

C

LRZ

CjLjRjZ

QR

CQRL 1

, 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

M

(tan)( 0

0

1

Q

QBW

0

12

1

2

1 2

0 QQL O

sfrequenciepowerHalf

Z

RGv

LEARNING EXAMPLE D t i th t f lit f t d


35/51

LEARNING EXAMPLE

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

LEARNING EXTENSION A series RLC circuit as the following properties:


36/51

LEARNING EXTENSION A series RLC circuit as the following properties:

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

Determine the values of L,C.

CRR

L

Q 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

C 6

620

20

1056.11016104

111

PROPERTIES OF RESONANT CIRCUITS: VOLTAGE ACROSS CAPACITOR


37/51

PROPERTIES OF RESONANT CIRCUITS: VOLTAGE ACROSS CAPACITOR

|||| 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


38/51

LEARNING EXAMPLE

mH50

F5

, max0

sr adLC

/2000)105)(105(

11620

CRR

L

Q 0

0 1

LC

1

0

20

maxmax

2

11

Qu

RQ

050.02000 2max 2

112000Q

R Q Wmax

50 2 18711 100 2000

Evaluated with EXCEL and rounded to zero decimals

Using MATLAB one can display the frequency response


39/51

R=50Low Q

Poor selectivity

R=1

High Q

Good selectivity

FILTER NETWORKS


40/51

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 on

PASSIVE filters

Simple low-pass filter


41/51

Simple low-pass filter

RCjCj

R

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


42/51

Simple high-pass filter

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

1L O

Simple band-pass filter


43/51

p p

Band-pass

CLjR

R

V

VGv

11

0

222 1)(

LCRC

RCM

11

LC

M 0)()0( MM

2

4/)/( 202

LRLRL O

LC

10

2

4/)/( 202 LRLR

H I

L

RBW L OH I

)(2

1)( H IL O MM

Simple band reject filter


44/51

Simple band-reject filter

0110

00

CLj

LC

10 VV circuitopenasactscapacitorthe0at

10 VV circuitopenasactsinductortheat

filterpass-band

theinasdeterminedareH IL O ,

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


45/51

can 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

Cj

V

V

S

C

1

1

0)(,10 S

C

S

C

V

V

V

VLow-pass

FCHLR 159,159,10 forplotBode

ACTIVE FILTERS


46/51

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


47/51

Basic Inverting Amplifier

0V

VVgainInfinite

0V

0 II-impedanceinputInfinite

1

1 2

0OVV

Z Z

21

1

O

ZV V

Z

Linear circuit equivalent

0

I

1

2

Z

ZG

1

11

Z

VI

EX MPLE

USING INVERTING MPLIFIER


48/51

LOW P SS FILTER

Basic Non-inverting amplifier


49/51

Basic Non-inverting amplifier

1V

1V

0I

1

1

2

10Z

V

Z

VV

1

1

120 V

Z

ZZV

1

21Z

ZG

01I

EX MPLE

USING NON INVERTING CONFIGUR TION


50/51

EX MPLE SECON ORDER FILTER


51/51

2( )V s

22 2 2

1 1 2 3

01/

OI NV VV V V V

R C s R R

2

2 2

01/

OVV

R C s

Documents

Variable frequency networks - basic engineering circuit analysis Irwin