L06_kvs_transfer_function_full.pdf

8/14/2019 L06_kvs_transfer_function_full.pdf

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ESc201: IntroductiontoElectronicsFrequencyDomainResponse

r. . . r vas avaDept. of Electrical Engineering

IIT Kanpur

1


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Time domain vs. Frequency domain analysis

Signal

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signalSpectrum Analyzer

spectrum

f

0.1KHz 1KHz 10KHz

signal spectrum

Fourier Analysis(Mathematical tool)

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Speech signal

4


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System What does this circuit do ?

5V

0V

5


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Suppose the capacitor is reduced to ~21pF.

It is hard to find out what impact the change in capacitor has on circuit behavior6


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Frequency domain analysisVO

7


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Measure response at many different frequencies for a constant input amplitude

f=1KHz

f=10KHz 8


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Measure response at many different frequencies for a constant input amplitude

f=100KHz

f=1000KHz

9


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Plot the amplitude and phase as a function of frequency

Amplitude as a function of frequency

One can clearly see the frequency selective (often called a filter) nature of the

c rcu

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Phase as a function of frequency

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Suppose the capacitor is reduced to ~21pF.

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Anal sis of si nals and s stems in fre uenc domain

often provides useful insight into their behavior.

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Frequency domain analysis

Transfer function is a useful tool for finding the frequency

response of a system

Linear System

Y

asor or asor or

rans er unct on: ( )X =

Transfer function has a magnitude and a phase14


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( )

Y

H

= ( )X

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Because of the wide dynamic range of frequency, plotting

frequency on log axis is often more revealing !

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Logarithmic frequency scale

1 decade1 octave

f (Hz)1041031021011 50

2 decades

210

1

No. of decades = log ( )f

210

flog ( )

2 12

1 10

No. of octaves = log ( ) =f log (2)

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Decibel scale often reveals more information about

behavior

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The magnitude of transfer function is often specified in decibels

210

1

10 log ( )dBGP

=

Because power is proportional to V2 or I2, voltage gain and

current ain in decibels is s ecified as

220lo VG = 2

1020log ( )dBI

G =1V 1

Decibel scale is more convenient for our perception of hearing

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Decibel Scale

10

1000 60

10 20

2 31 0

1/2 3

0.5

6

0.1 20

0.01 4020


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dB Scale

10.1

0.01

0.001

A plot of the decibel magnitude of transfer function versus

requency us ng a ogar m c sca e or requency s ca e a

Bode plot 21


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= 10

)(2

jjH

+

+

50001

1001

500

400

300

200

100

0 2 4 6 8 10

x 104 (rad/s) 22


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= 10

)(2

jjH

+

+

50001

1001

300

200

|H|

100

101

102

103

104

105

0

(rad/s)

23


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= 10

)(2

jjH

60 +

+

5000100

40

20

|(

dB)

0|H

-

-

101

102

103

104

105

(rad/s) 24


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How to determine the transfer function?

VO

VS 0 1/jC

)( 0

VjH = jH =

1)(

11

=

( )2

1 RC+25


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21

1)(

RCjH

+=

Plot Magnitude

( )210 1log20 RCH dB +=

21

3

10 1log20

+=dB

dBH

RC

dB3 =

dB3


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dB3

>>

dB

H10

log20

dB3

H3dB 0

3dB

100

3dB

40

-

110lo =

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RC

dB

13 =

3dB point

10HdB

corner frequency or half

power frequency

0 HdB 0

-10 103dB 20

1003dB 40

-20

-30 2

3

10 1log20

+=dB

dBH

101 102 103 104 105-40

(rad/s) 28


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Example10 3

3101 + j2

3dB

310 101log2020

+=

dBH

3

10

3

log20:10

>>

HdB

20

310-20dB/decade

410

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( )

1

H

j CR

=

+

0.1KHz 10KHz

3

3

( )1 10

110

Hj

j

= =+ +

4( ) 1 (100 ) 0.1 (10 )O

V t Sin t Sin t = +

10

-20dB/decade

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-20dB/decade

-40dB/decade

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Adding more RC stages, makes the characteristics sharper

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Sk t hi f T f f ti 10 1


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Sketching of Transfer function:

Bode Magnitude Plot10 1

( )

1 1

H

=

+ +10 10

2 2 20Lo 20 20 1 20 1H Lo Lo

= + +10 10

3

410-20dB/decade

-20dB/decade

-20dB/decade

-40dB/decade

34

Sketching of Transfer function


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Sketching of Transfer function

Bode Magnitude Plot 10 1 1

3 4 5

1 1 1

10 10 10j j j

+ + +

2 2 2

10 10 10 103 4 520Log ( ( ) ) 20 20 (1 ( ) ) 20 (1 ( ) ) 20 (1 ( ) )

10 10 10H Log Log Log

= + + +

( ) ( )H dB

3

410510

-20dB/decade

-20dB/decade-20dB/decade

-40dB/decade

-60dB/decade35


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( ) ( )NH j =Bode Magnitude Plot

10 1020Log ( ( ) ) 20 ( )H N Log =

( ) ( )H dB

1

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D t i t f f ti ?


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Determine transfer function?

1

vO(t) ( )

( )

O

S

HV

= j C

vS R ( )1

j CRHj CR

= +

3( / )dBj 1 13

1 ( / )

dBj + 3dB 3dBRC 2 RC

210 10 10

3 3

20Log ( ( ) ) 20 ( ) 20 (1 ( ) )dB dB

H log log

= +

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Bode Magnitude Plot

2

10 10 10

3 3

20Log ( ( ) ) 20 ( ) 20 (1 ( ) )dB dB

H log log

= +

20 dB/decade( ) ( )H dB

3dB

-20 dB/decade

High Pass Filter

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313dB

2 RC= =

High Pass Filter 39

Bode Plot segments


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Bode Plot segments

1 1n t

= 31 2 ( / ) {1 ( / )}o r mj j +

20log(K)

20t dB/decade

3

o 1

20n dB/decade-

-20r dB/decade

40

Example: 1 1


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Example: 1 1 ( ) 200

10 100H j

=

+ +

1 1 ( ) 0.2H j

=

10 100

+ +

( ) ( )H dB 10 10020 dB/decade +6 dB

-14 -20 dB/decade

.

-

-20dB/decade

-34

41


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Filter -pass a band of frequency and reject the remaining

|H(f)| H fLow pass High pass

ff

|H(f)||H(f)|

Band Stop

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C 1 FvS

vO

vO(t)

C

vS R

43

3dB Frequency of single capacitor fil ters


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=

3dB Frequency of single capacitor fil ters

v

vO 33dB

1= 10 /

RCrad s =

v

vO

1

3dB

1 2

=

R R C

2

CLinear Circuit3dB

eq

1 1=R C

=

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Bandpass Filter


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p

>

C1 R2

vS C2

vO(t)

2 1

1 1 2 2

f ; f

2 2R C R C

46

Example: Band Pass filter


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p

47

Bandstop Filter


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p

f1 f2>f1

48

What does this circuit do?


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Low f

High f

Vo ~Vin Vo ~Vin 49

What does this circuit do?


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Bandstop or Notch Filter

50

R-L Circuits (Filters) ( )V


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( ) ( ) O

VH

=

vO(t)S

3( / )

dBjj L

H

= =vS

31 ( / )

dBR j L j + +

R3dB

L=

High pass filter

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R-L Circuits ( )( ) O

VH


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( ) OHV

=j L

vS

vO(t)L

R1

( ) R

H = =3dB

R =

L

Low pass filter

52

Amplitude Modulated (AM) Radio


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.

For example, one may want to receive a 450KHz signal but reject 460KHz or

z

( ) ( )H dB 450KHz

460KHz

-60dB

2460log( ) 10 decades

This implies an attenuation of -6000 dB/decade !!

53

Resonance


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,

Washington, United States. The bridge opened on July 1, 1940 and fromthe start became notorious for its movement during windy days, earning

the nickname "Gallo in Gertie". The wind-induced colla se occurred on

November 7, 1940, due partially to a physical phenomenon known

as mechanical resonance..wikepedia54


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Nuclear ma netic resonance

55

Resonance


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A small disturbance leads to oscillatory behavior 56


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T = 1.1s57


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T = 0.9s58


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= s

The amplitude is 10 times larger even though input magnitude is same !59

Series Resonant Circuit

R i diti i hi h iti d i d ti


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Resonance is a condition in which capacitive and inductive

reactance cancel each other to give rise to a purely resistivecircuit

1eqZ R j L j C = +

vS C

Resonant frequency: 1 1

0O OO

j L jC LC

= =

12

Of

LC=

eqZ R=

Current and voltage are in phase (power factor is unity) and

current is maximum ! 60

LR( ) m

VI =


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( )I

vS Ci(t) ( )R L C +

1 and 2 are called half-power frequenciessince P I2

61

( )1

mV

I =


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2 21

C

1( )

1 2

m mV V

I

R

= =

1

1C

+

2

2 2

( )1 2

m mV VIR

= =

2C1 2 -

power frequencies since P I2

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= 1 2O2 1B

L = =

ua y ac or: arpness o resonance

Peak Stored Energy2Q =

21m

L IL

1 1

21

2 m O

RI R T

= =

O

OCRLC

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LR. j0.9K

vS C

-j1.1K

Z=0.1K-j0.2K

LR j1K0.1K

vS C -j1K Z=0.1KImpedance is in k

Not very large change in impedance as we approach resonance !64


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LR. Z=0.1K-j0.2meg

Im edance is in MvS C

-j1.1meg

LR j1meg0.1K = .

Impedance is in k

vS C -j1meg

very large change in impedance as we approach resonance !

Implying high quality factor 65

Quality factor Q


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LR j1K.

vS C -j1K Z=0.1K

LR j1meg

0.1K

vS C -j1meg Z=0.1K

O OQ or Q

R R

= =66

RB O

L


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B = =O=LR

O OQ

B = =

For high Q circuits:

67

R-L-C filters


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vO(t)

LR

vO(t)

R

vS C

j C

vS L

j L

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LC


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vO

RvS

j Lj C

How much Q do we need to pass 450KHz but reject 460KHz by

60dB?( )

( ) OV R

H

V

= =

( )C

+

Assumin V = 1V and notin that Q = L/R

1V =

For =O, VO = 1 so the signal simply passesthrough !

2 2

21 ( 1)

O

Q

+ 3 62 450 10 2.8 10 /O

rad s = = 70

2

1( )OV =


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2 221 (1 )OQ +

srad

sra

/1089.2104602

.

63

0

==

==

For an attenuation of -60dB or 10-3 at : Q=23,000

Example: for Q = 104 at 450KHz

3

Suppose 10 ; 0.28 ; 125L H R C pF

= = =o

Q R

=

71

Suppose 0.1 ; 28 ; 1.25L H R C pF= = =

Parallel Resonance


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C LIM 0 R 1 1=

eqR L

L

Resonant frequency: 1 10O OO

j C jL LC

= =

2

Of

LC

=eqZ R=

72

L+ 2 2( ) m

I RV =


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L

M

- C 21 ( )LL C +

73

2 22

( )1

mI R

VR C

=


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2

2L C

For high Q:74

What is the resonant frequency ?


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752

oO

f

=

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