l11.hegde.unlocked

Transfer  Func,ons  and  Frequency  Response                                                                              Part  2  

ESC201A  :    Introduc,on  to  Electronics  

rhegde  Dept.  of  Electrical  Engineering  

IIT  Kanpur  

Sketching of Transfer function: Bode Magnitude Plot

3 4

10 1 ( )1 1

10 10

Hj j

ωω ω

= ×+ +

ω

( ) ( )H dBω

20

310

-20dB/decade

410

-20dB/decade

-40dB/decade

2 210 10 103 4 20Log ( ( ) ) 20 20 (1 ( ) ) 20 (1 ( ) )

10 10H Log Logω ωω = − + − +

-20dB/decade

2  

Sketching of Transfer function Bode Magnitude Plot

3 4 5

10 1 1 ( )1 1 1

10 10 10

Hj j j

ωω ω ω

= × ×+ + +

ω

( ) ( )H dBω

20

310-20dB/decade

410

-20dB/decade

-40dB/decade

2 2 210 10 10 103 4 5 20Log ( ( ) ) 20 20 (1 ( ) ) 20 (1 ( ) ) 20 (1 ( ) )

10 10 10H Log Log Logω ω ωω = − + − + − +

-60dB/decade

510-20dB/decade

3  

( ) ( )NH jω ω=

10 10 20Log ( ( ) ) 20 ( )H N Logω ω= ×

1

20N dB/decade

( ) ( )H dBω

ω

4  

Bode  Magnitude  Plot  

Determine transfer function?

vS

vO(t)

R

C ( ) ( )( )

O

S

VHV

ωω

ω=

( )1j CRHj CRω

ωω

=+

1 j Cω

3

3

( / ) ( )1 ( / )

dB

dB

jHjω ω

ωω ω

=+ 3dB 3dB

1 1= ; f =RC 2 RC

ωπ

210 10 10

3 3

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

H log logω ωω

ω ω= − +

5  

ω3dB

20 dB/decade

( ) ( )H dBω

ω

210 10 10

3 3

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

H log logω ωω

ω ω= − +

-20 dB/decade

High Pass Filter 6  

Bode  Magnitude  Plot  

33dB

1 f 102

HzRCπ

= =

High Pass Filter 7  

Bode Plot segments

31 2

1 1 ( ) ( / ) {1 ( / )}( / ) {1 ( / )}

n to r mH K j j

j jω ω ω ω ω

ω ω ω ω= × × × +

+

( ) ( )H dBω

ω

20log(K)

20n dB/decade

oω2ω

-20m dB/decade

-20r dB/decade

1ω

20t dB/decade 3ω

8  

Example: 1 1 ( ) 20010 100

H jj j

ω ωω ω

= × × ×+ +

1 1 ( ) 0.21 1

10 100

H jj j

ω ωω ω

= × × ×+ +

( ) ( )H dBω

ω-14

10

-20 dB/decade

100

-20 dB/decade

-20dB/decade

1 20 dB/decade

9  

0.1+6 dB

-34

Filter -pass a band of frequency and reject the remaining

f

f f

f

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

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

Low pass High pass

Band pass Band Stop

10  

R =1K

C 1µFvS

vO

vS

vO(t)

R

C

11  

R =1K

C 1µFvS

vO

3dB Frequency of single capacitor filters

33dB

1= 10 /RC

rad sω =

C 1µFvS

vO

R1 =1K

R2 =1K

3dB1 2

1=R R C

ω

CLinear Circuit 3dB

eq

1 1=R C

ωτ=

12  

One can often tell the type of filter by looking at behavior at very low and very high frequencies and keeping in mind that capacitor offers very high impedance at low frequencies and very low impedance at high frequencies.

Bandpass filter

Low f High f

Vo ~0 Vo ~0

13  

Bandpass Filter

f2 f1>f2

vS R1

C1

C2

vO(t)R2

2 11 1 2 2

1 1 f ; f2 2RC R Cπ π

≅ ≅14  

Example: Band Pass filter

15  

Bandstop Filter

f1 f2>f1

Will this work?

16  

What does this circuit do?

Low f High f

Vo ~Vin Vo ~Vin 17  

Bandstop or Notch Filter

What does this circuit do?

18  

R-L Circuits (Filters)

R

vS

vO(t)

L

( ) ( )( )

O

S

VHV

ωω

ω=

3

3

( / ) ( )1 ( / )

dB

dB

jj LHR j L j

ω ωωω

ω ω ω= =

+ +j Lω

3dBRL

ω =

High pass filter

19  

R-L Circuits

vS

vO(t)L

R

( ) ( )( )

O

S

VHV

ωω

ω=

3

1 ( )1 ( / )dB

RHR j L j

ωω ω ω

= =+ +

j Lω

3dBRL

ω =

Low pass filter

20  

Amplitude Modulated (AM) Radio

Different radio channels are separated by very narrow frequency interval.

For example, one may want to receive a 450KHz signal but reject 460KHz or 440KHz

( ) ( )H dBω

ω

450KHz

460KHz

-60dB

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

2460log( ) 10450

decades−≅

21  

Resonance

Galloping Gertie, the first Tacoma Narrows Bridge in Tacoma Washington, United States. The bridge opened on July 1, 1940 and from the start became notorious for its movement during windy days, earning the nickname "Galloping Gertie". The wind-induced collapse occurred on November 7, 1940, due partially to a physical phenomenon known as mechanical resonance…..wikepedia

22  

Nuclear magnetic resonance

23  

Resonance

A small disturbance leads to oscillatory behavior 24  

T = 1.1µs 25  

T = 0.9µs 26  

T = 1µs

The amplitude is 10 times larger even though input magnitude is same ! 27  

Series Resonant Circuit

vS Ci(t)

LR

Resonance is a condition in which capacitive and inductive reactance cancel each other to give rise to a purely resistive circuit

1eqZ R j L j

Cω

ω= + −

Resonant frequency: 1 10O OO

j L jC LC

ω ωω

− = ⇒ =

12Of LCπ

=eqZ R=

Current and voltage are in phase (power factor is unity) and current is maximum ! 28  

vS Ci(t)

LR

2 2

( )1( )

mVIR L

C

ωω

ω

=

+ −

ω1 and ω2 are called half-power frequencies since P ∝ I2

29  

12 2

11

( )1 2( )

m mV VIRR L

C

ωω

ω

= =

+ −

22 2

22

( )1 2( )

m mV VIRR L

C

ωω

ω

= =

+ −ω1 and ω2 are called half-power frequencies since P ∝ I2

30  

2 2

( )1( )

mVIR L

C

ωω

ω

=

+ −

1 2Oω ωω=2 1

RBL

ω ω= − =

Quality (Q) factor: Sharpness of resonance Peak Stored Energy2

Energy dissipated in one period at resonanceQ π=

2

2

122 12

mO

m O

L I LQRI R T

ωπ

×= × =

×

1 1O

O

QCRLC

ωω

= ⇒ =

31  

vS C

LR

vS C

LR0.1K j0.9K

-j1.1K Z=0.1K-j0.2K

j1K

-j1K

0.1K

Z=0.1K

Not very large change in impedance as we approach resonance ! 32  

Impedance is in kΩ

Impedance is in kΩ

vS C

LR

vS C

LR0.1K j0.9meg

-j1.1meg

Z=0.1K-j0.2meg

j1meg

-j1meg

0.1K Z=0.1K

very large change in impedance as we approach resonance ! Implying high quality factor 33  

Impedance is in MΩ

Impedance is in kΩ

Quality factor Q

vS C

LR j1K

-j1K

0.1K

Z=0.1K

vS C

LR j1meg

-j1meg

0.1K

Z=0.1K

1O OL CQ or QR R

ω ω= =

34  

2 1RBL

ω ω= − =OLQR

ω=

O OQBω ω

ω= =

ΔHence Q represents sharpness of resonance

For high Q circuits: 35  

R-L-C filters

vS C

vO(t)LR

j Lω

j Cω−vS

vO(t)R

C

Lj Lω

j Cω−

36  

vO(t)L

R

C

vSj Lωj Cω−

vS

RvO(t)

L

C

j Lω

j Cω−

37  

vO(t)L

R

C

vSj Lωj Cω−

How much Q do we need to pass 450KHz but reject 460KHz by 60dB?

2 2

( )( )( ) 1( )

O

IN

V RHV

R LC

ωω

ωω

ω

= =

+ −

Assuming VIN = 1V and noting that Q = ωOL/R

22 2

2

1( )1 ( 1)

O

O

VQ

ωωω

=

+ −

For ω=ωO, VO = 1 so the signal simply passes through !

3 62 450 10 2.8 10 /O rad sω π= × × × = ×38  

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

Such a large value of Q is not easy to get !

22 2

2

1( )1 (1 )

O

O

VQ

ωωω

=

+ −

sradsrad

/1089.2104602

/10827.210450263

630

×=××=

×=××=

πω

πω

39  

Example: for Q = 104 at 450KHz

3Suppose 10 ; 0.28 ; 125L H R C pF−= ⇒ = Ω ⇒ =oLQRω

=

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

Parallel Resonance

C LIM 0 R

C LvS

R

1 1eqY j C j

R Lω

ω= + −

Resonant frequency: 1 10O OO

j C jL LC

ω ωω

− = ⇒ =

12Of LCπ

=eqZ R=

40  

LIM 0 R+V- C

2 22

2

( )11 ( )

mI RVR C LL C

ω

ωω

=

+ −

41  

2 22

2

( )11 ( )

mI RVR C LL C

ω

ωω

=

+ −

For high Q: 42  

What is the resonant frequency ?

43  2o

Ofωπ

=