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Circuit Theory

Chapter 14

Frequency Response

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

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Frequency ResponseChapter 14

14.1 Introduction

14.2 Transfer Function

14.3 Series Resonance

14.4 Parallel Resonance

14.5 Passive Filters

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What is FrequencyResponse of a Circuit?

It is the variation in a circuit’s

behavior with change in signal

frequency and may also be

considered as the variation of the gain

and phase with frequency.

14.1 Introduction (1)

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http://en.wikipedia.org/wiki/Tacoma_Narrows_Bridge_(1940)

http://lpsa.swarthmore.edu/Analogs/ElectricalMechanicalAnalogs.html

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14.2 Transfer Function (1)

• The transfer function H(ω) of a circuit is the

frequency-dependent ratio of a phasor output

Y(ω) (an element voltage or current ) to a phasor

input X(ω) (source voltage or current).

|)(H|

)(X

)(Y )(H

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14.2 Transfer Function (2)

• Four possible transfer functions:

)(V

)(V gain Voltage )(H

i

o

)(I

)(I gain Current )(H

i

o

)(I

)(V ImpedanceTransfer )(H

i

o

)(V

)(I AdmittanceTransfer )(H

i

o

|)(H|

)(X

)(Y )(H

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Example 1

For the RC circuit shown below, obtain the transfer function Vo/Vsand its frequency response. Let vs = Vmcosωt.

Solution:

The transfer function is

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,

The magnitude is2)/(1

1)(H

o

The phase iso

1tan

1/RCo

RC j1

1

C j1/ R

Cj

1

V

V)(H

s

o

Low Pass Filter

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Example 2

Obtain the transfer function Vo/Vs of the RL circuit shown below, assuming vs = Vmcosωt. Sketch its frequency response.

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14.2 Transfer Function (6)

Solution:

The transfer function is

,

The magnitude is2)(1

1)(H

o

The phase iso

1tan90

R/Lo

L j

R1

1

L jR

Lj

V

V)(H

s

o

High Pass Filter

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Ex3) Find the transfer function H(ω)=Vout/Vin. What type of filter is it?

11H(w)=1000/(jw+11000), Lowpass filter

Wolfram Alpha, add abs( ) for mag

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Plotting phase in Wolfram

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Magnitude of H(w)

Phase of H(w)

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Plot in Matlab:

w = [1:100:1e6];H = 1000./(j*w+11000)plot(w, abs(H) )

semilogx(w,abs(H)) magnitude

semilogx(w,angle(H)*180/pi) phase

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Ex 4) Use the Inverting Op-Amp formula to derive the transfer function. Find the cutoff frequency of the filter. Obtain the output voltage in s.s.s. when the input voltage is Vin=1cos(100t) V and

Vin=1cos(10,000t+90°) V.

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ωc=104 rad/s; vout, s.s.s=0.01cos(100t-90°)V, when vin=1cos(100t)V; vout, s.s.s=0.707cos(10,000t-45°)V, when vin=1cos(10,000t+90°)V

magnitude

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w = [1:100:1e6];H = 1./ (j*(10000./w) - 1);semilogx(w,abs(H))

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14.3 Series Resonance (1)Resonance is a condition in an RLC circuit in which the capacitive and inductive reactance are equal in magnitude, thereby resulting in purely resistiveimpedance.

Resonance frequency:

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14.3 Series Resonance (2)

The features of series resonance:

The impedance is purely resistive, Z = R;

• The supply voltage Vs and the current I are in phase, so

cos q = 1;

• The magnitude of the impedance Z(ω) is minimum;

• The inductor voltage and capacitor voltage can be much

more than the source voltage.

)C

1L ( jRZ

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14.3 Series Resonance (3)

Bandwidth B

The frequency response of the resonance circuit current is )

C

1L ( jRZ

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m

)C /1L (R

V|I|I

The average power absorbed by the RLC circuit is

RI2

1)(P 2

The highest power dissipated occurs at resonance: R

V

2

1)(P

2

mo

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14 3 Series Resonance (4)

Half-power frequencies ω1 and ω2 are frequencies at which the dissipated power is half the maximum value:

The half-power frequencies can be obtained by setting Z

equal to √2 R.

4R

V

R

)2/(V

2

1)(P)(P

2

m

2

m21

LC

1)

2L

R(

2L

R 2

1 LC

1)

2L

R(

2L

R 2

2 21 o

Bandwidth B12 B

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14.3 Series Resonance (5)

Quality factor, CR

1

R

L

resonanceat period onein

circuit by the dissipatedEnergy

circuit in the storedenergy Peak Q

o

o

• The quality factor is the ratio of its resonant frequency to its bandwidth.

• If the bandwidth is narrow, the quality factor of the resonant circuit must be high.

• If the band of frequencies is wide,the quality factor must be low.

The relationship between the B, Q and ωo:

CRQL

RB 2

oo

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Example 5

A series-connected circuit has R = 4 Ω

and L = 25 mH.

a. Calculate the value of C that will produce a quality factor of

50.

b. Find ω1 and ω2, and B.

c. Determine the average power dissipated at ω = ωo, ω1, ω2.

Take Vm= 100V.

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14.4 Parallel Resonance (1)

Resonance frequency:

HzoLC2

1for rad/s

LC

1o

)L

1C ( j

R

1Y

It occurs when imaginary part of Y is zero

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Summary of series and parallel resonance circuits:

14.4 Parallel Resonance (2)

LC

1

LC

1

RC

1or

R

L

o

o

ω

ωRCor

L

Ro

o

Q

o

Q

o

2Q )

2Q

1( 1 2 o

o

2Q )

2Q

1( 1 2 o

o

2

Bo

2

Bo

characteristic Series circuit Parallel circuit

ωo

Q

B

ω1, ω2

Q ≥ 10, ω1, ω2

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Example 6

A parallel resonant circuit has R = 100 kOhm, L = 20 mH, and C = 5 nF. Calculate wo, w1, w2, Q and B

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14.5 Passive Filters (1)

• A filter is a circuit that is designed to pass signals with desired frequencies and reject or attenuate others.

• Passive filter consists of only passive element R, L and C.

• There are four types of filters.

Low Pass

High Pass

Band Pass

Band Stop

Example 7

For the circuit in the figure below, obtain the transfer function Vo(ω)/Vi(ω).

Identify the type of filter the circuit represents and determine the corner

frequency. Take R1=100W =R2 and L =2mH.

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krad/s25Answer:

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EXAMPLE 8) a) Find the transfer function H(ω)=Vout/Vinb) At what frequency will the magnitude of H(ω) be maximum and what is the maximum value of the magnitude of H(ω)?c) At what frequency will the magnitude of H(ω) be minimum and what is the minimum value of the magnitude of H(ω)?

29a) H(ω)= -45455/(jω+4545.5)b) Hmax=10 as ω=0 rad/sc) Hmin=0 as ω goes to infinity.

HANDOUTS

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Example 1

For the RC circuit shown below, obtain the transfer function Vo/Vsand its frequency response. Let vs = Vmcosωt.

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Example 2

Obtain the transfer function Vo/Vs of the RL circuit shown below, assuming vs = Vmcosωt. Sketch its frequency response.

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Ex3) Find the transfer function H(ω)=Vout/Vin. What type of filter is it?

33H(w)=1000/(jw+11000), Lowpass filter

Magnitude of H(w)

Phase of H(w)

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Plot in Matlab:

w = [1:100:1e6];H = 1000./(j*w+11000)plot(w, abs(H) )

semilogx(w,abs(H)) magnitude

semilogx(w,angle(H)*180/pi) phase

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Ex 4)Find the cutoff frequency of the filter. Use voltage divider to derive the transfer function. Obtain the output voltage in s.s.s when the input voltage is Vin=1cos(100t) V and Vin=1cos(10,000t+90°) V.

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ωc=104 rad/s; vout, s.s.s=0.01cos(100t-90°)V, when vin=1cos(100t)V; vout, s.s.s=0.707cos(10,000t-45°)V, when vin=1cos(10,000t+90°)V

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Example 5

A series-connected circuit has R = 4 Ω

and L = 25 mH.

a. Calculate the value of C that will produce a quality factor of

50.

b. Find ω1 and ω2, and B.

c. Determine the average power dissipated at ω = ωo, ω1, ω2.

Take Vm= 100V.

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Example 7

For the circuit in the figure below, obtain the transfer function Vo(ω)/Vi(ω).

Identify the type of filter the circuit represents and determine the corner

frequency. Take R1=100W =R2 and L =2mH.

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krad/s25Answer:

EXAMPLE 8) a) Find the transfer function H(ω)=Vout/Vinb) At what frequency will the magnitude of H(ω) be maximum and what is the maximum value of the magnitude of H(ω)?c) At what frequency will the magnitude of H(ω) be minimum and what is the minimum value of the magnitude of H(ω)?

38a) H(ω)= -45455/(jω+4545.5)b) Hmax=10 as ω=0 rad/sc) Hmin=0 as ω goes to infinity.