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Alexander-Sadiku Alexander-Sadiku Fundamentals of Electric Fundamentals of Electric

CircuitsCircuits

Chapter 14Chapter 14

Frequency ResponseFrequency Response

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

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Frequency ResponseFrequency ResponseChapter 14Chapter 14

14.1 Introduction14.2 Transfer Function14.3 Series Resonance14.4 Parallel Resonance14.5 Passive Filters

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What is FrequencyResponse of a Circuit?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)14.1 Introduction (1)

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14.2 Transfer Function (1)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)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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14.2 Transfer Function (3)14.2 Transfer Function (3)Example 1

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

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14.2 Transfer Function (4)14.2 Transfer Function (4)

Solution:

The transfer function is

,

The magnitude is 2)/(1

1)(H

o

The phase iso

1tan

1/RCo

RC j1

1

C j1/ RCj

1

V

V)(H

s

o

Low Pass FilterLow Pass Filter

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14.2 Transfer Function (5)14.2 Transfer Function (5)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)14.2 Transfer Function (6)

Solution:

The transfer function is

,

The magnitude is 2)(1

1)(H

o

The phase iso

1tan90

R/Lo

L jR

1

1

L jR

Lj

V

V)(H

s

o

High Pass FilterHigh Pass Filter

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14.3 Series Resonance (1)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 resistive impedance.

)C

1L ( jRZ

Resonance frequency:

HzLC2

1f

rad/sLC

1

o

oro

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14.3 Series Resonance (2)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 = 1;• The magnitude of the transfer function H(ω) = 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)14.3 Series Resonance (3)

Bandwidth B

The frequency response of the resonance circuit current is )

C

1L ( jRZ

22

m

)C /1L (R

V|I|I

The average power absorbed by the RLC circuit is

RI2

1)(P 2

The highest power dissipated The highest power dissipated occurs at resonance:occurs at resonance: R

V

2

1)(P

2mo

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14 3 Series Resonance (4)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

2m

2m

21

LC

1)

2L

R(

2L

R 21

LC

1)

2L

R(

2L

R 22 21 o

Bandwidth BBandwidth B12 B

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14.3 Series Resonance (5)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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14.3 Series Resonance (6)14.3 Series Resonance (6)

Example 3

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)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 zeroIt occurs when imaginary part of Y is zero

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Summary of series and parallel resonance circuits:Summary of series and parallel resonance circuits:

14.4 Parallel Resonance (2)14.4 Parallel Resonance (2)

LC

1

LC

1

RC

1or

R

L

o

o

ω

ωRCor

L

Ro

o

Qo

Qo

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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14.4 Parallel Resonance (3)14.4 Parallel Resonance (3)Example 4

Calculate the resonant frequency of the circuit in the figure shown below.

rad/s2.1792

19AnswerAnswer::

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14.5 Passive Filters (1)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

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14.5 Passive Filters (2)14.5 Passive Filters (2)

Example 5

For the circuit in the figure below, obtain the transfer function For the circuit in the figure below, obtain the transfer function

Vo(ω)/Vi(ω). Identify the type of filter the circuit represents Vo(ω)/Vi(ω). Identify the type of filter the circuit represents

and determine the corner frequency. Take R1=100and determine the corner frequency. Take R1=100=R2 =R2

and L =2mH.and L =2mH.

krad/s25AnswerAnswer::