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EE221 Circuits II

Chapter 14Frequency Response

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Frequency ResponseChapter 14

14.1 Introduction14.2 Transfer Function14.3 Bode Plots14.4 Series Resonance14.5 Parallel Resonance14.6 Passive Filters14.7 Active filters

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What is Frequency Response 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

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14.2 Transfer Function

The transfer function H(ω) of a circuit is the frequency-dependent ratio of a phasor output Y(ω) (voltage or current ) to a phasor input X(ω)(voltage or current).

φωωωω ∠== |)(H|

)(X)(Y )(H

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14.2 Transfer Function

Four possible transfer functions:

)(V)(V gain Voltage )(H

i

o

ωωω ==

)(I)(I gain Current )(H

i

o

ωωω ==

)(I)(V ImpedanceTransfer )(H

i

o

ωωω ==

)(V)(I AdmittanceTransfer )(H

i

o

ωωω ==

φωωωω ∠== |)(H|

)(X)(Y )(H

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14.2 Transfer Function 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

Solution:

The transfer function is

,

The magnitude is 2)/(11)(H

oωωω

+=

The phase isoωωφ 1tan−−=

1/RC=oω

RC j11

C j1/ RCj

1

VV)(H

s

o

ωωωω

+=

+==

Low Pass FilterLow Pass Filter

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14.2 Transfer FunctionExample 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

Solution:

The transfer function is

,

The magnitude is 2)(1

1)(H

ωω

ωo+

=

The phase isoω

ωφ 1tan90 −−°∠=

R/L=oω

L jR1

1L jR

LjVV)(H

s

o

ωω

ωω+

=+

==

High Pass FilterHigh Pass Filter

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14.4 Bode Plots

Bode Plots are semilog plots of the magnitude (in dB) and phase (in deg.) of the transfer function versus frequency.

HHHeH

dB

j

10log20== φ

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Bode Plot of Gain K

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Bode Plot of a zero (jω)

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Bode plot of a zero

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Bode Plot of a quadratic pole

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Example 1

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Example 1

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Example 2

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14.4 Series Resonance

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

)C

1L ( jRZω

ω −+=

Resonance frequency:

HzLC2

1f

rad/sLC1

o π

ω

=

= oro

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14.4 Series Resonance

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 θ = 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 ( jRZω

ω −+=

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14.4 Series Resonance

Bandwidth B

The frequency response of the resonance circuit current is )

C 1L ( jRZ

ωω −+=

22m

)C /1L (RV|I|I

ωω −+==

The average power absorbed by the RLC circuit is

RI21)(P 2=ω

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

V21)(P

2m=oω

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14 4 Series Resonance

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.

4RV

R)2/(V

21)(P)(P

2m

2m

21 === ωω

LC1)

2LR(

2LR 2

1 ++−=ωLC1)

2LR(

2LR 2

2 ++=ω 21ωωω =o

Bandwidth BBandwidth B 12 B ωω −=

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14.4 Series Resonance

Quality factor, CR1

RL

resonanceat period onein circuit by the dissipatedEnergy circuit in the storedenergy Peak Q

o

o

ωω

===

• 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.

QLRB2ooωω===

The relationship between the B, Q and ωo:

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14.4 Series Resonance

Example 3

A series-connected circuit has R = 4 Ωand 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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14.5 Parallel Resonance

Resonance frequency:

Hzo LC21for rad/s

LC1

o πω ==

)L

1C ( jR1Y

ωω −+=

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:

14.5 Parallel Resonance

LC1

LC1

RC1or

RL

o

o

ωω RCor

LR

oo

ωω

Qoω

Qoω

2Q )

2Q1( 1 2 o

oωω ±+

2Q )

2Q1( 1 2 o

oωω ±+

2B

±oω 2B

±oω

characteristic Series circuit Parallel circuit

ωo

Q

B

ω1, ω2

Q ≥ 10, ω1, ω2

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14.5 Resonance Example 4

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

rad/s2.179219

==ωAnswerAnswer::

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14.6 Passive Filters • 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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Low Pass Filter (Passive)

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High Pass Filter (Passive)

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Band Pass Filter (Passive)

See Equation 14.33 for corner Frequencies ω1 and ω2

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Band Reject Filter (Passive)

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Low Pass Filter (Active)

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High Pass Filter (Active)

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Band Pass Filter (Active)

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Band Reject Filter (Active)

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Magnitude and Frequency Scaling

Example: 4th order low-pass filter

Corner Frequency: 1 rad/secResistance: 1Ω

Corner Frequency: 100π krad/secResistance: 10 kΩ