EE221 Circuits IIeebag/AlexanderCh14.pdf · 2010. 11. 16. · EE2003 Circuit Theory Author: EE...
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1 EE221 Circuits II Chapter 14 Frequency Response
Transcript of EE221 Circuits IIeebag/AlexanderCh14.pdf · 2010. 11. 16. · EE2003 Circuit Theory Author: EE...
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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
