lecture 11 ac circuits frequency response · resonance is a condition in an rlc circuit in which...
TRANSCRIPT
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Lecture 11 AC CircuitsFrequency Response
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Contents
• Introduction
• Transfer Function
• Series Resonance
• Parallel Resonance
• Passive 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.
Introduction
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Transfer Function• 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
)(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
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Transfer Function: Example
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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Solution (Example 1)
Solution:
The transfer function is
,
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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Transfer Function: Example
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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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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Solution (Example 2)
Solution:
The transfer function is
,
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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 resistive impedance.
ω Cω Lj R
1Z
Resonance frequency:
HzLC2
1f
rad/sLC
1
o
oro
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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 = 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.
ω Cω Lj RZ
1
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Series Resonance
Bandwidth B
The frequency response of the resonance circuit current is
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 occurs at resonance: R
V
2
1)(P
2
mo
ω Cω Lj RZ
1
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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.
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
R
V
2
1)(P
2
mo
Highest power
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Series Resonance: 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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Series Resonance
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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Parallel Resonance
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:
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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Parallel Resonance
Example 4
Calculate the resonant frequency of the circuit in the figure shown below.
rad/s2.1792
19Answer:
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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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Passive Filters
Example 5
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.
krad/s25Answer: