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CHAPTER IV
FREQUENCY DEPENDENT CIRCUITS
1. RESONANT CIRCUITS
2. FILTERS
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Objectives
• To define the resonance phenomenon.
• To calculate the resonance frequency of series and parallel
circuits.
• To identify the half power points and to write an expression for
the circuit bandwidth.
• To define the quality factor of the series and parallel resonant
circuits.
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Resonance In Electric Circuits
• Any passive electric circuit will resonate if it has an
inductor and capacitor.
• Resonance is characterized by the input voltage and current
being in phase and the driving point impedance (or
admittance) is completely real when this condition exists.
• In this chapter only series and parallel resonance
circuits are considered. Multiple resonance circuits are
not covered.
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Resonance
• Resonant circuits (series or parallel) are useful for constructing
filters, as their transfer functions can be highly frequency selective.
• They are used in many applications such as selecting the desired
stations in radio and TV receivers.
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Series Resonance
• Consider the series RLC circuit shown below.
• The input impedance is given by:
1
)
wC
• The magnitude of the circuit current is;
Z  R  j ( wL 
I | I |
Vm
R 2  ( wL 
1 2
)
wC
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Series Resonance
• Resonance results when the imaginary part of the transfer
function is zero, or
• The value of ω that satisfies this condition is called the
resonant frequency ω0. Thus, the resonance condition is
• This is an important equation to remember. It applies to
both series and parallel resonant circuits.
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Important Notes
• The impedance is purely resistive, thus, Z = R. In other
words, the LC series combination acts like a short circuit, and
the entire voltage is across R.
• The voltage Vs and the current I are in phase, so that the
power factor is unity.
• The magnitude of the impedance Z(ω) is minimum.
• The inductor voltage and capacitor voltage can be much
more than the source voltage.
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BandWidth
• The frequency response of the circuit’s current magnitude
Half power point
BW = w2 – w1
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Half Power Points
• The average power dissipated by the RLC circuit is
• The highest power dissipated occurs at resonance, when
I = Vm/R,
.
so that
• At certain frequencies ω = ω1, ω2, the dissipated power is half the
maximum value; that is,
•
Hence, ω1 and ω2 are called the half-power frequencies.
The half-power frequencies are obtained by setting Z equal to√2R.
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• After some insightful algebra one will find two frequencies
at which the previous equation is satisfied, they are:
2
R
1
 R 
and
w1  
 


2L
 2 L  LC
2
R
1
 R 
w2 
 


2L
 2 L  LC
• The two half-power frequencies are related to the resonant
frequency by
wo  w1w2
• The bandwidth of the series resonant circuit is given by;
BW  wb  w2  w1 
R
L
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Quality Factor
The Q (quality factor) of the circuit is defined as;
Q = (Reactive power of L or C at resonance) / (Active power at resonance)
wo L
1
1 L
Q


 
R
wo RC R  C 
Using Q, we can write the bandwidth as;
BW 
wo
Q
Quality Factor
• 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.
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An Observation:
• By using Q = woL/R in the equations for w1and w2 we have;
2
 1



1
w1  wo 
 
 1

2Q 
 2Q




2
 1



1
w2  wo 
 
 1

 2Q

 2Q 


and
Also;
• If Q > 10, one can safely use the approximation;
BW
w1  wo 
2
and
BW
w2  wo 
2
• These are useful approximations.
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Parallel Resonance
Consider the circuits shown below:
V
I
R
L
1
1 
I  V   jwC 

R
jwL


C
L
R
V
C
I

1 
V  I  R  jwL 

jwC


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Duality Between Series and Parallel Resonance
1
1 
I  V   jwC 

R
jwL



1 
V  I  R  jwL 

jwC


We notice the above equations are the same provided:
I
V
R
L
1
R
C
If we make the inner-change, then one equation becomes the same as
the other. For such case, we say the one circuit is the dual of the other.
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Duality Between Series and Parallel Resonance
Series Resonance
Parallel Resonance
1
LC
wL
Q O
R
1
LC
w 
O
w 
O
Q  w RC
o
BW  ( w  w )  w 
1
BW  w 
RC
2
1
BW
BW
R
L
 1
1 
 1 
w ,w  
 
 

2
RC
2
RC
LC




 R
1 
 R
w ,w  
   

2
L
2
L
LC
 


1

 1 
w ,w  w   
 1
 2Q 
 2Q

1

 1 
w ,w  w   
 1
 2Q 
 2Q

2
1
2
2
1
2
o
2
1
2
2
1
2
o
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Example 1
• Determine the resonant frequency of the circuit shown
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Example 2:
Determine the resonant frequency for the circuit below.
R
L
C
1
jwL ( R 
)
( w2 LRC  jwL )
jwC
ZI N 

2
1
(
1

w
LC )  jwRC
R  jwL 
jwC
At resonance, the phase angle of Z must be equal to zero.
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Analysis
(  w 2 LRC  jwL )
( 1  w 2 LC )  jwRC
For zero phase;
wL

wRC
(  w 2 LCR ) ( 1 w 2 LC
This gives;
w 2 LC  w 2 R 2 C 2 1
or
wo 
1
( LC  R 2C 2 )
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Example 3:
A series RLC resonant circuit has a resonant frequency admittance of 2x10-2 S(mohs).
The Q of the circuit is 50, and the resonant frequency is10,000 rad/sec.
Calculate the values of R, L, and C. Find the half-power frequencies and the bandwidth.
First R = 1/G = 1/(0.02) = 50 ohms.
Second, from
Q
Third, we can use
Fourth: We can use
w OL
R
Solve for L, knowing Q, R, and wo to find L = 0.25 H.
C
wBW
Q
50

100  F
wO R 10,000 x50
wo 1x104
 
 200 rad / sec
Q
50
Fifth: Use the approximations;
w1 = wo - 0.5BW = 10,000 – 100 = 9,900 rad/sec
w2 = wo - 0.5BW = 10,000 + 100 = 10,100 rad/sec