Fundamentals of
Electric Circuits
Chapter 14
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Overview
• This chapter will introduce the idea of the
transfer function: a means of describing the
relationship between the input and output of
a circuit.
• Bode plots and their utility in describing the
frequency response of a circuit will also be
introduced.
• The concept of resonance as applied to LRC
circuits will be covered as well
• Finally, frequency filters will be discussed.
2
Frequency Response
• Frequency response is the variation in a
circuit’s behavior with change in signal
frequency.
• This is significant for applications involving
filters.
• Filters play critical roles in blocking or
passing specific frequencies or ranges of
frequencies.
• Without them, it would be impossible to have
multiple channels of data in radio
communications.
3
Transfer Function
• One useful way to analyze the
frequency response of a
circuit is the concept of the
transfer function H(ω).
• It is the frequency dependent
ratio of a forced function Y(ω)
to the forcing function X(ω).
H   
Y  
X  
4
Transfer Function
• Four possible transfer functions:
H( )  Voltage gain 
Vo ( )
Vi ( )
H( ) 
Io ( )
H( )  Current gain 
Ii ( )
H( )  T ransferImpedance 
Vo ( )
Ii ( )
Y( )
 | H( ) | 
X( )
I o ( )
H( )  T ransferAdmittance
Vi ( )
5
Example 1
For the RC circuit shown below, obtain the transfer
function Vo/Vs and its frequency response.
Let vs = Vmcosωt.
6
Solution:
The transfer function is
1
V
1
jC
H( )  o 

Vs R  1/ j C 1  j RC
,
The magnitude is
The phase is
H( ) 
   tan1

o
o  1/RC
1
1  ( / o ) 2
Low Pass Filter
7
Example 2
Obtain the transfer function Vo/Vs of the RL circuit shown
below, assuming vs = Vmcosωt. Sketch its frequency
response.
8
14.2 Transfer Function (6)
Solution:
Vo
j L
1
H( ) 


Vs R  j L 1  R
j L
The transfer function is
High Pass Filter
,
The magnitude is
The phase is
1
H ( ) 
1 (
  90  t an1
o  R/L
o 2
)


o
9
Zeros and Poles
• To obtain H(ω), we first convert to frequency
domain equivalent components in the circuit.
• H(ω) can be expressed as the ratio of
numerator N(ω) and denominator D(ω)
polynomials.
N  
H   
D  
• Zeros are where the transfer function goes to
zero.
• Poles are where it goes to infinity.
• They can be related to the roots of N(ω) and
D(ω)
10
Decibel Scale
• We will soon discuss Bode plots.
• These plots are based on logarithmic scales.
• The transfer function can be seen as an
expression of gain.
• Gain expressed in log form is typically
expressed in bels, or more commonly
decibels (1/10 of a bel)
GdB  10log10
P2
P1
11
Bode Plots
• One problem with the transfer function is
that it needs to cover a large range in
frequency.
• Plotting the frequency response on a
semilog plot (where the x axis is plotted in
log form) makes the task easier.
• These plots are referred to as Bode plots.
• Bode plots either show magnitude (in
decibels) or phase (in degrees) as a function
of frequency.
12
Standard Form
• The transfer function may be written in terms
of factors with real and imaginary parts. For
example:
K  j  1  j / z  1  j 2  /    j /   


1
H   
2
1
1
k
1  j / p1  1  j 2 2 / n   j / n 
k
2


• This standard form may include the following
seven factors in various combinations:
– A gain K
– A pole (jω)-1 or a zero (jω)
– A simple pole 1/(1+jω/p1) or a simple zero
(1+jω/z1)
– A quadratic pole 1/[1+j22ω/ ωn+ (jω/ ωn)2] or zero
1/[1+j21ω/ ωn+ (jω/ ωk)2]
13
Bode Plots
• In a bode plot, each of these factors is
plotted separately and then added
graphically.
• Gain, K: the magnitude is 20log10K and the
phase is 0°. Both are constant with
frequency.
• Pole/zero at the origin: For the zero (jω), the
slope in magnitude is 20 dB/decade and the
phase is 90°. For the pole (jω)-1 the slope in
magnitude is -20 dB/decade and the phase is
-90°
14
Bode Plots
• Simple pole/zero: For the simple zero, the
magnitude is 20log10|1+jω/z1| and the phase
is tan-1 ω/z1.
• Where:
H dB
j
 20log10 1 
z1
 20log10
as  

z1
• This can be approximated as a flat line and
sloped line that intersect at ω=z1.
• This is called the corner or break frequency
15
Bode Plots
• The phase can be plotted as a series straight
lines
• From ω=0 to ω≤z1/10, we let =0
• At ω=z1 we let =45°
• For ω≥10z1, we let = 90°
• The pole is similar, except the corner
frequency is at ω=p1, the magnitude has a
negative slope
16
Bode Plots
• Quadratic pole/zero: The magnitude of the
quadratic pole 1/[1+j22ω/ ωn+ (jω/ ωn)2] is 20log10 [1+j22ω/ ωn+ (jω/ ωn)2]
• This can be approximated as:
H dB
0
as  0
 40log10
as  

n
• Thus the magnitude plot will be two lines,
one with slope zero for ω<ωn and the other
with slope -40dB/decade, with ωn as the
corner frequency
17
Bode Plots
• The phase can be expressed as:
 0 0
2 2 / n 
   tan 1
  90   n
2
2
1   / n
180   

• This will be a straight line with slope of 90°/decade starting at ωn/10 and ending at 10
ωn.
• For the quadratic zero, the plots are inverted.
18
Bode Plots
19
Bode Plots
20
Figure 14.14
Resonance
• The most prominent feature of the frequency
response of a circuit may be the sharp peak
in the amplitude characteristics.
• Resonance occurs in any system that has a
complex conjugate pair of poles.
• It enables energy storage in the firm of
oscillations
• It allows frequency discrimination.
• It requires at least one capacitor and
inductor.
22
Series Resonance
• A series resonant circuit
consists of an inductor and
capacitor in series.
• Consider the circuit shown.
• Resonance occurs when the
imaginary part of Z is zero.
• The value of ω that satisfies
this is called the resonant
frequency
0 
Z  R  j ( L 
1
)
C
1
rad/s
LC
23
Series Resonance
• At resonance:
–
–
–
–
The impedance is purely resistive
The voltage Vs and the current I are in phase
The magnitude of the transfer function is minimum.
The inductor and capacitor voltages can be much more than
the source.
24
Half-power frequencies ω1 and ω2 are frequencies at which the
dissipated power is half the maximum value:
1 (Vm / 2 ) 2 Vm2
P(1 )  P(2 ) 

2
R
4R
The half-power frequencies can be obtained by setting Z equal to √2
R.
1  
R
R
1
 ( )2 
2L
2L
LC
Bandwidth B
2 
R
R
1
 ( )2 
2L
2L
LC
o  12
B  2  1
25
Quality Factor
• The “sharpness” of the resonance is
measured quantitatively by the quality factor,
Q.
• It is a measure of the peak energy stored
divided by the energy dissipated in one
period at resonance.
Q
0 L
R

1
0CR
• It is also a measure of the ratio of the
resonant frequency to its bandwidth, B
B
R 0

L Q
26
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.
27
28
Parallel Resonance
• The parallel RLC circuit
shown here is the dual of the
series circuit shown
previously.
• Resonance here occurs when
the imaginary part of the
admittance is zero.
• This results in the same
resonant frequency as in the
series circuit.
29
It occurs when imaginary part of Y is zero
1
1
Y   j ( C 
)
R
L
Resonance frequency:
1
1
o 
rad/s or f o 
Hz
LC
2 LC
30
Series Resonance
• The relevant equations for the parallel
resonant circuit are:
2
1
1
 1 
1  
 


2 RC
 2 RC  LC
1
B
RC
2
1
1
 1 
2 
 


2 RC
 2 RC  LC
R
Q  0 RC 
0 L
31
Figure 14.29
Passive Filters
• A filter is a circuit that is designed to pass
signals with desired frequencies and reject
or attenuate others.
• A filter is passive if it consists only of
passive elements, R, L, and C.
• They are very important circuits in that many
technological advances would not have been
possible without the development of filters.
33
Passive Filters
• There are four types of filters:
– Lowpass passes only low
frequencies and blocks high
frequencies.
– Highpass does the opposite of
lowpass
– Bandpass only allows a range of
frequencies to pass through.
– Bandstop does the opposite of
bandpass
34
Lowpass Filter
• A typical lowpass filter is formed
when the output of a RC circuit
is taken off the capacitor.
• The half power frequency is:
c 
1
RC
• This is also referred to as the
cutoff frequency.
• The filter is designed to pass
from DC up to ωc
35
Highpass Filter
• A highpass filter is also
made of a RC circuit, with
the output taken off the
resistor.
• The cutoff frequency will be
the same as the lowpass
filter.
• The difference being that the
frequencies passed go from
ωc to infinity.
36
Bandpass Filter
• The RLC series resonant
circuit provides a bandpass
filter when the output is taken
off the resistor.
• The center frequency is:
0 
1
LC
• The filter will pass frequencies
from ω1 to ω2.
• It can also be made by feeding
the output from a lowpass to a
highpass filter.
37
Bandstop Filter
• A bandstop filter can be
created from a RLC circuit by
taking the output from the LC
series combination.
• The range of blocked
frequencies will be the same
as the range of passed
frequencies for the bandpass
filter.
38
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.
Answer:
  25 krad/s
39
40