Transcript ee2.cust.edu.tw

Fundamentals of
Electric Circuits
Chapter 14
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Overview
• 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.
• The concept of resonance as applied to
LRC circuits .
• Frequency filters.
2
Frequency Response
• Frequency response is the variation in
a circuit’s behavior with change in
signal frequency.
• 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
• There are four possible input/output
combinations:
H    Voltage gain 
H    Current gain 
Vo  
Vi  
I o  
I i  
H    Transfer impedance 
H    Transfer admittance 
Vo  
I i  
I o  
Vi  
5
Example
For the RC circuit, obtain the transfer function
Vo Vs and its frequency response. Let vs  Vm cos t
H( w) 
H 
Vo
1 jwC
1


Vs
R  1 jwC
1  jwRC
1
1   w wo 
2
,    tan 1
w
wo
wo 
1
RC
6
7
Example
For the RL circuit, obtain the transfer function
Vo Vs and its frequency response. Let vs  Vm cos t
Vo
jwL
H ( w) 

Vs R  jwL
H
wL
wL
,   90  tan
2
2 2
R
R w L
o
1
wo 
R
L
8
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)
9
Example
For the RL circuit, calculate the gain Io ( w)
its poles and zeros
Ii ( w) and
4  j2w
I o ( w) 
I i ( w)
4  j 2 w  1 j 0.5w
I o ( w)
4  2S
S ( S  2)

 2
, S  jw
I i ( w) 4  2 S  2 S S  2 S  1
10
Decibel Scale
• 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)
P2
V2
GdB  10log10  20log10
P1
V1
Properties of logarithms
1. log P1 P2  log P1  log P2 , 3. log P n  n log P
2. log P1 P2  log P1  log P2 , 4. log 1=0
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 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.
H  H   He j  ln H  ln H  ln e j = ln H  j
The real part of ln H is a function of the magnitude
while the imaginary part is the phase. In a Bode plot
H dB  20log10 H (the magnitude)
12
13
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]
14
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°
15
Bode plot for gain K
Bode plot for a zero(jw) at the origin
16
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
17
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
18
Bode plot of zero(1+ jw z1 )
19
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
20
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.
21
Bode plot of quadratic (1+ j 2 w wn  w2 wn2 )-1
22
Bode Plots
23
Bode Plots
24
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 requires at least one capacitor and
inductor.
25
Example
Construct the Bode plots for
200 jw
the transfer function H( w) 
( jw  2)( jw  10)
H( w) 

200 jw
10 jw

( jw  2)( jw  10) (1  jw 2)(1  jw 10)
10 jw
90o  tan 1 w 2  tan 1 w 10
1  jw 2 1  jw 10
 H dB  20log10 10  20log10 jw  20log10 1 
  90o  tan 1
jw
jw
 20log10 1 
2
10
w
w
 tan 1
2
10
26
27
Example
Draw the Bode plots for the
transfer function H ( w)  5( jw  2)
jw( jw  10)
28
Example
Draw the Bode plots for the
transfer function H ( w)  ( jw  10) 2
jw( jw  5)
29
Example
Draw the Bode plots for the
transfer function H ( s)  2000000( S  5) 2
( S  10)( S  100)
30
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 is called the
resonant frequency
1
0 
rad/s
LC
1

Hz
2 LC
Vs
1
1
Z  H ( w) 
 R  jwL 
 R  j ( wL 
)
I
jwC
wC
1
1
1
Im( Z)  wL 
 0  wo L 
 wo 
wC
woC
LC
31
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.
Vm
Vm
I  I 
R   wL  1 wC 
2
VL 
2

R
Vm
V
1
wo L  QVm  m
 VC
R
R woC
• There are two frequencies above and below
resonance where the dissipated power is half the
max:
2
2
R
1
R
1
 R 
 R 
wo  w1w2
1  
 





2



2L
 2L 
LC
2L
 2L 
LC
32
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
Peak energy stored in the circuit
 2
0CR
Energy dissipated by the ckt in one period at resonance
• It is also a measure of the ratio of the
resonant frequency to its bandwidth, B
B
R 0

 w2  w1
L Q
33
B
R 0

 w2  w1
L Q
It is said to be a high-Q circuit while Q  10.
w1
wo 
B
, w2
2
wo 
B
2
34
Example
In the circuit, R  2, L  1mH , C  0.4F
(1)Find wo , w1 and w2 . (2)Calculate Q and B
(3) Determine the amplitude of the current at
wo , w1 and w2
1
1

 50krad / s,
3
9
LC
10  0.4  10
wo L 50  103  10-3
Q
=
=25 10,
R
2
w
50
B
B o 
 2krad / s, w1  wo   50  1  49krad / s,
Q 25
2
B
w2  wo   50  1  51krad / s
2
V
20
At w  wo , I  m 
 10A
R
2
V
10
At w  w1 , w2 I  m 
 7.071A
2R
2
wo 
35
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.
36
Parallel Resonance
• The relevant equations for the parallel
resonant circuit are:
2
1
1
 1 
1  
 


2 RC
 2 RC  LC
1
B  w2  w1 
RC
2
1
1
 1 
2 
 


2 RC
 2 RC  LC
Q
wo
R
 0 RC 
B
0 L
B
B
2
w1  wo 1  1 2Q  
, w2  wo 1  1 2Q  
2Q
2Q
B
B
For high-Q circuit (Q  10), w1 wo  , w2 wo 
2
2
2
37
38
Example
Determine the resonant frequency of the
circuit.
1
1
2  j2w

 0.1  j0.1w+
10 2  j 2 w
4  4 w2
2w
At resonance Im( Y)  0  0.1w  0  w  wo  2 rad / s
2
4  4w
Y=j0.1w +
39
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.
40
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
41
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
Vo
H( w) 
H( wc ) 
Vi

1
1  jwRC
1
1  wc2 R 2C 2

1
1
 wc 
RC
2
42
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
V
ωc to infinity.
H( w) 

o
Vi
H( wc ) 
jwRC
1  jwRC
wc RC
1  wc2 R 2C 2

1
1
 wc 
RC
2
43
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
Vo
highpass filter.
H ( w) 

Vi
R
R  j ( wL  1 wC )
44
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.
Vo
R2
jw
H ( w) 

, highpass filter
Vi R1  R2 jw  wc
R1 R2
wc 
 25 krad/s
( R1  R2 ) L
45
Example
Determine what type of filter is shown in the
figure. Calculate the corner or cutoff freqency.
R  2k, L  2H, C  2 F
The corner frequency is the same as
the half-power frequency, i.e., H  1
H( s ) 

2
Vo
R //(1 sC )

Vi sL  R //(1 sC )
R (1  sRC )
R
 2
sL  R (1  sRC ) s RLC  sL  R
2
1
w L
2
2
H 
  2  1  wc LC    c 
2
2
 R 
R

w
RLC
 c   wc2 L2 2
R
2
 2  1  wc2  4  106    wc  103   2  1  4 wc2   wc2 , Let wc be krad/s
2
2
2
 16wc4  7 wc2  1  0  wc2  0.5509  wc  0.742krad / s  742rad / s
46
Example
Determine what type of filter is shown in the
figure. Calculate the corner or cutoff freqency.
R1  R2 =100, L  2mH
H ( w) 
Vo
R2
jw

, highpass filter
Vi R1  R2 jw  wc
R1 R2
wc 
 25 krad/s
( R1  R2 ) L
47
Example
If the bandpass filter in the figure is to reject a
200 Hz sinusoid while passing other freq.
,calculate the values of L, C and R  150 .
The bandwidth is 100 Hz
B  2 f  2  100  200 rad/s
R
R 150
B L 
 0.2387 H
L
B 200
wo  2 f o  2  200  400 rad/s
wo 
1
1
1
C  2 
 2.653 F
2
wo L  400   0.2387
LC
48
Active Filters
• Passive filters have a few drawbacks.
– They cannot create gain greater than 1.
– They do not work well for frequencies below the
audio range.
– They require inductors, which tend to be bulky
and more expensive than other components.
• It is possible, using op-amps, to create all the
common filters.
• Their ability to isolate input and output also
makes them very desirable.
49
First Order Lowpass
• The corner frequency will be:
c 
1
Rf C f
Zf
Vo
H ( w) 

Vi
Zi
Rf
1
Zi  Ri and Z f  R f //

jwC f 1  jwR f C f
H ( w)  
Zf
Zi

Rf
1
Ri 1  jwR f C f
Rf
1
 The dc gain is 
and the corner freq. wc 
Ri
Rf C f
50
First Order Highpass
• The corner frequency will be:
c 
1
Ri Ci
Zf
Vo
H( w) 

Vi
Zi
Zi  Ri 
H( w)  
1  jwRi Ci
1

and Z f  R f
jwCi
jwCi
Zf
Zi

 The gain is 
Rf
Ri  1 jwCi
Rf
Ri

jwR f Ci
1  jwRi Ci
as w   and the corner freq. wc 
1
Ri Ci
51
Bandpass
• To avoid the use of an inductor, it is possible
to use a cascaded series of lowpass active
filter into a highpass active filter.
• To prevent unwanted signals passing, their
gains are set to unity, with a final stage for
amplification.
52
Figure 14.45
The analysis of the bandpass filter
H ( w) 
R
R
Vo
1
jwRC2
1
jwRC2
 (
) (
)(  f )   f
Vi
1  jwRC1
1  jwRC2
Ri
Ri 1  jwRC1 1  jwRC2
The lowpass section sets the upper corner freq. as w2 
1
RC1
The highpass section sets the lower corner freq. as w1 
1
RC2
The center freq., bandwidth and quality factor
wo  w1w2 , B  w2  w1 , Q 
wo
B
Rf
Rf
j w w1
jww2
H( w)  

Ri (1  j w w2 )(1  j w w1 )
Ri ( w1  jw)( w2  jw)
Rf
R f w2
jww2
H( wo )  

, wo  w1w2
Ri ( w1  jw)( w2  jw) Ri w1  w2
The passband gain K 
Rf
w2
Ri w1  w2
54
Bandreject
• Creating a bandstop filter requires using a
lowpass and highpass filter in parallel.
• Both output are fed into a summing amplifier.
• It will function by amplifying the desired
signals compared to the signal to be
rejected.
55
Figure 14.47
The analysis of the bandreject filter

1
jwRC2  R f (1  j 2 w w1   jw  w1w1 )


 1  jwRC 1  jwRC   R (1  j w w )(1  j w w )
1
2 
i
2
1

1
The highpass section sets the upper corner freq. as w2 
RC1
2
R
V
H( w)  o   f
Vi
Ri
The lowpass section sets the lower corner freq. as w1 
1
RC2
The center freq., bandwidth and quality factor
wo  w1w2 , B  w2  w1 , Q 
wo
B
R (1  j 2 wo w1   jwo  w1w1 ) R f 2 w1
H( wo )  f

, wo  w1w2
Ri (1  j wo w2 )(1  j wo w1 )
Ri w1  w2
2
57
Example
Design a lowpass active filter with a dc
gain of 4 and a corner frequency of 500 Hz.
1
wc  2 f c  2  500 
Rf C f
The dc gain is H (0)  
Rf
Ri
 4
If we select C f  0.2  F , then
1
Rf 
 1.59k   1.6k 
6
2  500  0.2  10
Rf
Ri 
 397.5  400
4
58
Example
Design a highpass active filter with a high
frequency gain of 5 and a corner frequency
of 2kHz. Use a 0.1 F capacitor in your design
1
wc  2 f c  2  2000 
Ri Ci
1
 Ri 
 795.7  800
6
2  2000  0.1  10
Rf
The gain of H ( )  
 5  R f  5Ri  4k 
Ri
59