1
12. Variable frequency
network performance
EMLAB
2
Learning Goals
Variable-Frequency Response Analysis
Network performance as function of frequency.
Transfer function
Sinusoidal Frequency Analysis
Bode plots to display frequency response data
Resonant Circuits
The resonance phenomenon and its characterization
Scaling
Impedance and frequency scaling
Filter Networks
Networks with frequency selective characteristics:
low-pass, high-pass, band-pass
EMLAB
3
Variable frequency-response analysis
In AC steady state analysis the frequency is assumed constant (e.g., 60Hz).
Here we consider the frequency as a variable and examine how the performance
varies with the frequency.
Variation in impedance of basic components
Resistor
Z R  R  R0
EMLAB
4
Inductor
Z L  j L   L90
EMLAB
5
Capacitor
1
1
Zc 

  90
j C  C
EMLAB
6
Series RLC network
2
1
( j ) 2 LC  jRC  1  j RC  j ( LC  1)


Z eq  R  jL 


j
C
jC
jC
Simplification in notation
Driving Point functions
j  s
s 2 LC  sRC  1
Z eq ( s) 
sC
1  
(RC ) 2  (1   2 LC ) 2
| Z eq |
C
LC  1 

Z eq  tan 

RC


2
EMLAB
7
Pulse represented by sum of sinusoidal signal
7
EMLAB
8
Network functions
Nomenclature
When voltages and currents are defined at different terminal pairs we define the
ratios as Transfer Functions
INPUT
Voltage
Current
Current
OUTPUT
Voltage
Voltage
Current
TRANSFER FUNCTION
Voltage Gain
Transimpedance
Current Gain
SYMBOL
Gv(s)
Z(s)
Gi(s)
If voltage and current are defined at the same terminals we define
Driving Point functions (Impedance/Admittance)
Network functions for basic components
Z R ( s)  R, Z L ( s)  sL, ZC 
1
sC
General form
V ( s ) am s m  am1s m1  ...  a1s  a0
Z ( s) 

I ( s)
bn s n  bn 1s n 1  ...  b1s  b0
EMLAB
9
Nomenclature
Port #1
Port #2
I1
I2

V1

V2
Circuit


terminals
terminals
V2 or I2
Transfer function
Driving point function
{V2 , I 2 }
H ( s) 
{V1 , I1}
Vi
Ii
Z ( s)  , Y ( s) 
Ii
Vi
(i  1,2)
EMLAB
10
Example 12.1
sL
1
sC
R
sRC
VS  2
VS
R  sL  1/ sC
s LC  sRC  1
s  j
jRC
Vo 
VS
( j ) 2 LC  jRC  1
Vo ( s) 
R
j (15  2.53  103 )
Vo 
100
2
3
3
( j ) (0.1 2.53  10 )  j (15  2.53  10 )  1
MATLAB can be effectively used to compute frequency response characteristics
1
100
0.8
| V0 |
V0
0.6
0.4
0
-50
0.2
0
50
0
200
400
600
800
1000
-100
0
200
400
600
800
1000
EMLAB
11
Some Matlab command
 2 3 1

A  
 4 5 7
A= [2,3,1; 4, 5, 7]
A  1 3 5 7 9
A= [1:2:9]
AA
A= A’
T
x  [1 : 2 : 1.e3];
y  an x n  an 1 x n 1    a1 x1  a0
a  [a n , a n 1 ,, a1 , a 0 ];
y  polyval (a , x );
x  [1 : 2 : 1e3];
H
m 1
am x  am1 x    a1 x  a0
bn x n  bn 1 x n 1    b1 x1  b0
m
1
a  [a m , a m 1 ,, a1 , a 0 ];
b  [b n , b n 1 ,, b1 , b0 ];
num  polyval(a, x);
den  polyval(b, x);
H  num ./den;
EMLAB
12
Matlab example
1
0.8
| V0 |
0.6
0.4
0.2
0
0
200
400
600
800
1000
clear; % clear all variables.
clf; % clear current figure.
w = [1:1:1.e3];
b=[15*2.53*1e-3 0];
a=[0.1*2.53*1e-3 15*2.53*1e-3 1];
num = polyval(b,i*w);
den = polyval(a,i*w);
H = num./den;
mag = abs(H);
plot(w,mag);
EMLAB
Example
13
Find the pole and zero locations and the value of K0 for the
voltage gain
V ( s)
G( s)  o
VS ( s )
H ( s)  K 0
( s  z1 )( s  z2 )...( s  zm )
( s  p1 )( s  p2 )...( s  pn )
Zeros = roots of numerator
Poles = roots of denominator
For this case the gain was shown to be
H ( s) 
 
VO  sCin Rin  
1
s
  40,000 




VS 1  sCin Rin  1  sCo Ro   s  100   s  40,000 
zero : z1  0
poles : p1  50 Hz , p2  20,000 Hz
K 0  (4  107 )
EMLAB
14
1
0.9
0.8
0.7
0.6
|H |
0.5
clear;
0.4
clf;
0.3
w = logspace(1,6,200);
0.2
b = [1.e0 * 40e3*pi, 0];
0.1
0
a = [1, 40.1e3*pi, 4e6*pi*pi];
0
2
4
6
8
10
x 10
100
num = polyval(b,i*w);
den = polyval(a,i*w);
H = num./den;
80
60
%arg = angle(H)*180/pi;
H
40
20
mag = abs(H);
0
plot(w,mag);
-20
%semilogx(w,mag);
-40
-60
%freqs(num, den,w);
-80
-100
5
0
2
4
6
8
10
x 10
5
EMLAB
15
Poles and Zeros
(More nomenclature)
am s m  am 1s m 1  ...  a1s  a0
H ( s) 
bn s n  bn1s n1  ...  b1s  b0
; Arbitrary network function
Using the roots, every (monic) polynomial can be expressed as a
product of first order terms
H ( s)  K 0
( s  z1 )( s  z2 )...( s  zm )
( s  p1 )( s  p2 )...( s  pn )
z1 , z2 ,..., zm  zeros of the network function
p1 , p2 ,..., pn  poles of the network function
The network function is uniquely determined by its poles and zeros
and its value at some other value of s (to compute the gain)
EMLAB
Positive real function
16
am s m  am 1s m 1  ...  a1s  a0
H ( s) 
bn s n  bn1s n1  ...  b1s  b0
H ( s)  K 0
( s  z1 )( s  z2 )...( s  zm )
( s  p1 )( s  p2 )...( s  pn )
1. The coefficients of the polynomial are related to the values of R, L,
C and they are all real and positive.
2. At the same time, the real part of the poles are all negative if the
circuit contains only passive components.
EMLAB
17
Sinusoidal frequency analysis
A0e j ( t  )


B0 cos( t   )
H (s)

A0 H ( j )e j ( t  )

 B0 | H ( j ) | cos t    H ( j ) 
Circuit represented by
network function
To study the behavior of a network as a function of the frequency we analyze
the network function H ( j ) as a function of .
Notation
M ( ) | H ( j ) |
 ( )  H ( j )
H ( j )  M ( )e j ( )
Plots of M ( ),  ( ), as function of  are generally called
magnitude and phase characteri stics.
20 log10 (M ( ))
BODE PLOTS
vs log10 ( )

(

)

EMLAB
18
Transfer function
Vout ( s)  H ( s) Vin ( s)
+
Vin
-
H(s)
1
100
0.9
80
0.8
60
0.7
40
|H |
0.6
0.5
0
0.4
0.3
-40
0.2
-60
0.1
-80
-100
0
2
4
6
8
10
x 10
H (s)
H
20
-20
0
A0e j ( t  ) 

B0 cos( t   )
+
Vout
-
0
2
4
6
8
10
5
x 10
5

A0 H ( j )e j ( t  )

 B0 | H ( j ) | cos t    H ( j ) 
EMLAB
19
Drawing transfer function is difficult!
H ( s)  K 0
Bode plot
( s  z1 )( s  z2 )...( s  zm )
( s  p1 )( s  p2 )...( s  pn )
H ( s ) | H ( s ) | e jH ( s )
H1 ( s ) H 2 ( s ) | H1 ( s ) || H 2 ( s ) | e j [ H1 ( s )H 2 ( s )]
log | H1 ( s ) || H 2 ( s ) |  log | H1 ( s ) |  log | H 2 ( s ) |
Magnitude :
log 10 | H ( s ) | log 10 K 0
( s  z1 )( s  z2 )...( s  zm )
( s  p1 )( s  p2 )...( s  pn )
 log 10 | K 0 |  log 10 | s  z1 |  log 10 | s  z2 | ...  log 10 | s  zm |
 log 10 | s  p1 |  log 10 | s  p2 | ...  log 10 | s  pn |
Phase :

( s  z1 )( s  z2 )...( s  zm ) 
H ( s )    K 0

(
s

p
)(
s

p
)...(
s

p
)
1
2
n 

 K 0  ( s  z1 )  ( s  z2 )  ...  ( s  zm )
 ( s  p1 )  ( s  p2 )  ...  ( s  pn )
EMLAB
20
History of the Decibel
Originated as a measure of relative (radio) power
P2 |dB (over P1 )  10 log
P2
P1
V2
V22
I 22
PI R
 P2 |dB (over P1 )  10 log 2  10 log 2
R
V1
I1
2
V |dB  20 log10 | V |
By extension
I |dB  20 log10 | I |
G |dB  20 log10 | G |
Using log scales the frequency characteristics of network functions
have simple asymptotic behavior.
The asymptotes can be used as reasonable and efficient approximations
EMLAB
Standard form of a transfer function

s 
1  1 
z1 

H ( s)  K0

s 
1  1 
p1 

20 log 10 | H ( s ) | 20 log 10 | K 0 | 20 log 10 1 
 20 log 10 1 
21
s 
s 
...1  
z2  
zm 
s  
s 
...1  
p2  
pn 
s
s
s
 20 log 10 1 
 ...  20 log 10 1 
z1
z2
zm
s
s
s
 20 log 10 1 
 ...  20 log 10 1 
p1
p2
pm
On applying to log function, terms containing poles and zeroes can be
separated. Then, it becomes easier to draw the transfer function.
EMLAB
22
How to draw Bode plot of simple zero or poles
Bode plots can be drawn simply following the procedures below.
1. Draw first two extreme regions where  / z1  1 and  / z1  1 .
2. Connect those two regions with a smooth curve.
3. Mark the exact point where  / z1  1 .
Magnitude
   2 
s
y  20 log 10 1   10 log 10 1    
z1
  z1  
0 dB

  1 


 
s

  1  j   tan 1  
z1 
z1 

 z1 
90
y
3 dB
Phase
 20 dB/dec
45
log 10 z1
0
log 10 z1
EMLAB
23
Bode plot of quadratic poles
Drawing a transfer function containing quadratic polynomial deserves a special care.
Depending on the value of ζ, the shape of the transfer function curves changes
significantly.
H (s) 
(  1)
(  1)
(  1)
1
 s   s 
1  2     
 0   0 
2

1

s 
s 
1  1  
p1 
p2 

1. p1 and p2 are real numbers and have different values.
2. p1 and p2 are real numbers and have the same value.
3. p1 and p2 are complex number and have different values.
EMLAB
24
When the poles are real valued (   1)
H (s) 
1
 s   s 
1  2     
 0   0 
2

1
1


s 
s  
s 
s 
1  1       2  1      2  1  
p1 
p2  
0 
0 

(1) p1, p2 same
(2) p1, p2 distinct
20 log 10 | H ( j ) |
20 log 10 | H ( j ) |
log 10 p1
log 10 p1
 40 dB/dec
log 10 p2
 20 dB/dec
 40 dB/dec
EMLAB
25
When the poles are complex valued (   1)
H ( s) 
1
 s   s 
1  2     
 0   0 
2

1
1


s 
s  
s 
s 
1  1      j 1   2     j 1   2  
p1 
p2  
0 
0 


p1, 2  0   j 1   2
With smaller
3 dB
log 10 p1


3 dB
log 10 p1
0 dB
0 dB
 40 dB/dec
 40 dB/dec
EMLAB
Standard form of quadratic functions
2 
1
Q
H (s) 
26
1
2
 s 
1 s 
      1
 0  Q  0 
Instead of ζ, Q is more widely used and is called ‘Quality factor’.
The Q factor is related with the power loss of a passive component.
(Q  1 / 2)
(Q  1 / 2)
(Q  1 / 2)
1. p1 and p2 are real numbers and have different values.
2. p1 and p2 are real numbers and have the same value.
3. p1 and p2 are complex number and have different values.
EMLAB
27
General form of a network function
Poles/zeros at the origin
Frequency independent
K 0 ( j ) N (1  j1 )[1  2 3 ( j3 )  ( j3 )2 ]...
H ( j ) 
(1  ja )[1  2 b ( jb )  ( jb )2 ]...
log( AB )  log A  log B First order terms
log(
N
)  log N  log D
D
Quadratic terms for
complex conjugate poles/zeros
| H ( j ) |dB  20 log10 | H ( j ) |  20 log10 K 0  N 20 log10 | j |
 20 log10 | 1  j1 | 20 log10 | 1  2 3 ( j3 )  ( j3 ) 2 | ...
z1z2  z1  z2
z
 1  z1  z2
z2
 20 log10 | 1  ja | 20 log10 | 1  2 b ( jb )  ( jb ) 2 | ..
H ( j )  0  N 90
2 33
 tan 1 1  tan 1
 ...
1  (3 ) 2
 tan 1 a  tan 1
2 bb
 ...
2
1  (b )
Display each basic
term separately and
add the results to
obtain final answer
Let’s examine each basic term
EMLAB
28
a. Constant Term
the x - axis is log10
this is a straight line
b. Poles/Zeros at the origin
( j )
N
| ( j )  N |dB   N  20 log10 ( )

( j )  N   N 90

EMLAB
29
c. Simple pole or zero

| 1  j |dB  20 log10 1  ( ) 2
1  j  (1  j )  tan1 
(1  j )  0
 1  | 1  j |dB  0 low frequency asymptote
 1  | 1  j |dB  20 log10  high frequency asymptote (20dB/dec) (1  j )  90
The two asymptotes meet when   1(corner/break frequency)
Behavior in the neighborhood of the corner
corner
octave above
octave below
distance to
FrequencyAsymptoteCurve asymptote Argument
3dB
3
45
  1 0dB

 2
6dB
7db
1
63.4
  0 .5
0dB
1dB
1
26.6
Asymptote for phase
Low freq. Asym.
High freq. asymptote
EMLAB
30
Simple zero
Simple pole
EMLAB
d. Quadratic pole or zero
| t |  20 log 1  ( )   2 
2 2
2 dB
  1
  1
  1
t 2  [1  2 ( j )  ( j )2 ]  [1  2 ( j )  ( )2 ]
t 2  tan 1
2
10
2
1  ( )2
| t2 |dB  0 low frequency asymptote
t2  0
| t 2 |dB  20 log10 ( )2 high freq. asymptote
| t2 |dB  20 log10 (2 )
  1  2 2
40dB/dec
t2  180
t2  90
Corner/break frequency
| t2 |dB  20 log10 2 1   2
31
Resonance frequency
t 2  tan
1
1  2 2


2
2
These graphs are inverted for a zero
Magnitude for quadratic pole
Phase for quadratic pole
EMLAB
32
Example
Generate magnitude and phase plots
Draw asymptotes
for each term
Gv ( j ) 
10(0.1 j  1)
( j  1)(0.02 j  1)
Breaks/cor ners : 1,10,50
Draw composites
dB
40
20
10 |dB
20dB / dec
0
 20dB / dec
 20
90
45 / dec
 45 / dec
0.1
1
10
100
 90
1000
EMLAB
33
EMLAB
Generate magnitude and phase plots
Example
25( j  1)
Draw asymptotes for each
Gv ( j ) 
( j ) 2 (0.1 j  1)
Form composites
34
Breaks (corners) : 1, 10
dB
40
28dB
20
0
 40dB / dec
 20
90
45 / dec
 45
90
180
0.1
1
10
100
 270
EMLAB
35
A function with complex conjugate poles
Example
25 j
2
G
(
j

)

t 2  [1  2 ( j )  ( j ) ]
2  1 / 25
( j  0.5) ( j ) 2  4 j  100
    0.2
Put in standard form
  0.1 
0.5 j
G ( j ) 
( j / 0.5  1) ( j / 10) 2  j / 25  1
Draw composite asymptote




dB
40
20
  1 | t2 |dB  20 log10 (2 )
0
8dB
 20
90
90
0.01
0.1close to corner of1conjugate pole/zero
10
Behavior
is too dependent on damping ratio.
Computer evaluation is better
 270
100
EMLAB
Using MATLAB to compute Magnitude & Phase information
36
am s m  am 1s m 1  ...  a1s  a0
Vo ( s) 
bn s n  bn1s n1  ...  b1s  b0
 num  [am , am 1 ,..., a1 , a0 ];
 den  [bn , bn1 ,..., b1 , b0 ];
 freqs(num , den)
EXAMPLE
MATLAB commands required to display magnitude
and phase as function of frequency
NOTE: Instead of comma (,) one can use space to
separate numbers in the array
a1
j (15  2.53  103 )
Vo 
( j ) 2 (0.1 2.53  103 )  j (15  2.53  103 )  1
b2
b1
b0
» num=[15*2.53*1e-3,0];
» den=[0.1*2.53*1e-3,15*2.53*1e-3,1];
» freqs(num,den)
» num=[15*2.53*1e-3 0];
» den=[0.1*2.53*1e-3 15*2.53*1e-3 1];
» freqs(num,den)
Missing coefficients must
be entered as zeros
This sequence will also
work. Must be careful not
to insert blanks elsewhere
EMLAB
37
EMLAB
38
clear;
clf;
w = logspace(1,7,200);
num = [1.e0 * 40e3*pi, 0];
den = [1, 40.1e3*pi, 4e6*pi*pi];
freqs(num, den,w);
EMLAB
39
Evaluation of frequency response using MATLAB
G ( j ) 
25 j
( j  0.5) ( j ) 2  4 j  100


Using default options
» num=[25,0]; %define numerator polynomial
» den=conv([1,0.5],[1,4,100]) %use CONV for polynomial multiplication
den =
1.0000
4.5000 102.0000
50.0000
» freqs(num,den)
EMLAB
40
Determining Transfer function from Bode plot
This is the inverse problem of determining frequency characteristics. We will use only the composite
asymptotes plot of the magnitude to postulate a transfer function. The slopes will provide information on
the order
A. different from 0dB.
There is a constant Ko
A
B
C
K 0 |dB  20  K 0
D
E
K 0 |dB
 10 20
B. Simple pole at 0.1
( j / 0.1  1)1
C. Simple zero at 0.5
( j / 0.5  1)
D. Simple pole at 3
( j / 3  1)1
E. Simple pole at 20
G ( j ) 
10( j / 0.5  1)
( j / 0.1  1)( j / 3  1)( j / 20  1)
( j / 20  1)1
If the slope is -40dB we assume double real pole. Unless we are given more data
EMLAB
Determine a transfer function from the composite
magnitude asymptotes plot
Example
C
E
A
B
D
41
A. Pole at the origin.
Crosses 0dB line at 5
5
j
B. Zero at 5
C. Pole at 20
D. Zero at 50
E. Pole at 100
5( j / 5  1)( j / 50  1)
G ( j ) 
j ( j / 20  1)( j / 100  1)
EMLAB
42
12.3 Resonant circuits
Vout ( s)  H ( s) Vin ( s)
RS
+
Vin
-
H(s)
RL
+
Vout
-
A circuit that selects one
frequency is called a
resonant circuit.
EMLAB
43
Series resonance
Standard form : s 2 
0
Q
s  02  s 2  20 s  02
H(s)
L
C
+
Vout
-

Vout
R2
R2Cs

 2
Vin R  sL  1
s LC  R2Cs  1
2
sC
s
1
Q0


2
 s 0 
1 s  s 
  
1

Q
1
  
Q 0  0 
 0 s 
H ( s) 

  0 

 H ( s) 
1
1 

, Q0 
R2C 
LC
1
  
1  jQ  0 
 0  
EMLAB
44
Transfer function of a resonator
| H () |2
Magnitude
10
10
10
10
0
-1
-2
-3
10
H ( )
clear;
clf;
w0=10^2;
Q=10;
H0=1;
w = logspace(1,4,200);
num = [1/(Q*w0), 0];
den = [1/(w0*w0), 1/(Q*w0),1];
freqs(num, den,w);
1
2
3
2
3
10
10
Frequency (radians)
10
4
Phase (degrees)
100
50
0
-50
-100
1
10
10
10
Frequency (radians)
10
4
Spectrum analyzer
EMLAB
45
Half Power Band-width
1
| H ( s ) |
0.9
1
  
1  Q 2   0 
 0  
0.8
2
0.7
| H () |2
1
x  x  1  0,
Q
0.4
0.3
0.2
1
1
x

1
2Q
4Q 2
0.1
0
80
1
1
x  

 1 ( x  0)
2Q
4Q 2
 1
 x    
 0  Q
  
0
Q
 BW
0.6
0.5
  0 
1
    
Q
 0  
2
Power 가 1/2이 되는
지점.

Q  0  10

90
100
90
100
110
120
80
60
H ( )
40
20
0
-20
Half power band-width
-40
-60
-80
80
110
120
EMLAB
46
HPBW vs. Q
On a log scale, a point where |H| is lowered by 3dB
from the peak value is called a half power point.
3dB-BW
| H () |2
10
-1
-2
10
-3
10
1
2
3
10
10
Frequency (radians)
10
10
4
Phase (degrees)
0
-50
H ( )
-2
-3
1
2
3
10
10
Frequency (radians)
10
2
3
10
10
Frequency (radians)
Q2
10
4
0
-2
-4
10
1
2
3
2
3
10
10
Frequency (radians)
10
4
100
50
0
-50
-100
1
10
10
10
4
100
50
-100
1
10
10
-1
10
100
Phase (degrees)
10
0
Magnitude
10
10
Phase (degrees)
10
0
Magnitude
Magnitude
10
2
3
10
10
Frequency (radians)
10
4
Q  10
50
0
-50
-100
1
10
10
10
Frequency (radians)
10
4
Q  100
With large Q, BW gets narrow.
Q : Quality factor
EMLAB
47
Parallel resonance

H(s)
I in
sL ||
1
sC
I out
1


RC
1 
I in R  sL || 1
1  2  sL 

2
sC
L 
sC 
s
1
Q0


2
 s 0 
1 s  s 
 

1

Q
1
  
s 
Q 0  0 
 0
H ( s) 
I out
EMLAB
48
Resonant circuits
These are circuits with very special frequency characteristics. And resonance is a very
important physical phenomenon
Parallel RLC circuit
Y ( j )  G  jC 
Series RLC circuit
1
Z ( j )  R  jL 
jC
1
jL
The reactance of each circuit is zero when
L 
1
 0 
C
1
LC
The frequency at which the circuit becomes purely resistive is called
the resonance frequency
EMLAB
49
Properties of resonant circuits
At resonance the impedance/admittance is minimal
Z ( j )  R  jL 
| Z |2  R 2  (L 
1
j C
Y ( j )  G 
1 2
)
C
1
jL
| Y |2  G 2  (C 
 j C
1 2
)
L
Current through the serial circuit/
voltage across the parallel circuit can
become very large (if resistance is small)
Quality Factor : Q 
0 L
R

1
0CR
Given the similarities between series and parallel resonant circuits,
we will focus on serial circuits
EMLAB
50
Properties of resonant circuits
At resonance the power factor is unity

VR


j L

j
V1
L

I
VC   j
C

CIRCUIT
SERIES
PARALLEL
Phasor diagram for series circuit
GV1
BELOW RESONANCE
CAPACITIVE
INDUCTIVE
jCV1
ABOVE RESONANCE
INDUCTIVE
CAPACITIVE
Phasor diagram for parallel circuit
EMLAB
51
Example 12.7
Determine the resonant frequency, the voltage across each
element at resonance and the value of the quality factor
1
  0 L  50
 0C
I
VC 
1
j 0 C
I   j 50  5  250  90
Q
1
1

 2000rad / sec
3
6
LC
(25  10 H )(10  10 F )
At resonance Z  2
V
100
I S 
 5A
Z
2
0 
0 L  (2 103 )(25 103 )  50
VL  j0 LI  j50  5  25090 (V )
0 L
R

50
 25
2
At resonance
VS
 Q | VS |
R
| VC | Q | VS |
| VL |  0 L
1
1 2
s LC  sRC  1
R
sC
sC
2
1  s 
1 s  

      1
sC  0  Q  0  


VS  sL 
EMLAB
Example 12.8
0 
52
Given L = 0.02H with a Q factor of 200, determine the capacitor
necessary to form a circuit resonant at 1000Hz
1
1
 C  1.27  F
 2  1000 
0.02C
LC
What is the rating for the capacitor if the
circuit is tested with a 10V supply?
100V
At resonance
VS
 Q | VS |
R
| VC | 2000V
| VC | Q | VS |
| VL |  0 L
L with Q  200  200 
I
0 L
R
R
1
1


1  sRC  s 2 LC 
sC sC
2
 s  
1 
s

  
1 
sC  0Q  0  


Z  R  sL 
2 1000  0.02
 1.59
200
10
 6.28 A
1.59
The reactive power on the capacitor
exceeds 12kVA
EMLAB
Energy transfer in resonant circuits
i (t ) 
53
V
cos  t[ A]
R
m
O
Normalization
factor
EMLAB
Quality factor in terms of Energy
54
Energy dissipated in one cycle
1
WD  Ri 2  T
2
1
WL  Li 2
2
Energy stored in L
1 2
Li
2
WL
L
1 0 L Q




1
WD
2
Ri 2  T R  T 2 R
2
W
 Q  2 L
WD
EMLAB
A series RLC circuit as the following properties:
Example 12.10
55
R  4,0  4000rad / sec, BW  100rad / sec
Determine the values of L,C.
0 
1
LC
Q
0 L
R

1
 0CR
BW 
0
Q
1. Given resonant frequency and bandwidth determine Q.
2. Given R, resonant frequency and Q determine L, C.
Q
0
BW
L
QR
C
1
0


L 02
4000
 40
100
40  4
 0.040 H
4000

1
1

 1.56  106 F
2
6
 0 RQ 4  10  16  10
EMLAB
Example 12.12
56
The Tacoma Narrows Bridge
Opened: July 1, 1940
Collapsed: Nov 7, 1940
Likely cause: wind
varying at frequency
similar to bridge
natural frequency
0  2  0.2
EMLAB
57
Displacement vs. voltage

x
3.77'


F  ma  mx  kx  bx  F0 cos  t
ms2x  kx  bsx  F0
0.44’
1.07’
m 2
b
F0
s x  sx  x 
k
k
k
k
k
0 
, Q
m
0b
EMLAB
58
Example 12.15
Increasing selectivity by cascading low Q circuits
Single stage tuned amplifier
0 
Q
1

LC
0
BW

10
6

1
H 2.54  1012 F

 6.275  108 rad / s  99.9 MHz
1 C
C
R
G L
L
2.54  1012
 250 
 0.398
6
10
EMLAB
59
12.5 Filter networks
Networks designed to have frequency selective behavior
COMMON FILTERS
High-pass filter
Low-pass filter
We focus first on
PASSIVE filters
Band-reject filter
Band-pass filter
EMLAB
Examples : voice signal spectrum
Time domain signal
60
Frequency domain spectrum
EMLAB
Example : ADSL signal spectrum
61
EMLAB
62
RS
+
Vin
-
Filter
H(s)
RL
+
Vout
-
1-st order
Low-pass filter
EMLAB
63
Prototype low pass filter
H ( s) 
Type #1
1
sC
I out
1
1



I in R  1
1  sR 2 C 1  s
2
sC
p1

1 
 p1

R
C
2


y
Type #2
log 10 p1
0 dB
 3 dB
H ( s) 
Vout
R2
1
1



Vin R2  sL 1  sL 1  s
R2
p1
R 

 p1  2 
L 

 20 dB/dec
EMLAB
64
LPF-to-HPF transformation
s
p
 1
p1
s'
y
H ( s) 
1
s
1
p1
log 10 p1
0 dB
 3 dB
log 10 
y
substituti on
1
LPF→HPF
3 dB
1
H ( s' ) 
p1
s

1
1
p1
s
log 10 p1
0 dB
 20 dB/dec
 20 dB/dec
s
1
p1
s
1
p1
EMLAB
65
LPF→HPF
replace
p12
s
s'
p12 L1
1
sL1 

1
s'
s' 2
p1 L
1
s'
 2  s' L
sC
p1 C
C1 
1
p12 L1
L1 
1
p12C1
EMLAB
66
LPF-to-BPF transformation
replace
  s'  
s
 0   0 
p1
BW  0 s' 
H ( s' ) 
y
LPF→BPF
y
H ( s) 
1
log 100
3 dB
1
s
p1
0 dB
 20 dB/dec
log 10 p1
0 dB
 3 dB
1
  s'  
1  0   0 
BW   0 s' 
log 10 
 '  1
 20 dB/dec
'  2
  s ' 0 
s
 0 
  1
p1
BW  0 s ' 
 20 dB/dec
s '  j1
0  s ' 0 

    j 0
BW  0 s ' 
BW
 1  1 

   j
 0 
s '  j 2
0  s ' 0 
    1 
    j 0  2
 j
BW  0 s ' 
BW  0 
EMLAB
67
Replace
  s'  
s
 0   0 
p1
BW  0 s' 
sL1  L1
p1 0  s'  0   p1
1

    
L1  s'
BW   0 s'   BW 
 BW 

s' 
2
 p1 0 L1 
1
1
1
1
 

p1C 02 1
sC
C p1 0  s'  0  p1C
s'
  
BW
BW s'
BW   0 s' 
EMLAB
LPF-to-BRF transformation
Replace
s

p1
0
BW
H (s' ) 
1
y
LPF→BPF
y
H ( s) 
3 dB
1
1
s
p1
log 10 p1
0 dB
 3 dB
68
1
 s ' 0 
  
 0 s' 
1
1
 0  s'  0 

 
BW   0 s ' 
log 100
0 dB
 20 dB/dec
log 10 
 20 dB/dec
 '  1
 20 dB/dec
'  2
  s ' 0 
s
 0 
   1
p1
BW  0 s ' 
s'  j1
0  s' 0 
    2 
    j 0  1
j
BW  0 s' 
BW  0 
EMLAB
s
1
replace

p1
 0  s ' 0 
  
BW  0 s' 
p1
1
sL1  L1

 02 1
1
 0  s'  0 
s'

 
L1 p1 BW s'
BW   0 s'  L1 p1BW
1
1 0
 
sC
C p1BW
69
 s' 0 
02 1
1
   
s'

s
'
Cp
BW
Cp1BW s'
1
 0

EMLAB
Example 12.19
70
A simple notch filter to eliminate 60Hz interference
vin (t )  sin 2  60t   0.2 sin 2 1000t 
L  70.3mH , C  100 F
1
L
jC
C
ZR 

1
1
jL 
j (L 
)
jC
C
jL
V0 
Req
Req  Z R
Vin
1 
1 


Z R  



V



0
0
LC 
LC 


EMLAB
71
Active filter
Passive filters have several limitations
1. Cannot generate gains greater than one
2. Loading effect makes them difficult to interconnect
3. Use of inductance makes them difficult to handle
Using operational amplifiers one can design all basic filters, and more,
with only resistors and capacitors
The linear models developed for operational amplifiers circuits are valid, in a
more general framework, if one replaces the resistors by impedances
These currents are zero
Ideal Op-Amp
EMLAB
72
Basic Inverting Amplifier
I1 
V1
Z1
I  0
V  0
V  0
Infinite gain  V  V
Infinite input impedance  I -  I   0
V1 V2

0
Z1 Z 2
V2  
Linear circuit equivalent
Z2
V1
Z1
G
Z2
Z1
EMLAB
73
Basic Non-inverting amplifier
I1  0
V1
I  0
V1
V0  V1 V1

Z2
Z1
V0 
Z 2  Z1
V1
Z1
Basic Non-inverting Amplifier
G 1
Z2
Z1
Due to the internal op-amp circuitry, it has
limitations, e.g., for high frequency and/or
low voltage situations. The Operational
Transductance Amplifier (OTA) performs
well in those situations
EMLAB
Example 12.40 “BASS-BOOST” AMPLIFIER
(non-inverting op-amp)
74
DESIRED BODE PLOT
f 
P
500
2
Find the pole and zero locations and the value of Ko
for the voltage gain
OPEN SWITCH
(6dB)
EMLAB
Example 12.41
Original player response
75
TREBLE BOOST
Desired boost
Proposed boost circuit
Non-inverting amplifier
EMLAB