VARIABLE-FREQUENCY NETWORK
PERFORMANCE
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
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
Inductor
Z L  jL  L90
Capacitor
Zc 
1
1

  90
jC C
Frequency dependent behavior of 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
" Simplifica tion in notation" j  s
s 2 LC  sRC  1
Z eq ( s ) 
sC
| Z eq |
(RC )  (1   LC )
C
2
2
2
1  
LC  1 

Z eq  tan 
 RC 
2
Simplified notation for basic components
Z R ( s)  R, Z L ( s)  sL, ZC 
1
sC
For all cases seen, and all cases to be studied, the impedance is of the form
am s m  am 1s m 1  ...  a1s  a0
Z ( s) 
bn s n  bn1s n1  ...  b1s  b0
Moreover, if the circuit elements (L,R,C, dependent sources) are real then the
expression for any voltage or current will also be a rational function in s
LEARNING EXAMPLE
1
sC
R
sRC
VS  2
VS
R  sL  1/ sC
s LC  sRC  1
s  j
jRC
Vo 
VS
2
( j ) LC  jRC  1
Vo ( s) 
sL
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
USING MATLAB TO COMPUTE MAGNITUDE AND PHASE INFORMATION
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)
Missing coefficients must
be entered as zeros
This sequence will also
» num=[15*2.53*1e-3 0];
work. Must be careful not
» den=[0.1*2.53*1e-3 15*2.53*1e-3 1];
to insert blanks elsewhere
» freqs(num,den)
GRAPHIC OUTPUT PRODUCED BY MATLAB
Log-log
plot
Semi-log
plot
LEARNING EXAMPLE
A possible stereo amplifier
Desired frequency characteristic
(flat between 50Hz and 15KHz)
Log frequency scale
Postulated amplifier
Frequency Analysis of Amplifier
Vin ( s ) 
Rin
VS ( s )
Rin  1 / sC in
G ( s) 
Vo ( s) 
1 / sC o
[1000Vin ]
1 / sC o  Ro
Vo ( s ) Vin ( s ) Vo ( s )

VS ( s ) VS ( s ) Vin ( s )
Voltage Gain
Frequency domain equivalent circuit
s
 40,000 
 sC in Rin 

 

1

[
1000
]
G ( s)  
 s  40,000 
[1000]1  sC R   s  100 
1

sC
R




in in 
o o
C in Rin 
1

9

 100 
 3.18  10  10
Co Ro 1  79.58  109
6 1
1
 100 (50 Hz )
 40,000 (20kHz )
s
40,000
100 | s | 40,000  G ( s )  [1000]
s
40,000
Frequency dependent behavior is
caused by reactive elements
actual
required
NETWORK FUNCTIONS
Some nomenclature
When voltages and currents are defined at different terminal pairs we
define the ratios as Transfer Functions
INPUT
Voltage
Current
Current
Voltage
OUTPUT TRANSFER FUNCTION SYMBOL
Voltage
Voltage Gain
Gv(s)
Voltage
Transimpedance
Z(s)
Current
Current Gain
Gi(s)
Current
Transadmittance
Y(s)
If voltage and current are defined at the same terminals we define
Driving Point Impedance/Admittance
EXAMPLE
To compute the transfer functions one must solve
the circuit. Any valid technique is acceptable
I 2 ( s)  Transadmittance

V1 ( s )  Transfer admittance
V ( s)
Gv ( s)  2
Voltage gain
V1 ( s)
YT ( s ) 
LEARNING EXAMPLE
VOC ( s ) 
sL
V1 ( s )
sL  R1
The textbook uses mesh analysis. We will
use Thevenin’s theorem
1
sLR1
1

 R1 || sL 
sC sL  R1
sC
s 2 LCR1  sL  R1
ZTH ( s ) 
sC ( sL  R1 )
ZTH ( s) 
I 2 ( s)  Transadmittance

V1 ( s )  Transfer admittance
V ( s)
Gv ( s)  2
Voltage gain
V1 ( s)
YT ( s ) 
ZTH (s)

VOC (s)


sL
V1 ( s )
sL  R1
VOC ( s )
sC ( sL  R1 )
I 2 ( s) 


s 2 LCR1  sL  R1 sC ( sL  R1 )
R2  ZTH ( s )
R2 
sC ( sL  R1 )
I 2 ( s)
R2 V2 ( s )

s 2 LC
YT ( s)  2
s ( R1  R2 ) LC  s( L  R1R2C )  R1
Gv ( s ) 
Vs ( s ) R2 I 2 ( s )

 R2YT ( s )
V1 ( s )
V1 ( s)
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)
EXAMPLE
zeros : z1  1,
poles : p1  2  j 2, p2  2  j 2
H (0)  1
H ( s)  K 0
( s  1)
s 1
 K0 2
( s  2  j 2)( s  2  j 2)
s  4s  8
1
H (0)  K 0  1 
8
H ( s)  8
s 1
s 2  4s  8
LEARNING EXTENSION
Find the driving point impedance at VS (s)
I (s)
Z ( s) 
VS ( s )
I ( s)
KVL : VS ( s)  Rin I ( s ) 
Z ( s )  Rin 
1
I ( s)
sC in
1
100 
 1 
M
sC in 
s 
Replace numerical values
LEARNING EXTENSION Find the pole and zero locations and the value of K o
for the voltage gain G ( s ) 
H ( s)  K 0
Vo ( s )
VS ( s )
( 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
s
 40,000 
 sC in Rin 

 

1

[
1000
]
G ( s)  
 s  40,000 
[1000]1  sC R   s  100 
1

sC
R




in in 
o o
zero : z1  0
poles : p1  50 Hz, p2  20,000 Hz
K 0  (4  107 )
Variable
Frequency
Response
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 characteristics.
20 log10 (M ( ))
BODE PLOTS
vs log10 ( )

(

)

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
General form of a network function showing basic terms
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(
Quadratic terms for
complex conjugate poles/zeros
N
)  log N  log D
D
| 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 | ...
 20 log10 | 1  ja | 20 log10 | 1  2 b ( jb )  ( jb ) 2 | ..
z1z2  z1  z2 H ( j )  0  N 90
Display each basic term
z1
2


  z1  z2
1
1
separately and add the
3
3

tan


tan

...
1
z2
results to obtain final
1  (3 ) 2
2 bb
 tan a  tan
 ...
1  (b ) 2
1
1
answer
Let’s examine each basic term
Constant Term
the x - axis is log10
this is a straight line
Poles/Zeros at the origin
( j )
N
| ( j )  N |dB   N  20 log10 ( )

( j )  N   N 90


| 1  j |dB  20 log10 1  ( ) 2
1

j

Simple pole or zero


(1  j )  tan 1 

 1  | 1  j |dB  0 low frequency asymptote
(1  j )  0
 1  | 1  j |dB  20 log10  high frequency asymptote (20dB/dec)
The two asymptotes meet when   1(corner/break frequency)
(1  j )  90
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
Simple zero
Simple pole
Quadratic pole or zero t 2  [1  2 ( j )  ( j ) ]  [1  2 ( j )  ( )2 ]
2
| t 2 |dB  20 log10
1  ( ) 
2 2
t 2  tan 1
 2 
2
  1 | t2 |dB  0 low frequency asymptote
2
1  ( ) 2
t2  0
  1 | t2 |dB  20 log10 ( )2 high freq. asymptote 40dB/dec t2  180
  1 | t2 |dB  20 log10 (2 ) Corner/break frequency
t2  90
  1  2 2 | t2 |dB  20 log10 2 1   2 Resonance frequency
These graphs are inverted for a zero
Magnitude for quadratic pole
t 2  tan
1
1  2 2

Phase for quadratic pole

2
2
LEARNING EXAMPLE
Draw asymptotes
for each term
Generate magnitude and phase plots
Gv ( j ) 
10(0.1 j  1)
( j  1)(0.02 j  1)
Breaks/corners : 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
asymptotes
LEARNING EXAMPLE
Generate magnitude and phase plots
Draw asymptotes for each
Gv ( j ) 
Form composites
25( j  1)
( j ) 2 (0.1 j  1)
Breaks (corners) : 1, 10
dB
40
28dB
20
0
 40dB / dec
 20
90
45 / dec
 45
90
180
0.1
1
10
100
 270
Final results . . . And an extra hint on poles at the origin
 40
dB
dec
 20
dB
dec
 40
1
K0
 0     K 0 2
( j ) 2 dB
dB
dec
LEARNING EXTENSION
Sketch the magnitude characteristic
breaks : 2, 10, 100
104 ( j  2)
G ( j ) 
But the function is NOT in standard form
( j  10)( j  100)
20( j / 2  1)
We need to show about
Put in standard form G ( j ) 
4 decades
( j / 10  1)( j / 100  1)
dB
40
25 |dB
20
0
 20
90
1
10
100
1000
90
LEARNING EXTENSION
Sketch the magnitude characteristic
It is in standard form
100(0.02 j  1
G ( j ) 
( j ) 2
break at 50
Double pole at the origin
dB
40
20
0
 20
90
90
1
10
100
Once each term is drawn we form the composites
 270
1000
LEARNING EXTENSION
Put in standard form
j
G ( j ) 
( j  1)( j / 10  1)
Sketch the magnitude characteristic
G ( j ) 
10 j
( j  1)( j  10)
not in standard form
zero at the origin
breaks : 1, 10
dB
40
20
0
 20
 20dB / dec
20dB / dec
90
90
0.1
1
10
Once each term is drawn we form the composites
100
 270
LEARNING EXAMPLE
A function with complex conjugate poles
t 2  [1  2 ( j )  ( j )2 ]
Put in standard form
G ( j ) 
G ( j ) 
Draw composite asymptote
25 j
( j  0.5) ( j ) 2  4 j  100


0.5 j
( j / 0.5  1) ( j / 10) 2  j / 25  1


2  1 / 25
    0.2
  0.1 
dB
40
20
  1 | t2 |dB  20 log10 (2 )
0
8dB
 20
90
90
Behavior close to corner of conjugate pole/zero
is too dependent on damping ratio.
Computer evaluation is better
0.01
0.1
1
10
 270
100
Evaluation of frequency response using MATLAB
G ( j ) 
Using default options
25 j
( j  0.5) ( j ) 2  4 j  100


» 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)
Evaluation of frequency response using MATLAB User controlled
G ( j ) 
25 j
( j  0.5) ( j ) 2  4 j  100


>> clear all; close all %clear workspace and close any open figure
>> figure(1) %open one figure window (not STRICTLY necessary)
>> w=logspace(-1,3,200);%define x-axis, [10^{-1} - 10^3], 200pts total
>> G=25*j*w./((j*w+0.5).*((j*w).^2+4*j*w+100)); %compute transfer function
>> subplot(211) %divide figure in two. This is top part
>> semilogx(w,20*log10(abs(G))); %put magnitude here
>> grid %put a grid and give proper title and labels
>> ylabel('|G(j\omega)|(dB)'), title('Bode Plot: Magnitude response')
Evaluation of frequency response using MATLAB User controlled
Repeat for phase
Continued
USE TO ZOOM IN A SPECIFIC REGION OF INTEREST
>> semilogx(w,unwrap(angle(G)*180/pi)) %unwrap avoids jumps from +180 to -180
>> grid, ylabel('Angle H(j\omega)(\circ)'), xlabel('\omega (rad/s)')
>> title('Bode Plot: Phase Response')
No xlabel here to avoid clutter
Compare with default!
LEARNING EXTENSION
Sketch the magnitude characteristic
t 2  [1  2 ( j )  ( j ) ]
2
0.2( j  1)
G ( j ) 
j[( j / 12) 2  j / 36  1]
dB
40
  1 / 12
2  1 / 36    1 / 6
  1 | t2 |dB  20 log10 (2 )
 9.5dB
20
 20dB / dec
0
 20
0dB / dec
90
 40dB / dec
90
0.1
1
12
10
100
 270
G ( j ) 
0.2( j  1)
j[( j / 12) 2  j / 36  1]
» num=0.2*[1,1];
» den=conv([1,0],[1/144,1/36,1]);
» freqs(num,den)
DETERMINING THE TRANSFER FUNCTION FROM THE 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
LEARNING EXTENSION
Determine a transfer function from the composite
magnitude asymptotes plot
A. Pole at the origin.
Crosses 0dB line at 5
C
E
A
B
D
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)
Sinusoidal
RESONANT CIRCUITS - SERIES RESONANCE
 Im{ Z }  0

QUALITY FACTOR
RESONANT FREQUENCY
PHASOR DIAGRAM
RESONANT CIRCUITS
These are circuits with very special frequency characteristics…
And resonance is a very important physical phenomenon
Parallel RLC circuit
Series RLC circuit
Z ( j )  R  jL 
1
jC
Y ( j )  G  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
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
Properties of resonant circuits
At resonance the power factor is unity

VR


j L

j
V1
L

VC   j
I
C
GV1
jCV1

CIRCUIT
SERIES
PARALLEL
BELOW RESONANCE
CAPACITIVE
INDUCTIVE
Phasor diagram for series circuit
ABOVE RESONANCE
INDUCTIVE
CAPACITIVE
Phasor diagram for parallel circuit
LEARNING EXAMPLE
Determine the resonant frequency, the voltage across each
element at resonance and the value of the quality factor
I
1
  0 L  50
 0C
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
LEARNING EXAMPLE
0 
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?
At resonance
VS
 Q | VS |
R
| VC | 2000V
| VC | Q | VS |
| VL |  0 L
L with Q  200  200 
I
0 L
R
R
2 1000  0.02
 1.59
200
10
 6.28 A
1.59
The reactive power on the capacitor
exceeds 12kVA
LEARNING EXTENSION
Find the value of C that will place the circuit in resonance
at 1800rad/sec
0 
1
1
1
1800 
C 
0.1( H )  C
LC
0.1 18002
C  3.86 F
Find the Q for the network and the magnitude of the voltage across the
capacitor
Q
0 L
R
Q
1800  0.1
 60
3
At resonance
VS
 Q | VS | | V | 600V
C
R
| VC | Q | VS |
| VL |  0 L
M ( ) 
Resonance for the series circuit
Z ( j )  R  jL 
| Z |2  R 2  (L 
1
j C
1 2
)
C
1
1/ 2

0 2 
2 
1

Q
(

) 





0
BW 
0
Q
Claim : The voltage gain is
Gv 
VR
1

V1 1  jQ (    0 )
0
Gv 
At resonance :
 0 L  QR,  0C 
R
1
R  jL 
jC
1
QR
Z ( j )  R  j
Gv 
R
Z


R
Z ( j )
Half power frequencies
 ( )   tan 1 Q (


QR  j 0 QR
0


  
 R 1  jQ (  0 )
0  

M ( ) | Gv |,  ( ) | Gv
 LO
2
 1

 1 
  0 
 
  1
 2Q

 2Q 
 0
 )
0 
The Q factor
0 L
1
R
 0CR
For series circuit : High Q  Low R
For parallel circuit : High Q  High R (low G)
Q

dissipates
Stores as E
field
High Q  Small BW
M
Stores as M
field
Capacitor and inductor exchange stored
energy. When one is at maximum the
other is at zero
Q  2
Q can also be interpreted from an
energy point of view
WS
maximum energy stored
 2
WD
energy dissipated by cycle
2
WD  RI eff

2
0

1
2
2
RI mx

2
0
1 2
1
2
W S  LI mx
 CVmx
2
2
Ws
L 0
Q


WD 2  R 2
ENERGY TRANSFER IN RESONANT CIRCUITS
i (t ) 
V
cos  t[ A]
R
m
O
Normalization
factor
LEARNING EXAMPLE
Determine the resonant frequency, quality factor and
bandwidth when R=2 and when R=0.2
2
5 F
2mH
0 
0 
R
2
0.2
Q
1
LC
Q
0 L
R

1
(2  103 )(5  106 )
Q
10
100
10000  0.002
R
1
 0CR
BW 
0
Q
 104 rad / sec
R
2
0.2
Q
10
100
BW(rad/sec)
1000
100
BW  10000 / Q
Evaluated with EXCEL
LEARNING EXTENSION
A series RLC circuit as the following properties:
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
L
C
0
BW
QR
0
1
L 02

4000
 40
100

40  4
 0.040 H
4000

1
1
6


1
.
56

10
F
2
6
 0 RQ 4  10  16  10
LEARNING EXAMPLE
Find R, L, C so that the circuit operates as a band-pass filter
with center frequency of 1000rad/s and bandwidth of 100rad/s
Gv 
0 
R
R  jL 
1
LC
1
jC
Q

R
Z ( j )
0 L
R

1
 0CR
BW 
0
Q
dependent
Strategy:
1. Determine Q
2. Use value of resonant frequency and Q to set up two equations in the three
unknowns
3. Assign a value to one of the unknowns
0
1000
Q

 10
BW 100
1
1
0 
 (103 ) 2 
LC
LC
Q
0 L
R
 10 
1000 L
R
For example C  1 F  106 F
L  1H
R 100
PROPERTIES OF RESONANT CIRCUITS: VOLTAGE ACROSS CAPACITOR
At resonance
| V0 | Q | VR |
But this is NOT the maximum value for the
voltage across the capacitor
1
jC
V0
1


1
VS
1   2 LC  jCR
R  jL 
jC
0 L
2
1

V
Q

0
R
 0CR u   ; g  V
0
S
1
dg
2(1  u 2 )(2u)  2(u / Q )(1 / Q )
g ( u) 
1
2
2
0

2
(
1

u
)



2
2
2
du
2 2 u


Q
2
1

u





u
2
1

u

  
Q





 Q  

1
0 
LC

umax


1
 max  1 
0
2Q 2

gmax

1
Q2
Q | VS |


|
V
|

0
 1
1
1  1 1
1




2

4
2
4
1

4Q
4Q  Q
2Q 
4Q 2
LEARNING EXAMPLE
Determine 0 , max when R  50 and R  1
50mH
0 
5 F
1
LC
umax 
0 
Q
1
1

 2000rad / s
2
6
LC
(5  10 )(5  10 )
2000 0.050
R
 max  2000  1  1
R
50
1
Q
0 L
R

1
 0CR
 max
1
 1
0
2Q 2
2Q 2
Q Wmax
2
1871
100 2000
Evaluated with EXCEL and rounded to zero decimals
Using MATLAB one can display the frequency response
R=50
Low Q
Poor selectivity
R=1
High Q
Good selectivity
LEARNING EXAMPLE
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
Tacoma Narrows Bridge Simulator
Assume a low Q=2.39
31.66  F
20 H
9.5
Vinmx  42
(1V  1 ft )
1
At failure a 42mph wind caused 4' deflection.
For the model at resonance
v0
RB
4


vin RA  RB 42
3.77'
0.44’
1.07’
PARALLEL RLC RESONANT CIRCUITS
Impedance of series RLC
Admittance of parallel RLC
1
1
Y ( j )  G 
 j C
Notice
equivalenc
es
j C
jL
1 2 R  G, L  C , C  L
1 2
| Z |2  R 2  (L 
)
| Y |2  G 2  (C 
)
Z

Y
,
V

I
C
L
Z ( j )  R  jL 
Series RLC
I S  YVS
0 
1
LC
Q
0 L
R

1
 0CR
Parallel RLC
At resonance
1
0 
1
LC
Q
 0C
G

 0C 
Y G
G
0
0 L
IG  GVS  I S
BW

Series RLC
Y
Q
IG  I S
j C
0
I C  jCVS 
IS I  I
Parallel
RLC
BW

C
L
Y
Q

C
1
0
1
jL | I C | G | I S |  Q | I S |
IL 
VS 
IS
j L
Y
1
| I L |
|I |
 0 LG S
1
 0 LG
VARIATION OF IMPEDANCE AND PHASOR DIAGRAM – PARALLEL CIRCUIT
LEARNING EXAMPLE
If the source operates at the resonant frequency of the
network, compute all the branch currents
IG  0.011200  1.20( A)  I S
1
1
0 

 117.85rad / s

4
LC
0.120  (6  10 )
VS  1200, G  0.01S
C  600  F ,
IC  (190)  (117.85)  (600 106 ) 1200  8.4990( A)
L  120mH
At resonance
1
 0C 
Y G
G
0 L
IG  GVS  I S
Y
IG  I S
j C
I C  jCVS 
IS I  I
C
L
Y
 0C
1
|
I
|

| IS |  Q | IS |
C
1
j L
G
IL 
VS 
IS
j L
Y
1
| I L |
|I |
 0 LG S
I L  8.49  90( A)
I x  _______
Derive expressions for the resonant frequency, half power
frequencies, bandwidth and quality factor for the transfer
characteristic
Vout
LEARNING EXAMPLE
H
I in
2
G
1
G 
h  
 
 
2C
 2C  LC
Vout
YT  G  jC 
I
V
1
 in  H  out 
YT
I in YT
| H |
1
1
1
jL
1

2
2
Half power frequencies | H ( j h ) |2  0.5 | H |2max
2
Q
0
BW

G
C
1 C
C
R
G L
L
Replace and show
1 

G


C



jL
L 

1
1
Resonant frequency :  0 
| H max |  R
LC
G
G  jC 
BW   HI   LO 
1

1 
 G
  2G 2   hC 
G 2    hC 

L
h L 
h

Q
 LO
 0C
G

1
 0 LG
2
 1

 1 
  0 
 
  1
 2Q

 2Q 
LEARNING EXAMPLE
Increasing selectivity by cascading low Q circuits
Single stage tuned amplifier
0 
1

LC
10
6

1
H 2.54  1012 F

1 C
C
Q 0 
R
BW G L
L
2.54  1012
 250 
 0.398
6
10

 6.275  108 rad / s  99.9 MHz
Determine the resonant frequency, Q factor and bandwidth
LEARNING EXTENSION
R  2k, L  20mH , C  150 F
Parallel RLC
1
LC
0 
0 
Q
 0C
G


1
BW  0
Q
 0 LG
1
3
6
(20  10 )(150  10 )
577  150  106
Q
 173
1/ 2000
BW 
577
 3.33rad / s
173
 577 rad / s
LEARNING EXTENSION Determine L, C,
R  6k, BW  1000rad / s, Q  120
Parallel RLC
1
LC
0 
Q
 0C
G


1
BW  0
Q
 0 LG
0  Q  BW  120 1000  1.2 105 rad / s
C
Q
120

 0.167  F
R0 6000 1.2  105
L
R
6000

 417  H
Q0 120  1.2  105
Can be used to verify computations
0
PRACTICAL RESONANT CIRCUIT
The resistance of the inductor coils cannot be
neglected
I
. At resonance the voltage and impedance are maxima
Y
2
   2   L 2 
   R L 2 
R 2   R L
  R1   R   0  
Z MAX 
 R1  

  R  
   0   R  
R




Z MAX  RQ02
V  ZI 
Y ( j )  jC 
1
R  jL

R  jL R  jL
R  jL
R 2  (L) 2


R
L


Y ( j )  2

j
C

2
2
2

R  (L)
R  (L) 

How do you define a quality factor for
this circuit?
Y ( j )  jC 
L
1  R
Y real  C  2

0



 
R
LC  L 
R  (L)2
1
1
0 L



1

0 
, Q0 
 R
0
2
LC
R
Q0
2
LEARNING EXAMPLE
1
1
 L
, Q0  0   R   0 1  2
Q0
LC
R
0 
0 
Q0 
Determine both 0 ,  R for R  50, 5
1
3
6
(50  10 H )(5  10 F )
 2000rad / s
2000  0.050
1
,  R  2000 1  2
R
Q0
R
50
5
Q0
2
20
Wr(rad/s) f(Hz)
1732
275.7
1997
317.8
RESONANCE IN A MORE GENERAL VIEW
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
For series connection the impedance reaches maximum at resonance. For parallel
connection the impedance reaches maximum
Ys 
jC
( j ) 2 LC  jCR  1
Zp 
jL
( j ) 2 LC  jLG  1
In Bode plots the quadratic term was written as
( j )2  2 j  1
  LC 
series
2  CR  2   0CR 
A high Q circuit is highly
under damped
1
0
parallel
1
Q
Q
2  LG  2   0 LG 
1
2
1
Q
Resonance
SCALING
Scaling techniques are used to change an idealized network into a more
realistic one or to adjust the values of the components
Magnitude or impedance scaling
Frequency or time scaling
R'  K M R
ω'  K F ω
L'  K M L
Impedance of each component is unchanged
C '
C
KM
 ' L'  L,
LC  L' C '   0 
Q
 0 L  0 L'
R

1
1

LC
L' C '
R'
Magnitude scaling does not change the
frequency characteristics nor the quality
of the network.
1
1

 ' C ' C
R'  R
L' 
L
KF
C' 
C
KF
 0'  K F 0
Q' 
BW ' 
0' L'
 0'
Q'
R'
Q
Constant Q
networks
 K F ( BW )
LEARNING EXAMPLE
Determine the value of the elements and the characterisitcs
of the network if the circuit is magnitude scaled by 100 and
frequency scaled by 1,000,000
 0  2rad / s, Q 
 1H
R'  R
1
F
2
 2
Magnitude or impedance scaling
R'  K M R
BW ' 
C
C '
KM
R'  200
1
C'
F
200
L' '  100mH
C ''
C
KF
 0'  K F 0
L'  K M L
L'  100 H
R' '  200
L
L' 
KF
C' 
Q,0 are unchanged
 0'
Q'
2
, BW  2
2
1
F
200
0''  1.414 106 rad / s
 K F ( BW )
LEARNING EXTENSION
An RLC network with R  10, L  1H, C  2F is magnitude
scaled by 100 and frequency scaled by 10,000. Determine
the resulting circuit elements
Magnitude or impedance scaling
Frequency scaling
R'  K M R
R'  R
L'  K M L
C
C '
KM
R'  1000
L'  100 H
C '  0.02 F
L
L' 
KF
C
C 
KF
'
R' '  1k
L' '  0.01H
C ' '  2 F
Scaling
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
Simple low-pass filter
1
V
1
jC
Gv  0 

V1 R  1
1  jRC
jC
1
Gv 
;   RC
1  j
M ( ) | Gv |
1
1   
2
Gv   ( )   tan 1 
1 1

M max  1, M    

2

1
   half power frequency

BW 
1

Simple high-pass filter
Gv 
V0
R
jCR


V1 R  1
1  jCR
jC
Gv 
j
;   RC
1  j

M ( ) | Gv |
Gv   ( ) 
1   
2

2
 tan 1 
1 1

M max  1, M    

2

1
   half power frequency

 LO 
1

Simple band-pass filter
Band-pass
V
Gv  0 
V1
M ( ) 
 LO 
R
1 

R  j  L 

C 

RC
 HI 
RC 2   2 LC  1
2
1 

M  
  1 M (  0)  M (  )  0
LC 

0 
M ( LO ) 
1
LC
1
 M ( HI )
2
 ( R / L) 
 R / L2  4 20
2
( R / L) 
 R / L2  4 20
2
BW   HI   LO 
R
L
Simple band-reject filter

1
1 
  0
 j  0 L 
LC

C

0 
at   0 the capacitor acts as open circuit  V0  V1
0 
at    the inductor acts as open circuit  V0  V1
 LO ,  HI are determined as in the
band - pass filter
LEARNING EXAMPLE
Depending on where the output is taken, this circuit
can produce low-pass, high-pass or band-pass or bandreject filters
Band-reject filter
Band-pass
Bode plot for R  10, L  159H , C  159 F
VL

VS
VC

VS
jL
1 

R  j  L 

C 

1
jC
1 

R  j  L 

C 

VL
  0  0, VL (  )  1
VS
VS
VC
  0  1, VC (  )  0
VS
VS
High-pass
Low-pass
LEARNING EXAMPLE
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 


LEARNING EXTENSION
Sketch the magnitude characteristic of the Bode plot for Gv ( j )
1
1
jC
Gv ( j ) 

1
1  jRC
R
jC
Break/corner frequency : 5rad/s
low frequency asymptote of 0dB/dec
  RC  (10 103 )(20 106 F )  0.2rad / s
High frequency asymptote of - 20dB/dec
LEARNING EXTENSION
Sketch the magnitude characteristic of the Bode plot for Gv ( j )
20dB/dec. Crosses 0dB at  
Gv ( j ) 
  RC  (25 103 )(20 106 F )  0.5rad / s
R
R
1
jC

1

 2rad / s
jRC
1  jRC
Break/corner frequency : 2rad/s
low frequency asymptote of 0dB/dec
High frequency asymptote of - 20dB/dec
LEARNING EXTENSION
Sketch the magnitude characteristic of the Bode plot for Gv ( j )
Band-pass
20dB/dec. Crosses 0dB at  
Gv ( j ) 
1
0 
 1000
LC
1
 1000rad / s
RC
R
jRC

2
1
1

j

RC

(
j

)
LC
R
 jL
jC
 2  LC    106  103 ,
Break/corner frequency : 1000 rad/s
103
3
6
2  RC  10  10   
 0.5
2  103
High frequency asymptote of - 40dB/dec
 LO 
 HI 
 ( R / L) 
 R / L2  4 20
2
( R / L) 
 R / L2  4 20
2
low frequency asymptote of 0dB/dec
 618rad / s
 1618rad / s
 40dB / dec
ACTIVE FILTERS
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
Basic Inverting Amplifier
I1 
V1
Z1
I  0
V  0
V  0
Infinite gain  V  V
Infinite input impedance  I -  I   0
V1 VO

0
Z1 Z 2
VO  
Z2
V1
Z1
G
Z2
Z1
Linear circuit equivalent
EXAMPLE
USING INVERTING AMPLIFIER
LOW PASS FILTER
Basic Non-inverting amplifier
V1
I1  0
I  0
V1
V0  V1 V1

Z2
Z1
V0 
Z 2  Z1
V1
Z1
G 1
Z2
Z1
EXAMPLE
USING NON INVERTING CONFIGURATION
EXAMPLE
SECON ORDER FILTER
V V
V
V V V

 
0
R
1/ C s R
R
2
IN
2
1
1
V
V
 
0
R 1/ C s
2
V ( s)
2
2
2
O
2
2
2
O
3
Due to the internal op-amp circuitry, it has
Operational Transductance Amplifier (OTA)
limitations, e.g., for high frequency and/or
low voltage situations. The Operational
Transductance Amplifier (OTA) performs
well in those situations
Ideal OTA : Rin  R0  
COMPARISON BETWEEN OP-AMPS AND OTAs – PHYSICAL CONSTRUCTION
Comparison of Op-Amp and OTA - Parameters
Amplifier Type
Op-Amp
OTA
Ideal Rin Ideal Ro Ideal Gain Input Current input Voltage
0
0
0
gm
0
nonzero




Basic Op-Amp Circuit
Basic OTA Circuit
R0
gm v in
R0  RL
RL
v0 
Av vin
R0  RL
i0 
Rin
vin 
VS
RS  Rin
vin 
Rin
VS
RS  Rin
 RL   Rin 
v
A 0 
Av
VS  R0  RL   RS  Rin 
Gm 
i0  R0   Rin 

gm 

v in  R0  RL   RS  Rin 
Ideal OTA
Ideal Op - Amp
A  Av  
Gm  g m
Basic OTA Circuits
i0  gm v1
1t
v0   i0 ( x )dx  v0 (0)
C0
Integrator
gm t
v0 ( t ) 
v1 ( x )dx  v0 (0)

C 0
In the frequency domain
V0 
gm
V1
jC
i0   gm vin (notice polarity)
iin  i0  0
Simulated Resistor
vin
1
 Req 
iin
gm
OTA APPLICATION
gm1v1
gm1 v1  gm2 v2
gm 2v2
Basic OTA Adder
v0 
1
( gm1 v1  gm2 v2 )
gm 3
Simulated Resistor
Equivalent representation
Programmability of gm
 S
gm1  20  I ABC
 A
Typical values
gm  10mS
gm range : 3 - 7 decades
(e. g., gm 
10mS
)
7
10
Controlling transconductance
LEARNING EXAMPLE
Produce a 25k resistor
gm  4mS
4mS
 4  107 S
4
10
gm  20 I ABC
gm 
Simulated Resistor
vin
1
 Req 
iin
gm
25  103 
1
 gm  4  105 S  4  107 S
gm
S
4 105 S  20  I ABC ( A)
 A
I ABC  2 106 A  2A
LEARNING EXAMPLE
Floating simulated resistor
i0   gmvin
One grounded terminal
i01   gm1v1
i1  i01
i02  gm 2v1
i02  i1
For proper operation
gm1  gm 2
Produce a 10M resistor
gm  4mS
4mS
 4  107 S
4
10
gm  20 I ABC
1
7
gm 

10
S  4 107 S
6
10 10
gm 
The resistor cannot be produced
with this OTA!
LEARNING EXAMPLE
Select gm1 , gm 2 , gm 3 , to produce
a) v0  10v1  2v2
gm  4mS
4mS
7

4

10
S
4
10
gm  20 I ABC
gm 
b) v0  10v1  2v2
Case b
Reverse polarity of v2!
v0 
1
( gm1 v1  gm2 v2 )
gm 3
Case a
gm1
g
 10; m 2  2
gm 3
gm 3
Two equations in three unknowns.
Select one transductance
1
104 ( A)  5A
20
gm 2  0.2mS  I ABC 2  10A
gm 3  0.1mS  I ABC 3 
gm1  1mS  I ABC1  50A
ANALOG MULTIPLIER
Based on ‘modulating the control current
ASSUMES VG IS ZERO
AUTOMATIC GAIN CONTROL
For simplicity of analysis
we drop the absolute value
v small  v  Av
IN
O
v big  v 
IN
O
A
B
IN
OTA-C CIRCUITS
Circuits created using capacitors, simulated resistors, adders and integrators
resistor
Magnitude Bode plot
Gv 
V0
Vi1
integrator
Frequency domain analysis assuming
ideal OTAs
V0 
1
IC
jC
I01  gm1Vi1
V0 
IC  I01  I02
I 02   gm 2V0
1
gm1Vi1  gm 2V0 
jC
gm1
V0 
gm 2
Vi1
C
1  j
gm 2
g
Adc  m1
gm 2
gm 2
C
g
2 f C  m 2
C
C 
LEARNING EXAMPLE
gm  1mS
V
4
Desired : Gv  0 
Vi1 1  j
2 (105 )
1mS
 106 S
3
10
gm  20 I ABC
gm 
Find the transductances and biases
Adc 
gm1
gm 2
Adc  4 C  2 (105 )  fC  100kHz
gm 2
C
g
2 f C  m 2
C
C 
Two equations in three unknowns.
Select the capacitor value
C  25 pF
gm 2  2 (10 )(25 10
5
gm1  62.8S  I ABC1  3.14A
12
6
)  15.7 10 S
I ABC 2 
15.7
 0.785A
20
OK
TOW-THOMAS OTA-C BIQUAD FILTER
biquad ~ biquadratic
i03
V0 A( j ) 2  B ( j )  C

Vi ( j ) 2   0 ( j )   2
0
Q
V02 
V01  gm1
Vi1  V02
jC
I02  gm 2 (V01  Vi 2 )
1
( I 02  I 03 )
jC 2
I03  gm3 (Vi 3  Vo 2 )
Four equations and four unknowns (V01, V02 , I01, I02 )
 jC 2 gm 3 
 gm 3 
 jC1 
 jC1 gm 3 

V

V

Vi 3
i1
i2
 g



V

V

Vi 3
i1 
i2



gm 2 
gm 2 


m2
 gm1 
 gm1 gm 2 
V01 
V

02
 C1C 2 
2  gm 3C1 
 C1C 2 
2  gm 3C1 
 g g  ( j )  g g  ( j )  1
(
j

)
g g 
 g g  ( j )  1
 m1 m 2 
 m 2 m1 
 m1 m 2 
 m 2 m1 

g g

g
g g
C2
gm


0  m1 m 2 , 0  m 3 , Q  m12 m 2
 0
C
C1C 2
Q C2
C1
gm 3
gm1  gm 2   Q  gm
 
Filter Type
A
B
C
gm 3
C

C
1
2 
Low-pass
0
0
nonzero

Band-pass
0
nonzero
0
 BW  gm 3
High-pass nonzero
0
0

C
LEARNING EXAMPLE
Design a band - pass filter with center frequency of 500kHz,
bandwidth of 75kHz, and center frequency gain - 5.
Use the Tow - Thomas configurat ion and 50 - pF capacitors
C1  C2
gm  4mS
4mS
 4  107 S
4
10
gm  20 I ABC
gm 
Vi 3  0
BW
0 
gm1 gm 2 0 gm 3
,

,Q
C1C 2
Q C2
gm1 gm 2 C 2
C1
gm2 3
gm 3
 2  75 103  23.56S
12
50 10
g
| Gv ( j 0 ) | m 2  5  gm 2  117.8S
gm 3
BW 
 0 2  2  5 10

5 2
 jC1 
 jC1 gm 3 
Vi1  
V

 i 2  g g Vi 3
g
 m1 
 m1 m 2 
V02 
 C1C 2 
2  gm 3C1 
(
j

)
g g 
 g g  ( j )  1
 m1 m 2 
 m 2 m1 
 C1C 2 
( j ) 2  0

 gm1 gm 2 
  0  1  
gm1  117.8  106

 gm1  209.5S
(5  1013 ) 2
I ABC1  10.47 A
I ABC 2  5.89A
I ABC 3  1.18A
Bode plots for resulting amplifier
LEARNING BY APPLICATION
Using a low-pass filter to reduce 60Hz ripple
Using a capacitor to create a lowpass filter
1
VTH
1  jRTH C
| VTH |
| VOF |
2
1  RTH C 
VOF 
C 
1
RTH C
Design criterion: place the corner frequency
at least a decade lower
| VOF | 0.1 | VTH |
Thevenin equivalent for AC/DC
converter
500C 
1
 C  53.05 F
2  6
Filtered output
LEARNING EXAMPLE
Single stage tuned transistor amplifier
Select the capacitor for maximum
gain at 91.1MHz
Antenna Transistor
Voltage
Parallel resonant circuit
V0
4 


R || jL || 1
jC 
VA
1000 
4
1
j / C

1000 1  1  jC j / C
R jL
V0
4
j / C

VA
1000 ( j ) 2  j  1
RC LC
Band - pass with center frequency 1 / LC
1
2 91.1 106 
 C  3.05 pF
6
10 C
V0 
1 
4
R  100
 

VA 
LC  1000



Magnitude Bode plot for
V0
VA
LEARNING BY DESIGN
Anti-aliasing filter
Nyquist Criterion
When digitizing an analog signal, such as music, any frequency components
greater than half the sampling rate will be distorted
In fact they may appear as spurious components. The phenomenon is known as
aliasing.
SOLUTION: Filter the signal before digitizing, and remove all components higher
than half the sampling rate. Such a filter is an anti-aliasing filter
For CD recording the industry standard is to sample at 44.1kHz.
An anti-aliasing filter will be a low-pass with cutoff frequency of 22.05kHz
Single-pole low-pass filter
Resulting magnitude Bode plot
V01
1

Vin 1  jRC
C 
1
 2  22,050
RC
C  1nF  R  72.18k
Attenuation
in audio range
Improved anti-aliasing filter
Two-stage buffered filter
n - stage

V0 n
1

Vin 1  jRC n
v 01

V01
1

Vin 1  jRC
Four-stage
V02
1

V01 1  jRC
Two-stage
One-stage
LEARNING BY DESIGN
Notch filter to eliminate 60Hz hum
http://www.wiley.com/college/irwin/0470128690/animations/swf/12-38.swf
Notch filter characteristic
Magnitude Bode plot
Vamp
Vtape

Ramp
Ramp  Rtape  sL || 1 / sC 
C  10  F


L  0.704mH


Vamp
Ramp
s 2 LC  1



Vtape Ramp  Rtape  2

 
L
  1
 s LC  s



 Ramp  Rtape  
1
To design, pick one, e.g., C and determine the other
notch frequency 
LC
DESIGN EXAMPLE
ANTI ALIASING FILTER FOR MIXED MODE CIRCUITS
Signals of different
frequency and the same
samples
Visualization of aliasing
Ideally one wants to eliminate frequency components
higher than twice the sampling frequency and make
sure that all useful frequencies as properly sampled
Design specification
Simplifying assumption
Infinite input resistance (no load on RC circuit)
Design equation
 R  15.9k
http://www.wiley.com/college/irwin/0470128690/animations/swf/12-40.swf
DESIRED BODE PLOT
DESIGN EXAMPLE “BASS-BOOST” AMPLIFIER
(non-inverting op-amp)
f 
P
500
2
OPEN SWITCH
(6dB)
Switch closed??
DESIGN EXAMPLE
TREBLE BOOST
Original player response
http://www.wiley.com/college/irwin/0470128690/animations/swf/12-40.swf
Desired boost
Design equations
Proposed boost circuit
Non-inverting amplifier
Filters