Transcript No Slide Title

VARIABLE-FREQUENCY NETWORK
PERFORMANCE
Variable-Frequency Response Analysis
Network performance as function of frequency.
Transfer function
Sinusoidal Frequency Analysis
Bode plots to display frequency response data
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
" Simplification in notation" j  s
s 2 LC  sRC  1
Z eq ( s) 
sC
(RC)  ( LC  1)
| Z eq |
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
1
sC
sL
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) 
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
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) 
EXAMPLE
VOC ( s ) 
sL
V1 ( s )
sL  R1
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) 
V2 ( 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
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 log 10 (M ( ))
BODE PLOTS
vs log 10 ( )

(

)

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 log 10 | V |
By extension
I |dB  20 log 10 | I |
G |dB  20 log 10 | 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
Quadratic terms for
complex conjugate poles/zeros
N
)  log N  log D
D
| H ( j ) |dB  20 log 10 | H ( j ) |  20 log 10 K 0  N 20 log 10 | j |
log(
 20 log 10 | 1  j1 | 20 log 10 | 1  2 3 ( j 3 )  ( j 3 ) 2 | ...
 20 log 10 | 1  j a | 20 log 10 | 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 1  a  tan 1
 ...
1  ( b ) 2
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 log 10 ( )

( j )  N   N 90

2


|
1

j

|

20
log
1

(

)
dB
10
Simple pole or zero 1  j 

(1  j )  tan 1 

 1  | 1  j |dB  0 low frequency asymptote
(1  j )  0
 1  | 1  j |dB  20 log 10  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
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/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
asymptotes
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