Transcript No Slide Title

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
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
" Simplification 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
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) 
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
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 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
2
Quadratic pole or zero t2  [1  2 ( j )  ( j ) ]  [1  2 ( j )  ( ) ]
2

| t2 |dB  20 log 10 1  ( )
  2 
2 2
t 2  tan 1
2
  1 | t 2 |dB  0 low frequency asymptote
2
1  ( ) 2
t 2  0
  1 | t2 |dB  20 log10 ( )2 high freq. asymptote 40dB/dec t 2  180
  1 | t 2 |dB  20 log 10 (2 ) Corner/break frequency
t 2  90
  1  2 2 | t2 |dB  20 log10 2 1   2
t 2  tan
These graphs are inverted for a zero
Magnitude for quadratic pole
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/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
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
2
( j ) 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
26 |dB
20
0
 20
90
1
10
100
1000
90
LEARNING EXTENSION
Sketch the magnitude characteristic
It is in standard form
break at 50
Double pole at the origin
100(0.02 j  1
G ( j ) 
( j ) 2
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
G ( j ) 
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
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
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
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

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 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 (
 0
 )
0 


QR  j 0 QR
0


  
 R 1  jQ (  0 )
0  

M ( ) | Gv |,  ( ) | Gv
2
 1

 1 
 LO   0 
 
  1
 2Q

 2Q 
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
1
LC
Q
0 L
R

1
(2 103 )(5 106 )
Q
10
100
1
 0CR
BW 
0
Q
 104 rad / sec
R
2
0.2
Q
10
100
BW(rad/sec)
1000
100
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
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


2
1
VS
1


LC  jCR
R  jL 
jC
1
0 
LC
0 L
1

V0
Q

u

;
g

R
 0CR
0
VS
2
1
dg
2(1  u2 )( 2u)  2(u / Q )(1 / Q )
1
2
2
0

2
(
1

u
)



2
2
2
du
2 2 u


Q
2
1

u





u
2
1

u

  
Q





 Q  

 max
1
1
Q2
Q | VS |
umax 
 1  2 gmax 

|
V
|

0
0
2Q
 1
1
1  1 1
1




2

4
2
4
1

4Q
4Q  Q 2Q 
4Q 2
g ( u) 




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 2
0
2Q
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
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 ( ) 

M  

 LO 
R
1 

R  j L 

C 

RC
 HI 
RC 2   2 LC  1
2
1 
  1 M (  0)  M (  )  0
LC 
0 
M ( LO ) 
 ( R / L) 
1
LC
1
 M ( HI )
2
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
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