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EMLAB
1
12. Variable frequency network performance
EMLAB
2
Variable-Frequency Response AnalysisNetwork performance as function of frequency.Transfer function
Sinusoidal Frequency AnalysisBode plots to display frequency response data
Resonant CircuitsThe resonance phenomenon and its characterization
ScalingImpedance and frequency scaling
Filter NetworksNetworks with frequency selective characteristics:low-pass, high-pass, band-pass
Learning Goals
EMLAB
3
0RRZRResistor
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 performancevaries with the frequency.
Variation in impedance of basic components
EMLAB
4
90LLjZL
Inductor
EMLAB
5
Capacitor
9011
CCjZc
EMLAB
6
Series RLC network
Cj
RCjLCj
CjLjRZeq
1)(1 2
C
LCjRC
j
j
)1( 2
C
LCRCZeq
222 )1()(||
RC
LCZeq
1tan
21
sC
sRCLCssZeq
1)(
2
Driving Point functions
Simplification in notation sj
EMLAB
7
7
Pulse represented by sum of sinusoidal signal
EMLAB
8
General form
011
1
011
1
...
...
)(
)()(
bsbsbsb
asasasa
sI
sVsZ
nn
nn
mm
mm
sCZsLsZRsZ CLR
1,)(,)(
Network functions for basic components
Network functions
When voltages and currents are defined at different terminal pairs we define the ratios as Transfer Functions
If voltage and current are defined at the same terminals we defineDriving Point functions (Impedance/Admittance)
Nomenclature
INPUT OUTPUT TRANSFER FUNCTION SYMBOLVoltage Voltage Voltage Gain Gv(s)Current Voltage Transimpedance Z(s)
Current Current Current Gain Gi(s)
EMLAB
9
Nomenclature
CircuitCircuit
terminals
Port #1
terminals
Port #2
} ,{
},{)(
11
22
IV
IVsH
V2 or I2
Transfer function
)2,1()(,)( iV
IsY
I
VsZ
i
i
i
iDriving point function
1I2I
2V
1V
EMLAB
10
Example 12.1
sL
sC
1
R
So VsCsLR
RsV
/1)(
SV
sRCLCs
sRC
12
So VRCjLCj
RCjV
js
1)( 2
0101)1053.215()1053.21.0()(
)1053.215(332
3
jj
jVo
MATLAB can be effectively used to compute frequency response characteristics
0 200 400 600 800 10000
0.2
0.4
0.6
0.8
1
0 200 400 600 800 1000-100
-50
0
50
100
|| 0V 0V
EMLAB
11
Some Matlab command
754
132A A= [2,3,1; 4, 5, 7]
97531A A= [1:2:9]
TAA A= A’
01
11
1 axaxaxay nn
nn
);x,a(polyvaly
];a,a,,a,a[a
1.e3];:2:[1x
011nn
01
11
1
01
11
1
bxbxbxb
axaxaxaH
nn
nn
mm
mm
./den;numH
x);polyval(b,den
x);polyval(a,num
];b,b,,b,[bb
];a,a,,a,[aa
1e3];:2:1[x
011nn
011mm
EMLAB
12
0 200 400 600 800 10000
0.2
0.4
0.6
0.8
1
|| 0V
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);
Matlab example
EMLAB
13
Example
)104(
000,20,50 :poles
0 :zero
70
21
1
K
HzpHzp
z
)(
)()(
sV
sVsG
S
o
ooinin
inin
S
O
RsCRsC
RsC
V
VsH
1
1
1)(
000,40
000,40
100 ss
s
For this case the gain was shown to be
))...()((
))...()(()(
21
210
n
m
pspsps
zszszsKsH
Zeros = roots of numeratorPoles = roots of denominator
Find the pole and zero locations and the value of K0 for the voltage gain
EMLAB
14
0 2 4 6 8 10
x 105
-100
-80
-60
-40
-20
0
20
40
60
80
100
0 2 4 6 8 10
x 105
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
clear;
clf;
w = logspace(1,6,200);
b = [1.e0 * 40e3*pi, 0];
a = [1, 40.1e3*pi, 4e6*pi*pi];
num = polyval(b,i*w);
den = polyval(a,i*w);
H = num./den;
%arg = angle(H)*180/pi;
mag = abs(H);
plot(w,mag);
%semilogx(w,mag);
%freqs(num, den,w);
clear;
clf;
w = logspace(1,6,200);
b = [1.e0 * 40e3*pi, 0];
a = [1, 40.1e3*pi, 4e6*pi*pi];
num = polyval(b,i*w);
den = polyval(a,i*w);
H = num./den;
%arg = angle(H)*180/pi;
mag = abs(H);
plot(w,mag);
%semilogx(w,mag);
%freqs(num, den,w);
|| H
H
EMLAB
15
Poles and Zeros
(More nomenclature)
011
1
011
1
...
...)(
bsbsbsb
asasasasH n
nn
n
mm
mm
; Arbitrary network function
Using the roots, every (monic) polynomial can be expressed as aproduct of first order terms
))...()((
))...()(()(
21
210
n
m
pspsps
zszszsKsH
functionnetwork theof poles,...,,
functionnetwork theof zeros,...,,
21
21
n
m
ppp
zzz
The network function is uniquely determined by its poles and zerosand its value at some other value of s (to compute the gain)
EMLAB
16
011
1
011
1
...
...)(
bsbsbsb
asasasasH n
nn
n
mm
mm
Positive real function
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.
))...()((
))...()(()(
21
210
n
m
pspsps
zszszsKsH
EMLAB
17
Sinusoidal frequency analysis
)(sH
Circuit represented bynetwork function
)cos(0
)(0
tB
eA tj
)(cos|)(|
)(
0
)(0
jHtjHB
ejHA tj
)()()(
)()(
|)(|)(
Notation
jeMjH
jH
jHM
stics.characteri phase and magnitude
calledgenerally are offunction as ),(),( of Plots M
)(log)(
))(log2010
10
PLOTS BODE vs
(M
. of functiona as )( functionnetwork the
analyze wefrequency theof functiona asnetwork a ofbehavior study the To
jH
EMLAB
18
Transfer function
H(s)H(s)+
Vin-
+Vout
-
0 2 4 6 8 10
x 105
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
)(sH
)cos(0
)(0
tB
eA tj
)(cos|)(|
)(
0
)(0
jHtjHB
ejHA tj
)()()( sVsHsV inout
|| H
0 2 4 6 8 10
x 105
-100
-80
-60
-40
-20
0
20
40
60
80
100
H
EMLAB
19
Drawing transfer function is difficult!
))...()((
))...()(()(
21
210
n
m
pspsps
zszszsKsH
||log...||log||log
||log...||log||log||log
))...()((
))...()((log|)(|log
10210110
10210110010
21
2101010
n
m
n
m
pspsps
zszszsK
pspsps
zszszsKsH
|)(|log|)(|log|)(||)(|log
|)(||)(|)()(
|)(|)(
2121
)]()([2121
)(
21
sHsHsHsH
esHsHsHsH
esHsHsHsHj
sHj
)(...)()(
)(...)()(
))...()((
))...()(()(
21
210
21
210
n
m
n
m
pspsps
zszszsK
pspsps
zszszsKsH
Magnitude :
Phase :
Bode plot
EMLAB
20
History of the Decibel
Originated as a measure of relative (radio) power
1
2)2 log10(|P
PP dB 1Pover
21
22
21
22)2
22 log10log10(|
I
I
V
VP
R
VRIP dB 1Pover
By extension
||log20|
||log20|
||log20|
10
10
10
GG
II
VV
dB
dB
dB
Using log scales the frequency characteristics of network functionshave simple asymptotic behavior.The asymptotes can be used as reasonable and efficient approximations
EMLAB
21
Standard form of a transfer function
n
m
ps
ps
ps
zs
zs
zs
KsH
1...11
1...11
)(
21
210
m
m
p
s
p
s
p
s
z
s
z
s
z
sKsH
1log20...1log201log20
1log20...1log201log20||log20|)(|log20
102
101
10
102
101
1001010
On applying to log function, terms containing poles and zeroes can be separated. Then, it becomes easier to draw the transfer function.
EMLAB
22
2
110
110 1log101log20
zz
sy
1
1
11
tan11zz
jz
s
Magnitude Phase
0
90
110log z
45dB0
110log z
y
dB3 dB/dec20
How to draw Bode plot of simple zero or polesBode plots can be drawn simply following the procedures below.
1. Draw first two extreme regions where and .
2. Connect those two regions with a smooth curve.
3. Mark the exact point where .
1/ 1 z 1/ 1 z
1/ 1 z
EMLAB
23
Bode plot of quadratic polesDrawing 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.
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.
21
2
00
11
1
21
1)(
ps
psss
sH
)1( )1( )1(
EMLAB
24
0
2
0
2
21
2
00
11
1
11
1
21
1)(
ssps
psss
sH
(1) p1, p2 same
dB/dec40
|)(|log20 10 jH
dB/dec40
|)(|log20 10 jH
110log p110log p
210log p
dB/dec20
(2) p1, p2 distinct
When the poles are real valued )1(
EMLAB
25
When the poles are complex valued )1(
0
2
0
2
21
2
00
11
1
11
1
21
1)(
sj
sj
ps
psss
sH
dB0
110log pdB3
dB/dec40
dB0
110log pdB3
dB/dec40
With smaller
202,1 1 jp
EMLAB
26
Standard form of quadratic functions
11
1)(
0
2
0
s
Qs
sHQ
12
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.
)2/1( Q
)2/1( Q
)2/1( Q
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.
EMLAB
27
]...)()(21)[1(
]...)()(21)[1()()( 2
233310
bbba
N
jjj
jjjjKjH
General form of a network function
Frequency independentPoles/zeros at the origin
First order terms Quadratic terms for complex conjugate poles/zeros
..|)()(21|log20|1|log20
...|)()(21|log20|1|log20
||log20log20
21010
233310110
10010
bbba jjj
jjj
jNK
DND
N
BAAB
loglog)log(
loglog)log(
|)(|log20|)(| 10 jHjH dB
212
1
2121
zzz
z
zzzz
...)(1
2tantan
...)(1
2tantan
900)(
211
23
3311
1
b
bba
NjH
Display each basic Display each basic term separately and term separately and add the results to add the results to obtain final answerobtain final answer
Let’s examine each basic termLet’s examine each basic term
EMLAB
28
a. Constant Term
b. Poles/Zeros at the origin
90)(
)(log20|)(|)( 10
Nj
Njj
NdB
NN
linestraight a is this
log is axis- xthe 10
EMLAB
29
c. Simple pole or zeroj1
1
210
tan)1(
)(1log20|1|
j
j dB
asymptotefrequency low 0|1| dBj
(20dB/dec) asymptotefrequency high 10log20|1| dBj
frequency) akcorner/bre1 whenmeet asymptotes two The (
Behavior in the neighborhood of the corner
FrequencyAsymptoteCurvedistance to asymptote Argument
corner 0dB 3dB 3 45
octave above 6dB 7db 1 63.4
octave below 0dB 1dB 1 26.6
125.0
0)1( j
90)1( j
1
1
Asymptote for phase
High freq. asymptoteLow freq. Asym.
EMLAB
30
Simple zero
Simple pole
EMLAB
31
d. Quadratic pole or zero ])()(21[ 22 jjt ])()(21[ 2 j
222102 2)(1log20|| dBt 2
12
)(1
2tan
t
1 asymptotefrequency low 0|| 2 dBt 02t
1 asymptote freq. high 2102 )(log20|| dBt 1802t
1 )2(log20|| 102 dBt 902tCorner/break frequency221 2
102 12log20|| dBt
2
12
21tan
t
2
2Resonance frequency
Magnitude for quadratic pole Phase for quadratic pole
dB/dec40
These graphs are inverted for a zero
EMLAB
32
Example Generate magnitude and phase plots
)102.0)(1(
)11.0(10)(
jj
jjGv
Draw asymptotesfor each term 1,10,50 :nersBreaks/cor
Draw composites
40
20
0
20
dB
90
90
1.0 1 10 100 1000
dB|10
decdB /20
dec/45
decdB /20
dec/45
EMLAB
33
EMLAB
34Generate magnitude and phase plots
)11.0()(
)1(25)( 2
jj
jjGv 10 1, :(corners) Breaks
40
20
0
20
dB
90
270
90
1.0 1 10 100
Draw asymptotes for each
dB28
decdB /40
180
dec/45
45
Form composites
Example
EMLAB
35A function with complex conjugate poles
1004)()5.0(
25)( 2
jjj
jjG
Put in standard form
125/)10/()15.0/(
5.0)( 2
jjj
jjG
40
20
0
20
dB
90
90
01.0 1.0 1 10 100 270
1 )2(log20|| 102 dBt
2.01.0
25/12
])()(21[ 2
2 jjt
dB8
Draw composite asymptote
Behavior close to corner of conjugate pole/zerois too dependent on damping ratio.Computer evaluation is better
Example
EMLAB
36
Using MATLAB to compute Magnitude & Phase information
011
1
011
1
...
...)(
bsbsbsb
asasasasV n
nn
n
mm
mm
o
),(
];,,...,,[
];,,...,,[
011
011
dennumfreqs
bbbbden
aaaanum
nn
mm
MATLAB commands required to display magnitudeand phase as function of frequency
NOTE: Instead of comma (,) one can use space toseparate numbers in the array
1)1053.215()1053.21.0()(
)1053.215(332
3
jj
jVo
EXAMPLE
» num=[15*2.53*1e-3,0];» den=[0.1*2.53*1e-3,15*2.53*1e-3,1];» freqs(num,den)
1a
2b1b
0b
Missing coefficients mustbe entered as zeros
» num=[15*2.53*1e-3 0];» den=[0.1*2.53*1e-3 15*2.53*1e-3 1];» freqs(num,den)
This sequence will alsowork. Must be careful notto 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
1004)()5.0(
25)( 2
jjj
jjG
» num=[25,0]; %define numerator polynomial» den=conv([1,0.5],[1,4,100]) %use CONV for polynomial multiplicationden = 1.0000 4.5000 102.0000 50.0000» freqs(num,den)
Using default options
EMLAB
40
Determining Transfer function from Bode plotThis 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
A. different from 0dB.There is a constant Ko
B
B. Simple pole at 0.11)11.0/( j
C
C. Simple zero at 0.5
)15.0/( j
D
D. Simple pole at 3
1)13/( j
E
E. Simple pole at 20
1)120/( j
)120/)(13/)(11.0/(
)15.0/(10)(
jjj
jjG
20
|
00
0
1020|dBK
dB KK
If the slope is -40dB we assume double real pole. Unless we are given more data
EMLAB
41
Example Determine a transfer function from the composite magnitude asymptotes plot
A
A. Pole at the origin. Crosses 0dB line at 5
j5
B
B. Zero at 5
C
C. Pole at 20
D
D. Zero at 50
E
E. Pole at 100
)1100/)(120/(
)150/)(15/(5)(
jjj
jjjG
EMLAB
42
12.3 Resonant circuits
H(s)H(s)+
Vout-
)()()( sVsHsV inout
A circuit that selects one frequency is called a resonant circuit.
SR
LR+
Vin-
EMLAB
43
Series resonance
H(s)H(s)
L C
V_ACSRC1
RR2R=50 Ohm
CC1
LL1
RR1
ss
QssQ
Qs
CsRLCs
CsR
sCsLR
R
V
VsH
in
out
0
0
2
00
0
22
2
2
2
1
1
11
11)(
0
0
1
1)(
jQ
sH
+Vout
-
CRQ
LC 200
1,
1
200
220
02 2:form Standard sss
Qs
EMLAB
44
Spectrum analyzer
Transfer function of a resonator
101
102
103
104
-100
-50
0
50
100
Frequency (radians)
Pha
se (
de
gre
es)
101
102
103
104
10-3
10-2
10-1
100
Frequency (radians)
Ma
gni
tud
e
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);
2|)(| H
)(H
EMLAB
45
80 90 100 110 1200
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
2
0
0
21
1|)(|
Q
sH
BWQ
Qx
xQQ
x
QQxx
Qx
Q
0
0
2
22
0
0
1
)0(14
1
2
1
14
1
2
1,01
1
1
Half Power Band-width
80 90 100 110 120-80
-60
-40
-20
0
20
40
60
80
2|)(| H
100
Q
Half power band-width
Power 가 1/2 이 되는 지점 .
)(H
EMLAB
46
101
102
103
104
-100
-50
0
50
100
Frequency (radians)
Pha
se (
de
gre
es)
101
102
103
104
10-3
10-2
10-1
100
Frequency (radians)
Ma
gni
tud
e
101
102
103
104
-100
-50
0
50
100
Frequency (radians)
Pha
se (
de
gre
es)
101
102
103
104
10-4
10-2
100
Frequency (radians)
Ma
gni
tud
e
101
102
103
104
-100
-50
0
50
100
Frequency (radians)
Pha
se (
de
gre
es)
101
102
103
104
10-3
10-2
10-1
100
Frequency (radians)
Ma
gni
tud
e
2Q 10Q 100Q
On a log scale, a point where |H| is lowered by 3dB from the peak value is called a half power point.
HPBW vs. Q
With large Q, BW gets narrow.
3dB-BW
Q : Quality factor
2|)(| H
)(H
EMLAB
47
RR1
I_ACSRC1
RR2L
L1CC1
Parallel resonance
H(s)H(s) LL1
CC1
ss
QssQ
Qs
sCsL
LCR
sCsLR
sCsL
I
IsH
in
out
0
0
2
00
0
22
1
1
11
11
11
||
1||
)(
outIinI
EMLAB
48
Resonant circuits
These are circuits with very special frequency characteristics. And resonance is a very important physical phenomenon
CjLjRjZ
1
)(
circuit RLCSeries
LjCjGjY
1
)(
circuit RLCParallel
LCCL
11
whenzero iscircuit each of reactance The
0
The frequency at which the circuit becomes purely resistive is calledthe resonance frequency
EMLAB
49
Properties of resonant circuitsAt resonance the impedance/admittance is minimal
Current through the serial circuit/voltage across the parallel circuit canbecome very large (if resistance is small)
CRR
LQ
0
0 1
:FactorQuality
222 )1
(||
1)(
CLRZ
CjLjRjZ
222 )1
(||
1)(
LCGY
CjLj
GjY
Given the similarities between series and parallel resonant circuits, we will focus on serial circuits
EMLAB
50
Properties of resonant circuitsAt resonance the power factor is unity
CIRCUIT BELOW RESONANCE ABOVE RESONANCESERIES CAPACITIVE INDUCTIVEPARALLEL INDUCTIVE CAPACITIVE
Phasor diagram for series circuit Phasor diagram for parallel circuit
RV
C
IjVC
Lj
1GV1CVj
L
Vj
1
EMLAB
51
Example 12.7 Determine the resonant frequency, the voltage across eachelement at resonance and the value of the quality factor
LC
10 sec/2000
)1010)(1025(
163
radFH
I
AZ
VI S 5
2
010
2Z resonanceAt
50)1025)(102( 330L
)(902505500 VjLIjVL
902505501
501
0
00
jICj
V
LC
C
R
LQ 0 25
2
50
||||
|||| 0
SC
SS
L
VQV
VQR
VLV
resonanceAt
111
111
0
2
0
2
s
Q
s
sC
sRCLCssC
RsC
sLVS
EMLAB
52
Given L = 0.02H with a Q factor of 200, determine the capacitornecessary to form a circuit resonant at 1000Hz
R
L0200200Q withL
LC
10
C02.0
110002 FC 27.1
What is the rating for the capacitor if the circuit is tested with a 10V supply?
VVC 2000|| ||||
|||| 0
SC
SS
L
VQV
VQR
VLV
resonanceAt
59.1200
02.010002R
AI 28.659.1
10
The reactive power on the capacitorexceeds 12kVA
Example 12.8
V010
2
00
2
11
111
s
Q
s
sC
LCssRCsCsC
sLRZ
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53
Normalizationfactor
( ) cos [ ]m
O
Vi t t A
R
Energy transfer in resonant circuits
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54
2
2
1LiWL
TRiWD 2
2
1
D
L
D
L
W
WQ
Q
R
L
TR
L
TRi
Li
W
W
2
22
1
21
21
0
2
2
Energy stored in L
Energy dissipated in one cycle
Quality factor in terms of Energy
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55
Example 12.10 A series RLC circuit as the following properties:
sec/100sec,/4000,4 0 radBWradR
Determine the values of L,C.
CRR
LQ
0
0 1
LC
10
QBW 0
1. Given resonant frequency and bandwidth determine Q.2. Given R, resonant frequency and Q determine L, C.
40100
40000 BW
Q
HQR
L 040.04000
440
0
FRQL
C 662
020
1056.11016104
111
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56
The Tacoma Narrows Bridge Opened: July 1, 1940Collapsed: Nov 7, 1940
Likely cause: windvarying at frequencysimilar to bridgenatural frequency
2.020
Example 12.12
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x
tbkmamF cos0Fxxx
02 Fxxx bskms
0.44’
1.07’
'77.3
Displacement vs. voltage
b
kQ
m
k
ks
k
bs
k
m
00
02
,
F
xxx
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Increasing selectivity by cascading low Q circuits
Single stage tuned amplifier
MHzsradFHLC
9.99/10275.61054.210
11 8
1260
398.010
1054.2250 6
12
L
CR
L
C
GBWQ 10
Example 12.15
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12.5 Filter networksNetworks designed to have frequency selective behavior
COMMON FILTERS
Low-pass filterLow-pass filterHigh-pass filterHigh-pass filter
Band-pass filterBand-pass filter
Band-reject filterBand-reject filter
We focus first onPASSIVE filters
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60
Examples : Examples : voice signal spectrum
Time domain signal Frequency domain spectrum
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61
Example : ADSL signal spectrum
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62
Filter
H(s)
Filter
H(s)
+Vout
-
SR
LR+
Vin-
CC1
V_ACSRC1
RR1
RR2
Low-pass filterLow-pass filter
1-st order1-st order
LL1
RR2
V_ACSRC1
RR1
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63
Prototype low pass filter
RR1
V_ACSRC1
LL1
RR2
CC1
V_ACSRC1
RR1
RR2
12
2
2
1
1
1
1)(
ps
RsLsLR
R
V
VsH
in
out
1
22 1
1
1
11
1
)(
psCsR
sCR
sCI
IsH
in
out
CRp
21
1
L
Rp 2
1
dB0110log p
y
dB3
dB/dec20
Type #1
Type #2
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64
1
1
1)(
ps
sH
LPF-to-HPF transformation
dB0
y
dB3
dB/dec20
sp
sp
sH11 1
1
1
1)'(
dB0
y
dB3
dB/dec20
onsubstituti'1
1 s
p
p
s
10log110log p110log p
LPF→HPF
11
p
s1
1
p
s
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65
RR1
V_ACSRC1
LL1
RR2
CC1
LPF→HPF
'replace
21
s
ps
Lpss
LpsL
21
121
1 1'
1
' L
L11
21
1
1
LpC
LsCp
s
sC'
'121
CC1
LL1
CC1
RR2
V_ACSRC1
RR1
CC1
RR2
V_ACSRC1
RR1
LL1
R=L=1.0 nH
RR2
V_ACSRC1
RR1
121
1
1
CpL
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66
1
1
1)(
ps
sH
dB0
y
dB3
dB/dec20
''
1
1)'(
0
0
0
ss
BW
sH
dB0
y
dB3
dB/dec20
'
' replace 0
0
0
1 s
s
BWp
s
10log
010log
110log p
LPF→BPF
1'
' 0
0
0
1
s
s
BWp
s
jBW
js
s
BW
js
jBW
js
s
BW
js
0
1200
0
0
2
0
1100
0
0
1
'
'
'
'
'
'
dB/dec20
2' 1'
LPF-to-BPF transformation
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67
1201
110
0
0111
'
1'
'
'
LpBW
s
sLBW
p
s
s
BW
pLsL
LL1
'1
'
1
''
11120110
0
01
sBW
Cps
BWCp
ss
BW
pCsC
'
' Replace 0
0
0
1 s
s
BWp
s
LL1
CC1
CC1
RR1
V_ACSRC1
LL1
RR2 V_AC
SRC1
RR2R=50 Ohm
CC1
LL1
RR1
CC1
RR2
V_ACSRC1
RR1
CC1
LL1 R
R2V_ACSRC1
RR1
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1
1
1)(
ps
sH
dB0
y
dB3
dB/dec20
''
11
1)'(
0
0
0
ss
BW
sH
dB0
y
dB3
dB/dec20
''
1 Replace
0
0
01
ss
BW
p
s
10log
010log
110log p
LPF→BPF
1'
' 0
0
0
1
s
s
BWp
s
jBW
js
s
BW
js
0
2100
0
0
1
'
'
'
dB/dec20
LPF-to-BRF transformation
2' 1'
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69
'1
'1
1
''
11
20
11
0
0
0
111
sBWpLs
BWpLss
BW
pLsL
LL1
'
1'
1
'
'11
1
20
1
0
01
0
sBWCps
BWCps
s
BWpCsC
''
1 replace
0
0
01
ss
BW
p
s
LL1
CC1
CC1
RR1
V_ACSRC1
LL1
RR2
CC1
RR2
V_ACSRC1
RR1
CC1
LL1
RR2
V_ACSRC1
RR1
RR2
RR1
CC1
LL1
V_ACSRC1
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Example 12.19 A simple notch filter to eliminate 60Hz interference
)1
(1
1
CLj
CL
CjLj
CjLj
ZR
inReq
eq VZR
RV
0
LCZR
1 01
0
LCV
tttvin 10002sin2.0602sin)( FCmHL 100,3.70
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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 amore general framework, if one replaces the resistors by impedances
Ideal Op-Amp
These currents are zero
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Basic Inverting Amplifier
0V
VV gain Infinite
0V
0 II -impedanceinput Infinite
02
2
1
1 Z
V
Z
V
11
22 V
Z
ZV
Linear circuit equivalent
0I
1
2
Z
ZG
1
11 Z
VI
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Basic Non-inverting amplifier1V
1V
0I
1
1
2
10
Z
V
Z
VV
11
120 V
Z
ZZV
1
21Z
ZG
01 I
Basic Non-inverting Amplifier
Due to the internal op-amp circuitry, it haslimitations, e.g., for high frequency and/orlow voltage situations. The OperationalTransductance Amplifier (OTA) performswell in those situations
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(non-inverting op-amp)Example 12.40 “BASS-BOOST” AMPLIFIER
DESIRED BODE PLOT
OPEN SWITCH
(6dB)
500
2Pf
gain voltagefor the
of value theand locations zero and pole theFind oK
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Example 12.41 TREBLE BOOST
Original player responseDesired boost
Proposed boost circuit
Non-inverting amplifier
Recommended