ec010303 network theory-module5
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EC010 303 NETWORK THEORYMODULE-3
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Frequency Response, Bode Plots
1. Determine the output of a filter for a given input
consisting of sinusoidal components using the
filter’s transfer function.
3. Use circuit analysis to determine the transfer
functions of simple circuits.
4. Draw first-order lowpass or highpass filter circuits
and sketch their transfer functions.
5. Understand decibels, logarithmic frequency scales,
and Bode plots.
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.–
. , ,Functions 6.1.2 Filters
* Filters process the sinusoidal components of an input signal
differently depending of the frequency of each component.
Often, the goal of the filter is to retain the components incertain frequency ranges and reject components in other
frequency ranges.
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.–
. , ,Functions 6.1.3 Filters and Transfer Functions
* Since the impedances of inductances and capacitances change
with frequency, RLC circuits provide one way to realize electrical
filters.
* The transfer function of a two-port filter is defined as:
90- f 2
190-
C
1Z ,90 fL290 L Z C L
phase th e is - H(f)
magn i tude th e is H(f) whe r e
H(f) H(f)) f H
in ou t
in
ou t
VV
V
V
)(
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.–
. , ,Functions Example 6.1 – Using Transfer Function to Find Output
For the transfer functions shown, find the output signal,
given the input: )40t 2000cos( 2 )t ( vin
in
out 303 )1000( H
Hz 1000 f is signal input theof frequencyThe
V
V
402inV 706 402303* )1000( H inout V V
)702000cos(6)( t t vout
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.–
. , ,Functions Example 6.2 – Multi-input components, Superposition Principle
The input involves two components:
2in1in
in
)70t 4000cos( )t 2000( sco2 )t ( v
V V
306 02303 )1000( H
: principle itionsuperposuse We
1in1out V V
102701602 )0002( H 2in2out V V
)t ( v )t ( v )t ( v 2out 1out out
)10t 4000cos( 2 )30t 2000cos( 6 )t ( vout
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.–
. , ,Functions Example 6.2 – Multi-input components, Superposition Principle
)70t 4000cos( )t 2000( sco2 )t ( vin
)10t 4000cos( 2 )30t 2000cos( 6 )t ( vout
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6. Frequency Response – 6.2 First-Order Low-Pass Filters * Ideal Filters
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6. Frequency Response – 6.2 First-Order Low-Pass Filters 6.2 First Order Low-Pass Filters
A low-pass filter is designed to pass low-frequency components
and reject high-frequency components. In other words, for low
frequencies, the output magnitude is nearly the same as theinput; while for high frequencies, the output magnitude is
much less than the input.
6.2.1 Transfer Function
fC π j21 R
havewe ,V phasor ahaving
l sinusoidaais signal input the shown,as
filter pass-loworder - first theConsider
in
in
V I
RC f π j21 fC π j21 R fC π j2
1 inin
V V I V
fC π j2
1out
fRC π 2 j1
1 H(f)
in
out
V
V
frequency power" -half " the
frequency,break" " the RC π 2
1 f defineWe B
) f f j( 1
1 H(f)
B
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6. Frequency Response – 6.2 First-Order Low-Pass Filters 6.2.2 Magnitude and Phase Plots of the Transfer Function
)(1
1)(
B f f j f H
2)(1
1)(
B f f f H
)arctan()( B f
f f H
Power Half V P since ,V 2
1V ,2
1 H(f ) , f f As
90 H(f )also
rejected,components frequency-high 0 H(f ) , f f As
0 H(f )also
passed,components frequency-low 1 ) f ( H 0, f As
2
rmsrmsinrmsout B
B
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6. Frequency Response – 6.2 First-Order Low-Pass Filters Example 6.3 – Calculation of RC Low-pass Output
1000 f 0,5
100, f 0,5 10, f 0,5
)t 2000cos( 5 )t 200cos( 5 )t 20cos( 5 )t ( v
33in
22in11in
in
V
V V
) f f ( j1
1 ) f ( H
B
Hz RC
f B 10010*10*)21000(*2
1
2
16
71.59950.0 )10( H
457071.0 )100( H 29.840995.0 )1000( H
71.5975.4 )10( H 1in1out V V
)71.5t π 20cos( 975.4 )t ( v 1out
45535.3 )100( H 2in2out V V
)45t π 200cos( 535.3 )t ( v 2out
29.844975.0 )1000( H 3in3out V V
)29.84t π
2000cos( 4975.0 )t ( v 3out
d l
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6. Frequency Response – 6.2 First-Order Low-Pass Filters Example 6.3 – Calculation of RC Low-pass Output
1000 f 0,5
100, f 0,5 10, f 0,5
)t π 2000cos( 5 )t π 200cos( 5 )t π 20cos( 5 )t ( v
33in
22in11in
in
V
V V
(t)v )29.84t π 2000cos( 4975.0
(t)v )45t π 200cos( 535.3 (t)v )71.5t
π
20cos( 975.4 )t ( v
3out
2out
1out out
d l
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6. Frequency Response – 6.2 First-Order Low-Pass Filters Quiz – Exercise 6.4: Another First-Order Low-Pass Filter
This is also a low-pass filter
L R/2 f where
) j(f /f 1
1 H(f)
is functiontransfer thethat Show
B
Bin
out
V
V
6 F R 6 3 D ib l d h C d C i
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6. Frequency Response –
6.3 Decibels and the Cascade Connection 6.2 Decibels and the Cascade Connections
6.3.1 Decibels
* We usually express the ratio of voltage (or power) amplitude
in decibels.
power for ) f H( 10log ) f H( ) f H(
voltage for ) f ( H log 20 ) f ( H ) f H(
db
dB
6 F R 6 3 D ib l d h C d C i
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6. Frequency Response –
6.3 Decibels and the Cascade Connection 6.3.2 Cascade two-Port Networks
2in
2out
1in
1out
1out
2out
1in
1out
1in
2out
in
out ) f ( H V
V
V
V
V
V
V
V
V
V
V
V
)()()( 21 f H f H f H
dB2dB1dB ) f ( H ) f ( H ) f ( H
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6. Frequency Response – 6.4 Bode Plots 6.4 Bode Plots
2
)(1
1)(
B f f
f H
6 F R 6 4 B d Pl t
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6. Frequency Response –
6.4 Bode Plots 6.4 Bode Plots
2)(1
1)(
B f f f H
2 B
dB
) f f ( 1
1log 20 ) f ( H
BdB B
B
f
f log 20 ) f ( H f f For
dB0 H(f) f f For
B f
f f H arctan)(
90 H(f) , f 10 f For
0 H(f) /10, f f For
B
B
B f frequencybreak
6 F R 6 5 Fi t O d Hi h P Filt
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6. Frequency Response –
6.5 First-Order High-Pass Filters 6.5 First-Order High-Pass Filters
6.5.1 Transfer Function
)(1
)()(
B
B
in
out
f f j
f f j
V
V f H
RC f B
2
1
2
B
B
f f 1
f f ) f H(
B f
f f H arctan90)(
6 F R 6 5 Fi t O d Hi h P Filt
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6. Frequency Response –
6.5 First-Order High-Pass Filters 6.5.2 Bode Plots
21)(
B
B
f f
f f f H
2
1log10log20)( B B
dB f
f
f
f f H
0 f H , f 10 f For
90 f H /10, f f For
0 f H , f f For
f
f log 20 ) f ( H , f f For
B
B
dB B
BdB B
6 Frequency Response 6 5 First Order High Pass Filters
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6. Frequency Response –
6.5 First-Order High-Pass Filters Exercise 6.13– Another First-Order High-Pass Filter
L R/2 f where
) f f ( j1
) f f ( j
V
V ) f ( H
:iscircuit theof
functiontransfer thethat Show
B
B
B
in
out
6 Frequency Response 6 5 First Order Filters
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6. Frequency Response –
6.5 First-Order Filters First-Order Low-Pass Filters
First-Order High-Pass Filters
6 Frequency Response 6 8 Ideal and Second Order Filters
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6. Frequency Response –
6.8 Ideal and Second-Order Filters 6.8 Ideal and Second-Order Filters
6.8.1 Ideal Filters
6 Frequency Response 6 8 Ideal and Second Order Filters
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6. Frequency Response –
6.8 Ideal and Second-Order Filters 6.8.1 Ideal Filters
6 Frequency Response 6 8 Ideal and Second Order Filters
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6. Frequency Response –
6.8 Ideal and Second-Order Filters
6.8.2 Second-Order Low-Pass Filter
) f f f f ( jQ1
) f f ( jQ
V
V ) f ( H
00 s
0 s
in
out
LC π 2
1 f 0
R
L f π 2Q 0
s
CR f π 2
1
0
1Qchoose
passband theinconstant
elyapproximat beto gainthe
want we f ilter,adesign In
s
6 Frequency Response 6 8 Ideal and Second Order Filters
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6. Frequency Response –
6.8 Ideal and Second-Order Filters
6.8.2 Second-Order High-Pass Filter
6 Frequency Response 6 8 Ideal and Second Order Filters
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6. Frequency Response –
6.8 Ideal and Second-Order Filters
6.8.2 Second-Order Band-Pass Filter
6 Frequency Response 6 8 Ideal and Second Order Filters
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6. Frequency Response –
6.8 Ideal and Second-Order Filters
6.8.2 Second-Order Band-Reject (Notch) Filter
6 Frequency Response 6 8 Ideal and Second-Order Filters
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6. Frequency Response –
6.8 Ideal and Second-Order Filters
Example 6.7 – Filter Design
Design a second-order filter with L=50mH that passes
components higher in frequency than 1kHz, rejects
components lower than 1kHz.
We need a high-pass filter.
To obtain a approximately
constant transfer function
In the pass-band, we choose
LC 2
1 f since
1kHz f select and 1Q
0
0 s
F 507 .0 L f )2(
1
C havewe 2
0
2
1.314Q
L f 2 R and
s
0
6 Frequency Response 6 8 Ideal and Second-Order Filters
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6. Frequency Response –
6.8 Ideal and Second-Order Filters
* The Popular Sallen-Key Filters
6 Frequency Response 6 8 Ideal and Second Order Filters
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6. Frequency Response – 6.8 Ideal and Second-Order Filters
* Higher-order Filters using Cascade of 2nd -order Filters
6 Frequency Response 6 8 Ideal and Second Order Filters
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6. Frequency Response – 6.8 Ideal and Second-Order Filters
* Higher-order Filters using Cascade of 2nd -order Filters