Ec010303 Network Theory-module5

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EC010 303 NETWORK THEORYMODULE-3



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.



.–

. , ,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.



.–

. , ,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

)(



.–

. , ,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



.–

. , ,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



.–

. , ,Functions Example 6.2 – Multi-input components, Superposition Principle

)70t 4000cos( )t 2000( sco2 )t ( vin

)10t 4000cos( 2 )30t 2000cos( 6 )t ( vout



6. Frequency Response – 6.2 First-Order Low-Pass Filters ＊ Ideal Filters



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



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



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



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



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



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




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



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




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




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




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




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




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




6.8 Ideal and Second-Order Filters 6.8 Ideal and Second-Order Filters

6.8.1 Ideal Filters





6.8 Ideal and Second-Order Filters 6.8.1 Ideal Filters





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






6.8.2 Second-Order High-Pass Filter






6.8.2 Second-Order Band-Pass Filter






6.8.2 Second-Order Band-Reject (Notch) Filter

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





* The Popular Sallen-Key Filters




6. Frequency Response – 6.8 Ideal and Second-Order Filters

* Higher-order Filters using Cascade of 2nd -order Filters


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