frequency response - kalyana veluvoluncbs.knu.ac.kr/teaching/acs_files/ct2_freq_resp.pdf ·...

1

‹#›ELEC 24409: Circuit Theory 2 Dr. Kalyana Veluvolu

Frequency Response

Chapter Objectives:

Ø Understand the Concept of Transfer Functions.Ø Be Familiar with the Decibel Scale.Ø Learn how to make Bode Magnitude and Phase plots.Ø Know Different Types of Passive and Active Filters and their

Characteristics.


What is Frequency Response of a Circuit?

It is the variation in a circuit’s behavior with change in signal

frequency and may also be considered as the variation of the gain

and phase with frequency.

FREQUENCY RESPONSE

2


TRANSFER FUNCTIONØThe transfer function H(w) of a circuit is the is the frequency dependent ratio of the phasor output Y(w) to a phasor input X(w).

Ø Considered input and output may be either the current or the voltage variable.

Ø 4 types of possible transfer functions.

Y( )H( ) X( )

= H( ) |

wwww f

=

Ð

)(V)(V gain Voltage )(H

i

o

www == )(I

)(V ImpedanceTransfer )(Hi

o

www ==

)(I)(I gain Current )(H

i

o

www ==

)(V)(I AdmittanceTransfer )(H

i

o

www ==


Magnitude plot for a low-pass filter

TRANSFER FUNCTION of Low-pass RC Circuit

R=20 kΩC=1200 pF

At low frequencies At high frequencies

3


Phase plot for a low-pass filter

TRANSFER FUNCTION of Low-pass RC Circuit

At low frequencies At high frequencies

R=20 kΩC=1200

pF


TRANSFER FUNCTION of High-pass RC Circuit

Magnitude plot for a high-pass filter

At high frequencies At low frequencies

R=20 kΩC=1200 pF

4


TRANSFER FUNCTION of High-pass RC Circuit

Magnitude plot for a high-pass filter

Phase plot for high-pass filter

At high frequencies At low frequencies

R=20 kΩC=1200 pF


TRANSFER FUNCTION of a Band-pass RC Circuit

5


Frequency Response of the RC Circuit

2

1

1( ) 1( ) Transfer Function1( ) 1

Magnitude Response

Phase Res

1( )1 ( )

( ) ( ) tan

1Where

ponse

o

o

o

o

s

H

H

V j CHV j RCR C

C

j

R

w

w www w

ww

wf w ww

w

w

-

= = =++

=+

= Ð = -

=

a) Time Domain RC Circuit b) Frequency Domain RC Circuit


Drawing Frequency Response of RC Circuit

a) Amplitude Response b) Phase Response

2

1( )1 ( )

o

H www

=+

1( ) ( ) tan o

H wf w ww

-= Ð = -

Low Pass Filter

ØThe frequency value of wo is of special interest.

Ø Because output is considerable only at low values of frequency, the circuit is also called a LOW PASS FILTER.

6


HIGH Pass Filter


TRANSFER FUNCTIONØ The transfer function H(w) can be expressed in terms of its numerator polynomial N(w) and its denominator polynomial D(w).

( )( )( )NHDwww

=

Ø The roots of N(w)=0 are called ZEROS of H(w) (jw=z1, z2, z3, ….).Similarly The roots of D(w)=0 are called POLES of H(w) (jw=p1, p2, p3, ….). A zero as a root of the numerator polynomial, results in a zero value of the transfer function. A pole as a root of the denominator polynomial, results in an infinite value of the transfer function.

21 1

1

21

1

2( ) 1 1( )( )( ) 21 1

k k

n n

jj jK j zNHD jj j

p

V ww ww w wwww V ww w

w w

± é ùæ öæ ö+ + +ç ÷ ç ÷ê úè ø è øë û= =é ùæ öæ ö+ + +ç ÷ ç ÷ê úè ø è øë û

K

K

7


s=jw


Vx

0.5Vx0.5VxVx

8


DECIBEL SCALEØ The DECIBEL provides us with a unit of less magnitude.

210

1

10log is defined as Decibel GaindBPGP

=

22 2

2 2 2 110 10 10 102

11 1 2

1

2 2 210 10 10 2 1

1 1 1

10log 10log 10log 10log

20log 10log 20log if

dB

VP R V RG

VP V RR

V R V R RV R V

æ ö= = = +ç ÷

è ø

= - = =

2 210 10

1 1

10log 20logdBP VGP V

= =


DECIBEL SCALE

210 10

1

20log 20logdBVH HV

= =Magnitude H Decibel Value HdB

0.001 -600.01 -400.1 -200.5 -6

1/√2 -31 0

√2 32 6

10 2020 26

100 40

1 2 1 2

11 2

2

log log log

Some Properties of Logar

log log log

log loglog

ith s

1 0

m

n

PP P P

P P PP

P n P

= +

æ ö= -ç ÷

è ø=

=

Ø The DECIBEL value is a logarithmic measurement of the ratio of one variable to another of the same type.

ØDecibel value has no dimension.

Ø It is used for voltage, current and power gains.

9


Typical Sound Levels and Their Decibel Levels.


SEMILOG SCALE

10


BODE PLOTSØ Bode plots are APPROXIMATE semilog plots of the magnitude (in Decibels) and phase (in degrees) of a transfer function versus frequency. They are much easier to plot.

High pass filter circuit magitude response.

Bode plots are APPROXIMATE plots of the magnitude and phase responses.


BODE PLOTS

Actual Response

BODE PLOT

11


BODE PLOTS

Bode plot and actual magnitude and phase responses of the RC high pass filter circuit.


BODE PLOTSØ Bode plots are semilog plots of the magnitude (in Decibels) and phase (in degrees) of a transfer function versus frequency.

Bode plots carry same information. They are much easier to plot.

jH He ff= Ð =H

Logarithmic axis

Decibel values

Total Magnitude Response

12


BODE PLOTS

21 1

1

21

1

2( ) 1 1( )( )( ) 21 1

k k

n n

jj jK j zNHD jj j

p

V ww ww w wwww V ww w

w w


K

K

Ø This representation is called the STANDARD FORM. It has several different factors.We can draw the Bode plots by plotting each of the terms of the transfer function separately and then adding them.The different factors of the transfer function are

1.) Gain term K.2.) A pole (jw)-1 or a zero (jw) at the origin.

3.) A simple pole or zero

4.) A quadratic pole or zero

11 j

pwæ ö+ç ÷

è ø 11 j

zwæ ö+ç ÷

è ø

2121

n n

j jV w ww w

é ùæ ö+ + ç ÷ê úè øë û

2121

k k

j jV w ww w

é ùæ ö+ + ç ÷ê úè øë û

Ø A transfer function may be written in terms of factors that have real and imaginary parts.


BODE PLOTS, The Decade

Ø A DECADE is an interval between 2 frequencies with a ratio of 10 (between 10 Hz and 100 Hz or between 500 Hz and 5000 Hz). 20 dB/decade means that magnitude changes 20 dB whenever the frequency changes tenfold or one decade.

The DC value (ω→ 0) does not appear on Bode plots ( Log0 = - ∞).

Ø Slopes are expressed in dB/decade.

One Decade

20 dB

13


BODE PLOTS

2

0.4(1STANDARD F

10)( )(1 5)

ORMjH

j jww

w w+

=+

Ø To plot the Bode plots of a given transfer function.

1.) Put the transfer function in STANDARD FORM.2.) Write the Magnitude and phase equations from the STANDARD FORM.3.) Plot the magnitude of each term separately.4.) Add all magnitude terms to obtain the magnitude transfer function.5.) Repeat 2-4 for the phase response.6.) The total magnitude response in Decibel units is the summation and subtraction of the responses of different terms.7.) The total phase response in degrees is the summation and subtraction of the phase responses of different terms.

10

10

10

10

20log 0.420log 1 10

20log

40log 1 5

dbHj

j

j

w

w

w

= +

+ -

-

+

-1

-1

0 tan ( 10)90 2 tan ( 5)

f w

w

= °+

- °-


BODE PLOTSØ We examine how to plot different terms that may appear in a transfer function.

The total response will be obtained by adding all the responses.

CONSTANT TERM 1020log K

21 1

1

21

1

2( ) 1 1( )( )( ) 21 1

k k

n n

jj jj zNHD jj j

p

K V ww ww w wwww V ww w

w w


K

K

14


BODE PLOTSZERO AT THE ORIGIN ( )jw

1020log w

POLE AT THE ORIGIN ( ) 1jw -

-20 -90°


SIMPLE ZERO ( )11 j zw+

1010 1

10 110 1

20log 1 0 0 0 dB 020log 1

20log 0 20 log dBH j zzz

w ww

w ww w= ® £ì ì

= + = í í ³»

®¥ îî

11

1

0, =0tan 45 , =z

90 , z

wwf w

w

-

ìæ ö ï= = °íç ÷è ø ï ° ®¥î

Ø Approximate the magnitude response of a simple zero by two linear curves before and after ω=z1Approximate the phase response of a simple zero by three linear curves before ω=0.1z1 after ω=10z1 and between ω=0.1z1 and ω=10z1

CORNER FREQUENCY

BREAK FREQUENCY

3 dB FREQUENCY

15


SIMPLE POLE ( )11 j pw+

-20 -90°

10 120log 1 1 j pw+

Ø Approximate the magnitude response of a simple pole by two linear curves before and after ω=p1Approximate the phase response of a simple pole by three linear curves before ω=0.1p1after ω=10p1 and between ω=0.1p1 and ω=10p1

ØNOTICE the pole and zero responses are in opposite directions.

CORNER FREQUENCY

11

1

0, =0tan 45 , =p

90 , p

wwf w

w

-

ìæ ö ï= - = - °íç ÷è ø ï- ° ®¥î


EXACT RESPONSE OF QUADRATIC POLE

21

10

21

220log 1

1( )21

dB

n

n

n

n

j jH

Hj j

V w ww

w ww w

w

wV

æ ö= - +

=é ùæ ö+ + ç ÷ê ú

+ ç

ø

è

è û

÷ø

ë

1

12

2

2

tan1

n

n

V wwf

ww

-= --

BODE PLOTS

16


BODE PLOT OF QUADRATIC ZERO

21

1010

0 as 0220log 1 40log as dBn n

n

j jHwV w www w ww

»®ìïæ ö= + + ç ÷ í ®¥è ø ïî

1

12

2

2 0, 0tan 90 ,

1 180 ,

nn

n

V w wwf w w

www

-

®ìï= » ° =í

- ï ° ®¥î

Ø The EXACT responses can be approximated by BODE plots in terms of the corner frequency ωn


BODE PLOT OF QUADRATIC POLE 2

110

10

0 as 0220log 1 40log as dBn n

n

j jHwV w w

ww w ww»

®ìïæ ö= - + + ç ÷ í- ®¥è ø ïî

1

12

2

2 0, 0tan 90 ,

1 180 ,

nn

n

V w wwf w w

www

-

®ìï= - » - ° =í

- ï- ° ®¥î

17


• Express transfer function in Standard form.

• Find the Pole and Zero frequencies.

• Express the magnitude and phase responses.

• Sketch each term of the magnitude and phase responses.

• Add each term of magnitude response to find total magnitude response.

• Add each term of phase response to find total phase response.

Another Procedure• Zeros cause an increase and poles cause a decrease in the slope .

• Start with the lowest frequency of the Bode plot.

• Move along the frequency axis and increase or decrease the slope at each corner frequency.

•Repeat the same procedure for both the magnitude and the phase.

Procedure for Bode plot Construction




• Two corner frequencies at ω=2, 10 and a zero at the origin ω=0.

• Sketch each term and add to find the total response.

EXAMPLE 14.3 Construct Bode plots for

( )( )200( )2 10jH

j jww

w w=

+ +

10STANDARD FORM ( )(1 2)(1 10)

jHj j

www w

=+ +

10

-1 -1

10 10 1020log 10 20log 20log 1 2 20log 1

90 tan ( 2)

10

tan ( 10)dbH j j j

f

w

w

w

w

w

= ° -

= +

-

- + - +

18


EXAMPLE 14.3 Construct Bode plots for ( )( )

200( )2 10jH

j jww

w w=

+ +

10 10 10 1020log 10 20log 20log 1 2 20log 1 10dbH j j jw w w= + - + - +

-1 -190 tan ( 2) tan ( 10)f w w= °- -

1020log 2 6dB=

26 dB

X X

XX


EXAMPLE 14.3 Continued: Let us calculate |H| and f at w=50 rad/sec graphically.

26 dB

10(50) (10) 20log (50 /10) 26 20 0.7 26 14 12H H dB= - = - ´ = - =

10 10 10(50) 90 45 log (1/ 0.2) 90 log (20 /1) 45 log (50 / 20)90 45 0.7 90 1.3 45 0.4 90 31.5 117 18 76.5

f = °- °´ - °´ - °´

= °- °´ - °´ - °´ = °- °- °- ° = - °

w=50

w=50

19


EXAMPLE 14.4 Construct Bode plots for ( )210( )

5jH

j jww

w w+

=+

2

0.4(1STANDARD F

10)( )(1 5)

ORMjH

j jww

w w+

=+

10 10 10 1020log 0.4 20log 1 10 20log 40log 1 5dbH j j jw w w= + + - - +-1 -10 tan ( 10) 90 2 tan ( 5)f w w= °+ - °-



• Two corner frequencies at ω=5, 10 and a zero at ω=10.

• The pole at ω=5 is a double pole. The slope of the magnitude is -40 dB/decade and phase has slope -90 degree/decade.

• Sketch each term and add to find the total response.


EXAMPLE 14.4 Construct Bode plots for ( )210( )

5jH

j jww

w w+

=+

10

10

10

10

20log 0.420log 1 10

20log

40log 1 5

dbHj

j

j

w

w

w

= +

+ -

-

+

-1

-1

0 tan ( 10)90 2 tan ( 5)

f w

w

= °+

- °-

X O

X O

20


PRACTICE PROBLEM 14.6 Obtain the transfer function for the Bode plot given.

2 2

2

2

A zero at 0.5,

A pole at 1,

Two poles at 10,

1 0.5 (1 0.5)(0.5 ) 200( 0.5)( )(1 1)(1 10) (1 100)(1 )(10 ) ( 1)( 10)

1 0.51

1 11

(1 10)

j

j

j j sHj j j j s s

j

w

w

w

w www

w

w

w w

w

w

=

=

=

+ + += = =

+ + + + + +

+

+

+


21


frequency response - kalyana veluvoluncbs.knu.ac.kr/teaching/acs_files/ct2_freq_resp.pdf ·...

Documents