Frequency Response and Filter Design By Poles and Zeros

Positioning

Dr. Mohamed Bingabr

University of Central OklahomaSlides For Lathi’s Textbook Provided by Dr. P. Cheung

Frequency Response of a LTI System

We have seen that LTI system response to x(t)=est is H(s)est. We represent such input-output pair as:

Instead of using a complex frequency, let us set s = j, this yields:

It is often better to express H(j) in polar form:

Therefore

( )st ste H s e

( )j t j te H j e cos Re( ) Re[ ( ) ]j t j tt e H j e

( )( ) ( ) j H jH j H j e

cos ( ) cos[ ( )]t H j t H j

Frequency

Response

Amplitude Response

Phase Response

h(t) = 3e-5t

5

3)(

ssH ts

ii

iesHth

)()(

|H(s)|

Frequency Response Example (1)

Find the frequency response of a system with transfer function:

Then find the system response y(t) for input x(t)=cos2t and x(t)=cos(10t-50)

Substitute s=j

0.1( )

5

sH s

s

0.1( )

5

jH j

j

21 1

2

0.01( ) and ( ) ( ) tan tan

0.1 525H j H j j

Frequency Response Example (2)

2

2

0.01( )

25H j

1 1( ) ( ) tan tan

0.1 5j H j

Frequency Response Example (3)

For input x(t)=cos2t, we have:

Therefore

2

2

2 0.01( 2) 0.372

2 25H j

1 12 2( 2) tan tan 65.3

0.1 5j

( ) 0.372cos(2 65.3 )y t t

Frequency Response Example (4)

For input x(t)=cos(10t-50), we will use the amplitude and phase response curves directly:

Therefore

( 10) 0.894H j

( 10) ( 10) 26j H j

( ) 0.894cos(10 50 26 ) 0.894cos(10 24 )y t t t

Frequency Response of delay of T sec

H(s) of an ideal T sec delay is:

Therefore

That is, delaying a signal by T has no effect on its amplitude.

It results in a linear phase shift (with frequency), and a gradient of –T.

The quantity:

is known as Group Delay.

( ) (Time-shifting property)sTH s e

( ) 1 and ( )j TH j e j T

( ) =Tg

d

d

Frequency Response of an ideal differentiator

H(s) of an ideal differentiator is:

Therefore

This agrees with:

That’s why differentiator is not a nice component to work with – it amplifies high frequency component (i.e. noise!).

/ 2( ) and ( ) jH s s H j j e

( ) and ( )2

H j H j

(cos ) sin cos( / 2)d

t t tdt

Frequency Response of an ideal integrator

H(s) of an ideal integrator is:

Therefore

This agrees with:

That’s why integrator is a nice component to work with – it suppresses high frequency component (i.e. noise!).

/ 21 1 1( ) and ( ) jj

H s H j es j

1( ) and ( )

2H j H j

1 1cos sin cos( / 2)t dt t t

Filter Design By Poles and Zeros Positioning

Effects of Poles & Zeros on Frequency Response (1)

Consider a general system transfer function:

The value of the transfer function at some complex frequency s = p is:

1 2zeros at , ,...., Nz z z

1 2poles at , ,...., Np p p

i

ir

Effects of Poles & Zeros on Frequency Response (2)

Therefore the magnitude and phase at s = p are given by:

Effects of Poles & Zeros on Frequency Response (3)

Frequency Response of a system is obtained by evaluating H(s) along the y-axis (i.e. taking all value of s=j).

Consider the effect of two complex system poles on the frequency response. 1

( )( ( ))( ( ))

H ss j s j

( )K

H jdd

1 2( ) ( )H j

Near to a pole ENHANCES

amplitude

Effects of Poles & Zeros on Frequency Response (4)

Consider the effect of two complex system zeros on the frequency response.

( ) ( ( ))( ( ))H s s j s j

( ) 'H j rr 1 2( ) ( )H j

Near to a zero SUPPRESSES

amplitude

Poles & Low-pass Filters

Use the enhancement and suppression properties of poles & zeros to design filters.

Low-pass filter (LPF) has maximum gain at =0, and the gain decreases with .

Simplest LPF has a single pole on real axis, say at (s=-c). Then

To have a “brickwall” type of LPF (i.e. very sharp cut-off), we need a WALL OF POLE as shown, the more poles we get, the sharper the cut-off.

( ) and ( )c c

c

H s H js d

Chebyshev Filter

x

x

xx

x

x

x

jc

-jc

Re0

x

x

x

x

x

x

x

Butterworth Filter

N = 2

Poles & Band-pass Filter

Band-pass filter has gain enhanced over the entire passband, but suppressed elsewhere.

For a passband centred around 0, we need lots of poles opposite the imaginary axis in front of the passband centre at 0.

Notch Filter

Notch filter could in theory be realised with two zeros placed at ±j0. However, such a filter would not have unity gain at zero frequency, and the notch will not be sharp.

To obtain a good notch filter, put two poles close to the two zeros on the semicircle as shown. Since both pole/zero pair are equal-distance to the origin, the gain at zero frequency is exactly one. Same for =±.

Notch Filter Example

Design a second-order notch filter to suppress 60 Hz hum in a radio receiver.

Make 0 =120. Place zeros are at s = ±j0 , and poles at - 0 cos ± j0

sin. We get:

0 0

0 0 0 0

2 2 20

2 2 20 0

( )( )( )

( cos sin )( cos sin )

142122.3

(2 cos ) (753.98cos ) 142122.3

s j s jH s

s j s j

s s

s s s s

HW7_Ch4: 4.8-1(a,b), 4.8-2 (a,b,d), 4.10-1, 4.10-3, 4.10-5, 4.10-8, 4.10-9