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# Frequency Response of an LSI System

Module by: Richard Baraniuk. E-mail the author

Summary: An introduction to the frequency response of an LSI system.

H H: circulent matrix

h h: H H's 0th column

H H: DFT of h h; H=FHh H F H h contain equivalent info!

Effect of LSI system on an input x x is easy to describe in the Fourier domain (frequency domain)...

ie: Yk=HkXk Y k H k X k ; 0kN1 0 k N 1

ie: pointwise multiplication

## Example 1

2 point smoother

Compute frequency response H H if LSI system...

Hk=1N n =0N1hne(i2πNkn)=1N n =0N112e(i2πNkn)=1N(1+ei2πNk)=1N(e(i2πNk2)+ei2πNk2)ei2πNk2=2NcosπNkeiπNk H k 1 N n N 1 0 h n 2 N k n 1 N n N 1 0 1 2 2 N k n 1 N 1 2 N k 1 N 2 N k 2 2 N k 2 2 N k 2 2 N N k N k
(1)
|Hk|=2N|cosπNk| H k 2 N N k
(2)
where 0 is low frequency, π is high frequency, 2π 2 is low frequency, and w k =2πnk w k 2 n k

Therefore it is a lowpass filter

ie: smoothing lowpass filtering

## Example 2

ie: h h is a "square box".

The Frequency response is (using answer from test 1):

Hk=1NsinMπNksinπNke(iπN(M1)k) H k 1 N M N k N k N M 1 k
(3)
Where sinMπNksinπNk M N k N k is the "Dirichlet Kernel."

H0= H 0 ?

### Note:

sin0=0 0 0 L'Hopitâl's rule to the rescue...
Hk| k =0=dd k numerator | k =0dd k denominator | k =0=1N(MπN)cosMπNk| k =0(πN)cosπNk| k =0e0=1NM=MN k 0 H k k 0 k numerator k 0 k denominator 1 N k 0 M N M N k k 0 N N k 0 1 N M M N
(4)

## Example 3: Edge Detector

Hk=1N n =0N1hne(i2πNkn)=1N(1e(i2πNkn))=1N(ei2πNk2e(i2πNk2))e(i2πNk2)=1N2isinπNke(iπNk) H k 1 N n 0 N 1 h n 2 N k n 1 N 1 2 N k n 1 N 2 N k 2 2 N k 2 2 N k 2 1 N 2 N k N k
(5)

## Example 4: Circular Shift

Hk=1N n =0N1hne(i2πNkn)=1Ne(i2πNkm) H k 1 N n 0 N 1 h n 2 N k n 1 N 2 N k m
(6)
|Hk|=1N H k 1 N , Hk=(2πNkm) H k 2 N k m That is, delay "linear phase shift." (slope =(m2πN) m 2 N

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