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# Eigenanalysis of LSI Systems

Module by: Richard Baraniuk. E-mail the author

LSI systems circulent (normal) matrix multiplication have eigenvectors and eigenvalues.

## Exercise 1

Find eigenvectors and eigenvalues of LSI system

### Solution

v k =kth DFT basis vector v k k th DFT basis vector

v k n=12Nei2πknN v k n 1 2 N 2 k n N

Complex sinusoid!!

## Proof

Easiest with circular convolution formula

yn=l=0N1x(nl)modNhl=l=0N1ei2πk(nl)modNNhl=l=0N1ei2πknNe(i)2πklNhl=ei2πknN kl =0N1hle(i)2πklN=Hkei2πknN y n l 0 N 1 x n l N h l l 0 N 1 2 k n l N N h l l 0 N 1 2 k n N 2 k l N h l 2 k n N k l 0 N 1 h l 2 k l N H k 2 k n N
(1)

## Note:

kl =0N1hle(i)2πklN k l 0 N 1 h l 2 k l N is the DFT of h Hk H k

## Note:

Hk H k is the eigenvalue and ei2πknN 2 k n N is the eigenvector

In other words, the DFT basis complex sinusoids are eigenvectors of LSI systems. Corresponding eigenvalues are the DFT elements of impulse response.

## Note:

call Hk H k the frequency responce

H=FDFH H F D F , where H=( H0000 0H100 00...0 000HN1 ) H H 0 0 0 0 0 H 1 0 0 0 0 ... 0 0 0 0 H N 1 Yk=HkXk Y k H k X k where 0kN1 0 k N 1

where

Y=FHy Y F y

H=FHh H F h

X=FHx X F x

In other words, circular convolution in time domain is equivalent to multiplication in frequency domain.

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