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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.

Figure 1
Figure 1 (fig1.png)
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)...

Figure 2
Time Domain
Time Domain (fig2.png)
Figure 3
Frequency Domain
Frequency Domain (fig3.png)
ie: Yk=HkXk Y k H k X k ; 0kN1 0 k N 1

ie: pointwise multiplication

Figure 4
Figure 4 (fig4.png)

Example 1

2 point smoother

Figure 5
Figure 5 (fig5.png)
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)
Figure 6
Figure 6 (fig6.png)
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

Figure 7
Figure 7 (fig7.png)
ie: smoothing lowpass filtering

Example 2

Figure 8
Figure 8 (fig8.png)
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)
Figure 9
Figure 9 (fig9.png)
Figure 10
Figure 10 (fig10.png)
Figure 11
Figure 11 (fig11.png)
Figure 12
Figure 12 (fig12.png)
Figure 13
Figure 13 (fig13.png)

Example 3: Edge Detector

Figure 14
Figure 14 (fig14.png)
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)
Figure 15
High Pass Filter
High Pass Filter (fig15.png)

Example 4: Circular Shift

Figure 16
Figure 16 (fig16.png)
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
Figure 17
Alll Pass Filter
Alll Pass Filter (fig17.png)
That is, delay "linear phase shift." (slope =(m2πN) m 2 N

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