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**1. Homework Statement**

Determine the output if this signal is processed by a filter with the following transfer

functions:

[itex]u= \sum_{k=-\infty}^\infty c_n*exp(j*\pi*n*x)= \sum_{k=-\infty}^\infty rect(x-2k-.5)[/itex]

c_n = 0 for even

= 4/(j*[itex]\pi[/itex]*n) for odd

Determine the output if this signal is processed by a filter with the following transfer

functions:

a) H([itex]\xi[/itex])=exp{[itex]\phi[/itex]([itex]\xi[/itex])}

[itex]\phi[/itex]([itex]\xi[/itex])= [itex]\pi[/itex]/2 ; [itex]\xi[/itex]>0

0; [itex]\xi[/itex]=0

-[itex]\pi[/itex]/2 ; [itex]\xi[/itex]<0

## Homework Equations

f(x)= 2*rect(x-.5) [itex]\otimes[/itex] .5 comb(x/2)

[itex]\otimes[/itex]- convolution.

## The Attempt at a Solution

f(x) is a series of rect functions of width 1 and centered at .5, 2.5,-1.5 etc

In the frequency domain F([itex]\xi[/itex]) is a shifted since envelope with delta functions at .5 intervals.

F([itex]\xi[/itex])=2* sinc([itex]\xi[/itex])* exp(-i*pi*[itex]\xi[/itex]).comb( 2*[itex]\xi[/itex])

The filter H([itex]\xi[/itex]) is a phase filter which is equal to 1 at [itex]\xi[/itex]=0, i for [itex]\xi[/itex] >0 and -i for [itex]\xi[/itex]<0

G([itex]\xi[/itex])= H([itex]\xi[/itex])* F([itex]\xi[/itex])

I cant figure out what G([itex]\xi[/itex]) and g(x) {inverse fft} looks like. Please help me

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