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ppoonamk
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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
f(x)= 2*rect(x-.5) [itex]\otimes[/itex] .5 comb(x/2)
[itex]\otimes[/itex]- convolution.
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 can't figure out what G([itex]\xi[/itex]) and g(x) {inverse fft} looks like. Please help me
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 can't figure out what G([itex]\xi[/itex]) and g(x) {inverse fft} looks like. Please help me
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