# Filtering problem. Help with the output

1. Homework Statement

Determine the output if this signal is processed by a filter with the following transfer
functions:
$u= \sum_{k=-\infty}^\infty c_n*exp(j*\pi*n*x)= \sum_{k=-\infty}^\infty rect(x-2k-.5)$

c_n = 0 for even
= 4/(j*$\pi$*n) for odd
Determine the output if this signal is processed by a filter with the following transfer
functions:
a) H($\xi$)=exp{$\phi$($\xi$)}
$\phi$($\xi$)= $\pi$/2 ; $\xi$>0
0; $\xi$=0
-$\pi$/2 ; $\xi$<0

## Homework Equations

f(x)= 2*rect(x-.5) $\otimes$ .5 comb(x/2)
$\otimes$- 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($\xi$) is a shifted since envelope with delta functions at .5 intervals.

F($\xi$)=2* sinc($\xi$)* exp(-i*pi*$\xi$).comb( 2*$\xi$)

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

G($\xi$)= H($\xi$)* F($\xi$)
I cant figure out what G($\xi$) and g(x) {inverse fft} looks like. Please help me

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uart
1. Homework Statement

Determine the output if this signal is processed by a filter with the following transfer
functions:
$u= \sum_{k=0}^\infty a_n*exp{j*\pi*n*x}= \sum_{k=-\infty}^\infty rect(x-2k-.5)$

a_n = 0 for even
= 4/(j*$\pi$*n)
Before you start you need to correct some mistakes in the question. The given Fourier series is incorrect for given rectangular pulse train (I'm assuming that it is a rectangular pulse train and that the sum from k=infinity to infinity was a typo - which I corrected in the quote).

Also

1. If you're going to use the complex exponential form of the series on that real time function then you'll need to sum from a_n from -infinity to infinity, not from zero to infinity.

2. $a_n = 0$ for n even, except for a_0 which equals 1/2.

3. $a_n = 1/(j n \pi)$ for n odd.

4. We usually denote the coefficients of the complex series as c_n (a_n is usually reserved for the real cosine series).

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Before you start you need to correct some mistakes in the question. The given Fourier series is incorrect for given rectangular pulse train (I'm assuming that it is a rectangular pulse train and that the sum from k=infinity to infinity was a typo - which I corrected in the quote).

Also

1. If you're going to use the complex exponential form of the series on that real time function then you'll need to sum from a_n from -infinity to infinity, not from zero to infinity.

2. $a_n = 0$ for n even, except for a_0 which equals 1/2.

3. $a_n = 1/(j n \pi)$ for n odd.

4. We usually denote the coefficients of the complex series as c_n (a_n is usually reserved for the real cosine series).
Hi,

Thank you for correcting the typo. The series can alternatively be expressed in terms of rect and that is what I have been working on. Could you help me with the output of the filter. I know the phase change should the change the sines into cosines. I can understand what is happening but I cant figure out the math behind it

uart
Hi,

Thank you for correcting the typo. The series can alternatively be expressed in terms of rect and that is what I have been working on. Could you help me with the output of the filter. I know the phase change should the change the sines into cosines. I can understand what is happening but I cant figure out the math behind it
Yes I agree that the series can alternatively be represented as a single rectangular pulse convolved with an impulse train. However since the problem gave the Fourier series I thought perhaps that was how they wanted you do deal with it.

You realize that with the Fourier series your "phase filter" (also known as a Hilbert transform filter) simply turns each sine component to an equivalent cosine component.

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uart
BTW. There is also a typo in your filter definition.

H($\xi$)=exp{$\phi$($\xi$)}

In my previous response I've assumed that you meant to write:

H($\xi$)=exp{$j \phi$($\xi$)}

H(ξ)=exp{jϕ(ξ)}

Yes you are right. Sorry about having so many errors. This was the first time I was using latex.
A gentle bounce. Can anyone help me?

uart
Look at how your filter behaves, term (pair) by term (pair), on the Fourier series.

Input term: $\frac{\exp(j k \pi x) - \exp(-j k \pi x)}{j k} = \frac{2 \sin(k \pi x)} {k \pi}$

Output term: $\frac{\exp(j k \pi x + j \pi/2) - \exp(-j k \pi x - j \pi/2)}{j k \pi} = \frac{2 \cos(k \pi x)} {k \pi}$

The easiest way to express the output of that filter is in terms of its Fourier series.

Here's a plot of the first 50 terms of the FS of both input and output.

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