Filtering problem. Help with the output

[ppoonamk]

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 can't figure out what G(\xi) and g(x) {inverse fft} looks like. Please help me

 

[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).

 

[ppoonamk]

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 can't 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 can't 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.

 

[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)}

 

[ppoonamk]

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.