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Filtering problem. Help with the output

  • Thread starter ppoonamk
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  • #1
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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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Answers and Replies

  • #2
uart
Science Advisor
2,776
9
1. Homework Statement

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

a_n = 0 for even
= 4/(j*[itex]\pi[/itex]*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. [itex]a_n = 0[/itex] for n even, except for a_0 which equals 1/2.

3. [itex]a_n = 1/(j n \pi)[/itex] 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).
 
Last edited:
  • #3
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0
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. [itex]a_n = 0[/itex] for n even, except for a_0 which equals 1/2.

3. [itex]a_n = 1/(j n \pi)[/itex] 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
 
  • #4
uart
Science Advisor
2,776
9
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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  • #5
uart
Science Advisor
2,776
9
BTW. There is also a typo in your filter definition.

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

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

H([itex]\xi[/itex])=exp{[itex]j \phi[/itex]([itex]\xi[/itex])}
 
  • #6
28
0
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?
 
  • #7
uart
Science Advisor
2,776
9
Look at how your filter behaves, term (pair) by term (pair), on the Fourier series.

Input term: [itex]\frac{\exp(j k \pi x) - \exp(-j k \pi x)}{j k} = \frac{2 \sin(k \pi x)} {k \pi}[/itex]

Output term: [itex]\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}[/itex]

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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