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Thread: Finding output eqn using DTFT and inverse DTFT knowing impulse and input eqns!

  1. #1
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    Apr 2010
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    Finding output eqn using DTFT and inverse DTFT knowing impulse and input eqns!

    Hi,

    The question is apparently simple, but I am very stuck and have been working on it for way too long, I must be missing something

    Any help is greatly appreciated.

    I am given an impulse response:

    $\displaystyle h(n) = (-1/2)^n u(n) $

    and input:

    $\displaystyle x(n) = (1/4)^n u(n) $

    so to calculate $\displaystyle y(n) $

    I know that:

    $\displaystyle y(n) = x(n) * h(n) $ and $\displaystyle Y(f) = X(f) \times H(f) $

    so my strategy (and following the questions instructions) was to find $\displaystyle X(f)$ then multiply it with $\displaystyle H(f)$ to get $\displaystyle Y(f)$ and finally use the inverse DTFT to find $\displaystyle y(n)$.

    so...

    $\displaystyle X(f) = \sum\limits_{n=0}^\infty x(n) e^{-(j2 \pi fn)} $

    $\displaystyle = \sum\limits_{n=0}^\infty (1/4) e^{-(j2 \pi fn)} $

    simplifying this summation...

    $\displaystyle X(f) = \frac {1}{1 - \frac{1}{4} e^{-(j2 \pi f)}} $

    similarly

    $\displaystyle H(f) = \frac {1}{1 + \frac{1}{2} e^{-(j2 \pi f)}} $

    Now, when I multiply these two together, I have no idea how to find the inverse DTFT.

    I have got:

    $\displaystyle Y(f) = \frac{1}{(1 - \frac{1}{4} e^{-(j2 \pi f)}) \times (1 + \frac{1}{2} e^{-(j2 \pi f)}) } $

    how can I find the DTFT of something like this? I don't know the period of it and I have no idea how to simplify it into something to apply the inverse DTFT which I know as:



    like I said, any advice would be greatly appreciated.

    I have just found myself completely lost with this problem and doing loops trying to simplify my $\displaystyle Y(f) $ is absolutely killing me.

    Thanks a lot!
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  2. #2
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    Joined
    Apr 2010
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    So looking further at my notes I've found that apparently the period of my functions x(n) and h(n), the period will always be 1.

    So I then find:

    $\displaystyle x(n) = \int\limits_{-0.5}^{0.5} X(f) e^{j2 \pi fn} df $

    so from my $\displaystyle Y(f) $ I get:

    $\displaystyle Y(n) = \int\limits_{0.5}^{0.5} \frac{e^{j2 \pi fn}}{ (1 - \frac{1}{4} e^{-j2 \pi f}) \times (1+ \frac{1}{2}e^{-j2 \pi f} ) } df$

    I have no idea where to start when evaluating this integral...
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