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

**Dint** · April 22nd 2010, 03:16 AM

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:

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

and input:

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

so to calculate $y(n)$

I know that:

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

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

so...

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

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

simplifying this summation...

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

similarly

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

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

I have got:

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

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 $Y(f)$ is absolutely killing me.

Thanks a lot!

**Dint** · April 22nd 2010, 03:43 AM

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:

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

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

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

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

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