I am trying to use Fourier transforms to get the discrete time response of an LTI system.

I am given: $$h[n] = (0.5)^n u[n],\space{}x[n]=\cos(\pi*n/2)$$

I am trying to find $y[n]$ via Fourier transforms.

So far, I have:

$$H(\Omega) = 1/(1-0.5\cdot e^{-j\Omega})$$

$$X(\Omega) = \pi\sum_{m=-\infty}^{\infty}[\delta(\Omega-\pi/2 - 2\pi m)+\delta(\Omega+\pi/2-2\pi m)]$$

$$Y(\Omega) = H(\Omega)X(\Omega)= 1/(1-0.5\cdot e^{-j\Omega}) \cdot \pi\sum_{m=-\infty}^{\infty}[\delta(\Omega-\pi/2 - 2\pi m)+\delta(\Omega+\pi/2-2\pi m)]$$

Now I am trying to simplify $Y(\Omega)$ to more easily convert back via transforms to discrete time $y[n]$.

I think I need to use the fact that DTFTs have a period of $2\pi$ to evaluate the $X(\Omega)$, but I am not sure where to begin.



Hint: Use the fact that $\delta(x-x_0)f(x)=\delta(x-x_0)f(x_0)$ to arrive at $$Y(\Omega) = \pi\sum_{m=-\infty}^{\infty}\frac{\delta(\Omega-\pi/2-2\pi m)}{1+0.5j}+\frac{\delta(\Omega+\pi/2-2\pi m)}{1-0.5j}. \tag 1$$ Afterwards, use $$\frac{1}{1+0.5j}=\frac{1-0.5j}{1.25}, \tag 2$$ and $$\frac{1}{1-0.5j}=\frac{1+0.5j}{1.25}. \tag 3$$

  • $\begingroup$ How did you eliminate the $e^{-j\Omega}$ in $H(\Omega)$? $\endgroup$ – sbdchd Nov 7 '17 at 22:09
  • 1
    $\begingroup$ @sbdchd Use $\delta(x-x_0)f(x)=\delta(x-x_0)f(x_0)$ to establish $$\frac{\delta(\Omega-\pi/2-2\pi m)}{1-0.5e^{-j \Omega}}=\frac{\delta(\Omega-\pi/2-2\pi m)}{1+0.5j}.$$ $\endgroup$ – Math Lover Nov 7 '17 at 22:20

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