Difference between revisions of "ECE438 HW3 Solution" - Rhea

Revision as of 13:07, 18 September 2010

Inverse Z-transforms

$ x[n] = \begin{cases} 1, & n = 4 \\ 2, & n = 5 \\ 3, & n = 2 \\ 0, & \mbox{else} \end{cases} $

This is equivalent to

$ \begin{align} x[n] &= \delta[n-4] + 2\delta [n-5] + 3\delta [n-2], \\ X(z) &= z^{-4} + 2z^{-5} + 3z^{-2}, (\text{converges for all } z \neq 0) \\ &= \left[z^{-4} + 2z^{-5} + 3z^{-2}\right] \sum_{n=-\infty}^{\infty} \delta[n]z^{-n} \\ &= \sum_{n=-\infty}^{\infty} \delta[n]z^{-n-4} + 2\sum_{n=-\infty}^{\infty} \delta[n]z^{-n-5} + 3\sum_{n=-\infty}^{\infty} \delta[n]z^{-n-2} \end{align} $

Using a change in variables to bring equation to the right form,

$ \begin{align} j = n+4 \\ k = n+5\\ l = n+2 \\ X(z) &= \sum_{j=-\infty}^{\infty} \delta[j-4]z^{-j} + 2\sum_{k=-\infty}^{\infty} \delta[k-5]z^{-k} + 3\sum_{l=-\infty}^{\infty} \delta[l-2]z^{-l} \\ x[n] &= \mathcal{Z}^{-1}(X(z)) \text{ and } X(z) = \sum_{n=-\infty}^{\infty}x[n]z^{-n}\\ x[n] &= \delta[n-4] + 2\delta [n-5] + 3\delta [n-2] \end{align} $

$ \begin{align} x[n] &= - a^n u[-n-1], \left |z \right| < \left | a \right| \\ X(z) &= \frac{1}{1-az^{-1}}\\ &= \frac{1}{1-a/z}\\ &= \frac{z}{z-a}\\ \end{align} $

Multiplying and dividing X(z) by -1,

$ \begin{align} X(z) &= \frac{-z}{a-z}\\ &= \frac{-z}{a(1-z/a)}\\ &= \frac{-z}{a}\frac{1}{1-(z/a)} \\ &= \frac{-z}{a}\sum_{n=0}^{\infty}(z/a)^n \\ &= - \sum_{n=0}^{\infty}(z/a)^{n+1} \\ &= - \sum_{n=-\infty}^{\infty}(z/a)^{n+1}u[n] \end{align} $

Changing variable using n+1 = -k

$ \begin{align} X(z) &= - \sum_{k=-\infty}^{\infty}(z/a)^{-k}u[-k-1] \\ &= - \sum_{k=-\infty}^{\infty}a^{k}u[-k-1](z)^{-k} \\ x[n] &= \mathcal{Z}^{-1}(X(z)) \text{ and } X(z) = \sum_{n=-\infty}^{\infty}x[n]z^{-n}\\ x[n] &= -a^n u[-n-1] \end{align} $

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