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<math>
 
<math>
 
\begin{align}
 
\begin{align}
x[n] &= u[n-4] + 2u[n-5] + 3u[n-2] \\
+
x[n] &= \delta[n-4] + 2\delta [n-5] + 3\delta [n-2], \\
& {\color{blue} \text{I think you mean } \delta[n-4] + 2\delta [n-5] + 3\delta [n-2]}, \\
+
 
{\color{blue} \text{and thus } X(z)} &{\color{blue}= z^{-4} + 2z^{-5} + 3z^{-2} , \text{ which converges for any finite }z\neq 0.}\\
+
X(z) &= z^{-4} + 2z^{-5} + 3z^{-2} \\
 +
&= \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}
 
\end{align}
 
</math>
 
</math>
 +
 +
Using a change in variables to bring equation to the right form,
 +
 +
<math>
 +
\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}
 +
</math>
 +
 +
  
 
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Revision as of 10:31, 18 September 2010

$ 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} \\ &= \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} $


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Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood