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&= \left[z^{-4} + 2z^{-5} + 3z^{-2}\right] \sum_{n=-\infty}^{\infty} \delta[n]z^{-n} \\ | &= \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} | &= \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} | ||
| + | &= \sum_{n=-\infty}^{\infty} | ||
| + | |||
\end{align} | \end{align} | ||
</math> | </math> | ||
| Line 34: | Line 36: | ||
---- | ---- | ||
| + | <math> | ||
| + | \begin{align} | ||
| + | x[n] &= - a^n u[-n-1] \\ | ||
| + | X(z) &= \frac{1}{1-az^{-1}}, \left |z \right| < \left | a \right| \\ | ||
| + | &= \frac{1}{1-a/z}\\ | ||
| + | &= \frac{z}{z-a}\\ | ||
| + | \end{align} | ||
| + | </math> | ||
| + | Multiplying and dividing X(z) by -1, | ||
| + | |||
| + | <math> | ||
| + | \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} | ||
| + | </math> | ||
| + | |||
| + | Changing variable using n+1 = -k | ||
| + | |||
| + | <math> | ||
| + | \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} | ||
| + | </math> | ||
| + | |||
| + | ---- | ||
| + | |||
| + | <math> | ||
| + | \begin{align} | ||
| + | x[n] &= a^n (u[n-2] + u[n])\\ | ||
| + | X(z) &= \frac{z^2 + a^2}{z^2 - az}, \left |z \right| > \left | a \right| \\ | ||
| + | \end{align} | ||
| + | </math> | ||
| + | |||
| + | This [[Z-transforms_and_inverse_z-transforms_ECE438F10|page]] has a discussion about calculating this particular X(z). | ||
| + | |||
| + | <math> | ||
| + | \begin{align} | ||
| + | &= \frac{z^2}{z^2(1-a/z)} + \frac{a^2}{z^2(1-a/z)}\\ | ||
| + | &= \frac{1}{1-a/z} + \frac{a^2}{z^2(1-a/z)}\\ | ||
| + | &= \sum_{n=0}^{\infty} (a/z)^n + (a/z)^2 \sum_{n=0}^{\infty} (a/z)^n\\ | ||
| + | &= \sum_{n=-\infty}^{\infty} a^n u[n]z^{-n} + \sum_{n=-\infty}^{\infty} a^{n+2}u[n]z^{-(n+2)} | ||
| + | \end{align} | ||
| + | </math> | ||
| + | |||
| + | Making a change of variable using n+2 = k | ||
| + | |||
| + | <math> | ||
| + | \begin{align} | ||
| + | X(z) &= \sum_{n=-\infty}^{\infty} a^n u[n]z^{-n} + \sum_{k=-\infty}^{\infty} a^{k}u[k-2]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] + a^n u[n-2])\\ | ||
| + | &= a^n (u[n-2] + u[n]) | ||
| + | \end{align} | ||
| + | </math> | ||
| + | |||
| + | ---- | ||
| + | |||
| + | <math> | ||
| + | \begin{align} | ||
| + | x[n] &= a^n u[n-2] + b^n u[n-1]\\ | ||
| + | X(z) &= \frac{a^2z^{-2}}{1-az^{-1}} + \frac{bz^{-1}}{1-bz^{-1}}, \left |z \right| > \left | a \right| \left |z \right| > \left | b \right|\\ | ||
| + | &= (a/z)^2\frac{1}{1-(a/z)} + (b/z)\frac{1}{1-(b/z)}\\ | ||
| + | &= \sum_{n=0}^{\infty}(a/z)^{n+2} + (b/z)\sum_{n=0}^{\infty}(b/z)^n+1 \\ | ||
| + | &= \sum_{n=-\infty}^{\infty}(a/z)^{n+2}u[n] + \sum_{n=-\infty}^{\infty}(b/z)^{n+1}u[n]\\ | ||
| + | \end{align} | ||
| + | </math> | ||
| + | |||
| + | Change of variable using n+2 = j, n+1 = k | ||
| + | |||
| + | <math> | ||
| + | \begin{align} | ||
| + | X(z) &= \sum_{j=-\infty}^{\infty}(a/z)^{j}u[n-2] + \sum_{k=-\infty}^{\infty}(b/z)^{k}u[k-1]\\ | ||
| + | X(z) &= \sum_{j=-\infty}^{\infty}a^ju[n-2]z^-j + \sum_{k=-\infty}^{\infty}b^ku[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-2] + b^n u[n-1]\\ | ||
| + | |||
| + | \end{align} | ||
| + | </math> | ||
| + | |||
| + | ---- | ||
[[2010_Fall_ECE_438_Boutin|Back to 438 main page]] | [[2010_Fall_ECE_438_Boutin|Back to 438 main page]] | ||
Latest revision as of 16:42, 29 November 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} &= \sum_{n=-\infty}^{\infty} \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] \\ X(z) &= \frac{1}{1-az^{-1}}, \left |z \right| < \left | a \right| \\ &= \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} $
$ \begin{align} x[n] &= a^n (u[n-2] + u[n])\\ X(z) &= \frac{z^2 + a^2}{z^2 - az}, \left |z \right| > \left | a \right| \\ \end{align} $
This page has a discussion about calculating this particular X(z).
$ \begin{align} &= \frac{z^2}{z^2(1-a/z)} + \frac{a^2}{z^2(1-a/z)}\\ &= \frac{1}{1-a/z} + \frac{a^2}{z^2(1-a/z)}\\ &= \sum_{n=0}^{\infty} (a/z)^n + (a/z)^2 \sum_{n=0}^{\infty} (a/z)^n\\ &= \sum_{n=-\infty}^{\infty} a^n u[n]z^{-n} + \sum_{n=-\infty}^{\infty} a^{n+2}u[n]z^{-(n+2)} \end{align} $
Making a change of variable using n+2 = k
$ \begin{align} X(z) &= \sum_{n=-\infty}^{\infty} a^n u[n]z^{-n} + \sum_{k=-\infty}^{\infty} a^{k}u[k-2]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] + a^n u[n-2])\\ &= a^n (u[n-2] + u[n]) \end{align} $
$ \begin{align} x[n] &= a^n u[n-2] + b^n u[n-1]\\ X(z) &= \frac{a^2z^{-2}}{1-az^{-1}} + \frac{bz^{-1}}{1-bz^{-1}}, \left |z \right| > \left | a \right| \left |z \right| > \left | b \right|\\ &= (a/z)^2\frac{1}{1-(a/z)} + (b/z)\frac{1}{1-(b/z)}\\ &= \sum_{n=0}^{\infty}(a/z)^{n+2} + (b/z)\sum_{n=0}^{\infty}(b/z)^n+1 \\ &= \sum_{n=-\infty}^{\infty}(a/z)^{n+2}u[n] + \sum_{n=-\infty}^{\infty}(b/z)^{n+1}u[n]\\ \end{align} $
Change of variable using n+2 = j, n+1 = k
$ \begin{align} X(z) &= \sum_{j=-\infty}^{\infty}(a/z)^{j}u[n-2] + \sum_{k=-\infty}^{\infty}(b/z)^{k}u[k-1]\\ X(z) &= \sum_{j=-\infty}^{\infty}a^ju[n-2]z^-j + \sum_{k=-\infty}^{\infty}b^ku[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-2] + b^n u[n-1]\\ \end{align} $
