# Convolution using z transform

1. Dec 3, 2013

### cutesteph

x_1(n) = (!/4)^n u(n-1) and x_2(n) = [1- (1/2)^n] u(n)

X_1(z) = (1/4)z^-1 / (1-(!/4)z^-1 and X_2(z) = 1/(1-z^-1) + 1/(1-(1/2)z^-1)

Y(z) = X_1(z) X_2(z) = (-4/3) /(1-(1/4)z^-1 + (1/3) / (1-z^-1) + 1/(1-(1/2)z^-1

Last edited: Dec 3, 2013
2. Dec 4, 2013

### rbj

may i suggest that you try using the $\LaTeX$ pasteup provided by physicsforums? it makes it much easier to read. also try to use the convention of square brackets (instead of parenths) for discrete-time or discrete-frequency signals. like

$$x[n] = \frac{1}{N} \sum_{k=0}^{N-1} X[k] e^{j 2 \pi k n / N}$$

as opposed to

$$x(t) = \int_{-\infty}^{+\infty} X(f) e^{j 2 \pi f t} df$$

$z$ is a continuous variable, BTW. but $n$ is discrete.