Convolution using z transform

  • Thread starter cutesteph
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  • #1
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Main Question or Discussion Point

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
 
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Answers and Replies

  • #2
rbj
2,226
8
may i suggest that you try using the [itex]\LaTeX[/itex] 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

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

as opposed to

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

[itex] z [/itex] is a continuous variable, BTW. but [itex]n[/itex] is discrete.
 

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