Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Convolution using z transform

  1. Dec 3, 2013 #1
    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. jcsd
  3. Dec 4, 2013 #2


    User Avatar

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook