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Homework Help: Unit Sample Response Question

  1. Jul 8, 2015 #1
    What's j2 in this problem?

    1. The problem statement, all variables and given/known data


    2. Relevant equations

    3. The attempt at a solution

    a. What's j2 in this problem?

    b. How did we use the fact that we have a unit impulse?

    c. What are our first thoughts and strategies when seeing the problem ?

    What is our strategy? It looks like j2 is 4R/2 almost
  2. jcsd
  3. Jul 13, 2015 #2


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    I think the [itex] j2 [/itex] is merely two times the unit imaginary number. In other words, it's the same as [itex] \left( 2 \right) \left( \sqrt{-1} \right) [/itex] which is the same thing as saying [itex] \sqrt{-4} [/itex].

    You'll have to forgive me for not being more skilled with using the operator expressions. But maybe I can add some insight, albeit just a little bit.

    If you're wondering where the [itex] \frac{Y}{X} = \frac{3+4R}{1+R^2} [/itex] came from, it might help by rearranging the original difference equation.
    [tex] y[n] + y[n-2] = 3x[n] +4x[n-1] [/tex]
    Now transform the equation such that each delay corresponds an [itex] R [/itex]. [itex] [n] [/itex] gets no [itex] R [/itex], [itex] [n-1] [/itex] corresponds a single [itex] R [/itex], and [itex] [n-2] [/itex] corresponds to [itex] RR = R^2 [/itex]. Now solve for [itex] \frac{Y}{X} [/itex].

    The step where [itex] \frac{Y}{X} = \frac{3+4R}{1+R^2} = \frac{\frac{3}{2} - j2}{1-jR} + \frac{\frac{3}{2} + j2}{1+jR}[/itex] is the result obtained from partial fraction decomposition. Note that [itex] 1+R^2 [/itex] doesn't factor using real numbers. That's where imaginary numbers come in.
  4. Jul 13, 2015 #3


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    By the way, it looks like what's being done here is solving the problem by using what's more commonly called the "z-transform." I've never seen it done with [itex] R[/itex]s as the notation, which is what threw me. It's more commonly notated with zs. Anyway, if you want to investigate further, do some online research on the "z-transform."

    [Edit: Oh, and the approach your coursework is using uses positive exponents of [itex] R, R^2, [/itex] etc., where the z transform uses negative exponents, [itex] z^{-1}, z^{-2}, [/itex] etc. So the approach that your coursework is using is a little different than the standard z-transform (although it's conceptually equivalent; just substitute [itex] R^n \Leftrightarrow z^{-n} [/itex]). So that's something else to keep in mind.]
    Last edited: Jul 13, 2015
  5. Jul 17, 2015 #4

    rude man

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    Transwform the finite-difference equation into z transform form:
    Y(z) + z-2Y(z) = αX(z) + βz-1X(z)
    which gives you Y/X (z).
    What is the z transform of the given unit impulse function?
    Now use partial fractions to decompose the final Y(z) into manageable terms for the purpose of inverting each term.

    BTW I haven't done that myself and I suspect their derivation has a math error in it also.
  6. Jul 19, 2015 #5

    rude man

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    EDIT: nope, the given answer for the last 2 terms is correct:

    y[n] = 3δ[n] + (3/2 - j2) jn + (3/2 + j2) (-j)n.

    But the omission of the impulse term remains. In fact, given that Y(z) is not a "strictly proper" fraction (numerator and denominator order of z same) automatically requires such a term in the finite-difference equation solution.

    I also have never seen R used in lieu of z-1 but it seems OK. Not sure what the merit of it is; every z transform table I've encountered is in z, not in R = z-1. To use these tables the substitution R = z-1 and then multiplying num. & denom. by z would have to be made.
    Last edited: Jul 19, 2015
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