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How can I simplify this Convolution through a limit change?

  1. Feb 12, 2013 #1

    ElijahRockers

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


    Compute the convolution y[n] = x[n] * h[n]

    if discrete signals:
    [itex] x[n] = \alpha^nu[n][/itex]
    [itex] h[n] = \beta^nu[n][/itex]

    Where [itex]\alpha \neq \beta[/itex].

    2. Relevant equations

    [itex]y[n] = \sum_{k=-\infty}^{\infty} h[n-k]x[k][/itex]

    3. The attempt at a solution

    I plugged the two equations into the above convolution formula and got

    [itex]y[n] = \sum_{k=-\infty}^{\infty} \beta^{h-k}u[n-k]\alpha^ku[k][/itex]

    Simplifying, I get

    [itex]y[n] = \beta^n \sum_{k=-\infty}^{\infty} (\frac{\alpha}{\beta})^ku[n-k]u[k][/itex]

    And that's as far as I understand. The solution is supposed to be:

    [itex]y[n] = \beta^n \sum_{k=0}^n(\frac{\alpha}{\beta})^k , n\geq0[/itex]
    [itex]y[n] = (\frac{\beta^{n+1} - \alpha^{n+1}}{\beta - \alpha})u[n] , \alpha \neq \beta[/itex]

    But I'm not sure how to get there. I see that the change in limits somehow facilitates this, but I just can't wrap my head around how to arrive at that, especially the part where α≠β.

    I see that for each n, k goes through all possible values and sums each result, but I don't see why the lower limit for the solution is 0, or the upper limit is n.

    There's obviously something fundamental about convolutions that I'm not getting.
     
  2. jcsd
  3. Feb 12, 2013 #2

    Ray Vickson

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    Are the equations [itex] x[n] = \alpha^nu[n], \; y[n] = \beta^nu[n][/itex] supposed to hold for all integer n from -∞ to +∞, or are they supposed to hold only for integer n ≥ 0? (The answer makes a BIG difference.)
     
  4. Feb 12, 2013 #3

    rude man

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    Why? The u(n) says n > = 0 only, does it not?
    .
     
  5. Feb 12, 2013 #4

    rude man

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    Z transforms, anyone? Hint: anu[n] → 1/(1 - az-1).
     
  6. Feb 12, 2013 #5

    jbunniii

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    It seems to me that you're almost there. You just need to take into account the values of ##k## for which ##u[k]## and ##u[n-k]## are nonzero, and change the limits on the sum accordingly. Assuming the definition
    $$u[k] = \begin{cases}
    1 & \textrm{ if }k \geq 0 \\
    0 & \textrm{ otherwise} \\
    \end{cases}$$
    then for what values of ##k## is ##u[n-k]## nonzero? And then what about the product ##u[k]u[n-k]##?
     
  7. Feb 12, 2013 #6

    ElijahRockers

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    u[n-k] = 1 when k is ≤ n
    u[k] is 1 when k ≥ 0

    ok, so k only matters from 0≤k≤n, so i see where the limits are.... and 1 times 1 is 1....

    ok. its clicking.

    but why is the restriction on n≥0?

    Also, I used a formula for the sum when A/B ≠ 1 to get

    [itex]\beta^n(\frac{1-(\frac{\alpha}{\beta})^{n+1}}{1-\frac{\alpha}{\beta}})[/itex]

    But... is it just a matter of simplifying this to get the expression given as the solution?
     
  8. Feb 12, 2013 #7

    jbunniii

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    This is because if ##n < 0##, there are no values of ##k## which satisfy ##0 \leq k \leq n##, or putting it another way, ##u(k)u(n-k) = 0## for all ##k##. Therefore ##y[n] = 0## if ##n < 0##. This is why there is a ##u[n]## in the expression for the solution.
    Yes. Try multiplying through by ##\beta^n##, and clearing denominators.
     
  9. Feb 12, 2013 #8

    ElijahRockers

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    Ok, after grinding my gears for a bit, I get why n>0 for the first expression, duh, the sum upper limit cant be less than the lower limit,

    but where does the u[n] come from in the second expression?
     
  10. Feb 12, 2013 #9

    jbunniii

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    By the way, as you noted, your expression is invalid if ##\alpha = \beta##. But there is a valid solution if ##\alpha = \beta##. For completeness, you might consider finding a separate formula for that special case.
     
  11. Feb 12, 2013 #10

    ElijahRockers

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    That's part B), already figured that one out,

    y[n] = (n+1)αn...

    but the solution has u[n] also, just like in part A).... not sure where either of those comes from.
     
  12. Feb 12, 2013 #11

    jbunniii

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    See the remark I made above regarding why the expression only holds for ##n \geq 0##:
     
  13. Feb 12, 2013 #12

    ElijahRockers

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    Ok, that makes some kind of sense... it's kind of hard to visualize what's actually happening but I see where you are coming from.

    Thanks for your help!
     
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