convolution sum

Discussion in 'Homework Help' started by jut, Sep 20, 2009.

  1. jut

    Thread Starter Senior Member

    Aug 25, 2007
    224
    2
    Find the convolution of x(n) and h(n).
    x(n) = (-1/2)^n u(n-4)
    h(n) = 4^n u(2-n)

    where u(n)=\begin{cases}1,n\geq0\\0,n<0\end{cases}

    where the convolution is y(n)=x(n)\ast y(n)=\sum_{k=-\infty}^{\infty}x(k)h(n-k)
    note the \ast symbol is not multiply, it is the convolution operator. Also y(n) is causal, which means y(n) is only defined for n\geq 0, where n=0,1,2,3....

    Now,
    x(k) = (-1/2)^k u(k-4)
    h(k) = 4^{n-k} u(2-n+k)

    Plugging in,
    x(n)\ast y(n)=\sum_{k=-\infty}^{\infty} (-1/2)^k u(k-4) 4^{n-k} u(2-n+k)


    Simplifying,
    x(n)\ast y(n)=4^n\sum_{k=-\infty}^{\infty}(-1/2)^k 4^{-k} u(k-4) u(2-n+k)

    The argument passed into the unit step must be \geq 1 or else the value of the unit step function is 0. So if I need to change my limits of summation so that the two unit step functions resolve to 1. That is,

    if k-4\geq 0 then u(k-4)=1.
    if k+2-n\geq 0 then u(k+2-n)=1.

    So the limits on k would be,
    if k\geq 4
    if k\geq n-2

    I am bamboozled by the two lower limits on k. How can I do a summation with two lower limits? Would the summation diverge?
     
    Last edited: Sep 20, 2009
  2. steinar96

    Active Member

    Apr 18, 2009
    239
    4
    Signals and systems eh. Chapter 2 proplem :p. Funny thing is i got stuck on the same thing yesterday. The summation limits change with each n chosen which makes calculating the summation to a closed form a bit more tedious.

    Most likely it needs to be solved by looking at different scenarios when you plug in different "ranges" of n.
    K needs to be larger or equal to 4 so that u[k-4] be nonzero. From this we could say that we could start summing from k = 4.

    however let's say we have y[7]. Which means that u[2-n+k] won't be nonzero till k = 5.
    So with increasing n we can't start summing till k = n-2.

    So for n between 0 and 5 you can start summing from k = 4. While for n >= 6 you have to begin the summation at K = n-2.
     
  3. jut

    Thread Starter Senior Member

    Aug 25, 2007
    224
    2
    Hi, thanks for the reply.

    I was able to solve it shortly after posting. It's like you said, the limits depend on values of n. So the ultimate solution is piecewise.
     
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