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Scaled unit impulse/step sequences

  1. Nov 10, 2013 #1
    1. The problem statement, all variables and given/known data

    [itex]b_{k} = {4, 1, 1, 4}[/itex]

    x[n] = 2u[n]

    Write your answer using scaled unit impulse sequences and scaled unit step sequences. Write explicitly.

    2. Relevant equations



    3. The attempt at a solution

    Code (Text):


    4114
    2222222....
    ------------
    8228
     8228
      8228
       8228
        8228
          .....
    ------------
    8, 10, 12, 20, 20, 20, ...

     
    [itex]x[n] = 8\delta[n] + 10\delta[n-1] + 12\delta[n-2] + \sum^{\infty}_{k=3} 20 \delta[n-k][/itex]


    [itex]
    \delta[n] = \left\{\begin{matrix}
    0 & n<0\\
    8 & n = 0\\
    10 & n = 1 \\
    12 & n = 2\\
    20 & n \geq 3
    \end{matrix}\right.
    [/itex]

    Is this how you would answer this question?
     
    Last edited: Nov 10, 2013
  2. jcsd
  3. Nov 10, 2013 #2

    rude man

    User Avatar
    Homework Helper
    Gold Member

    You haven't stated the problem fully. What is the difference equation, exactly?


    In any case, u[n] is always 1.
     
  4. Nov 10, 2013 #3
    The general difference equation for a causal FIR is:

    [itex]y[n] = \sum^{\infty}_{k=0} b_{k} x[n-k][/itex]

    and then

    [itex]
    \delta[n] = \left\{\begin{matrix}
    0 & n<0\\
    8 & n = 0\\
    10 & n = 1 \\
    12 & n = 2\\
    20 & n \geq 3
    \end{matrix}\right.
    [/itex]
     
    Last edited: Nov 10, 2013
  5. Nov 10, 2013 #4

    rude man

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    Homework Helper
    Gold Member

    Your expression for x[n] is correct. But I don't quite understand your table, probably because I can't make out the column after the 1st equal sign.

    But δ[n] = 1, n = 0
    = 0, n > 0 always.

    BTW your equation is for a non-recursive filter, which is not necessarily an FIR filter.
    FIR filters can also be recursive, and IIR filters can be non-recursive. However, saying FIR → non-recursive and IIR → recursive is almost universal, if misleading.
     
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