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Unit-Pulse Response for Discrete Time System

  1. Feb 2, 2008 #1
    The question is: Compute the unit-impulse response h[n] for n=0,1,2,3 for each of the following discrete-time systems.

    y[n+1] + y[n] = 2x[n]

    I am trying to figure out how to solve this equation. I understand the example in the book but I don't understand what to do when it calls a future value (n+1)

    I rewrote the equation as:

    When n=0 delta[n] is 1 so:
    y[0]=2*1-y[1]<-----This is where I am getting confused. Doesn't y[1] refer to my answer when I use the value n=1? How can I get a solution if each equation will refer to the next future equation? The example in the book uses y[n-1] so for each value of n it refers to the previous answer for y[n].

    Any help would be much appreciated!!
  2. jcsd
  3. Feb 3, 2008 #2


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    It is exactly the contrary of what you did. You should write y[n+1] as a function of y[n] and x[n].
    By definition, the impulse response of a system is the zero state response of that system when the input is an impulse, so you have y[0] = 0.
    Now substitute x[n] by the impulse function and solve iteratively for y[1], y[2], y[3].
  4. Feb 21, 2008 #3
    Why don't you use Z transforms. That'll provide you with some more insight.
  5. Aug 13, 2009 #4

    Does this same reasoning apply if the equation is:

    y[n+2] + y[n+1] + y[n] = x[n+1] - x[n]

    if so, i too am lost.

    is there another way to describe it?
    or could you just go through it step by step
  6. Aug 13, 2009 #5


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    The reasoning is the same. Start with n = -1.
    x[n] = x[-1] = 0
    x[n+1] = x[0] = 1
    x[n+2] = x[n+3] = ... = 0
    y[n] = y[-1] = 0
    y[n+1] = y[0] = 0
    Or, as unplebeian suggested, use the z-transform
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