Discrete-time unit impulses help!

Discussion in 'Homework Help' started by felixshin8, Sep 19, 2011.

  1. felixshin8

    Thread Starter New Member

    Sep 18, 2011
    1
    0
    Hello, I've been trying to figure out how to answer this question.

    Represent x[n] = n^2 as a linear combination of time-shifted discrete-time unit impulses delta[n-k]'s

    Any help would be appreciated
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    The function

    y=x^2

    may be integrated between successive intervals 0-1,1-2,2-3 and so on, to give the required equivalent impulse area.

    The integral result then becomes

    Area=[\frac{x^3}{3}]^{L_1}_{L_2}

    For interval 0 to 1

    Area=[\frac{x^3}{3}]^{1}_{0}=\frac{1}{3}

    For interval 1 to 2

    Area=[\frac{x^3}{3}]^{2}_{1}=\frac{7}{3}

    For interval 2 to 3

    Area=[\frac{x^3}{3}]^{3}_{2}=\frac{19}{3}

    and so on

    so

    x[n]=\frac{1}{3}\mu_0(n)+\frac{7}{3}\mu_0(n-1)+\frac{19}{3}\mu_0(n-2)+\frac{37}{3}\mu_0(n-3)+....

    or perhaps it should be

    x[n]=\frac{1}{3}\mu_0(n-1)+\frac{7}{3}\mu_0(n-2)+\frac{19}{3}\mu_0(n-3)+\frac{37}{3}\mu_0(n-4)+....

    since the function is zero for n=0

    or maybe the answer is just

    x[n]=\mu_0(n-1)+4\mu_0(n-2)+9\mu_0(n-3)+16\mu_0(n-4)+....

    since the function isn't continuous
     
    Last edited: Sep 20, 2011
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