# Discrete-time unit impulses help!

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#### felixshin8

Joined Sep 18, 2011
1
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

#### t_n_k

Joined Mar 6, 2009
5,455
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

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