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I am trying to convolve the two discrete sequences

$$\left(\frac34\right)^nu(n-2)$$

and $$2^nu(-n-5)$$

where $u(n)$ is the unit step function.

Here is how I approached the problem ,using the convolution sum

$$\sum_{k=-\infty}^\infty x(k)h(n-k)$$

in this case ,

$$\sum_{k=-\infty}^\infty\left(\frac34\right)^ku(k-2)2^{n-k}u(-(n-k)-5)$$

which becomes

$$\sum_{k=-\infty}^\infty\left(\frac34\right)^ku(k-2)\frac{2^n}{2^k}u(-n+k-5)$$

now $u(k-2)$ in non zero when $k \ge 2$ and $u(-n+k-5)$ in non zero when $k \ge n+5$ (this is exactly where I am confused on how to set the limits of k based on the second condition)

if I use $k \ge 2$ for the sum , then the sum becomes

$$2^n\sum_{k=2}^\infty\left(\frac38\right)^k$$

using GP sum , it comes out to

$$2^n\left(\frac85 - 1-\frac38 \right)= 2^n\frac{9}{40}$$

But the solution given in the book by the author is

$$y(n) = \frac{1}{20}\left(\frac34\right)^{n+5}u(n+2)+\frac{9}{40}2^nu(-n-3)$$

This solution doesn't even remotely resemble mine.

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  • $\begingroup$ Please edit your question with Latex formatting. As is, it’s pretty unreadable. You can use this question as an example. Click “edit” on the question linked, and you’ll see how to nicely format things! $\endgroup$
    – Jdip
    Commented Jan 29, 2023 at 7:35
  • $\begingroup$ I am new to Latex formatting , if someone can help me with formatting this question , would be grateful , especially with symbols like sigma , upper and lower ,limits of sums ..,exponents ..etc.- Thanks. $\endgroup$ Commented Jan 29, 2023 at 7:42
  • $\begingroup$ I managed to format the question reasonably , hope its ok now - Thanks. $\endgroup$ Commented Jan 29, 2023 at 8:32
  • $\begingroup$ corrected the same now , hope this is good. $\endgroup$ Commented Jan 29, 2023 at 9:11

1 Answer 1

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Your first steps are correct, but then you need to distinguish two cases. Note that both conditions $k\ge 2$ and $k\ge n+5$ must be satisfied for the sum to be non-zero:

$$y[n]=2^n\sum_{k=\max\{2,n+5\}}^{\infty}\left(\frac38\right)^k$$

For $n\le -3$ we have the lower index $k=2$, and your result is correct. However, for $n\ge -2$ the lower limit becomes $k=n+5$, which will give you the other term mentioned in the book's solution. For the final result, you need to combine both terms, multiplied with appropriately shifted step sequences.

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