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I have two finite signals $x[n]$ and $h[n]$, I want to convolve these signals enter image description here

using the method defined in Example 2.2 of Signals and Systems by Oppenheim. I am getting this answer

$$ y[-3]=4, y[-2]=11, y[-1]=12, y[0]=8, y[1]=0, y[2]=-2, y[3]=-2, y[4]=-1 $$

and when I tried to solve this question using the method defined in example 2.1 of Signals and Systems by Oppenheim, I am getting different answer. Shouldn't it be the same? $$ y[-5]=4, y[-4]=11, y[-3]=12, y[-2]=8, y[-1]=0, y[0]=-2, y[1]=-2, y[2]=-1 $$ its 2 unit shifted right side

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Your solution based on Example 2.2 of A.Oppenheim's Signals & Systems is wrong.

In that example, the output $$y[n] = \sum_{k=-\infty}^{\infty} {x[k]h[n-k]}$$

is computed by the graphical method (a.k.a. flip-and-drag method).

Based on the nonzero range of input $x[n]$, this becomes: $$y[n] = \sum_{k=-2}^{2} { x[k]h[n-k] }$$

Also based on the nonzero range of $h[n]$, it can be seen that $y[n]$ has nonzero samples in the range from $n=-5$ to $n=2$.

In the flip & drag method, it's best to draw the shifted functions $h[n-k]$, to guide the computation process.

The key is to recognise the nonzero range for the flipped & dragged function $h[n-k]$ ; time reversed $h[-k]$ shifted by "$n$", drawn on an axis of $k$ and thus considered as a function of $k$.

You can see that the overlap between the functions $x[k]$ and $h[n-k]$ begins at the shift of $n = - 5$, from the left side of $x[k]$ (the sample $x[-2]$) and continues until the shift for $n = 2$ at the right side of $x[n]$ (sample $x[2]$). For any other "$n$" value, there is no overlap between $x[k]$ and shifted $h[n-k]$ and output $y[n]$ becomes zero.

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  • $\begingroup$ Can you please tell me the solution using example 2.2 $\endgroup$ – Shinning Eyes Dec 26 '15 at 6:13

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