# Question related to Discrete signal convolution

I have two finite signals $x[n]$ and $h[n]$, I want to convolve these signals using the method defined in Example 2.2 of Signals and Systems by Oppenheim. I am getting this answer

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

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? $$y[-5]=4, y[-4]=11, y[-3]=12, y[-2]=8, y[-1]=0, y=-2, y=-2, y=-1$$ its 2 unit shifted right side

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 $$x$$). For any other "$$n$$" value, there is no overlap between $$x[k]$$ and shifted $$h[n-k]$$ and output $$y[n]$$ becomes zero.

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