Is graphical method the best way to solve convolution questions whether they be discrete or continuous?

I was given a question:

$$x[n]=1$$ $$0\leq n \leq 4 $$

$$h[n]=\alpha^n$$ $$0\leq n \leq 6$$

for all other values $n$ is $0$.

I solved this question using graphical method and I was successful but when I solved the following question using graphical method, I was unable to:

$$x[n]=2^n u[-n]$$ $$h[n]=u[n]$$

Please tell me how can I solve this question?

  • $\begingroup$ Could you include a concrete example of what you're talking about? "Convolution questions" is a very broad topic, and "graphical method" too. Please edit your question (there's an Edit button) and include an example. $\endgroup$ – g6kxjv1ozn Oct 5 '18 at 13:22
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    $\begingroup$ The quick answer is no for most practical real world applications. Graphical methods of convolution help in show in very simple cases the underlying process but the equations exist to be solved. $\endgroup$ – Dan Boschen Oct 5 '18 at 13:27
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    $\begingroup$ Would it be possible for you, Ahmad, to confirm if @DanBoschen 's answer was satisfactory? If it was, would it be possible for you, Dan, to write it up as an answer and then Ahmad to accept it, so that we have this question closed? The alternative, Ahmad, is for you to delete the question (?) if you think that there is no reason for it anymore. $\endgroup$ – A_A Oct 7 '18 at 7:39
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    $\begingroup$ Ahmad- Are you familiar with the equation for convolution and how to do the math involved? $\endgroup$ – Dan Boschen Oct 7 '18 at 11:32

Graphical evaluation of convolution (flip n drag) is a very useful, helpful and indipensible method which aids in a very quick visual anticipation of the output, in terms of the input sequences. Indeed even if you don't use specifically the graphical method, you would still benefit from drawing a plot of input sequences and the rough sketch of the expected output in any case. Yet no method is the best for all types of problems.

For this purpose let me solve this problem without graphical method. Given the input sequences $$x[n] = 2^n u[-n] ~~~, ~~~\text{ for } -\infty <n \leq 0 $$ and $$h[n] = u[n] ~~~, ~~~\text{ for } 0 \leq n < \infty$$ then the convolution sum is : $$y[n] = x[n] \star h[n] = \sum_{k=-\infty}^{\infty} x[k] h[n-k]$$

First, observe (graphically) that output will extend from $n=-\infty$ to $n=\infty$. This is because a right sided sequence is convolved with a left sided sequence of both semi infinite lengths.

Then, for each $n$, look for the valid range of summing index $k$ that yields nonzero signal values according to their arguments. It can be seen from the definitions of nonzero ranges of $x[n]$ and $h[n]$ that the summing index $k$ should satisfy the following two relations due to $x[k]$ and $h[n-k]$ as:

$$ \{-\infty \leq k \leq 0 \} \cap \{0 \leq n-k < \infty \} $$

or equivalently $$ \{-\infty < k \leq 0 \} \cap \{-\infty < k \leq n \} $$

the intersection of which is: $$ \max\{ -\infty, -\infty \} < k \leq \min \{0 , n \} $$ $$ -\infty < k \leq \min \{0 , n \} ~~~,~~~\text{ for all } ~~~ n$$

then the convolution sum becomes: $$y[n] = \sum_{k=-\infty}^{\min\{0,n\}} 2^k$$

using the summation formula yields: $$y[n] = \frac{ 2^{-\infty} - 2^{\min\{0,n\}+1} }{1-2} = 2^{\min \{0,n\} + 1}~~~,~~~\text{ for all } ~~~ n $$

then according to whether $n<0$ or $n \geq 0$ the output $y[n]$ becomes: $$ y[n] = \begin{cases} 2^{n+1} ~~~ &, n < 0 \\ 2 ~~~&, n \geq 0\\ \end{cases}$$

Note that the graphical method should provide this answer in a less number of steps (due to visual aids that manipulate the indexing without hassle)...


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