0
$\begingroup$

A linear system S has the relationship $y[n]=\sum_k {x[k]*g[n-2k]}$, k ranging from $-\infty$ to $\infty$, between its input $x[n]$ and output $y[n]$, $g[n]=u[n]-u[n-4]$.

Determine $y[n]$ when $x[n]=\delta[n-1]$.

I know that the convolution operation can only be applied when both the operands have the same value of $n$ and that :

$$ y[n]=x[n]*h[n]=\sum_k {x[k]h[n-k]} $$

I don't understand how $x[k]$ and $g[n-2k]$ can be convolved with each other, when they do not have the same inputs. How do I perform discrete convolution on $x[k]$ and $g[n-2k]$?

$\endgroup$
  • 2
    $\begingroup$ Hint1: don't think about it as convolution but as a "recipe" to calculate the output $\endgroup$ – Hilmar Jan 29 at 15:37
  • $\begingroup$ Hint2: just manually calculate the first few output samples for x[n] = delta[n-1] by simply applying the recipe as written $\endgroup$ – Hilmar Jan 29 at 15:39
1
$\begingroup$

Hint:

For which values of $k$ is $x[k]=\delta[k-1]$ not equal to zero? How does that simplify the given sum?

$\endgroup$
0
$\begingroup$

The problem is in what you "know", viz.

the convolution operation can only be applied when both the operands have the same value of $n$ and that: $$ y[n]=x[n]*h[n]=\sum_k {x[k]h[n-k]} $$

The middle part $x[n]*h[n]$ and that both the arguments must be the same for the rightmost calculation to be valid is sheer unadulterated crap no matter how highly prestigious are the author(s) of whichever text you got the above from. Convolution is an operation that takes two input sequences ($x$ and $h$ in this instance and note the complete absence of square brackets and $n$'s) and produces an output sequence $y$ (again, notice the complete absence of square brackets and $n$'s), and is best written as $$y = x \star h$$ where I have used $\star$ to denote the convolution operation to avoid overloading $*$n since we might want to use it to denote complex conjugation or even multiplication in a computer language. Now, if you want to find the $n$-th term of the sequence $y$, it is customary to denote this term as $y[n]$ (though some might prefer $y_n$) and write $$y[n] = x\star h\big\vert_n$$ and then write down what what the convolution operation produces as the $n$-th term in the output sequence $y$, namely, $$y[n] = x\star h\big\vert_n = \sum_{k=-\infty}^\infty x[k]h[n-k]$$


So, what to do with $$y[n] = \sum_{k=-\infty}^\infty x[k]g[n-2k]$$ which looks a bit like a convolution sum but isn't quite right? Well, we have that for $n$ even, say $n=2m$, \begin{align} y[n] &= \sum_{k=-\infty}^\infty x[k]g[n-2k]\\ &= \sum_{k=-\infty}^\infty x[k]g[2m-2k] \tag{1} \end{align} and so if we define a new sequence $g_e$ as the even-numbered elements of the sequence $g$, that is, $g_e[\ell] = g[2\ell]$, and a new sequence $y_e$ as the even-numbered elements of the sequence $y$, that is, $y_e[m] = y[2m]$, then $y_e$ can be expressed as the result of the convolution of $x$ and $g_e$, that is, $$y_e = x \star g_e\\ y_e[m] = x \star g_e\big\vert_m = \sum_{k=-\infty}^\infty x[k]g_e[m-k] = \sum_{k=-\infty}^\infty x[k]g[2m-2k] =y[2m]. $$ Similarly, for $n$ odd, say $n=2m+1$, \begin{align} y[n] &= \sum_{k=-\infty}^\infty x[k]g[n-2k]\\ &= \sum_{k=-\infty}^\infty x[k]g[2m+1-2k] \tag{2} \end{align} and so if we define a new sequence $g_o$ as the odd-numbered elements of the sequence $g$, that is, $g_o[\ell] = g[2\ell+1]$, and a new sequence $y_o$ as the odd-numbered elements of the sequence $y$, that is, $y_o[m] = y[2m+1]$, then $y_o$ can be expressed as the result of the convolution of $x$ and $g_o$, that is, $$y_o = x \star g_o\\ y_o[m] = x \star g_o\big\vert_m = \sum_{k=-\infty}^\infty x[k]g_o[m-k] = \sum_{k=-\infty}^\infty x[k]g[2m-2k+1] =y[2m+1]. $$ Thus, the sequence $y$ can be viewed as the multiplexing or interleaving of the sequences $y_e$ and $y_o$ that are the result of convolving $x$ with $g_e$ and $g_o$ respectively.


Finally, with $g[n] = u[n]-u[n-4]$, we have that $g_e = g_o = [1~1]$ and so we only need to convolve $x$ with $[1~1]$, interleave the result with itself, and we are done. For $x[n] = \delta[n]$, the calculation is even simpler.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.