Having these two sequences:
$x_n = u_n u_{-n}$
$y_n = (n+1) u_n u_{3-n}$
Then the convolution of $ x_n * y_n $ equals...
The options are:
A. $x_n $
B. $y_n $
C. $(n+1) u_{n-4} u_{7-n}$
D. $(n-3) u_{n-4} u_{7-n}$
Note: $u_n = 1$ if $n \geq 0$ and $u_n = 0$ if $n < 0$
Since $x_n$ seems to be always 0, then I'd say option would be A but apparently the correct option is B??
Update: $x_n$ is always 0 except when n = 0.
Ok so by definition Convolution would be something like:
$\int_{-\infty}^\infty x_t * y_{n-t} dn $
$\int_{-\infty}^\infty x_n y_{n-t} = \int_{-\infty}^\infty u_n u_{-n} (n-t+1) u_{n-t} u_{3-(n-t)} dt $
But since $x_t$ is only 1 when n = 0. Then I would only evaluate
$\int_{0}^0 (n-t+1) dt = 0 $
What am I doing wrong??
Convolution of sequences is computed using a sum, not an integral. Specifically, if $x_n$ and $y_n$ are sequences, then their convolution is $$\sum_{k=-\infty}^\infty x_k y_{n-k}$$ In your case, $x_k = u_k u_{-k}$, which as you pointed out, is zero for $k \neq 0$, and $x_0 = u_0^2 = 1$. In other words, $$x_n = \begin{cases}1 & \text{ if $n = 0$ } \\ 0 & \text{ otherwise } \end{cases}$$ which is often written as $x_n = \delta_n$.
This means that only the $k=0$ term in the convolution sum is nonzero: $$\sum_{k=-\infty}^\infty x_k y_{n-k} = x_0 y_{n-0} = y_n$$ So (b) is the correct answer. Note that this choice of $x_n$ satisfies $x_n * y_n = y_n$ for any sequence $y_n$. We say that $x_n = \delta_n$ is an identity with respect to the convolution operation.