Sean las señales $x[n] = \alpha^{n} (u[n+a] - u[n-b])$ y $h[n] = \frac{28}{7}(u[-n] - u[-n+c])$ con $a=27$, $b=7$, y $c=25$. Si $y[n] = x[n] * h[n]$, ¿en qué casos $y[n] = 0$?

Steps to Solve

1. Simplify $x[n]$ and $h[n]$

First, let's simplify the expressions for $x[n]$ and $h[n]$ using the given values for $a$, $b$, and $c$. We have: $$ x[n] = \alpha^{n} (u[n+27] - u[n-7]) $$ $$ h[n] = \frac{28}{7}(u[-n] - u[-n+25]) = 4(u[-n] - u[-n+25]) $$

2. Understand the signals' ranges For the unit step functions to be non-zero, their arguments must be greater than or equal to zero.

For $x[n]$: $u[n+27] = 1$ for $n \geq -27$ and $u[n-7] = 1$ for $n \geq 7$. Therefore, $x[n]$ is non-zero for $-27 \leq n \leq 6$.

For $h[n]$: $u[-n] = 1$ for $-n \geq 0 \implies n \leq 0$ and $u[-n+25] = 1$ for $-n+25 \geq 0 \implies n \leq 25$. Therefore, $h[n]$ is non-zero for $-25 \leq n \leq -1$.

3. Compute the convolution sum The convolution $y[n] = x[n] * h[n]$ is defined as: $$ y[n] = \sum_{k=-\infty}^{\infty} x[k]h[n-k] $$

4. Determine the non-zero range for $y[n]$ For $y[n]$ to be non-zero, the ranges of $k$ for which $x[k]$ and $h[n-k]$ are non-zero must overlap. $x[k]$ is non-zero for $-27 \leq k \leq 6$ and $h[n-k]$ is non-zero for $-25 \leq n-k \leq -1$, or equivalently $n+1 \leq k \leq n+25$. Thus we require that $-27 \leq k \leq 6$ and $n+1 \leq k \leq n+25$, so $y[n]$ is only potentially non-zero when the two ranges overlap.

5. Find the range of $n$ where $y[n] = 0$

We want to find when the ranges $-27 \leq k \leq 6$ and $n+1 \leq k \leq n+25$ do not overlap.

• The left end of range 2 is greater than the right end of range 1: $n+1 > 6 \implies n > 5$.
• The right end of range 2 is less than the left end of range 1: $n+25 < -27 \implies n < -52$.

Therefore, $y[n] = 0$ for $n > 5$ and for $n < -52$.