To determine the output signal $y(t)$ for a system with input $x(t)$ and impulse response $h(t)$, we can use the convolution integral:

$y(t) = x(t) * h(t) = \int_{-\infty}^{\infty} x(\tau) h(t - \tau) \, d\tau$

Step 1: Analyzing $x(t)$ and $h(t)$

1. Input Signal $x(t)$:

   Given: $x(t) = u(t) - 2u(t - 1) + u(t - 2)$

   • $u(t)$: a unit step function that is 1 for $t \geq 0$ and 0 otherwise.
   • $u(t - 1)$: shifts the step by 1 unit to the right.
   • $u(t - 2)$: shifts the step by 2 units to the right.

   So, $x(t)$ will have values:

   • 1 for $t \geq 0$,
   • drops by 2 units at $t = 1$,
   • adds back 1 unit at $t = 2$.

2. Impulse Response $h(t)$:

   Given: $h(t) = u(t + 1) - u(t - 1)$

   This represents a pulse from $t = -1$ to $t = 1$.

Step 2: Convolution $y(t) = x(t) * h(t)$

The convolution process will involve evaluating the integral: $y(t) = \int_{-\infty}^{\infty} x(\tau) h(t - \tau) \, d\tau$

Given $h(t) = u(t + 1) - u(t - 1)$, it acts as a window for $x(t)$ from $t - 1 \leq \tau \leq t + 1$.

Step 3: Breaking Down the Convolution by Intervals

To simplify, we consider the piecewise nature of $x(t)$:

• For $t < 0$: $x(t) = 0$.
• For $0 \leq t < 1$: $x(t) = 1$.
• For $1 \leq t < 2$: $x(t) = -1$.
• For $t \geq 2$: $x(t) = 0$.

Then, by integrating each interval over $x(\tau)$ and $h(t - \tau)$ based on where $t$ lies, we find $y(t)$.

Final Convolution Result

After performing the convolution integrals for each segment, the resulting $y(t)$ can be found as a piecewise function depending on $t$.

Would you like me to walk through the detailed integration steps for each interval or provide a general graph of $y(t)$?

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1. What is the purpose of convolution in signal processing?
2. How does the unit step function $u(t)$ simplify piecewise functions?
3. What is the significance of an impulse response in a system?
4. How do delays (like $u(t - 1)$) affect signals in convolution?
5. How does convolution differ from other types of signal operations?

Tip: Visualizing the signals $x(t)$ and $h(t)$ can make convolution steps clearer, as you can see the overlap regions easily.