The problem is asking for the convolution of two signals:

1. $x_1(t) = e^{-3t} u(t)$
2. $x_2(t) = e^{-2t} u(t - 2)$

where $u(t)$ represents the unit step function.

Solution Approach

To find the convolution $y(t) = x_1(t) * x_2(t)$, we will use the convolution integral: $y(t) = \int_{-\infty}^{\infty} x_1(\tau) x_2(t - \tau) \, d\tau$

Since both signals are multiplied by the unit step function, we only need to evaluate the integral where both signals are non-zero, i.e., for $t \geq 0$ and $t \geq 2$ due to the shifted step function in $x_2(t)$.

Step-by-Step Solution

1. Set up the integral: $y(t) = \int_{-\infty}^{\infty} e^{-3\tau} u(\tau) \cdot e^{-2(t - \tau)} u(t - \tau - 2) \, d\tau$

2. Simplify the expression: Since $u(\tau)$ and $u(t - \tau - 2)$ restrict the limits of integration, we consider only the range where both are non-zero:

   • $\tau \geq 0$ (from $u(\tau)$)
   • $t - \tau \geq 2 \Rightarrow \tau \leq t - 2$ (from $u(t - \tau - 2)$)

   So, the limits of integration become $0 \leq \tau \leq t - 2$.

3. Evaluate the integral: Substitute $x_1(\tau) = e^{-3\tau}$ and $x_2(t - \tau) = e^{-2(t - \tau)}$: $y(t) = \int_0^{t-2} e^{-3\tau} \cdot e^{-2(t - \tau)} \, d\tau$ Simplify the exponent: $y(t) = \int_0^{t-2} e^{-3\tau} \cdot e^{-2t + 2\tau} \, d\tau = e^{-2t} \int_0^{t-2} e^{-\tau} \, d\tau$

4. Compute the integral: $y(t) = e^{-2t} \left[ -e^{-\tau} \right]_0^{t-2}$ Substitute the limits: $y(t) = e^{-2t} \left( -e^{-(t-2)} + 1 \right)$ Simplify: $y(t) = e^{-2t} \left( 1 - e^{-(t-2)} \right)$

5. Final answer: $y(t) = e^{-2t} - e^{-3t + 2}$

This is the convolution result $y(t) = x_1(t) * x_2(t)$.

Would you like further details or explanations on any step?

Related Questions

1. How do we apply the convolution theorem to simplify convolutions using Laplace transforms?
2. What is the role of the unit step function in time-domain analysis?
3. Can we perform convolutions with signals involving multiple time shifts?
4. How would the convolution result change if both signals were not multiplied by unit step functions?
5. What happens if we convolve a signal with a delta function?

Tip

When convolving signals with time shifts, always account for the shift in the limits of integration. This helps avoid unnecessary computation and provides clearer insights into the range where the signals interact.