Convolution Of Ramp And Exponential Signals

Problem Statement

Find the convolution $y(t) = x_1(t) * x_2(t)$ for the following signals: $$ x_1(t) = r(t) $$ $$ x_2(t) = e^{-2t} u(t) $$ Where $r(t)$ is the ramp function and $u(t)$ is the unit step function.

1. Definitions and Convolution Formula

First, let's define the functions involved:

• The unit step function, $u(t)$, is defined as:
  $u(t) = \begin{cases} 1, & t \ge 0 \\ 0, & t < 0 \end{cases}$
• The ramp function, $r(t)$, is defined as:
  $r(t) = t \cdot u(t) = \begin{cases} t, & t \ge 0 \\ 0, & t < 0 \end{cases}$

The convolution integral is given by the formula: $$ y(t) = x_1(t) * x_2(t) = \int_{-\infty}^{\infty} x_1(\tau) x_2(t-\tau) d\tau $$

2. Setting up the Integral

We need to express the signals in terms of the integration variable $\tau$:

• $x_1(\tau) = r(\tau) = \tau u(\tau)$
• $x_2(t-\tau) = e^{-2(t-\tau)} u(t-\tau)$

Substituting these into the convolution integral: $$ y(t) = \int_{-\infty}^{\infty} [\tau u(\tau)] [e^{-2(t-\tau)} u(t-\tau)] d\tau $$

3. Determining the Limits of Integration

The integrand is non-zero only when both unit step functions are non-zero.

• $u(\tau) \neq 0 \implies \tau \ge 0$
• $u(t-\tau) \neq 0 \implies t-\tau \ge 0 \implies \tau \le t$

For the integral to be non-zero, both conditions must be met simultaneously: $0 \le \tau \le t$. This implies that there is no overlapping region if $t < 0$. Therefore, we have two cases for the value of $t$.
Case 1: $t < 0$ If $t < 0$, the conditions $\tau \ge 0$ and $\tau \le t$ cannot both be true. There is no overlap, so the integral evaluates to zero. $$ y(t) = 0 \quad \text{for } t < 0 $$
Case 2: $t \ge 0$ If $t \ge 0$, the limits of integration become $0$ to $t$. The unit step functions are equal to $1$ within this range, so they can be removed from the integrand. $$ y(t) = \int_{0}^{t} \tau e^{-2(t-\tau)} d\tau $$

4. Solving the Integral

Now, we solve the integral for $t \ge 0$.
First, simplify the integrand by expanding the exponential term: $$ y(t) = \int_{0}^{t} \tau e^{-2t} e^{2\tau} d\tau $$ Since $e^{-2t}$ is a constant with respect to $\tau$, we can pull it out of the integral: $$ y(t) = e^{-2t} \int_{0}^{t} \tau e^{2\tau} d\tau $$ This integral requires Integration by Parts. The formula is $\int u \, dv = uv - \int v \, du$. Let:

• $u = \tau \implies du = d\tau$
• $dv = e^{2\tau} d\tau \implies v = \int e^{2\tau} d\tau = \frac{1}{2} e^{2\tau}$

Applying the formula: $$ \int \tau e^{2\tau} d\tau = \tau \left(\frac{1}{2}e^{2\tau}\right) - \int \frac{1}{2}e^{2\tau} d\tau $$ $$ = \frac{1}{2}\tau e^{2\tau} - \frac{1}{2} \int e^{2\tau} d\tau $$ $$ = \frac{1}{2}\tau e^{2\tau} - \frac{1}{2} \left(\frac{1}{2}e^{2\tau}\right) = \frac{1}{2}\tau e^{2\tau} - \frac{1}{4}e^{2\tau} $$ Now, we evaluate the definite integral from $0$ to $t$: $$ \int_{0}^{t} \tau e^{2\tau} d\tau = \left[ \frac{1}{2}\tau e^{2\tau} - \frac{1}{4}e^{2\tau} \right]_{0}^{t} $$ $$ = \left( \frac{1}{2}t e^{2t} - \frac{1}{4}e^{2t} \right) - \left( \frac{1}{2}(0)e^{0} - \frac{1}{4}e^{0} \right) $$ $$ = \left( \frac{1}{2}t e^{2t} - \frac{1}{4}e^{2t} \right) - \left( 0 - \frac{1}{4} \right) $$ $$ = \frac{1}{2}t e^{2t} - \frac{1}{4}e^{2t} + \frac{1}{4} $$ Substitute this result back into the expression for $y(t)$: $$ y(t) = e^{-2t} \left( \frac{1}{2}t e^{2t} - \frac{1}{4}e^{2t} + \frac{1}{4} \right) $$ Distribute the $e^{-2t}$ term: $$ y(t) = \frac{1}{2}t e^{-2t}e^{2t} - \frac{1}{4}e^{-2t}e^{2t} + \frac{1}{4}e^{-2t} $$ $$ y(t) = \frac{1}{2}t - \frac{1}{4} + \frac{1}{4}e^{-2t} $$

5. Final Result

Combining the results from both cases:

• For $t < 0$, $y(t) = 0$.
• For $t \ge 0$, $y(t) = \frac{1}{2}t - \frac{1}{4} + \frac{1}{4}e^{-2t}$.

We can write this as a single expression using the unit step function $u(t)$: $$ y(t) = \left( \frac{1}{2}t - \frac{1}{4} + \frac{1}{4}e^{-2t} \right) u(t) $$ $$ \boxed {y(t) = \left( \frac{1}{2}t - \frac{1}{4} + \frac{1}{4}e^{-2t} \right) u(t)} $$