Laplace transform of ramp function

Steps to Solve

1. Define the Ramp Function The ramp function, denoted as $r(t)$, is typically defined as: $$ r(t) = \begin{cases} 0 & \text{if } t < 0 \ t & \text{if } t \geq 0 \end{cases} $$

2. Set up the Laplace Transform Integral The Laplace transform $\mathcal{L}{r(t)}$ is given by the integral: $$ \mathcal{L}{r(t)} = \int_0^\infty e^{-st} r(t) , dt $$

3. Substitute the Ramp Function into the Integral Since $r(t) = t$ for $t \geq 0$, we substitute: $$ \mathcal{L}{r(t)} = \int_0^\infty e^{-st} t , dt $$

4. Evaluate the Integral To compute the integral, we can use integration by parts, where we let:

• ( u = t ) so ( du = dt )
• ( dv = e^{-st} dt ) which gives us ( v = -\frac{1}{s} e^{-st} )

Using integration by parts: $$ \int u , dv = uv - \int v , du $$ This becomes: $$ \int_0^\infty t e^{-st} dt = \left[-\frac{1}{s} t e^{-st} \right]_0^\infty + \frac{1}{s} \int_0^\infty e^{-st} dt $$

5. Evaluate the Bounds of the Integral Evaluating the first term:

• As $t \to \infty$, $t e^{-st} \to 0$ (dominated by the exponential decay)
• At $t = 0$, the term is $0$.

So, the first term is: $$ 0 - 0 = 0 $$

6. Evaluate the Second Integral Now we need to evaluate $\frac{1}{s} \int_0^\infty e^{-st} dt$. The integral $\int_0^\infty e^{-st} dt = \frac{1}{s}$ (for $s > 0$), hence we have: $$ \int_0^\infty t e^{-st} dt = 0 + \frac{1}{s^2} $$

Thus: $$ \mathcal{L}{r(t)} = \frac{1}{s^2} $$