Give the Laplace transform of the unit-step function, us(t).
The Correct Answer and Explanation is :
The Laplace transform of the unit-step function us(t)u_s(t) (also denoted u(t)u(t) or H(t)H(t)) is given by: L{u(t)}=1s,fors>0\mathcal{L}\{u(t)\} = \frac{1}{s}, \quad \text{for} \quad s > 0
Explanation:
The unit-step function u(t)u(t), also known as the Heaviside function, is defined as: u(t)={0,t<01,t≥0u(t) = \begin{cases} 0, & t < 0 \\ 1, & t \geq 0 \end{cases}
This function is used to model a sudden change or a “switch” that turns on at t=0t = 0 and remains on thereafter.
The Laplace transform of a time-domain function f(t)f(t) is defined as: L{f(t)}=∫0∞f(t)e−st dt\mathcal{L}\{f(t)\} = \int_{0}^{\infty} f(t) e^{-st} \, dt
For the unit-step function, we have f(t)=u(t)f(t) = u(t), so: L{u(t)}=∫0∞u(t)e−st dt\mathcal{L}\{u(t)\} = \int_{0}^{\infty} u(t) e^{-st} \, dt
Since u(t)=1u(t) = 1 for t≥0t \geq 0, the integral simplifies to: L{u(t)}=∫0∞e−st dt\mathcal{L}\{u(t)\} = \int_{0}^{\infty} e^{-st} \, dt
To solve this, we use the standard integral: ∫0∞e−st dt=1s,fors>0\int_{0}^{\infty} e^{-st} \, dt = \frac{1}{s}, \quad \text{for} \quad s > 0
Thus, the Laplace transform of the unit-step function is: L{u(t)}=1s,fors>0\mathcal{L}\{u(t)\} = \frac{1}{s}, \quad \text{for} \quad s > 0
Key Points:
- The unit-step function represents a sudden switch or step at t=0t = 0.
- The Laplace transform of u(t)u(t) is 1s\frac{1}{s}, which is a simple rational function in the frequency domain.
- The transform is valid for s>0s > 0 to ensure the integral converges.
This result is fundamental in systems and control theory, where it is used to analyze systems that “start” at a given time t=0t = 0.