Determine the Laplace transform of the real exponential signal, e^−at u(t) from the definition. Substitute a = 0 in the transform obtained and get the Laplace transform of the unit-step signal, u(t).

Question

Determine the Laplace transform of the real exponential signal,[latex]e^{−at} u(t)[/latex] from the definition. Substitute a = 0 in the transform obtained and get the Laplace transform of the unit-step signal, u(t).

Accepted Answer

[latex]X(s)=\int\limits_{0-}^{\infty }{e^{-at}u(t)e^{-st}dt } =\int\limits_{0-}^{\infty }{e^{-at}e^{-st}dt }[/latex]

[latex]=\int\limits_{0-}^{\infty }{e^{-(s+a)t}dt }=\left.-\frac{e^{-(s+a)t}}{s+a}\right|^{\infty}_{0^{-} } =\left.\frac{1}{s+a}-\frac{e^{-(s+a)t}}{s+a} \right|_{t=\infty }[/latex]
While the Laplace transform is applicable to a larger class of signals, for some signals the transform does not exist. Therefore, the condition, called the region of convergence, has to be mentioned for each transform. The Laplace transform pair for the exponential signal becomes
[latex]e^{-at}u(t)\leftrightarrow \frac{1}{s+a},Re(s)>-a. [/latex]
For complex-valued a, the convergence condition is Re(s) > Re(−a). Substituting a = 0, we get the transform pair for the unit-step signal u(t) as
[latex]u(t)\leftrightarrow \frac{1}{s},Re(s)>0.[/latex]
With a = ∓jω0,
[latex]e^{jw_{0}t }u(t)\leftrightarrow \frac{1}{s-jw_{0}}[/latex] and [latex]e^{-jw_{0}t }u(t)\leftrightarrow \frac{1}{s+jw_{0}},Re(s)>0.[/latex]
Then,
[latex]sin(ω0t)u(t) = −0.5j (e^{jw_{0}t }u(t)-e^{-jw_{0}t }u(t))\leftrightarrow\frac{w_{0}}{s^{2}+w^{2}_{0}} ,Re(s)>0. [/latex]

Question 8.7: Determine the Laplace transform of the real exponential sign...

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