Question

Determine the Laplace transform of the function:
f(t)=
{cos(t-(2π/3)) t>2π/3
{ 0 t<2π/3

Answer 1

Answer to Question #84502 – Math – Differential Equations

Question

Determine the Laplace transform of the function:

$$f(t) = \begin{cases} \cos(t - 2\pi/3), & t > 2\pi/3 \\ 0, & t < 2\pi/3 \end{cases}$$

Solution

$$Y(s) = \int_{0^{-}}^{\infty} \cos t \, e^{-st} \, dt$$ $$\cos t = \frac{e^{it} + e^{-it}}{2}$$ $$Y(s) = \int_{0^{-}}^{\infty} \frac{e^{it} + e^{-it}}{2} e^{-st} \, dt = \frac{1}{2} \int_{0^{-}}^{\infty} e^{it} e^{-st} \, dt + \frac{1}{2} \int_{0^{-}}^{\infty} e^{-it} e^{-st} \, dt$$ $$Y(s) = \frac{1}{2(s - i)} + \frac{1}{2(s + i)} = \frac{s + i + s - i}{2(s^2 + 1)} = \frac{s}{s^2 + 1}$$ $$\cos t \leftrightarrow \frac{s}{s^2 + 1}$$ $$\cos \left(t - \frac{2\pi}{3}\right) \leftrightarrow e^{-\frac{2\pi}{3} s} Y(s) = e^{-\frac{2\pi}{3} s} \frac{s}{s^2 + 1}$$

Answer: $\cos \left(t - \frac{2\pi}{3}\right) \leftrightarrow \frac{se^{-\frac{2\pi}{3} s}}{s^2 + 1}$.

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