Question

# a) Find the Laplace transformation of the following function by using definition of Laplace transformation: f(t) = cos(pt) text{ where } t geq 0

Laplace transform
a) Find the Laplace transformation of the following function by using definition of Laplace transformation:
$$f(t) = \cos(pt) \text{ where } t \geq 0$$

2020-10-28
Step 1
Given $$f(t)=\cos(pt)$$
Using the definition of Laplace Transform we have,
$$L\left[f(t)\right]=F(s)=\int_0^\infty e^{-st}f(t)dt$$
$$\Rightarrow L\left[\cos pt\right]=\int_0^\infty e^{-st}\cos pt dt$$
Now we will use Integration by parts to solve it
$$\text{Let } I=\int_0^\infty e^{-st} \cos pt dt$$
$$I=\cos pt \int e^{-st}dt-\int(-p \sin pt)\frac{e^{-st}}{-s}dt$$
$$I=\cos pt \left[\frac{e^{-st}}{-s}\right]-\frac{p}{s} \int \sin pt e^{-st}dt$$
$$I=-\frac{1}{s} \cos pt e^{-st} - \frac{p}{s}\left[\sin pt\left(\frac{e^{-st}}{-s}\right)-\int p \cos pt\left(\frac{e^{-st}}{-s}\right)\right]$$
$$I=\left[-\frac{1}{s} \cos pt e^{-st} - \frac{p}{(s^2)}\sin pt e^{-st}- \frac{(p^2)}{(s^2)}I\right]_0^\infty$$
$$\Rightarrow I\left[1+\frac{(p^2)}{(s^2)}\right]=-\frac{1}{s}\left[\cos pt e^{-st}+\frac{p}{s} \sin pt e^{-st}\right]_0^\infty$$
$$\Rightarrow I\left[\frac{(s^2+p^2)}{s^2}\right]=-\frac{1}{s}[(0+0)-(1+0)]$$
$$\Rightarrow I\left[\frac{(s^2+p^2)}{s^2}\right]=\frac{1}{s}$$
$$\Rightarrow I=\frac{s}{(s^2+p^2)}$$
Step 2
Hence,
$$I=\int_0^\infty e^{-st} \cos pt dt=L\left[\cos pt\right] =\frac{s}{(s^2+p^2)}$$
$$\text{Rightarrow By, definition of Laplace Transform of } (f(t)=\cos pt) \text{ is } \frac{s}{(s^2+p^2)}$$