The Laplace Transform

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Differential Equations › The Laplace Transform

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Find the Laplace transform of the periodic function.

$S(t)=\left{\begin{matrix} 1,& 0\leq2\ 0, & 2\leq4 \end{matrix}\right.$

$\frac{1}{s(1+e^{-2s})(1+e^{-s})}$

$\frac{1}{s(1-e^{-2s})(1+e^{-s})}$

$\frac{1}{s(1+e^{-2s})(1-e^{-s})}$

$\frac{1}{s(1+e^{-2s})(1+e^{s})}$

$\frac{1}{s(1+e^{2s})(1+e^{-s})}$

Explanation

This particular piecewise function is called a square wave. The period of this function is the length at which it takes the function to return to its starting point.

For this particular function

$S(t)=\left{\begin{matrix} 1,& 0\leq2\ 0, & 2\leq4 \end{matrix}\right.$

it has a period of

and furthermore,

Using the Transform of a Periodic Function Theorem which states,

$\mathcal{L}\begin{Bmatrix} f(t) \end{Bmatrix}=\frac{1}{1-e^{-sT}}\int_0^T e^{-sT}f(t)dt$

the problem can be solved as follows.

$\mathcal{L}\begin{Bmatrix} S(t) \end{Bmatrix}=\frac{1}{1-e^{-4s}}\int_0^4 e^{-st}S(t)dt$

$\=\frac{1}{1-e^{-4s}}\left[ \int_0^2 e^{-st}\cdot 1 dt+\int_2^4 e^{-st}\cdot 0dt \right] \\=\frac{1}{1-e^{-4s}}\frac{1-e^{-s}}{s}\\\leftarrow 1-e^{-4s}=(1+e^{-2s})(1-e^{-2s})\\\leftarrow (1-e^{-2s})=(1+e^{-s})(1-e^{-s}) \\=\frac{1-e^{-s}}{s(1+e^{-2s})(1-e^{-s})(1+e^{-s})} \\=\frac{1}{s(1+e^{-2s})(1+e^{-s})}$

Find the Laplace transform of the periodic function.

$S(t)=\left{\begin{matrix} 1,& 0\leq2\ 0, & 2\leq4 \end{matrix}\right.$

$\frac{1}{s(1+e^{-2s})(1+e^{-s})}$

$\frac{1}{s(1-e^{-2s})(1+e^{-s})}$

$\frac{1}{s(1+e^{-2s})(1-e^{-s})}$

$\frac{1}{s(1+e^{-2s})(1+e^{s})}$

$\frac{1}{s(1+e^{2s})(1+e^{-s})}$

Explanation

This particular piecewise function is called a square wave. The period of this function is the length at which it takes the function to return to its starting point.

For this particular function

$S(t)=\left{\begin{matrix} 1,& 0\leq2\ 0, & 2\leq4 \end{matrix}\right.$

it has a period of

and furthermore,

Using the Transform of a Periodic Function Theorem which states,

$\mathcal{L}\begin{Bmatrix} f(t) \end{Bmatrix}=\frac{1}{1-e^{-sT}}\int_0^T e^{-sT}f(t)dt$

the problem can be solved as follows.

$\mathcal{L}\begin{Bmatrix} S(t) \end{Bmatrix}=\frac{1}{1-e^{-4s}}\int_0^4 e^{-st}S(t)dt$

Find the Laplace Transform for the following function.

$\mathcal{L}\begin{Bmatrix} e^{-5t} \end{Bmatrix}$

$\mathcal{L}\begin{Bmatrix} e^{-5t} \end{Bmatrix}=\frac{1}{s+5}, s>-5$

$\mathcal{L}\begin{Bmatrix} e^{-5t} \end{Bmatrix}=\frac{1}{s-5}, s>-5$

$\mathcal{L}\begin{Bmatrix} e^{-5t} \end{Bmatrix}=\frac{1}{s+3}, s>-5$

$\mathcal{L}\begin{Bmatrix} e^{-5t} \end{Bmatrix}=\frac{1}{s+5}, s>5$

$\mathcal{L}\begin{Bmatrix} e^{-5t} \end{Bmatrix}=\frac{1}{s+5}, s<-5$

Explanation

To find the Laplace Transform for the following function

$\mathcal{L}\begin{Bmatrix} e^{-5t}\end{Bmatrix}$

use the transforms for basic functions which state,

$\mathcal{L}\begin{Bmatrix} 1 \end{Bmatrix}=\frac{1}{s}$

$\mathcal{L}\begin{Bmatrix} e^{at} \end{Bmatrix}=\frac{1}{s-a}$

$\mathcal{L}\begin{Bmatrix} e^{-5t} \end{Bmatrix}=\int_0^\infty e^{-st}e^{-5t}dt=\int_0^\infty e^{-(s+5)t} dt$

$\=\frac{-e^{-(s+5)t}}{s+5}\bigg|_0^\infty \\=\frac{1}{s+5}, s>-5$

Find the Laplace Transform for the following function.

$\mathcal{L}\begin{Bmatrix} e^{-5t} \end{Bmatrix}$

$\mathcal{L}\begin{Bmatrix} e^{-5t} \end{Bmatrix}=\frac{1}{s+5}, s>-5$

$\mathcal{L}\begin{Bmatrix} e^{-5t} \end{Bmatrix}=\frac{1}{s-5}, s>-5$

$\mathcal{L}\begin{Bmatrix} e^{-5t} \end{Bmatrix}=\frac{1}{s+3}, s>-5$

$\mathcal{L}\begin{Bmatrix} e^{-5t} \end{Bmatrix}=\frac{1}{s+5}, s>5$

$\mathcal{L}\begin{Bmatrix} e^{-5t} \end{Bmatrix}=\frac{1}{s+5}, s<-5$

Explanation

To find the Laplace Transform for the following function

$\mathcal{L}\begin{Bmatrix} e^{-5t}\end{Bmatrix}$

use the transforms for basic functions which state,

$\mathcal{L}\begin{Bmatrix} 1 \end{Bmatrix}=\frac{1}{s}$

$\mathcal{L}\begin{Bmatrix} e^{at} \end{Bmatrix}=\frac{1}{s-a}$

$\mathcal{L}\begin{Bmatrix} e^{-5t} \end{Bmatrix}=\int_0^\infty e^{-st}e^{-5t}dt=\int_0^\infty e^{-(s+5)t} dt$

$\=\frac{-e^{-(s+5)t}}{s+5}\bigg|_0^\infty \\=\frac{1}{s+5}, s>-5$

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