Transforms of Periodic Functions - Differential Equations

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Question

Find the Laplace transform of the periodic function.

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

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Answer

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.

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

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

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Question

Find the Laplace transform of the periodic function.

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

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Answer

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.

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

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

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Question

Find the Laplace transform of the periodic function.

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

Tap to reveal answer

Answer

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.

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

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

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Question

Find the Laplace transform of the periodic function.

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

Tap to reveal answer

Answer

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.

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

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

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