Q36E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

The unit triangular pulse $\mathit{\Lambda }\mathbf{\left(}\mathit{t}\mathbf{\right)}$is defined by 

$\mathit{\Lambda }\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\{\begin{array}{l}0,t<0\\ 2t,0<t<1/2\\ 2-2t,1/2<t<1\\ 0,t>1\end{array}$
1.sketch the graph of  why is it named? why is its symbol appropriate. 

2. show that  $\mathit{\Lambda }\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}{\mathbf{\int }}_{\mathbf{-}\mathbf{\infty }}^{\mathbf{t}}\mathbf{2}\left\{{\Pi }_{0,1/2}\left(\tau \right)-{\Pi }_{1/2,1}\left(\tau \right)\right\}\mathbf{d}\mathit{t}$

3 Find the Laplace transform of $\mathit{\Lambda }\mathbf{\left(}\mathit{t}\mathbf{\right)}$

 

 

Step-by-Step Solution

Answer

(a). The graph looks like a triangle and the symbol is the shape of the graph.

(b). Using the definition of the rectangular window function, we obtain the required result.

(c).$\mathcal{L}\left\{\Lambda \left(t\mathbf{\right)}\right\}\left(s\right)=\frac{2+2e^{-\frac{s}{2}}(2-s)+2e^{-s}(1-s)}{s^2}$

Step 1: Define rectangular window function

The rectangular window function can be denoted as $\mathbf{\prod }_{\mathbf{a}\mathbf{,}\mathbf{b}}\mathbf{\left(}\mathit{t}\mathbf{\right)}$and it is defined by,

$$\begin{gathered} \mathbf{\prod }_{\mathbf{a}\mathbf{,}\mathbf{b}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathit{u}\mathbf{(}\mathit{t}\mathbf{-}\mathit{a}\mathbf{)}\mathbf{-}\mathit{u}\mathbf{(}\mathit{t}\mathbf{-}\mathit{b}\mathbf{)} \\ \mathbf{\{}\begin{array}{l}\mathbf{0}\mathbf{,}\mathbf{t}\mathbf{<}\mathbf{a}\mathbf{,}\\ \mathbf{1}\mathbf{,}\mathbf{a}\mathbf{<}\mathbf{t}\mathbf{<}\mathbf{b}\mathbf{,}\\ \mathbf{0}\mathbf{,}\mathbf{b}\mathbf{<}\mathbf{t}\end{array} \end{gathered}$$

Step 2: (a) Sketch the graph of &#923; ( t ) and explain the name unit triangular pulse

Hence, the function is named unit triangular pulse because the graph looks like triangle, and the symbol is the shape of the graph. 

Since, the graph can be plotted as,

Step 3: (b) Use the rectangular window function to prove the given function .

Consider the given statement,

$\Lambda \left(t\right)={\int }_{-\infty }^{t}2\left(\prod _{0,{\displaystyle \frac{1}{2}}}\left(\tau \right)-\prod _{1,{\displaystyle \frac{1}{2}}}\right)dt$

Thus, the rectangular window function can be defined by,

$$\begin{gathered} \prod _{a,{\displaystyle b}}\left(t\right)=u(\tau -a)-u(\tau -b) \\ =\left\{\begin{array}{l}0,\tau <a\\ 1,a<\tau <b\\ 0,b<\tau \end{array}\right. \end{gathered}$$

Hence, from the given function we have.

$$\begin{gathered} \prod _{0,{\displaystyle \frac{1}{2}}}\left(t\right)=u\left(\tau \right)-u\left(\tau -\frac{1}{2}\right) \\ =\left\{\begin{array}{l}0,\tau <0\\ 1,0<\tau <\frac{1}{2}\\ 0,\frac{1}{2}<\tau \end{array}\right. \end{gathered}$$

And $a = 1/2$ $a = 1,b = 1/2$

$$\begin{gathered} -\prod _{1,{\displaystyle \frac{1}{2}}}\left(t\right)=u\left(\tau -\frac{1}{2}\right)-u(\tau -1) \\ =\left\{\begin{array}{l}0,\tau <\frac{1}{2}\\ 1,\frac{1}{2}<\tau <1\\ 0,1<\tau \end{array}\right. \end{gathered}$$

Therefore,

$2\left\{\prod _{0,{\displaystyle \frac{1}{2}}}\left(t\right)-\prod _{1,{\displaystyle \frac{1}{2}}}\left(t\right)\right\}=\left\{\begin{array}{l}0,\tau <0\\ 2,0<\tau <\frac{1}{2}\\ 2,\frac{1}{2}<\tau <1\\ 0,1<\tau \end{array}\right.$

Thus, we have

$$\begin{gathered} {\int }_{-\infty }^{t}2\left\{\prod _{0,{\displaystyle \frac{1}{2}}}\left(t\right)-\prod _{1,{\displaystyle \frac{1}{2}}}\left(t\right)\right\}dt=\left\{\begin{array}{l}{\int }_0^{t}0,\tau <0\\ {\int }_0^t\left(-2\right),0<\tau <\frac{1}{2}\\ {\int }_0^{t}0d\tau ,1<\tau \end{array}\right. \\ =\left\{\begin{array}{l}0,t<0\\ (2\tau {)}_0^t,0<t<\frac{1}{2}\\ (-2\tau {)}_t^1,\frac{1}{2}<t<1\\ 01<t\end{array}\right. \\ =\left\{\begin{array}{l}0,t<0\\ 2t,0<t<\frac{1}{2}\\ 2-2t,\frac{1}{2}<t<1\\ 0,1<t\end{array}\right. \end{gathered}$$

Hence,

 $\Lambda \left(t\right)={\int }_{-\infty }^{t}2\left(\prod _{0,{\displaystyle \frac{1}{2}}}\left(\tau \right)-\prod _{1,{\displaystyle \frac{1}{2}}}\right)dt$

Step 4: (c) Find the Laplace transform of &#923; ( t )

Consider the given function,

$\Lambda \left(t\right)=\left\{\begin{array}{l}0,t<0\\ 2t,0<t<1/2\\ 2-2t,1/2<t<1\\ 0,t>1\end{array}\right.$

Since,

data-custom-editor="chemistry" $$\begin{gathered} 2t\left(u\left(t\right)-u\left(t-\frac{1}{2}\right)\right)\mathit{=}\{\begin{array}{l}\mathrm{0}\mathrm{,}t\mathrm{<}\mathrm{0}\\ \mathrm{2}t\mathrm{,}\mathrm{0}\mathrm{<}t\mathrm{<}\frac{\mathrm{1}}{\mathrm{2}}\\ \mathrm{0}\mathrm{,}\frac{\mathrm{1}}{\mathrm{2}}\mathrm{<}t\end{array} \\ (2-2t)\left(u\left(t\mathrm{-}\frac{\mathrm{1}}{\mathrm{2}}\right)\mathrm{-}u\mathrm{(}t\mathrm{-}\mathrm{1}\mathrm{)}\right)=\left\{\begin{array}{l}0,t<\frac{1}{2}\\ 2-2t,\frac{1}{2}<t<1\\ 0,1<t\end{array}\right. \end{gathered}$$

 

Hence, the unit triangular pulse can be expressed as,

$$\begin{gathered} \Lambda \left(t\right)\backslash =2t\left(u\left(t\right)-u\left(t-\frac{1}{2}\right)\right)+(2-2t)\left(u\left(t-\frac{1}{2}\right)-u(t-1)\right) \\ =2tu\left(t\right)-2tu\left(t-\frac{1}{2}\right)+(2-2t)u\left(t-\frac{1}{2}\right)-(2-2t)u(t-1) \end{gathered}$$

Use the linearity of Laplace transform,

$$\begin{gathered} \mathcal{L}\left\{2tu\right(t\left)\right\}=\frac{2}{s^2} \\ -\mathcal{L}\left\{2tu\left(t-\frac{1}{2}\right)\right\}=-\frac{2e^{-\frac{s}{2}}}{s^2} \\ \mathcal{L}\left\{(2-2t)u\left(t-\frac{1}{2}\right)\right\}=\frac{2e^{-\frac{s}{2}}}{s}-\frac{2e^{-\frac{s}{2}}}{s^2} \\ -\mathcal{L}\left\{\right(2-2t\left)u\right(t-1\left)\right\}=\frac{2e^{-s}}{s^2}-\frac{2e^{-s}}{s} \end{gathered}$$

Thus, we get

$\mathcal{L}\left(\Lambda \left(t\right)\right)\left(s\right)=\frac{2+2e^{-\frac{k}{2}}(2-s)+2e^{-s}(1-s)}{s^2}$

Therefore, the Laplace transform of $\Lambda \left(t\right)$ is $\mathcal{L}\left(\Lambda \left(t\right)\right)\left(s\right)=\frac{2+2e^{-\frac{k}{2}}(2-s)+2e^{-s}(1-s)}{s^2}$