2.2 Discussion, week 3

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2.2.1 Questions

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2.2.2 Problem 1

Solution

Folding $h\left(\tau \right)$ to becomes $h(-\tau )$. Therefore, when $1+t<-1$ or $t<-2$, then $y\left(t\right)=0$ since there is no overlap.

When $-1<1+t<1$, or $-2<t<0$, then there is partial overlap. In this case$$\begin{array}{rl}y\left(t\right)& ={\int }_{-1}^{1+t}\cos \left(\pi \tau \right)d\tau -2<t<0\\ & =\frac{1}{\pi }{[\sin \left(\pi \tau \right)]}_{-1}^{1+t}\\ & =\frac{1}{\pi }[\sin \left(\pi (1+t)\right)-\sin (-\pi )]\\ & =\frac{1}{\pi }\sin \left(\pi (1+t)\right)\end{array}$$

When $1<1+t<3$, or $0<t<2$, then there is partial overlap. In this case$$\begin{array}{rl}y\left(t\right)& ={\int }_{t-1}^1\cos \left(\pi \tau \right)d\tau 0<t<2\\ & =\frac{1}{\pi }{[\sin \left(\pi \tau \right)]}_{t-1}^1\\ & =\frac{1}{\pi }[\sin \left(\pi \right)-\sin \left(\pi (t-1)\right)]\\ & =\frac{-1}{\pi }\sin \left(\pi (t-1)\right)\end{array}$$

When $3<1+t$ or $t>2$ then $y\left(t\right)=0$ since there is no overlap any more. Hence solution is$$y\left(t\right)=\{\begin{array}{ccc}0& & t\le -2\\ \frac{1}{\pi }\sin \left(\pi (1+t)\right)& & -2<t\le 0\\ \frac{-1}{\pi }\sin \left(\pi (t-1)\right)& & 0<t\le 2\\ 0& & t>2\end{array}$$ The following is a plot of $y\left(t\right)$

Figure 2.10:Plot of $y(t)$

2.2.3 Key solution

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