2.1 Discussion, second week

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2.1.1 Questions

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2.1.2 Problem 4.1

Solution

2.1.2.1 part a

We need to find linear combination of \(x_{1}\left [ n\right ] ,x_{2}\left [ n\right ] ,x_{3}\left [ n\right ] \) which gives \(x_{4}\left [ n\right ] \). In other words, looking at samples at \(n=0,1,2\) and adding corresponding samples gives\begin{align*} a+b+c & =1\\ b+c & =-1\\ c & =1 \end{align*}

But from second equation \(b=-1-1=-2\) and from first equation \(a=1-b-c=1+2-1=2\). Hence\[ 2x_{1}\left [ n\right ] -2x_{2}\left [ n\right ] +x_{3}\left [ n\right ] =x_{4}\left [ n\right ] \]

2.1.2.2 part b

Therefore by linearity\[ 2y_{1}\left [ n\right ] -2y_{2}\left [ n\right ] +y_{3}\left [ n\right ] =y_{4}\left [ n\right ] \] Hence \[ y_{4}\left [ n\right ] =2\delta \left [ n+1\right ] +\delta \left [ n\right ] +2\delta \left [ n-1\right ] -2\delta \left [ n-2\right ] \]

Figure 2.1:Plot of \(y[n]\)

2.1.2.3 part c

System is time invariant if shifted input gives same output but also shifted by the same amount as the input is shifted by. Let us consider \(x_{1}\left [ n\right ] \). By shifting it to the right by one, then the output should \(y_{1}\left [ n\right ] \) but shifted to the right by one which is \(y_{1}\left [ n-1\right ] \)

Figure 2.2:Plot of \(y_1[n-1]\)

Shifting \(x_{1}\left [ n\right ] \) by 2 now the output should be \(y_{1}\left [ n-2\right ] \,\)

Figure 2.3:Plot of \(y_1[n-2]\)

But adding \(x_{1}\left [ n\right ] +x_{1}\left [ n-1\right ] +x_{1}\left [ n-2\right ] \) gives \(x_{3}\left [ n\right ] \). Which has the output shown. Let us now add \(y_{1}\left [ n\right ] +y_{1}\left [ n-1\right ] +y_{1}\left [ n-2\right ] \) and see if this gives same as \(y_{3}\left [ n\right ] \)

Figure 2.4:Plot of all shifted inputs of \(x_1[x]\)

Since the above is not the same as \(y_{3}\left [ n\right ] \) then the system is not time invariant.

2.1.3 Problem 4.2

Solution

2.1.3.1 Part a

By folding \(x\left [ n\right ] \) and shifting to the right, we see that  \(y\left [ 0\right ] =2,y\left [ 1\right ] =2+2=4,y\left [ 2\right ] =2+2+2=6,y\left [ 3\right ] =8,y\left [ 4\right ] =6,y\left [ 5\right ] =4,y\left [ 6\right ] =2,y\left [ 7\right ] =0\) and \(y\left [ n\right ] =0\) for all other values.

Figure 2.5:\(y[n]\)

2.1.3.2 Part b

By folding \(x\left [ n\right ] \) and shifting to the right, we see that  \(y\left [ 0\right ] =0,y\left [ 1\right ] =0,y\left [ 2\right ] =0.5,y\left [ 3\right ] =1,y\left [ 4\right ] =1.5,y\left [ 5\right ] =1,y\left [ 6\right ] =0.5,y\left [ 7\right ] =0\) and \(y\left [ n\right ] =0\) for all other values.

Figure 2.6:\(y[n]\)

2.1.4 Problem 4.3

Solution

2.1.4.1 Part a

By folding \(x\left ( t\right ) \) and shifting, we see that for \(t<0\) that \(y\left ( t\right ) =0\). And for \(\,0<t<4\) the integral becomes\begin{align*} y\left ( t\right ) & =\int _{0}^{t}h\left ( \tau \right ) d\tau \qquad 0<t<4\\ & =\int _{0}^{t}1d\tau \\ & =t \end{align*}

And for \(\,0<t-4<4\) or \(4<t<8\)\begin{align*} y\left ( t\right ) & =\int _{t-4}^{4}h\left ( \tau \right ) d\tau \qquad 4<t<8\\ & =\int _{t-4}^{4}1d\tau \\ & =4-\left ( t-4\right ) \\ & =8-t \end{align*}

And for \(\,4<t-4\) or \(t>8\)\[ y\left ( t\right ) =0 \] Hence \(y\left ( t\right ) \) is\[ y\left ( t\right ) =\left \{ \begin{array} [c]{ccc}0 & & t<0\\ t & & 0<t<4\\ 8-t & & 4<t<8\\ 0 & & t>8 \end{array} \right . \]

Figure 2.7:\(y(t)\)

2.1.4.2 Part b

By folding \(h\left ( t\right ) \) and shifting, we see that for \(t<0\) that \(y\left ( t\right ) =0\). And for \(\,t>0\) the integral becomes\begin{align*} y\left ( t\right ) & =\int _{1}^{1+t}h\left ( \tau \right ) d\tau \qquad t>0\\ & =\int _{1}^{1+t}e^{-\left ( \tau -1\right ) }d\tau \\ & =\left [ \frac{e^{-\left ( \tau -1\right ) }}{-1}\right ] _{1}^{1+t}\\ & =-\left [ e^{-\left ( \tau -1\right ) }\right ] _{1}^{1+t}\\ & =-\left [ e^{-\left ( \left ( 1+t\right ) -1\right ) }-e^{-\left ( 1-1\right ) }\right ] \\ & =-\left [ e^{-t}-1\right ] \\ & =1-e^{-t} \end{align*}

Hence \(y\left ( t\right ) \) is

Figure 2.8:\(y(t)\)

2.1.4.3 Part c

By folding \(h\left ( t\right ) \) and shifting, we see that for \(-2+t<-1\) or \(t<1\) that \(y\left ( t\right ) =0\). And for \(\,\,-1<-2+t<3\) or \(1<t<5\) the integral becomes \(x\left ( t\right ) \) itself (i.e. original \(x\left ( t\right ) \) but shifted to right by 2). And for \(3<-2+t\) or \(t>5\) then \(y\left ( t\right ) =0\). Hence

\[ y\left ( t\right ) =\left \{ \begin{array} [c]{ccc}0 & & t<1\\ x\left ( t-2\right ) & & 1<t<5\\ 0 & & t>5 \end{array} \right . \]

Figure 2.9:\(y(t)\)

2.1.5 key solution

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