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I am stuck at question number 2.8

I am having a problem in the question number 2.8

This is how I have gone about solving it:

I have calculated $y(t)$ by convolving $x(t)$ with $h(t)$ using the fact that $x(t)$ convolved with an impulse at $t=t_0$ is just the same signal $x(t)$ time shifted by $t_0$. Hence,

$$y(t)= x(t+2) + 2x(t+1)$$

Now, according to the answer in the book, signal $y(t)$ is $t+3$ on $-2 < t \leq -1$. But, according to my answer, if we add $x(t+2) + 2x(t+1)$, the value of $y(t)$ at $t=-1$ is $4$. Why Please help me.

Also, why is it that in the answer in the book they haven't included $t=-2$ in the range?

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$$x(t)=\begin{cases} t+1\qquad & 0\leq t\leq1 \\ 2-t\qquad & 1<t \leq2 \\ 0\qquad & \mathrm{elsewhere} \end{cases}$$


$$f_1(t)=x(t+2)=\begin{cases} t+2+1\qquad & 0\leq t+2\leq1 \\ 2-(t+2)\qquad & 1<t+2 \leq2 \\ 0\qquad & \mathrm{elsewhere} \end{cases}$$

$$f_1(t)=x(t+2)=\begin{cases} t+3\qquad &-2\leq t\leq-1 \\ -t\qquad & -1<t \leq0 \\ 0\qquad &\mathrm{elsewhere} \end{cases}$$


$$f_2(t)=2x(t+1)=\begin{cases} 2(t+1+1)\qquad & 0\leq t+1\leq1 \\ 2(2-(t+1))\qquad & 1<t+1 \leq2 \\ 0\qquad &\mathrm{elsewhere} \end{cases}$$

$$f_2(t)=2x(t+1)=\begin{cases} 2t+4\qquad & -1\leq t\leq0\\ 2-2t\qquad & 0<t \leq1 \\ 0\qquad &\mathrm{elsewhere} \end{cases}$$


Now add them $$y(t)=x(t+2)+2x(t+1)=\begin{cases} t+3\qquad & -2\leq t<-1 \\ 3t+7\qquad &t=-1\\ t+4\qquad & -1<t \leq0 \\ 2-2t\qquad &0<t\leq1\\ 0\qquad &\mathrm{elsewhere} \end{cases}$$

Now check $f_1(-1)=2$,$f_2(-1)=2$,also $y(-1)=f_1(-1)+f_2(-1)=4$

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  • $\begingroup$ But isn't the value of x(t+2) and 2x(t+1) both 2 at t=-1. Hence should'nt y(t) be 2+2=4 at t=-1? $\endgroup$ Oct 22, 2017 at 10:24
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    $\begingroup$ @Rohit I have added a minus sign in the last system, I hope it's consistent $\endgroup$ Oct 22, 2017 at 11:34
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    $\begingroup$ @Fat32 LOL....So you are repeating my all words. :-)))...and this $0\geq t \leq 1$ LOL $\endgroup$
    – Rohit
    Oct 22, 2017 at 14:46
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    $\begingroup$ @YOGENDRASINGH Oh..I got you...you can include -2 too.. sorry my mistake $\endgroup$
    – Rohit
    Oct 22, 2017 at 17:27
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    $\begingroup$ @YOGENDRA SINGH If your doubt cleared from the answer then you should "accept the answer". $\endgroup$
    – Rohit
    Oct 25, 2017 at 12:18

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