2
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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


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

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


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

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


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 $y(-1)=f_1(-1)+f_2(-1)=4$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.