Question

Question 10: Signals -

The following signal x(t) is given, Sketch \(\large \color{green}{x(1 - \frac{t}{2})}\)

Answer 1

I have got :

Answer 2

The amplitude is 1 there..

Answer 3

@bradely are you trying it??

Answer 4

@Miracrown , your eyes are needed... I mean their "one look".. :P

Answer 5

I think it is right. You can substitute significant points to x(t) and see if it matches

Answer 6

And how can I do that??

Give me an example.. :)

Answer 7

I saw you typing.. :P What were you typing?? :P

Answer 8

Assume that the graph you derived is \(x'\left(1-\frac t2\right)\), the point at t=2 is a significant point where triangle ends and square begins in the graph. So when t=2, \(x'\left(1-\frac22\right)=x'(0)=0\). Look what x(0) shows in the original graph. The value of the original graph (0) is same as the second graph. Thus \(x'(0)=x(0)\) like that take few points in both graphs and check whether they are equal.

Answer 9

Sorry, I am not getting you @BAdhi .. Why have you written \(x'\) there??

Answer 10

How \(x'(0)\) is \(0\) ??

Answer 11

Sorry I was trying to imply that, take \(x'(t) \) as the new derived graph that you are trying to compare. Since \(x'(2)\) is 0, Consider t=2 to x(1-t/2) which is also zero. Like that compare points such as t=4, t=0, t=-2. If for all those values \(x'(t)\) and \(x(1-\frac t 2)\) gives the same answer we can conclude that \(x'(t)=x\left( 1-\frac t 2\right) \)

Answer 12

Another way to approach this problem: Write \(x(t)\) as a piecewise function:
\[x(t)=
\begin{cases}
1&\text{for }-1< t<0\\
t&\text{for }0< t<1\\
1&\text{for }1< t<2\\
0&\text{otherwise}
\end{cases}\]
Now if we replace \(t\) with \(1-\dfrac{t}{2}\), we have
\[x\left(1-\dfrac{t}{2}\right)=
\begin{cases}
1&\text{for }-1<1-\dfrac{t}{2}<0\\\\
1-\dfrac{t}{2}&\text{for }0< 1-\dfrac{t}{2}<1\\\\
1&\text{for }1< 1-\dfrac{t}{2}<2\\\\
0&\text{otherwise}
\end{cases}\]
and simplifying this gives
\[x\left(1-\dfrac{t}{2}\right)=
\begin{cases}
1&\text{for }2< t< 4\\\\
1-\dfrac{t}{2}&\text{for }0< t<2\\\\
1&\text{for }-2<t<0\\\\
0&\text{otherwise}
\end{cases}\]
and this definition agrees with your graph.

Answer 13

Yeah I got this.. :)

Answer 14

Thank you guys.. :) It helped a lot..