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So let's suppose this is the $x(t)$ signal:

enter image description here

This is what i know

  • $x(t+1)$ should be shifted by 1 on $y$ axis , so the signal will start from $-1$ and end to $1$ instead of $0$ to $2$
  • $x(-t)$ will reflect the signal
  • $x(1.5t)$ will suppress the signal by $1.5$

How can i correctly calculate and make a graphical representation of $x(-1.5t+1)$ ?

Note: I'm new with with this kind of mathematics and expressions , any edit would be appriciated!

Community
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Phill Alexakis
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1 Answers1

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If you just want the graph, then you can use the following sequence: $$x(t) \xrightarrow{\text{suppress}} x(1.5t) \xrightarrow{\text{reflect}}x(-1.5t) \xrightarrow{\text{shift by} \frac{2}{3}}x(-1.5t+1).$$

If you want algebraically then, you first need to find $x(t)$ from the graph. From the graph given, $$x(t)= \begin{cases} 0 & \text{ if } t \leq 0 \text{ or } t \geq 2\\ 1 & \text{ if } 0 <t \leq 1\\ 2-t & \text{ if } 1 < t <2. \end{cases}$$ Now you can compute the composite function $x(-1.5t+1)$.

Anurag A
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    Thank you !! is there any particular reason why we follow the sequence you mentioned and not in another way? – Phill Alexakis Aug 28 '19 at 09:27
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    @PhillAlexakis We could have approached as follows as well:$$x(t) \xrightarrow{\text{reflect}} x(-t) \xrightarrow{\text{suppress}}x(-1.5t) \xrightarrow{\text{shift}}x(-1.5t+1).$$ – Anurag A Aug 28 '19 at 09:29
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    Just make sure when you draw the graph that $x(-1.5t+1)$'s graph is not obtained by shifting $x(-1.5t)$'s graph left by $1$ unit. Since $x(-1.5t+1) = x\left(-1.5\left(t - \frac{1}{1.5}\right)\right)$, we actually have that $x(-1.5t+1)$'s graph is obtained by shifting $x(-1.5t)$'s graph right by $\frac{1}{1.5} =\frac{2}{3}$ units. – Minus One-Twelfth Aug 28 '19 at 09:30
  • Oh now i get it! thank you! – Phill Alexakis Aug 28 '19 at 09:32