At first I had trouble proving the time variance for this as I would do the usual mistake as in question Is the system represented by the equation $y(t) = x(2t)$ time invariant?. The solution has graphical representation of all the shifts in the process and they are the following. enter image description here

I get the shift from the first picture to the second. It gets compressed, I know that property. I also understand what happens in the third picture. The $x_1(t)$signal is shifted to the right because of $t-2$ . The fifth picture is fine as well, it's just a shift to the right again from the second picture.

What I'm having trouble with is the 4th picture. I know that $y_2(t)=x_2(t)=x_1(2t-t_0)$ I forgot to mention, all the graphs are for $t_0=2$.

  • Shouldn't the 4th picture be a shift of 2 to the right?
  • And the last one maybe $(2t-2t_0)$ so $2t_0=4$ units to the right?

1 Answer 1


The signals plotted on the figure are correct.

As your provided link already shows: $$y_d(t) = \mathcal{T}\{x_d(t)\} = \mathcal{T}\{x(t-d)\} = \mathcal{T}\{x(t-2)\} = x(2t-2) \not= y(t-2) $$ where $$ ~ y(t)=\mathcal{T}\{x(t)\} = x(2t) \longrightarrow y(t-2) = x(2(t-2)) = x(2t-4) $$

Looking at the progression on the provided figure, $x_1(t)$ is a pulse of domain $~ [-2,2] ~$ then $y_1(t)=\mathcal{T}\{x_1(t)\} = x_1(2t)$ will be a pulse of domain $~ [-1,1] ~$ which is obtained by compressing $x(t)$ by $2$ about the origin $~t=0~$

$x_2(t)$ is obtained from $x_1(t)$, delaying it by $2$ units to the right, hence $x_2(t)$ has a domain of $~[0,4]$. Therefore $y_2(t)=\mathcal{T}\{x_2(t)\} = x_2(2t)$ will be a pulse of domain $~ [0,2] ~$ which is obtained by compressing $x_2(t)$ by $2$ about the origin $~t=0~$

On the other hand shifting the pulse of $y_1(t)$ with the domain $[-1,1]$ by $2$ units to the right creates a new pulse with domain $[1,3]$ which is not equivalent to $[0,2]$; the domain of $y_2(t)$, from which we deduce, also graphically, that $y_1(t-2) \not = y_2(t)$


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