# Is the system represented by the equation $y(t) = x(2t)$ time invariant?

I came across this problem in the text book Signals and Systems - Oppenheim (Example-1.16).

To solve this, I followed the following algorithm (described in the book earlier for a separate problem):

\begin{align} y(t) &=x(2t)\\ y_1(t)&=x_1(2t)\end{align}

Let $x_2(t)=x_1(t-t_0)$ and $y_2(t)=x_2(2t)$

\begin{align} \implies y_2(t) &= x_1(2(t - t_0))\\ &= x_1(2t-2t_0)\end{align}

Now, $y_1(t-t_0) = x_1(2(t-t_0)) = x_1(2t - 2t_0)$.

Since $y_2(t)$ and $y_1(t-t_0)$ are equivalent, the system should be time-invariant.

However, the book takes a different(graphical) approach, wherein the time-shifted $y(t)$, $y(t - t _0)$ is $x(2t - t_0)$ and the resulting system is time-varying.

I would like to understand why is there this inconsistency between the mathematical and the graphical approach.

• why on earth do you do things like $y_2$$(t) and y_1$$(t-t_0)$ ?? and should be plaintext, and it's not only superfluous to put every single math mode element in $, it's also bad for readability! correcting that. – Marcus Müller Jan 27 '17 at 18:41 ## 3 Answers From your solution: I followed the following algorithm: $$y(t) =x(2t)$$ $$y_1(t) = x_1(2t)$$ Let $$x_2(t) = x_1(t-t_0) ~~~\text{and}~~~ y_2(t) = x_2(2t)$$ On this following step (time sampling of the shifted argument) you make the usual mistake: $$\implies y_2(t) = x_1(2(t - t_0))$$ which should instead be : $$\implies y_2(t) = x_2(2t) = x_1(t - t_0)|_{t=2t} = x_1(2t - t_0)$$ The remaining parts follow as usual to show that the time scaler is a time-varying system... there is not such inconsistency between the mathematical and the graphical approach, because the correct mathematical approach should be $$x_2(t) = x_1(t-t_0)$$ $$y_2(t) = x_2(2t) = x_1(2t-t_0)$$ as the formula says, the factor only multiplies the variable, it is like saying$ f(x) = x + 2 $so$ f(2x) $will be equal to$ 2(x) + 2 $and not$2(x+2) $I hope it answered your question • It did. You explained it really well. Thanks. – Hrinyaksh Jan 27 '17 at 18:51 1. Apply delay, then apply system function$x(t)\rightarrow$delay$t_0\rightarrowx(t-t_0)\rightarrow$apply system function (which doubles the time variable)$\rightarrowx(2t-t_0)$1. Apply system function, then apply delay$x(t)\rightarrow$apply system function$\rightarrowx(2t)\rightarrow$delay$t_0\rightarrowx(2(t-t_0)) = x(2t-2t_0)\$

Since the outputs don't match, the system is time variant.