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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.

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    $\begingroup$ 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. $\endgroup$ – Marcus Müller Jan 27 '17 at 18:41
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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...

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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

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  • $\begingroup$ It did. You explained it really well. Thanks. $\endgroup$ – Hrinyaksh Jan 27 '17 at 18:51
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  1. Apply delay, then apply system function

$x(t)$ $\rightarrow$ delay $t_0$ $\rightarrow$ $x(t-t_0)$ $\rightarrow$ apply system function (which doubles the time variable) $\rightarrow$ $x(2t-t_0)$

  1. Apply system function, then apply delay

$x(t)$ $\rightarrow$ apply system function $\rightarrow$ $x(2t)$ $\rightarrow$ delay $t_0$ $\rightarrow$ $x(2(t-t_0)) = x(2t-2t_0)$

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

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