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I was reading out a book and it said to prove that \$y(t) = sin (t) x(t-2)\$ is time variant, so far of all the inputs I have tried, as well as the general input of giving a shift of T, the system seems to be time invariant.

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This system is time variant because plugging in \$x(t-a)\$ does not equal \$y(t-a)\$.

For the first case you get:
\$y_1(t)=sin(t)x(t-a-2)\$

However, if you offset the output by a you get:
\$y_2(t)=y(t-a)=sin(t-a)x(t-a-2)\$

Since \$y_1(t)\$ does not equal \$y_2(t)\$ the system is time variant. Look at the first example on this page if you are confused: https://en.wikipedia.org/wiki/Time-invariant_system

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The system is time variant. Forget about the t-2 and think about what happens with sin(t) when you start the input at time t=0 as apposed to time t=pi/2 .

In both of these cases, let the input be a dirac delta function. In the first case, you should get nothing out, and in the second you would get the delta function back out.

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  • \$\begingroup\$ I had thought the same but the example of an RC circuit came in my mind in which a delayed sinusoidal input gives a delayed sinusoidal output \$\endgroup\$ – Syed Mohammad Asjad Jul 20 '17 at 13:36
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    \$\begingroup\$ oh hang on, now I get it :) thankyou \$\endgroup\$ – Syed Mohammad Asjad Jul 20 '17 at 13:48

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