Note: I posted this a month ago in the homework forum, but never got a reply. It really is an elementary question, and I think someone here might know the answer. (I hope I don't get an infraction for the re-post! If it's inappropriate, please delete.)(adsbygoogle = window.adsbygoogle || []).push({});

1. The problem statement, all variables and given/known data

Show that

[tex] y(t) = \frac{d}{dt}\left[e^{-t}x(t)\right][/tex]

is time invariant.

2. Relevant Information

I don't think this is TI!! I'm told it is TI, but I think I proved that it isnotTI! My proof is below. Am I wrong or is the question wrong in assuming that the system is TI?

3. The attempt at a solution

Let [itex]y_1[/itex] be the output when [itex]x(t+t_0)[/itex] is the input, then:

[tex] y_1(t) = \frac{d}{dt}\left[e^{-t}x(t+t_0)\right] [/tex]

but

[tex] y(t+t_0) = \frac{d}{dt}\left[e^{-(t+t_0)}x(t+t_0)\right] = e^{-t_0}\frac{d}{dt}\left[e^{-t}x(t+t_0)\right] [/tex]

Therefore [itex]y_1(t) \neq y(t+t_0)[/itex] and the system isnottime invariant.

[tex]\Box[/tex]

Also, just to make sure I wasn't missing some subtlety because of the differentiation I tried to prove this another way.

Since x(t) is arbitrary, I assumed x(t)=t, so that:

[tex]y(t)=\frac{d}{dt}[te^{-t}] = e^{-t}-te^{-t}[/tex]

Now I time shift the system by 2:

[tex]y(t+2) = e^{-(t+2)}-(t+2)e^{-(t+2)}=e^{-2}\left[e^{-t}-(t+2)e^{-t}\right][/tex]

Now I let [itex]y_1(t)[/itex] be the output when the input is [itex]x(t+2)=t+2[/itex]:

[tex]y_1(t) = \frac{d}{dt}[(t+2)e^{-t}] = e^{-t}-(t+2)e^{-t}[/tex]

Clearly, then, [itex]y(t+2)\neq y_1(t)[/itex] and the system is not TI for x(t)=t, and hence cannot be TI for arbitrary x(t).

[tex]\Box[/tex]

So, if itTI, what am I doing wrong? And how would I prove that it is TI?is

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# Time Invariance

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