Proving Time Invariance: How to Analyze and Confirm Time Invariance in a System

[WolfOfTheSteps]

Homework Statement

Show that

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

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 is not TI! My proof is below. Am I wrong or is the question wrong in assuming that the system is TI?

The Attempt at a Solution

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

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

but

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]

Therefore y_1(t) \neq y(t+t_0) and the system is not time invariant.

\Box

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:

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

Now I time shift the system by 2:

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

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

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

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

\Box

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

 

[WolfOfTheSteps]

Anyone know of a forum where this type of question is likely to get answered? I'm still desperate to get an answer to this. This is like basic systems... someone's got to know.

Can I move this to the EE forum?