# Proving Time Invariance

1. Jul 25, 2007

### WolfOfTheSteps

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

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?

3. 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?

2. Aug 2, 2007

### 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 gotta know.

Can I move this to the EE forum?