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?

 

[piercas]

First, let's define time invariance in a system. A system is considered time invariant if a time shift in the input signal results in an equivalent time shift in the output signal. In other words, if the input signal is delayed or advanced by a certain amount of time, the output signal should also be delayed or advanced by the same amount of time.

Now, let's look at your proof. You correctly showed that the output signal y(t) is not equal to y(t+t0), indicating that the system is not time invariant. However, your proof is based on the assumption that x(t) is an arbitrary function. This is not necessarily true. In fact, the given equation shows that x(t) is multiplied by e^-t, which is a time-dependent function. Therefore, the function x(t) itself is not arbitrary and cannot be used to prove time invariance.

To prove time invariance, we need to consider a specific input signal and show that the output signal follows the definition of time invariance. Let's take the input signal x(t) = t, as you did in your second attempt. In this case, the output signal y(t) = e^-t - te^-t. Now, if we time shift the input signal by t0, we get x(t+t0) = t+t0. Plugging this into the original equation, we get y(t+t0) = e^-(t+t0) - (t+t0)e^-(t+t0). This is equivalent to y(t) multiplied by e^-t0, which shows that the output signal is indeed delayed by t0, just as the input signal was delayed.

In summary, your proof was based on an assumption that x(t) is an arbitrary function, which is not true for the given equation. By considering a specific input signal, we can see that the output signal follows the definition of time invariance, proving that the system is indeed time invariant.