- #1
WolfOfTheSteps
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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?
