NB: Not a homework question, I am doing old exams in preparation for my own work, but I have no solutions for this, so I am a bit lost with this.
Question: The impulse response of a transfer function is shown in the graph below:
And I have four alternatives.
$$A= \frac{1}{s+1}$$ Initial value theorem: 0 Final value theorem: 1
$$B= \frac{1-2s}{1+2s}$$ Initial value theorem: -1 Final value theorem: 1
$$C= \frac{1}{(s+1)^2}$$ Initial value theorem: 0 Final value theorem: 1
$$D= \frac{-s}{s+1}$$ Initial value theorem: -1 Final value theorem: 0
I did the theorems, that is not in the question.
So none of them match the graph.
I have not worked with impulse before. I understand it is zero for all $t$ except $t=1$, where it is 1.
How do I approach this? I have done a TON of these with unit step, but not with impulse.
Let $F(s)$ denote the (one-sided) Laplace transform of $f(t)$, a continuous function defined on $[0,+\infty)$. Then the initial value theorem says that
$$f(0)=\lim_{s\rightarrow +\infty}sF(s)$$
and the final value theorem says that, if all the poles of $F$ are in the open left hand plane, then
$$\lim_{t\rightarrow+\infty}f(t)=\lim_{s\rightarrow 0}sF(s).$$
Using the two above and the fact the Laplace transform of the impulse response of a transfer function is the transfer function itself you can rule out all the above except for $A$.