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:

enter image description here

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.

  • 1
    $\begingroup$ I will need to see the picture. I have little idea of what you are describing. And I do know what an impulse is. $\endgroup$ – Ron Gordon May 4 '13 at 10:50
  • $\begingroup$ I've uploaded the graph, would appreciate any input on this! If it was a step function input I would multiply 1/s to the transfer function, but of course this is impulse. $\endgroup$ – DrOnline May 4 '13 at 11:16
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    $\begingroup$ And what exactly is the question? Are you to find the Laplace transform of the impulse? And what are $A$, $B$, etc.? $\endgroup$ – Ron Gordon May 4 '13 at 11:24
  • $\begingroup$ The question is: Which of these four transfer functions, given a dirac pulse input, would behave as the graph shows. A,B,C,D are multiple choice. I am confused because none of the 4 options match initial/final value theorem. Naturally, this is due to the input. However, I have no experience with dirac input, and so I need help understanding how it affects the graph. $\endgroup$ – DrOnline May 4 '13 at 11:37

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 $A$.

  • $\begingroup$ Initial value theorem (for A): 0, Final value theorem (for A): 1. Ok so you say - if you also account for the dirac pulse, it matches. But how. And why. Surely, at least the final value has to be 0...? $\endgroup$ – DrOnline May 4 '13 at 12:27
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    $\begingroup$ $lim_{s\rightarrow\infty}\frac{s}{s+1}=1$. $\endgroup$ – jkn May 4 '13 at 12:28
  • $\begingroup$ Ah, but you are looking at alternative D, also: the whole expression is negative in D. this, it's respective initial and final values should be -1 and 0. I think..? $\endgroup$ – DrOnline May 4 '13 at 12:43
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    $\begingroup$ I'm not, I'm looking at $A$, the theorem's look at limits involving $sT(s)$ not $T(s)$, that is the transfer function multiplied by $s$. $\endgroup$ – jkn May 4 '13 at 12:53
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    $\begingroup$ I'm multiplying with $s$ because the theorems ask you to do so. Have a more careful read, $F(s)$ is the Laplace transform of the function $f$ you are interested in. In your case $f$ is the output of a SISO LTI system with some input (whose Laplace transform is $T(s)U(s)$, with $T$ being the transfer function of the system and $U$ being the Laplace transform of the input). The Laplace transform of a dirac delta is simply $1$ while that of a unit step is $\frac{1}{s}$. Does this clear things up? $\endgroup$ – jkn May 4 '13 at 13:14

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