# Impulse response of transfer function

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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 $A$.

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### DrOnline

Updated on August 01, 2022

• DrOnline over 1 year

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.

• Ron Gordon over 10 years
I will need to see the picture. I have little idea of what you are describing. And I do know what an impulse is.
• DrOnline over 10 years
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.
• Ron Gordon over 10 years
And what exactly is the question? Are you to find the Laplace transform of the impulse? And what are $A$, $B$, etc.?
• DrOnline over 10 years
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.
• DrOnline over 10 years
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...?
• jkn over 10 years
$lim_{s\rightarrow\infty}\frac{s}{s+1}=1$.
• DrOnline over 10 years
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..?
• jkn over 10 years
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 $s$.
• DrOnline over 10 years
Ah, I think we are getting to the core of my confusion. you are multiplying the transfer function with s BECAUSE of the dirac pulse? Because if if the input was a step function we'd multiply with 1/s, which I am comfortable with. Can you explain/confirm your reason for multiplying with s, is it due to the signal being dirac pulse? OR: you are saying the theorem by DEFINITION looks at s*T(s), and the only reason I have never considered that s, is due to having worked exclusively with step functions removing it?
• jkn over 10 years
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?
• DrOnline over 10 years
Yes, it clears it up perfectly. 1000 vote ups for you my friend. The reason for my confusion is this: I've done like 100 such tasks with step response which of course removes that s introduced by the theorems. Now, the laplace transform of dirac is 1, I have to account for it, which solves everything. Wow.. we were kind of operating on auto pilot, it only worked each time (until now) because it was always step response. Thank you so much!