Engineering How would I solve this using Laplace transformation?

AI Thread Summary
The discussion revolves around solving for the parameter A in the transfer function H(s) = (s-1)/(1-2(s^2-s)-As-A/2) to ensure the impulse response h(t) includes terms of the form te^(at)σ(t). The user initially struggles with performing the inverse Laplace transformation and seeks guidance on partial fraction decomposition, which requires identifying the zeros of the denominator. Completing the square in the denominator is suggested as a key technique, leading to the conclusion that setting A to 2 or 6 simplifies the expression appropriately. This approach allows for the desired form of the impulse response to be achieved, emphasizing the importance of the term te^(at).
arhzz
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Homework Statement
A transfer function is given how A must be chosen so that the impulse response associated with H has components with te^{at}\sigma(t)
Relevant Equations
Laplace Transoformation
Hello!

Consider this transferfunction H(s);

$$ H(s) =\frac{s-1}{1-2(s^2-s)-As-\frac{A}{2}} $$

Now I need to determine A (note that A is coming from R) so that the impulse response h(t) (so in time domain) so that it contains components with $$te^{at} \sigma(t) $$.

Now I honestly really have no idea how to solve this.We are susposed to use the Laplace Transformation,so I tried a few things but it got me nowhere to be honest.

What I tried is; since the transferfunction is given in the frequency domain (s) and we need the impulse response in time domain (t) I was thinking of reverting the H(s) into h(t) (Inverse Laplace transformation) and then seeing what would that bring me; But the problem is how can I revert the function in the time domain? For that I need partial fraction decomposition,to get it in a form where I could use the standard Laplace Transformations (the one that are given on a sheet,the most common ones). And I can't do the decomposition when I can't find the zeros of the denominator.

I tried using the integral of the inverse Laplace transformation and it didnt bring me very far.Any insight would be great;

Also the solution should be ;

A = 2 and A = 6

Thanks
 
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arhzz said:
Homework Statement:: A transfer function is given how A must be chosen so that the impulse response associated with H has components with te^{at}\sigma(t)
Relevant Equations:: Laplace Transoformation

Hello!

Consider this transferfunction H(s);

$$ H(s) =\frac{s-1}{1-2(s^2-s)-As-\frac{A}{2}} $$

Now I need to determine A (note that A is coming from R) so that the impulse response h(t) (so in time domain) so that it contains components with $$te^{at} \sigma(t) $$.

Now I honestly really have no idea how to solve this.We are susposed to use the Laplace Transformation,so I tried a few things but it got me nowhere to be honest.

What I tried is; since the transferfunction is given in the frequency domain (s) and we need the impulse response in time domain (t) I was thinking of reverting the H(s) into h(t) (Inverse Laplace transformation) and then seeing what would that bring me; But the problem is how can I revert the function in the time domain? For that I need partial fraction decomposition,to get it in a form where I could use the standard Laplace Transformations (the one that are given on a sheet,the most common ones). And I can't do the decomposition when I can't find the zeros of the denominator.

I tried using the integral of the inverse Laplace transformation and it didnt bring me very far.Any insight would be great;

Also the solution should be ;

A = 2 and A = 6

Thanks
For this transfer function, ## H(s) =\frac{s-1}{1-2(s^2-s)-As-\frac{A}{2}}##, I believe the key is to complete the square in the denominator.

If I haven't made any errors, what I get for the denominator is ##-2[s + \frac 1 2(A/2 - 1)]^2 + \frac 1 8(A - 6)(A - 2)##
 
Mark44 said:
For this transfer function, ## H(s) =\frac{s-1}{1-2(s^2-s)-As-\frac{A}{2}}##, I believe the key is to complete the square in the denominator.

If I haven't made any errors, what I get for the denominator is ##-2[s + \frac 1 2(A/2 - 1)]^2 + \frac 1 8(A - 6)(A - 2)##
Ohhh okay,completing the square...Didnt really think about that.I get the same result,and if I am not mistaken,the next step would be taking A = 6 and A = 2 so that the right fraction becomes 0,and from that should give me the form which I want ( the one I mentioned in my #1 post)

Think that should be about it right?
 
Did you understand the significance of ##te^{at}##, as opposed to say just ##e^{at}##?
 
arhzz said:
Ohhh okay,completing the square...Didnt really think about that.
Keep that in mind, since that's a technique that can be used a lot in finding the inverse Laplace transform.

arhzz said:
I get the same result,and if I am not mistaken,the next step would be taking A = 6 and A = 2
If A = 2, the denominator simplifies to ##-2s^2##. See what you get when A = 6.
 

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