How would I solve this using Laplace transformation?

[arhzz]

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

 

[Mark44]

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 $-2[s + \frac 1 2(A/2 - 1)]^2 + \frac 1 8(A - 6)(A - 2)$

 

[arhzz]

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?

 

[vela]

Did you understand the significance of $te^{at}$, as opposed to say just $e^{at}$?

 

[Mark44]

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.

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.