# How would I solve this using Laplace transformation?

• Engineering
• arhzz
In summary, the transfer function H(s) is given as $$H(s) =\frac{s-1}{1-2(s^2-s)-As-\frac{A}{2}}$$. The task is to determine the value of A so that the impulse response h(t) contains components with $$te^{at} \sigma(t)$$. To solve this, we can use the Laplace Transformation and complete the square in the denominator. This gives us the form ##-2[s + \frac 1 2(A/2 - 1)]^2 + \frac 1 8(A - 6)(A - 2)##. Setting A = 6 and A = 2, we can simplify
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

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.

## 1. How do I determine which function to use for the Laplace transformation?

The function used for the Laplace transformation depends on the type of problem you are trying to solve. For example, if you are solving a differential equation, you would use the Laplace transform of the derivative. If you are solving an integral, you would use the Laplace transform of the integral. It is important to understand the properties of the Laplace transform and how they can be applied to different types of problems.

## 2. What are the steps involved in solving a problem using Laplace transformation?

The general steps for solving a problem using Laplace transformation are as follows:
1. Take the Laplace transform of the given function.
2. Use algebraic manipulation and the properties of Laplace transform to simplify the transformed function.
3. Inverse Laplace transform the simplified function to get the solution in the time domain.
4. Check the solution for accuracy by substituting it back into the original problem.

## 3. Can Laplace transformation be used for all types of problems?

No, Laplace transformation is mainly used for solving linear differential equations with constant coefficients. It can also be used for solving integral equations and some other types of problems, but it is not applicable to all types of problems. It is important to understand the limitations and assumptions of Laplace transformation before applying it to a problem.

## 4. How do I know if my solution using Laplace transformation is correct?

You can check the accuracy of your solution by substituting it back into the original problem. If the solution satisfies the initial conditions and the differential equation, then it is considered correct. It is also important to double-check for any algebraic errors or mistakes in the application of Laplace transform properties.

## 5. Are there any alternative methods to Laplace transformation for solving differential equations?

Yes, there are other methods such as the Fourier transform, the Z-transform, and the method of undetermined coefficients that can be used to solve differential equations. Each method has its own advantages and limitations, and the choice of method depends on the specific problem at hand. However, Laplace transformation is a commonly used and powerful tool for solving many types of differential equations.

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