Inverse Laplace transform Signals and systems

In summary, the problem is that the student was not given ω correctly, so they were not able to solve for y(t). I am going to edit my first post a couple of times after I first submitted it, so you may want to reread it.
  • #1
mattbrrtt
11
0

Homework Statement



Compute Y(t).


Homework Equations



Y(s)= (s(2s+11)+ω(8s+4ω^2))/((s^2+ω^2)(s+9)(s+2))

The Attempt at a Solution



(s(2s+11)+ω(8s+4ω^2 ))/((s^2+ω^2)(s+9)(s+2))= A/(s^2+ω^2 )+B/(s+9)+C/(s+2)

Every example I have looked at does not have the ω variable, but I am not sure that is the problem.

This is a problem from a lab for a signals and systems course. The problem starts by giving h(t) =[cos 2t + 4 sin 2t]u(t), and asking for the impulse response. Then an input of x(t) = (5/7)(e^-t)-(12/7)(e^-8t) is provided. The output response is then calculated, and that is where this question picks up.

Thank you for any assistance.
Matt
 

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  • #2
The numerator of the first fraction should be of the form As+B because the denominator is a quadratic.

I assume X(s) and H(s) are supposed to be the Laplace transforms for the given x(t) and h(t). If so, you need to recalculate X(s) as it isn't correct.

ω is a constant. It's the frequency in the cosine and sine terms, so you can set it to 2 in H(s).
 
Last edited:
  • #3
I am guessing that the mistake made with the X(s) and H(s) terms is that I simplified and ended up with unfavorable numerators. I am going to try using :

H(s)=1/(s+ω^2 )+4/(s^2+ω)
X(s)=1/(s+2)+1/(s+9)

and calculate from there.

Is this correct?

Thank you.
 
  • #4
mattbrrtt said:
I am guessing that the mistake made with the X(s) and H(s) terms is that I simplified and ended up with unfavorable numerators. I am going to try using :

H(s)=1/(s+ω^2 )+4/(s^2+ω)
X(s)=1/(s+2)+1/(s+9)

and calculate from there.

Is this correct?
I edited my first post a couple of times after I first submitted it, so you may want to reread it.

Your original H(s) was correct. You just have to set ω=2. It's the angular frequency which appears in the terms of x(t).

X(s), however, remains a mystery. I'm not sure how you're getting 1/(s+2) and 1/(s+9) terms in it.

It turns out if you get X(s) and H(s) correct, the algebra simplifies a lot, and you should find it relatively easy to find y(t).
 
  • #5
Thanks.
The X(s) comes from the formula that was given in the assignment.
x(t)=(5/7)(e^-t)-(12/7)(e^-8t)

When I was writing my explanation, I think I found an error. I will post back.
 

Related to Inverse Laplace transform Signals and systems

1. What is an inverse Laplace transform?

An inverse Laplace transform is a mathematical operation that takes a function in the Laplace domain and converts it into the time domain. This allows us to understand the behavior of a system over time.

2. How is the inverse Laplace transform different from the Laplace transform?

The Laplace transform is a mathematical tool used to convert a function in the time domain into the Laplace domain. The inverse Laplace transform is the opposite operation, converting a function in the Laplace domain back into the time domain.

3. What are some applications of inverse Laplace transforms in signal and systems analysis?

Inverse Laplace transforms are commonly used in the analysis of linear systems, such as electrical circuits and mechanical systems. They are also used in control theory to model and analyze the behavior of dynamic systems.

4. Are there different methods for finding the inverse Laplace transform?

Yes, there are several methods for finding the inverse Laplace transform, including the use of partial fraction decomposition, the method of residues, and the convolution integral method. The choice of method depends on the complexity of the function in the Laplace domain.

5. Can the inverse Laplace transform be used to solve differential equations?

Yes, the inverse Laplace transform can be used to solve differential equations. By converting a differential equation into the Laplace domain, it can be transformed into an algebraic equation, which can then be solved using inverse Laplace transforms. This is a useful tool in engineering and physics for solving complex systems of differential equations.

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