What is the Inverse Laplace Transform for V(s)=\frac{2s}{(s^{2}+4)^{2}}?

In summary, the conversation discusses finding the value of v(t) given the equation V(s)=\frac{2s}{(s^{2}+4)^{2} and provides the necessary equations and attempts at a solution. The final answer is v(t)=\frac{1}{2}tsin2tu(t). The conversation also mentions the need to account for the imaginary term and suggests using the quadratic form for the denominator.
  • #1
photonsquared
15
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1. Find [tex]v(t)[/tex] if [tex]V(s)=\frac{2s}{(s^{2}+4)^{2}}[/tex]

Ans: [tex]v(t)=\frac{1}{2}tsin2tu(t)[/tex]

2. Homework Equations :

[tex]V(s)=\frac{a_{n}}{(s-p)^{n}}+\frac{a_{n-1}}{(s-p)^{n-1}}+\cdots+\frac{a_{1}}{(s-p)}[/tex]
[tex]a_{n-k}=\frac{1}{k!}\frac{d^{k}}{ds^{k}}[(s-p)^{n}V(s)]_{s=p}[/tex]

3. Attempt at a solution:

[tex]V(s)=\frac{2s}{(s^{2}+4)^{2}}[/tex]

[tex]V(s)=\frac{2s}{(s^{2}+4)^{2}}=\frac{A}{(s^{2}+4)^{2}}+\frac{B}{(s^{2}+4)}[/tex]

[tex]A=\left[2s-B(s^{2}+4)\right]_{s=2i}[/tex]

[tex]A=4i[/tex]

[tex]B=\frac{d}{ds}\left[2s-B(s^{2}+4)\right]_{s=2i}[/tex]

[tex]B=2[/tex]

[tex]V(s)=\frac{4i}{(s^{2}+4)^{2}}+\frac{2}{(s^{2}+4)}[/tex]

I am not sure what to do with the imaginary term, but it does not translate to 1/2t, which is what is required for the answer.

[tex]?+sin2tu(t)[/tex]






 
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  • #2
photonsquared said:
1. Find [tex]v(t)[/tex] if [tex]V(s)=\frac{2s}{(s^{2}+4)^{2}}[/tex]

Ans: [tex]v(t)=\frac{1}{2}tsin2tu(t)[/tex]

2. Homework Equations :

[tex]V(s)=\frac{a_{n}}{(s-p)^{n}}+\frac{a_{n-1}}{(s-p)^{n-1}}+\cdots+\frac{a_{1}}{(s-p)}[/tex]
[tex]a_{n-k}=\frac{1}{k!}\frac{d^{k}}{ds^{k}}[(s-p)^{n}V(s)]_{s=p}[/tex]

3. Attempt at a solution:

[tex]V(s)=\frac{2s}{(s^{2}+4)^{2}}[/tex]

[tex]V(s)=\frac{2s}{(s^{2}+4)^{2}}=\frac{A}{(s^{2}+4)^{2}}+\frac{B}{(s^{2}+4)}[/tex]
No. Since the denominator, [itex]s^2+ 4[/itex] is quadratic you need
[tex]\frac{2s}{(s^2+4)^2}= \frac{Ax+ B}{(x^2+4)^2}+ \frac{Cx+ D}{x^2+4}[/tex]
 
  • #3
Thanks, I'll attempt again.
 

FAQ: What is the Inverse Laplace Transform for V(s)=\frac{2s}{(s^{2}+4)^{2}}?

What is the Inverse Laplace Transform?

The Inverse Laplace Transform is a mathematical operation that converts a function from the Laplace transform domain back to its original form in the time domain. It is the inverse of the Laplace transform and is used to solve differential equations and analyze systems in engineering and physics.

How is the Inverse Laplace Transform calculated?

The Inverse Laplace Transform is calculated using the formula ƒ(t) = (1/2πi) * ∫(c-i∞)^(c+i∞) F(s) * e^(st) ds, where ƒ(t) is the function in the time domain, F(s) is the function in the Laplace transform domain, and c is a constant chosen based on the properties of the function.

What are the properties of the Inverse Laplace Transform?

The properties of the Inverse Laplace Transform include linearity, time shifting, differentiation, integration, convolution, and initial and final value theorems. These properties allow for the manipulation and simplification of functions in the time domain using the Inverse Laplace Transform.

What are the applications of the Inverse Laplace Transform?

The Inverse Laplace Transform has many applications in engineering and physics, such as solving differential equations, analyzing electrical circuits, and understanding the behavior of systems over time. It is also used in control theory and signal processing to model and predict the behavior of dynamic systems.

What are some common misconceptions about the Inverse Laplace Transform?

One common misconception is that the Inverse Laplace Transform can be used to solve any type of function. In reality, it is only applicable to functions that have a Laplace transform. Another misconception is that it can only be used for linear systems, when in fact it can also be applied to nonlinear systems using the convolution property. Additionally, some may think that the Inverse Laplace Transform is just an algorithm, but it is actually a powerful mathematical tool used in various fields of science and engineering.

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