Laplace Transform Help: Solving for Unknown Variables in Electrical Circuits

In summary, the conversation is about a person who attempted a Laplace transform but got stuck and is asking for help. They provide the relevant equations and values, and attempt to solve for the inverse transform. They are then given a suggestion to find constants $\alpha$ and $\omega$ in order to find the inverse transform. The expert confirms that their working up to that point is correct.
  • #1
Sophie1
3
0
Evening All

I have had a go at a laplace transform and got stuck.

$$\frac{d^2v}{dt^2}+\frac R L \d v t+\frac 1{LC}v=\frac 1{LC}V_0$$

$$R=12 \Omega, L=0.16H, C=10^{-4}F, V_0=6V, v(0)=0, v'(0)=0$$

so subbing these in i get
$$\mathscr L \left[ \frac {d^2v}{dt^2}+75\d v t+62500 v \right]=\mathscr L[375000]$$

$$S^2X-Sx(0)-x'(0)+75(SX-x(0))+62500v=\frac{375000}{S}$$

subbing $v(0)=0, x'(0)=0$

$$S^2X+75SX+62500X=\frac{375000}{S}$$

$$X(S^2+75S+62500)=\frac{375000}{S}$$

$$X= \frac{375000}{S(S^2+75S+62500)}$$

not I'm stuck as i can't an inverse form to change it back into.

can someone help?
 
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  • #2
Sophie said:
Evening All

I have had a go at a laplace transform and got stuck.

$$\frac{d^2v}{dt^2}+\frac R L \d v t+\frac 1{LC}v=\frac 1{LC}V_0$$
$$R=12 \Omega, L=0.16H, C=10^{-4}F, V_0=6V, v(0)=0, v'(0)=0$$
(snip)
$$X= \frac{375000}{S(S^2+75S+62500)}$$

not I'm stuck as i can't an inverse form to change it back into.

Hi Sophie! Welcome to MHB! ;)

Suppose we could write $X$ as:
$$X = \frac A S + \frac {BS + C}{(s+\alpha)^2 + \omega^2}$$
for some yet to be determined $A,B,C,\alpha,\omega$.
Could you then find the inverse transform? (Wondering)
 
  • #3
could you give me a pointer of where to begin. I have not used or even see α,ω. With respect to laplace transform.
 
  • #4
Sophie said:
could you give me a pointer of where to begin. I have not used or even see α,ω. With respect to laplace transform.

From wiki:
$$\mathcal L^{-1}\left[\frac{s+\alpha}{(s+\alpha)^2+\omega^2}\right] = e^{-\alpha t} \cos(\omega t) \cdot u(t)$$

So if we can find such constants $\alpha,\omega$ we can find the inverse transform.
And we have that:
$$S^2+75S+62500 = (S+\alpha)^2 + \omega^2$$
so we just need to figure out what these $\alpha,\omega$ are... (Thinking)
 
  • #5
Would you be able to confirm that my working out upto the point i asked is correct as I'm not finding the answer. I will carry on trying.
 
  • #6
Sophie said:
Would you be able to confirm that my working out upto the point i asked is correct as I'm not finding the answer. I will carry on trying.

Your working up to:
$$X= \frac{375000}{S(S^2+75S+62500)}$$
is correct (disregarding some inconsistencies in $v$, $V$, and $x$ in the intermediate steps).
(Nod)
 

FAQ: Laplace Transform Help: Solving for Unknown Variables in Electrical Circuits

What is a Laplace transform and why is it important in science?

A Laplace transform is a mathematical operation that converts a function of time into a function of complex frequency. It is important in science because it allows us to solve differential equations and analyze systems in the frequency domain, which can provide valuable insights and predictions about their behavior.

How do I perform a Laplace transform?

To perform a Laplace transform, you need to have a function of time and use the appropriate integral formula. You can also use tables or software to assist with the calculations. It is important to follow the proper steps and consider the conditions for convergence to ensure accurate results.

What are the applications of Laplace transform in different fields of science?

Laplace transform has a wide range of applications in various fields of science, including physics, engineering, and mathematics. It is commonly used in circuit analysis, control systems, signal processing, and vibration analysis, among others.

How is Laplace transform related to Fourier transform?

Laplace transform and Fourier transform are closely related, as they both involve converting a function from the time domain to the frequency domain. However, Laplace transform is more general and can handle a wider range of functions, while Fourier transform is limited to periodic functions.

What are the benefits of using Laplace transform in scientific research?

The use of Laplace transform in scientific research allows for a more efficient and accurate analysis of systems and phenomena. It helps to simplify complex problems and provides a powerful tool for modeling and predicting the behavior of various systems, making it an essential tool for scientists and researchers in many fields.

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