# Laplace Transform

1. Oct 23, 2004

### rishid

Hello,

Anyone good source on how to do Laplace transforms? My teacher is really really bad and I have a test next week on it. I have no clue on how to do any of them. Example problems attached.

Thanks for any help at all.

#### Attached Files:

• ###### laplace.pdf
File size:
21.5 KB
Views:
98
2. Oct 23, 2004

### Galileo

The idea is to take the Laplace transform on both sides of the differential equation and use the properties of the laplace transform of derivatives to turn your differential equation into an algebraic equation.
Then, if you can find the inverse laplace transform of f(t) you have solved the equation.
You don't need to integrate if you may use the integral/laplace transform-tables.

3. Oct 23, 2004

### rishid

Why do I need to find the inverse laplace transform?

You got any idea what the I represents in problem 1?

Thanks for the help.

Last edited: Oct 23, 2004
4. Oct 24, 2004

### ReyChiquito

When you take the transform, you are no longer solving the differential equation for $x(t)$, you are now solving a differential equation for the Laplace transform. After you solve that equation, you need to reverse the transform to find $x(t)[/tex], wich is the original problem. The Laplace transform is given by the formula $$\tilde{f}(s)=\int_{0}^{\infty}e^{-st}f(t)dt$$ for example, the laplace transform for the derivative is given by $$\widetilde{f'}(s)=\int_{0}^{\infty}e^{-st}f'(t)dt=-sf(0)+s\int_{0}^{\infty}e^{-st}f(t)dt$$ so $$\widetilde{f'}(s)=-sf(0)+s\tilde{f}(s)$$ using this result $$\widetilde{f''}(s)=-sf'(0)+s\widetilde{f'}(s)=-sf'(0)-s^{2}f(0)+s^{2}\tilde{f}(s)$$ now, you only need to transform the whole equation, solve for [itex]\tilde{x}(s)$ and find the inverse transform as Galileo said (using tables). For further reference check Boyce DiPrima ode's book.

Ah... and your equation represents an harmonic oscilator under the influence of an external force $f(t)$. The $I$ must be a constant. The amplitude of a pulse.

Lets do 1. for example.

given $x(0)=0,x'(0)=0$ implies that

$$\widetilde{x'}(s)=s\tilde{x}(s),\widetilde{x''}(t)=s^{2}\tilde{x}(s)$$

and

$$\mathcal{L}[\delta(t-c)]=\int_{0}^{\infty}e^{-st}\delta(t-c)dt$$

wich by definition of the delta function $f(y)=\int f(x)\delta(x-y)dx$ yields

$$\mathcal{L}[\delta(t-c)]=e^{-sc}$$

so $\mathcal{L}[\delta(t)]=1$. Now, transforming the whole equation,

$$\widetilde{mx''(t)+kx(t)}=\widetilde{I\delta}(t)$$

due linearity

$$\widetilde{mx''(t)+kx(t)}=m\widetilde{x''}(t)+k\tilde{x}(t)=I\widetilde{\delta}(t)$$

substituting

$$(ms^{2}+k)\tilde{x}(s)=I$$

so

$$\tilde{x}(s)=\frac{I}{m}\frac{1}{s^{2}+\omega^{2}}$$

and given the fact that (try calculating this one, good exercise)

$$\mathcal{L}[sen(\omega t)]=\frac{\omega}{s^{2}+\omega^{2}}$$

implies

$$\tilde{x}(s)=\mathcal{L}[x(t)]=\frac{I}{m\omega}\mathcal{L}[sen(\omega t)]$$

Inverting

$$x(t)=\frac{I}{m\omega}sen(\omega t)$$

There is some details missing, but im too tired and i think this show the main point. You need to use your initial conditions and the homogeneous solution. Anyone here is willing to complete the exercise?

Last edited: Oct 24, 2004