Find the laplace transformation of the following:
I don't even know where to start< please Help>
Again I will only give hints:
You better do this using the definition of Laplace Transform.
To do it using properties you need three things
1)$\displaystyle L[e^{-3t}] = \delta(t - 3)$
2)$\displaystyle L[\int_{0}^{t} y(\tau) d\tau] = \frac{L[y(t)]}{s}$
3) Trickiest one $\displaystyle L[\frac{\sin 2t} {t}]$. This is the sinc function.Trying this separately is a good idea
Isomorphism has given some good hints. I'll give some different ones:
1. $\displaystyle LT \left[ e^{at} f(t) \right] = F(s - a)$ where F(s) = LT[f(t)]. This is one of the famous shift theorems.
In your problem, a = -3 and $\displaystyle f(t) = \int_0^t \frac{\sin (2\tau)}{\tau} \, d \tau$.
2. Same as Isomorphism's hint. In your problem $\displaystyle y(t) = \frac{\sin (2t)}{t}$.
3. $\displaystyle LT\left[ \frac{g(t)}{t} \right] = \int_{s}^{\infty} G(\tau) \, d \tau$ where G(s) = LT[g(t)]. In your problem $\displaystyle g(t) = \sin(2t)$.
4. $\displaystyle LT[\sin (2t)] = \frac{2}{s^2 + 2^2}$.
5. Therefore $\displaystyle LT\left[ \frac{\sin (2t)}{t}\right] = \int_{s}^{\infty} \frac{2}{\tau^2 + 2^2} \, d \tau = ..... $
I will point out that all of these results should be known to you. I can't help wondering if the learning curve of your self-study is too steep. Have you gone through - thoroughly - the basics?
Yes, I thought of 2 sinc(2x) later
But, I think getting it's transform from basics would be difficult for this member. And I don't think the transform of sinc (x) would be on the tables this member is using, which would make the time compression property unusable. Using the appropriate operational theorem is the most expeditious approach, I think.
What would be interesting is to see what tables this member will be using in the exam ......
By the way, please don't take this reply as a criticism of your suggestions.