# Easy laplace transform question

#### ihaveaquestion

Joined May 1, 2009
314
The laplace transform of f(t) = (2t+4)u(t) = 2/s^2 + 4/s

I can see that the laplace of f(t) = (2t) = (1/2)*(1/(s/2)^2) = 2/s^2

but I'm stumped at what I do with the 4... I see nothing like it on the laplace transform table... matlab gives the laplace of (t+4), a broken down version, to be 1/s^2 + 4/s. Again I see the 1/s^2 part from the t, but I don't se how they get 4/s for the 4 part... the laplace of u(t) is 1/t, is this why? Some explanation would be appreciated, thanks.

#### t_n_k

Joined Mar 6, 2009
5,455
The Laplace transform of a constant k is simply k/s. So the LT of a constant 4 is 4/s as you have indicated.

For the purposes of performing the Laplace transform, the Heavyside Function u(t) constrains the value of the time domain function it acts upon such that, for t<0 the function is zero and for t>=0 the function is the multiplicand of u(t). Presumably this is something that a rigorous application of the underlying mathematics demands - maybe it's all a bit academic for us ordinary folk.

#### ihaveaquestion

Joined May 1, 2009
314

#### t_n_k

Joined Mar 6, 2009
5,455
Laplace Transform of sin(3t) is 3/(s^2+9) - not what you have written.

Laplace Transform of cos(3t) is s/(s^2+9) - again not what you wrote.

Where did you get those forms you wrote - from a table of transfrm pairs?

Laplace Transform of f(t)=6sin(3t)+8cos(3t) is

F(s) = 6*3/(s^2+9) + 8*s/(s^2+9)
= 18/(s^2+9) +8s/(s^2+9)
= (8s+18)/(s^2+9)

#### ihaveaquestion

Joined May 1, 2009
314

#### t_n_k

Joined Mar 6, 2009
5,455
You made a fairly simple error near the end. Should be like this .....

1/2 X (s-2+s+6)/(s^2+4s-12) = 1/2X(2s+4)/(s^2+4s-12)=(s+2)/(s^2+4s-12)