Question

find the laplace trasfrom of

1. ft(t) = h(t-1)cos(t-1)

find the function laplace trasfrom of

2. f(s) = 1/s-2 - 4s/(s^2+2^2)^2

Answer 1

Use that:
\[\mathcal{L} \left[ f(t-t_0)\right]=e^{-s t_0}F(s); F(s)=\mathcal{L} \left\{ f(t) \right\}\]

Answer 2

For the first one that is.

Answer 3

hi , for1. i did a=1 gcos(t-1) = which gives me e^-s*se^-s/s^3+s is that correct?

Answer 4

Then finding the inverse laplace of the second one use:
\[\mathcal{L}(e^{\lambda t})=\frac{1}{s-\lambda}\]
And:
\[\mathcal{L}(\cos(\beta t))=\frac{s}{s^2+\beta^2}; \mathcal{L}(\sin(\beta t))=\frac{\beta}{s^2+\beta^2}; \]\[\mathcal{L}^{-1}(F(s)G(s))=g(t)*f(t)=\int\limits_0^t g(t-\lambda)f(\lambda)d \lambda\]

Answer 5

Is h a constant?

Answer 6

what confuses me is the cos(t-1) ? does is become g(t-1) or g(cos(t-1))?

Answer 7

yes h is constant

Answer 8

Then what you have is:
\[f(t)=t;g(t)=\cos(t) \implies f(t-1)g(t-1)=(t-1)\cos(t-1)\]

Answer 9

Pull the h out front of course.

Answer 10

Then you can use the shifting theorem to say that's just:
\[e^{-s} \mathcal{L}(f(t)g(t))\]

Answer 11

Then find the laplace of f(t)g(t)

Answer 12

so i was right it was cos(t-1) =g(t) , so i need to take the laplace of this first?

Answer 13

Either way you slice it you're going to have to take the laplace transform of:
\[\int\limits_0^{\infty}t \cos(t)e^{-st}dt\]

Answer 14

right laplace of (t-1)= e^-s/s and cos= s/s^2+1 do i mulitply these togther to get gT is it correct?

Answer 15

Well:
\[\mathcal{L}(f(t)g(t)) \ne F(s) G(s)\]

Answer 16

But you don't need the laplace of cos(t-1)
You need the laplace of (t-1)cos(t-1)