need help on Laplace Transform!

Hi all. I'm doing a laplace transform question using Heaviside functions. I can't solve the problem as it involves square of cosine when i try to integrate it.

The question is as below :
f(t) = [cos 2(t-4)]u(t-4)

I make it to become f(t) = [2cos(t-4)cos(t-4) - 1]u(t-4) and i cant proceed.

Anyone can give me some hints on this? Thanks.

kelvintc said:

Hi all. I'm doing a laplace transform question using Heaviside functions. I can't solve the problem as it involves square of cosine when i try to integrate it.

The question is as below :
f(t) = [cos 2(t-4)]u(t-4)

I make it to become f(t) = [2cos(t-4)cos(t-4) - 1]u(t-4) and i cant proceed.

Anyone can give me some hints on this? Thanks.

Click to expand...

Use the trig identity:

\(\displaystyle \large \cos(2A)=2 (\cos(A))^2-1\),

to reduce the square of the cosine, to a cosine of twice the argument.

RonL

ya i did that..
but the problem is that i dont know how to do the Laplace transform of square of cosine times unit step function.
cost*cost*u(t)

[/code]

kelvintc said:

ya i did that..
but the problem is that i dont know how to do the Laplace transform of square of cosine times unit step function.
cost*cost*u(t)

[/code]

Click to expand...

You don't do it that way, try something like:

\(\displaystyle \large \mathcal{L} \{ \cos(2(t-4)) u(t-4) \} =
\int\limits_0^\infty {\cos (2(t - 4))\,u(t - 4)e^{ - st} dt} \\
\ \ \ = \int\limits_4^\infty {\cos (2(t - 4))\,e^{ - st} dt} \\
\ \ \ = \int\limits_0^\infty {\cos (2\tau )\,e^{ - s(\tau + 4)} d\tau } \\
\ \ \ = e^{ - 4s} \int\limits_0^\infty {\cos (2\tau )\,e^{ - s\tau } d\tau}\\
\ \ \ = e^{ - 4s} \;\mathcal{L}(\cos (2t))\)

Which you can probably sort cut using known properties of the LT.

RonL

oh.. it looks so easy now.. thanks alot