1. ## Laplace/ heaviside help

My problem is regarding this function:

$\displaystyle H(t-pi/2) * cos(3t)$

and Im suppose to write the Laplace transformation of this product using the theorem:
$\displaystyle L[H(t-c) f(t-c)] = e^{-cs}*L(f)$

I understand the principle behind this problem, and I know the answer is $\displaystyle e^{-pi*s/2}$ * L(cos(3t) in terms of t-pi/6). Its just I can't figure out how to rewrite $\displaystyle cos(3t)$ in terms of $\displaystyle t-pi/2$. I tried to do cos$\displaystyle (2t + t),$ use the addition property, and end up getting stuck with a $\displaystyle 1-4sin(t)^2$ that I dont know what to do with. Can someone help me out?

2. Originally Posted by mike10003
My problem is regarding this function:

$\displaystyle H(t-pi/2) * cos(3t)$

and Im suppose to write the Laplace transformation of this product using the theorem:
$\displaystyle L[H(t-c) f(t-c)] = e^{-cs}*L(f)$

I understand the principle behind this problem, and I know the answer is $\displaystyle e^{-pi*s/2}$ * L(cos(3t) in terms of t-pi/6). Its just I can't figure out how to rewrite $\displaystyle cos(3t)$ in terms of $\displaystyle t-pi/2$. I tried to do cos$\displaystyle (2t + t),$ use the addition property, and end up getting stuck with a $\displaystyle 1-4sin(t)^2$ that I dont know what to do with. Can someone help me out?
Recall that $\displaystyle \sin\!\left(\tfrac{\pi}{2}-\theta\right)=\cos\theta$.

So $\displaystyle \cos\!\left(3t\right)=-\sin\!\left(3t-\tfrac{\pi}{2}\right)$

Now note that $\displaystyle \sin\!\left(3t-\tfrac{\pi}{2}\right)=\sin\!\left(3t-\tfrac{3\pi}{2}+\pi\right)=-\sin\!\left(3t-\tfrac{3\pi}{2}\right)=-\sin\!\left(3\left[t-\tfrac{\pi}{2}\right]\right)$

Therefore $\displaystyle \cos(3t)=\sin\!\left(3\left[t-\tfrac{\pi}{2}\right]\right)=\left.\sin(3t)\right|_{t\to t-\frac{\pi}{2}}$

Does this make sense?

3. Yes that is a great help! Didn't think to use those angle identities.

So the other problem related to this is:

$\displaystyle H(t-pi/6)*sin(2t)$

Using your guide as a reference, I did this
$\displaystyle sin(2t)=cos(pi/2-2t)$
$\displaystyle =cos(-2t+pi/2)$
$\displaystyle =cos(-2t+pi/3 + pi/6)$
$\displaystyle =cos(-2(t-pi/6)+pi/6))$

Here I wasnt sure what to do, so I split it up using the addition formula.

$\displaystyle =cos(-2(t-pi/6))cos(pi/6) - sin(-2(t-pi/6))sin(pi/6)$
$\displaystyle =\sqrt(3)/2*cos(-2(t-pi/6)) - 1/2*sin(-2(t-pi/6))$

Then converting to laplace, I get this

$\displaystyle =\sqrt3 /2*(s/s^{2}+4) - 1/2*(-2/s^{2}+4)$
$\displaystyle =(\sqrt3 s+2)/(2*(s^{2}+4))$
Then multiply that by the L of the heaviside, $\displaystyle e^{-pi*s/6}$

However, the answer in the back of the book is
$\displaystyle (1/2)e^{-pi*s/6}*(4+\sqrt3 s^{2})/(s^{2}+4)$

Am I doing something wrong? I dont know where they got a s^2 in the numerator, since the laplace of cos is s/(s^2+a^2) and sin is a^2/(s^2+a^2).