1
$\begingroup$

I've got a sinusodial signal:

\begin{equation} \Delta I=\cos(\Delta \omega t + \varphi)). \end{equation}

and I would like to rewrite it as the sum of a cosine and a sine signal(without a phaseterm):

\begin{equation} \Delta I = \mathcal{M}* \sin(\Delta \omega t) + \mathcal{N}* \cos(\Delta \omega t) \end{equation}

My book ("Quantum Optics" by J.C. Garrison p. 293) does it, but I can't see how they got there.

If I perform a sin/cos transform I get the period of the signal as a factor: \begin{align} M(t) &= \mathcal{S}(\Delta I(\Delta\omega)) \\ &= 4E_L E_P \int_{-T/2+t}^{T/2+t}cos(\Delta \omega \tau + \varphi))\sin(\Delta\omega)d\tau\\ &= -4E_L E_P T \sin(\varphi) \\ \end{align}

I'm fairly certain, that I have to integrate over \begin{equation} \cos(\Delta \omega \tau + \varphi))\sin(\Delta\omega) = \frac{1}{2}\left[ \sin(\varphi) + \sin(2\Delta\omega + \varphi) \right] \end{equation} and $\cos(\Delta \omega \tau + \varphi))\cos(\Delta\omega)$ in order to remove the rapidly oscillating term $\sin(2\Delta\omega t)$

$\endgroup$
3
$\begingroup$

Why don't you just use the relation $\cos(x+y)=\cos x \cos y - \sin x \sin y$? So

$$\cos(\Delta \omega t + \varphi))=\cos(\Delta\omega t)\cos\varphi-\sin(\Delta\omega t)\sin\varphi$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.