Multiplying some signal, a function of time, $m(t)$ by a cosine $\cos{\omega' t}$ causes a shift in frequency of $m(t)$, by $\pm\omega'$.
But what about multiplication by a sine wave, such $m(t)\cdot\sin{\omega' t}$ – or, what is the same, if the cosine multiplier was some $\cos(\omega' t + \phi)$?
The $\pm\omega'$ shift comes from the Fourier transform $\mathfrak{F}:\{\cos\omega'\}\rightarrow \pi\big(\delta(\omega-\omega')+\delta(\omega+\omega')\big)$, so if there was a non-zero phase shift $\phi$, this would be further multiplied by $e^{j\omega\phi}$?
I'm having trouble with the physical interpretation of this, doesn't it mean that the frequency shift is the same, but that there is a scale in magnitude, that is a function of the frequency as well as the phase? So the $\pm$ shifted frequency components now have different magnitudes?
Best Answer
If you write $x(t)=\cos(\omega_0t+\phi)$ as
$$x(t)=\frac12\left[e^{j\omega_0t}e^{j\phi}+e^{-j\omega_0t}e^{-j\phi}\right]$$
it's easy to see that its Fourier transform is
$$X(\omega)=\pi\left[e^{j\phi}\delta(\omega-\omega_0)+e^{-j\phi}\delta(\omega+\omega_0)\right]$$
If $M(\omega)$ is the Fourier transform of $m(t)$, we have
$$\mathcal{F}\{m(t)x(t)\}=\frac{1}{2\pi}M(\omega)*X(\omega)=\\ =\frac12\left[e^{j\phi}M(\omega-\omega_0)+e^{-j\phi}M(\omega+\omega_0)\right]$$
So you always get the shifted spectra, but they are multiplied with constant factors (which are complex conjugates of each other). Note that for $\phi=-\pi/2$ these factors are $\pm j$, which is what you get when multiplying the original signal with $\sin\omega_0t$.