What do we do at the receiver? How can we demodulate this signal? Once way would be to pass it through a amplitude modulator using a complex exponential $e^{-i\omega_0 t}$, and then we would get the signal shifted over to the left. We can then pass it through a low-pass filter with a pass band amplification of 2 to get the original height back. But, what if we had an oscillator using a cosine function at the receiving end instead of a complex exponential?

The following scheme is called \emph{synchronous demodulation}, and is when we have an identical cosine at the receiving end. Looking at what happened upon transmission, the signal was shifted $\pm \omega_0$ and the amplitude was cut in half. Applying this again, we would get a signal $q(t) = y(t)\cos(\omega_0 t)$, where we shift the shifted signal $\pm \omega_0$. This gives us three disjoint triangular components in the frequency spectrum. The first shift to the left $\omega_0$ gives us two triangles with a height of $1/4$ centered at $-2\omega_0$ and $0$. The shift to the right gives us two triangles with a height of $1/4$ centered at $2\omega_0$ and $0$. The two triangles centered at 0 add to form the original triangle halfed. So we can now put this signal $q(t)$ through a low-pass filter and amplify the signal by 2 to retrieve the original signal.

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At the receiving end the local oscillator has potentially a phase difference and frequency difference relative to the transmitter. This demodulation scheme assumes the same frequency and phase at the sender and receiver. What happens if at the receiver we have $q(t) = y(t)\cos((\omega_0+\Delta\omega)t + \theta)$?

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\begin{align*}
y(t) &= x(t)\cos(\omega_0 t) \\
r(t) &= y(t)\cos((\omega_0+\Delta\omega) t + \theta) \\
r(t) &= x(t)\cos(\omega_0 t)\cos((\omega_0+\Delta\omega) t + \theta) \\
\end{align*}

We can use the trigonometric identity:

$$ \cos\alpha \cos\beta = \frac{1}{2}\lr{\cos\lr{\alpha+\beta}+\cos\lr{\alpha-\beta}} $$

\begin{align*}
r(t) &= x(t)\cos(\omega_0 t)\cos((\omega_0+\Delta\omega) t + \theta) \\
r(t) &= \frac{x(t)}{2} \lr{\cos((2\omega_0 + \Delta\omega)t + \theta) + \cos(\Delta \omega t + \theta)} \\
\end{align*}

After we pass this signal through a low pass filter, we get rid of the first term containing the frequency $\omega_0+\Delta\omega)$. So if $g(t)$ denotes our signal after the low pass filter, then $g(t) = x(t)\cos(\Delta\omega t + \theta)$. We hope this is equal to $x(t)$.

Lets consider the different cases:

1) no frequency jitter ($\Delta\omega = 0$), such that $g(t) = x(t)\cos(\theta)$


If $\theta \approx \frac{\pi}{2}$, you get an extremely weak baseband. When $\theta = 0$, you have a synchronous demodulation scheme.


2) the phase is zero, but there is frequency jitter, then $g(t) = x(t)\cos(\Delta\omega t)$.


This is amplitude modulation with a very slow carrier signal, where we will have overlap in the frequency domain, causing echos and other problems.


Amplitude Modulation - Receiving