Q.Due to economic reasons, only the upper sideband of an AM wave is transmitted, but at the receiving station, there is a facility for generating the carrier. Show that if a device is available which can multiply two signals, then it is possible to recover the modulating signal at the receiver station.

Setup. Suppose the original message is $m(t)=A_m\cos(\omega_m t)$ and only the upper side band (USB) of the AM signal is transmitted:

where $\omega_c$ is the carrier's angular frequency (the carrier itself is NOT transmitted, only the USB).

Step 1 — Locally generate the carrier.

At the receiver, generate a local signal at exactly the carrier frequency: $\cos(\omega_c t)$.

Step 2 — Multiply the received USB signal by this local carrier.

Step 3 — Apply the product-to-sum identity $\cos X \cos Y = \tfrac{1}{2}[\cos(X-Y)+\cos(X+Y)]$, with $X=(\omega_c+\omega_m)t$ and $Y=\omega_c t$:

Step 4 — Low-pass filter.

The first term, $\frac{A}{2}\cos(\omega_m t)$, oscillates at the (low) message frequency $\omega_m$. The second term oscillates at approximately $2\omega_c$, far above $\omega_m$ (since $\omega_c \gg \omega_m$ for a radio carrier). Passing the product through a low-pass filter with cutoff between $\omega_m$ and $2\omega_c$ removes the high-frequency term entirely, leaving

$\frac{A}{2}\cos(\omega_m t)$ …

Multiply the received USB term $\cos[(\omega_c+\omega_m)t]$ by a locally regenerated carrier $\cos(\omega_c t)$; expand with the product-to-sum identity to get a low-frequency term at $\omega_m$ and a high-frequenc …

• Forgetting that the local carrier must match the transmitter's carrier frequency exactly (and ideally phase) — this is why it's called synchronous detection.
• Skipping the low-pass filter step and leaving both the $\omega_m$ and $2\omega_c$ terms in the 'recovered' signal. …