Amplitude modulation sidebands (${\omega }_c\pm \Omega$)

Apply the product-to-sum formula to the voltage model:The carrier wave component has peak voltage amplitude $A$; each AM sideband has peak voltage amplitude $Am/2$. These are temporal frequency spectrum components, unlike the spatial sidebands used in pattern stability. The separation $\Omega >0$ makes the three positive angular frequencies distinct, and $\Omega <{\omega }_c$ keeps the lower one positive. The span between sidebands is $2\Omega$, or $2f_m$ in hertz when $f_m=\Omega /(2\pi )$; this is the single-tone span, not a universal signal bandwidth for arbitrary messages.

A one-sided cosine-amplitude display assigns $A$ to the carrier line. A two-sided complex-exponential expansion instead assigns $A/2$ at each of $\pm {\omega }_c$, and $Am/4$ at each of the four sideband locations. The signal is identical; confusing the conventions creates a factor-of-two error. An ideal infinite-duration sinusoid gives a spectral line. A finite recording and finite analyser resolution broaden the displayed peak.

For a voltage applied across an ideal resistor of electrical resistance $R>0$, a component of peak voltage $B$ has root mean square voltage $B/\sqrt{2}$ and mean power $B^2/(2R)$ by Ohm's law. ThusThe proof of additivity is to average $s(t{)}^2/R$ over a long time: cross-products of distinct frequencies have zero mean. Consequently the whole signal's root mean square voltage is $A\sqrt{(1+m^2/2)/2}$, not simply its maximum wave envelope divided by $\sqrt{2}$. The carrier-component $A/\sqrt{2}$ remains unchanged as $m$ varies.

With $A=2\mathrm{V}$, $R=50\Omega$ and $m=0.5$, the carrier supplies $40\mathrm{mW}$ and each $0.5\mathrm{V}$ peak sideband supplies $2.5\mathrm{mW}$, giving $45\mathrm{mW}$ total. At $m=1.4$ each sideband is $1.4\mathrm{V}$ peak and supplies $19.6\mathrm{mW}$, giving $79.2\mathrm{mW}$ total. Neither case changes the carrier-component power.

The fraction of mean power in the message-bearing sidebands is $m^2/(2+m^2)$. It is $1/9$ at $m=0.5$ and $1/3$ at $m=1$. Conventional AM spends the remaining power on a carrier that enables an inexpensive envelope detector. Increasing $m$ beyond $1$ increases this fraction in the linear model but breaks ordinary envelope recovery. Removing the carrier gives suppressed-carrier amplitude modulation and changes the receiver requirement.