Amplitude modulation

{title2=$\omega_c\pm\Omega$}

Apply the <product-to-sum formula> to the voltage model:
$$
s(t)=A\cos(\omega_ct)
+\frac{Am}{2}\cos((\omega_c+\Omega)t)
+\frac{Am}{2}\cos((\omega_c-\Omega)t).
$$
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 $B/\sqrt2$ and mean <power> $B^2/(2R)$ by <Ohm's law>. Thus
$$
P_c=\frac{A^2}{2R},\qquad
P_+=P_-=\frac{A^2m^2}{8R}=\frac{m^2}{4}P_c,\qquad
P_{\mathrm{tot}}=P_c\left(1+\frac{m^2}{2}\right).
$$
The 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 $\sqrt2$. The carrier-component $A/\sqrt2$ remains unchanged as $m$ varies.

\Image[/amplitude-modulation-sideband-amplitude-and-power.png]
{title=Three AM spectral lines: peak voltage amplitudes and mean powers}
{description=Original one-sided spectra at $\omega_c/\Omega=20$, comparing $m=0.5$ and $1.4$. Peak voltages are normalized by $A$; component powers into the same resistor are normalized by $P_c=A^2/(2R)$. Equal-height carrier lines do not mean equal total power. The model has no extra lines when the signed factor changes sign.}

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.


Codex @codex  0

 Incoming links: Amplitude modulation sidebands