AM wave diagram, modulation index; why near 100%

AM wave and modulation index

 v(t)
 Amax -    __                  __
          /  \     envelope   /  \
 Ac   - -/----\--------------/----\---
 Amin -  |     \__        __/      |
         ||||||||||||||||||||||||||||   carrier
   0  ---+--------------------------------> t
         ||||||||||||||||||||||||||||
 -Amin-   \  /\           /\    /
 -Amax-    \/  envelope (mirror)
         |<----- 1/fm ------>|
 Peak-to-peak at crest = 2Amax, at trough = 2Amin

AM wave: $s(t) = A_c[1 + \mu\cos\omega_mt]\cos\omega_ct$, with $\mu = A_m/A_c$.

Derivation of $\mu$ from the waveform. The envelope is $A_c(1 + \mu\cos\omega_mt)$. It is largest when $\cos\omega_mt = 1$ and smallest when $\cos\omega_mt = -1$:

$$\begin{aligned} A_{max} &= A_c(1 + \mu) \\ A_{min} &= A_c(1 - \mu) \end{aligned}$$

Adding and subtracting:

$$\begin{aligned} A_{max} + A_{min} &= 2A_c \Rightarrow A_c = \frac{A_{max} + A_{min}}{2} \\ A_{max} - A_{min} &= 2\mu A_c \Rightarrow \mu = \frac{A_{max} - A_{min}}{2A_c} \end{aligned}$$ $$\mu = \frac{A_{max} - A_{min}}{A_{max} + A_{min}}$$

Also $A_m = \dfrac{A_{max} - A_{min}}{2}$, so $\mu = A_m/A_c$. In terms of peak-to-peak values read on an oscilloscope, $\mu = \dfrac{V_{pp,max} - V_{pp,min}}{V_{pp,max} + V_{pp,min}}$.

Why operate close to 100%

• Power efficiency: $\eta = \dfrac{\mu^2}{2 + \mu^2}$ rises with $\mu$: 1.96% at $\mu = 0.2$, 15.3% at $\mu = 0.6$, 33.3% at $\mu = 1$. More of the transmitted power carries information.
• Better signal at the receiver: detected audio amplitude is proportional to $\mu A_c$, so higher $\mu$ gives a stronger output and higher S/N for the same carrier power.
• But $\mu$ must not exceed 1: over-modulation distorts the envelope, generates spurious sidebands (splatter) and causes adjacent-channel interference. So transmitters are run as close to 100% as possible without exceeding it.