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Short note: Modulation index

Short answer

In AM, the modulation index μ = Am/Ac = (Amax−Amin)/(Amax+Amin), and efficiency η = μ²/(2+μ²). In FM and PM, β = Δf/fm or β = kpAm, setting the number of sidebands and bandwidth by Carson’s rule. In AM, transmitters aim for μ near 100% to maximise sideband power without overmodulation.

Write a short note on modulation index.

The modulation index tells how much a carrier parameter is changed by the message.

In AM

Ratio of message amplitude to carrier amplitude:

μ=AmAc=Amax−AminAmax+Amin\mu = \frac{A_m}{A_c} = \frac{A_{max} - A_{min}}{A_{max} + A_{min}}
  • μ<1\mu < 1: under-modulation, envelope follows the message.
  • μ=1\mu = 1: 100% modulation, maximum efficiency (33.3%) without distortion.
  • μ>1\mu > 1: over-modulation; envelope distortion, extra sidebands, envelope detector fails.

For several tones, μt=μ12+μ22+…\mu_t = \sqrt{\mu_1^2 + \mu_2^2 + \dots}.

Power relation: Pt=Pc(1+μ2/2)P_t = P_c(1 + \mu^2/2). A higher μ\mu puts more power in the sidebands and gives a stronger, clearer received signal.

Example: Amax=8A_{max} = 8 V, Amin=2A_{min} = 2 V gives μ=6/10=0.6\mu = 6/10 = 0.6 (60%).

In FM and PM

β=Δffm (FM),β=kpAm (PM)\beta = \frac{\Delta f}{f_m} \ (\text{FM}), \qquad \beta = k_pA_m \ (\text{PM})

β\beta is the peak phase deviation in radians; it decides the number of significant sidebands and the bandwidth (Carson's rule B=2(β+1)fmB = 2(\beta + 1)f_m). Unlike AM, β\beta can exceed 1. Example: Δf=75\Delta f = 75 kHz, fm=15f_m = 15 kHz gives β=5\beta = 5.

✓ Answer checked by Claude Opus 5.5 (high effort)

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