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Double-tone AM expression, bandwidth and modulation index

Short answer

A double-tone AM signal is s(t) = A_c[1+μ₁cosω₁t+μ₂cosω₂t]cosω_c t. Its bandwidth is twice the highest modulating frequency, B = 2f_max. Modulation indices are μ₁=A_{m1}/A_c, μ₂=A_{m2}/A_c, and the net index is μ_t = √(μ₁²+μ₂²).

Derive the expression of double-tone AM, and define BW and modulation indices.

Double-tone AM is amplitude modulation in which the message consists of two sinusoids:

m(t)=Am1cos⁡ω1t+Am2cos⁡ω2t,f1<f2≪fcm(t)=A_{m1}\cos\omega_1t+A_{m2}\cos\omega_2t,\qquad f_1<f_2\ll f_c

Expression

With carrier Accos⁡ωctA_c\cos\omega_ct, the standard AM wave is

s(t)=[Ac+Am1cos⁡ω1t+Am2cos⁡ω2t]cos⁡ωct=Ac[1+μ1cos⁡ω1t+μ2cos⁡ω2t]cos⁡ωct\begin{aligned} s(t)&=[A_c+A_{m1}\cos\omega_1t+A_{m2}\cos\omega_2t]\cos\omega_ct\\ &=A_c[1+\mu_1\cos\omega_1t+\mu_2\cos\omega_2t]\cos\omega_ct \end{aligned}

where μ1=Am1/Ac\mu_1=A_{m1}/A_c and μ2=Am2/Ac\mu_2=A_{m2}/A_c are the individual modulation indices. Expanding:

s(t)=Accos⁡ωct+μ1Ac2[cos⁡(ωc+ω1)t+cos⁡(ωc−ω1)t]+μ2Ac2[cos⁡(ωc+ω2)t+cos⁡(ωc−ω2)t]\begin{aligned} s(t)&=A_c\cos\omega_ct\\ &+\frac{\mu_1A_c}{2}\left[\cos(\omega_c+\omega_1)t+\cos(\omega_c-\omega_1)t\right]\\ &+\frac{\mu_2A_c}{2}\left[\cos(\omega_c+\omega_2)t+\cos(\omega_c-\omega_2)t\right] \end{aligned}

So the spectrum has five lines: the carrier at fcf_c, and side frequencies at fc±f1f_c\pm f_1 and fc±f2f_c\pm f_2.

                         Ac
                         |
                         |
   mu2Ac/2   mu1Ac/2     |     mu1Ac/2   mu2Ac/2
      |         |        |        |         |
 -----+---------+--------+--------+---------+-----> f
    fc-f2     fc-f1      fc     fc+f1     fc+f2

Bandwidth

The bandwidth is set by the highest modulating frequency:

B=(fc+f2)−(fc−f2)=2f2=2fmaxB=(f_c+f_2)-(f_c-f_2)=2f_2=2f_{max}

Modulation indices

  • Individual indices: μ1=Am1Ac\mu_1=\frac{A_{m1}}{A_c}, μ2=Am2Ac\mu_2=\frac{A_{m2}}{A_c}.
  • Net (total, effective) index: the single-tone index that gives the same total sideband power:
Pt=Pc(1+μ122+μ222)=Pc(1+μt22) ⇒ μt=μ12+μ22P_t=P_c\left(1+\frac{\mu_1^2}{2}+\frac{\mu_2^2}{2}\right)=P_c\left(1+\frac{\mu_t^2}{2}\right)\ \Rightarrow\ \mu_t=\sqrt{\mu_1^2+\mu_2^2}

For no over-modulation, μt≤1\mu_t\le1 (strictly, μ1+μ2≤1\mu_1+\mu_2\le1 guarantees that the envelope never goes negative).

Example: μ1=0.3\mu_1=0.3, μ2=0.4\mu_2=0.4 gives μt=0.5\mu_t=0.5.

✓ Answer checked by Claude Opus 5.5 (high effort)

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