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Two-tone AM: equation, index, power, spectrum

Short answer

AM signal: s(t)=[20+10cos(2π1000t)+15cos(π1000t)]cos(3π10⁶t). Modulation indices μ1=0.5, μ2=0.75; total μt=√(0.5²+0.75²)=0.901. Carrier power Pc=200W (1Ω), total power Pt=Pc(1+μt²/2)=281.25W. Spectrum: carrier at 1.5MHz, sidebands at ±1kHz (5V) and ±500Hz (7.5V).

The signals m(t) = 10cos2×10³πt + 15cos1×10³πt, c(t) = 20cos3×10⁶πt are applied to AM. Find a) The equation of the resulting signal. b) Modulation index c) Total power d) Draw spectrum

Given:

m(t)=10cos⁡(2π 1000 t)+15cos⁡(2π 500 t),c(t)=20cos⁡(2π×1.5×106t)m(t) = 10\cos(2\pi\,1000\,t) + 15\cos(2\pi\,500\,t), \quad c(t) = 20\cos(2\pi\times1.5\times10^6t)

So Am1=10A_{m1} = 10 V at fm1=1f_{m1} = 1 kHz, Am2=15A_{m2} = 15 V at fm2=500f_{m2} = 500 Hz, Ac=20A_c = 20 V, fc=1.5f_c = 1.5 MHz. Assume standard AM with ka=1k_a = 1: s(t)=[Ac+m(t)]cos⁡ωcts(t) = [A_c + m(t)]\cos\omega_ct, and R=1 ΩR = 1\ \Omega.

a) Equation of AM signal

s(t)=[20+10cos⁡(2π103t)+15cos⁡(π103t)]cos⁡(3π106t)=20[1+0.5cos⁡(2π103t)+0.75cos⁡(π103t)]cos⁡(3π106t)\begin{aligned} s(t) &= [20 + 10\cos(2\pi10^3t) + 15\cos(\pi10^3t)]\cos(3\pi10^6t) \\ &= 20[1 + 0.5\cos(2\pi10^3t) + 0.75\cos(\pi10^3t)]\cos(3\pi10^6t) \end{aligned}

Expanded:

s(t)=20cos⁡2π(1.5M)t+5cos⁡2π(1.501M)t+5cos⁡2π(1.499M)t+7.5cos⁡2π(1.5005M)t+7.5cos⁡2π(1.4995M)t\begin{aligned} s(t) = {} & 20\cos 2\pi(1.5\text{M})t \\ & + 5\cos2\pi(1.501\text{M})t + 5\cos2\pi(1.499\text{M})t \\ & + 7.5\cos2\pi(1.5005\text{M})t + 7.5\cos2\pi(1.4995\text{M})t \end{aligned}

b) Modulation index

μ1=1020=0.5,μ2=1520=0.75\mu_1 = \frac{10}{20} = 0.5, \quad \mu_2 = \frac{15}{20} = 0.75 μt=0.52+0.752=0.8125=0.9014\mu_t = \sqrt{0.5^2 + 0.75^2} = \sqrt{0.8125} = 0.9014

c) Total power

Pc=Ac22=4002=200 WPt=Pc(1+μt22)=200(1+0.81252)=281.25 W\begin{aligned} P_c &= \frac{A_c^2}{2} = \frac{400}{2} = 200\ \text{W} \\ P_t &= P_c\left(1 + \frac{\mu_t^2}{2}\right) = 200\left(1 + \frac{0.8125}{2}\right) = 281.25\ \text{W} \end{aligned}

(Efficiency =81.25/281.25=28.9%= 81.25/281.25 = 28.9\%.)

d) Spectrum

 A (V)
 20 |                  |
    |                  |
7.5 |         |        |        |
  5 |   |     |        |        |     |
    +---+-----+--------+--------+-----+---> f (MHz)
     1.499 1.4995    1.5     1.5005 1.501

Bandwidth =2×1= 2 \times 1 kHz =2= 2 kHz.

Answer: μt=0.901\mu_t = 0.901, Pt=281.25P_t = 281.25 W.

✓ Answer checked by Claude Opus 5.5 (high effort)

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