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Two-tone AM: spectrum, net index, power, efficiency

Short answer

DSB-FC with two tones: μ₁=0.3 at 1 kHz, μ₂=0.4 at 200 Hz. Net modulation index μ = √(0.3²+0.4²) = 0.5. Carrier power Pc = 5000 W, total power Pt = 5625 W, efficiency η = 11.11%. Spectrum: carrier 100 V at 1 MHz; sidebands at ±1 kHz (15 V) and ±200 Hz (20 V).

An Amplitude modulated wave is given by s(t) = 100cos(2π×10⁶t) + 30cos(2π×10⁶t)cos(2π×10³t) + 40cos(2π×10⁶t)cos(4π×10²t) Volt a) Draw the frequency spectrum of modulated wave b) Net modulation index c) Total modulated power d) Efficiency

Write in standard form:

s(t)=100[1+0.3cos⁡(2π103t)+0.4cos⁡(2π 200 t)]cos⁡(2π106t) Vs(t) = 100\left[1 + 0.3\cos(2\pi10^3t) + 0.4\cos(2\pi\,200\,t)\right]\cos(2\pi10^6t)\ \text{V}

since 30/100=0.330/100 = 0.3, 40/100=0.440/100 = 0.4 and 4π102t=2π(200)t4\pi10^2t = 2\pi(200)t.

Ac=100A_c = 100 V, fc=1f_c = 1 MHz; μ1=0.3\mu_1 = 0.3 at 1 kHz; μ2=0.4\mu_2 = 0.4 at 200 Hz. Assume R=1 ΩR = 1\ \Omega.

a) Frequency spectrum

Side-frequency amplitudes =μAc/2= \mu A_c/2: 30/2=1530/2 = 15 V (1 kHz tone), 40/2=2040/2 = 20 V (200 Hz tone).

FrequencyAmplitude
999 kHz15 V
999.8 kHz20 V
1000 kHz100 V
1000.2 kHz20 V
1001 kHz15 V
  A(V)
 100 |                |
     |                |
  20 |        |       |       |
  15 |   |    |       |       |    |
     +---+----+-------+-------+----+---> f (kHz)
       999 999.8    1000  1000.2  1001

b) Net modulation index

μt=μ12+μ22=0.09+0.16=0.25=0.5\mu_t = \sqrt{\mu_1^2 + \mu_2^2} = \sqrt{0.09 + 0.16} = \sqrt{0.25} = 0.5

c) Total power

Pc=Ac22R=10022=5000 WPt=Pc(1+μt22)=5000(1+0.125)=5625 W\begin{aligned} P_c &= \frac{A_c^2}{2R} = \frac{100^2}{2} = 5000\ \text{W} \\ P_t &= P_c\left(1 + \frac{\mu_t^2}{2}\right) = 5000(1 + 0.125) = 5625\ \text{W} \end{aligned}

d) Efficiency

η=μt22+μt2=0.252.25=0.1111=11.11%\eta = \frac{\mu_t^2}{2 + \mu_t^2} = \frac{0.25}{2.25} = 0.1111 = 11.11\%

Answer: μt=0.5\mu_t = 0.5; Pt=5625P_t = 5625 W; η=11.11%\eta = 11.11\%.

✓ Answer checked by Claude Opus 5.5 (high effort)

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