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AM numerical: s(t) = 5[1 + 0.6cos(3140t)]cos(2π10³t)

Short answer

Given s(t)=5[1+0.6cos(3140t)]cos(2π10³t), modulation index μ=0.6 → 60%. Max amplitude = Ac(1+μ)=8V, min = Ac(1-μ)=2V. Carrier power Pc=Ac²/(2R)=12.5mW, total power Pt=Pc(1+μ²/2)=14.75mW across 1kΩ. fm=3140/(2π)≈500Hz, fc=1000Hz; USB=fc+fm≈1.5kHz, LSB=fc-fm≈500Hz.

An AM wave is represented by s(t) = 5[1 + 0.6cos(3140t)].cos(2π10³t) volts, then find the followings: a) Modulation percentage, b) Maximum and minimum amplitude of AM wave, c) Power dissipated across 1K ohm resistor and d) Frequency of USB and LSB.

Compare with s(t)=Ac[1+μcos⁡ωmt]cos⁡ωcts(t) = A_c[1 + \mu\cos\omega_mt]\cos\omega_ct:

Ac=5A_c = 5 V, μ=0.6\mu = 0.6, ωm=3140\omega_m = 3140 rad/s, ωc=2π×103\omega_c = 2\pi\times10^3 rad/s.

fm=31402π=499.75 Hz≈500 Hz,fc=1000 Hzf_m = \frac{3140}{2\pi} = 499.75\ \text{Hz} \approx 500\ \text{Hz}, \qquad f_c = 1000\ \text{Hz}

a) Modulation percentage

% modulation=μ×100=0.6×100=60%\%\,\text{modulation} = \mu \times 100 = 0.6 \times 100 = 60\%

b) Maximum and minimum amplitude

Amax=Ac(1+μ)=5(1.6)=8 VAmin=Ac(1−μ)=5(0.4)=2 V\begin{aligned} A_{max} &= A_c(1 + \mu) = 5(1.6) = 8\ \text{V} \\ A_{min} &= A_c(1 - \mu) = 5(0.4) = 2\ \text{V} \end{aligned}

Check: μ=8−28+2=0.6\mu = \dfrac{8-2}{8+2} = 0.6.

c) Power across 1 kΩ

Pc=Ac22R=522×1000=12.5 mWPt=Pc(1+μ22)=12.5(1+0.18)=14.75 mW\begin{aligned} P_c &= \frac{A_c^2}{2R} = \frac{5^2}{2\times1000} = 12.5\ \text{mW} \\ P_t &= P_c\left(1 + \frac{\mu^2}{2}\right) = 12.5(1 + 0.18) = 14.75\ \text{mW} \end{aligned}

Each sideband: μ24Pc=0.09×12.5=1.125\frac{\mu^2}{4}P_c = 0.09 \times 12.5 = 1.125 mW.

d) USB and LSB frequencies

fUSB=fc+fm=1000+499.75=1499.75 Hz≈1.5 kHzfLSB=fc−fm=1000−499.75=500.25 Hz≈0.5 kHz\begin{aligned} f_{USB} &= f_c + f_m = 1000 + 499.75 = 1499.75\ \text{Hz} \approx 1.5\ \text{kHz} \\ f_{LSB} &= f_c - f_m = 1000 - 499.75 = 500.25\ \text{Hz} \approx 0.5\ \text{kHz} \end{aligned}

Answer: 60%; Amax=8A_{max} = 8 V, Amin=2A_{min} = 2 V; Pt=14.75P_t = 14.75 mW; USB ≈ 1.5 kHz, LSB ≈ 500 Hz.

✓ Answer checked by Claude Opus 5.5 (high effort)

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