AM numerical: 15 sin2π(1500t) audio on 60 sin2π(100,000t)

Given: message $m(t)=15\sin2\pi(1500t)$, so $A_m=15$ V, $f_m=1.5$ kHz. Carrier $c(t)=60\sin2\pi(100{,}000t)$, so $A_c=60$ V, $f_c=100$ kHz.

i) Modulated wave

For standard AM, the carrier amplitude varies as $A_c+m(t)$:

$$\begin{aligned} s(t)&=[A_c+m(t)]\sin2\pi f_ct\\ &=[60+15\sin2\pi(1500t)]\sin2\pi(100{,}000t)\\ &=60[1+0.25\sin2\pi(1500t)]\sin2\pi(100{,}000t) \end{aligned}$$

Expanding with $\sin A\sin B=\frac{1}{2}[\cos(A-B)-\cos(A+B)]$:

$$s(t)=60\sin2\pi(10^5t)+7.5\cos2\pi(98{,}500t)-7.5\cos2\pi(101{,}500t)$$

The envelope varies between $60+15=75$ V and $60-15=45$ V.

ii) Modulation index and percent modulation

$$\begin{aligned} \mu&=\frac{A_m}{A_c}=\frac{15}{60}=0.25\\ \text{Percent modulation}&=0.25\times100=25\% \end{aligned}$$

Check: $\mu=\frac{E_{max}-E_{min}}{E_{max}+E_{min}}=\frac{75-45}{75+45}=0.25$.

iii) Frequencies in the spectrum

Side-frequency amplitude $=\mu A_c/2=0.25\times60/2=7.5$ V. Bandwidth $=2f_m=3$ kHz.

          60 V
           |
   7.5 V   |   7.5 V
     |     |     |
 ----+-----+-----+-----> f (kHz)
   98.5   100  101.5

(Extra: on a 1 $\Omega$ load, $P_c=60^2/2=1800$ W, each sideband $28.125$ W, total $1856.25$ W.)

iv) Sketch of audio and carrier waves

Audio period $T_m=1/1500=0.667$ ms; carrier period $T_c=1/100{,}000=10\ \mu$s, so about 66.7 carrier cycles fit in one audio cycle.

 Audio: 15 sin(2*pi*1500 t), peak 15 V
  15 |    .--.
     |  /      \
   0 +-'--------'\--------/----> t
     |            \      /
 -15 |             '--'
     |<---- 0.667 ms ---->|

 Carrier: 60 sin(2*pi*1e5 t), peak 60 V
  60 | /\  /\  /\  /\  /\  /\
   0 +/--\/--\/--\/--\/--\/--\--> t
 -60 |
     |<>| 10 us

 AM wave: envelope between 45 V and 75 V
  75 |   /\/\/\
  45 |/\/      \/\/\      /\/
   0 +-------------------------> t
 -45 |\/\      /\/\/\    /\/\
 -75 |   \/\/\/      \/\/