---
title: "AM numerical: 750 kHz carrier, 110 V and 90 V amplitudes"
subject: "Communication Systems"
chapter: 2
key: "semester_6/communication_systems:am-750khz-110v-90v"
source: https://www.bctengineer.com/q/semester_6/communication_systems/am-750khz-110v-90v
site: BCT Engineering
---

# AM numerical: 750 kHz carrier, 110 V and 90 V amplitudes

Communication Systems · Chapter 2: Amplitude Modulation

- Asked in: 2075 Baisakh (CS I), 2+3+2 marks
- Video: [AMPLITUDE MODULATION  (SOLVED PROBLEMS)](https://www.youtube.com/watch?v=7f3Y2dvfnqI) — Prof. Barapate's Tutorials
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/am-750khz-110v-90v
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-2

## Question

A cosine carrier of frequency 750 kHz is amplitude modulated by another cosine wave of frequency 325 Hz resulting in maximum and minimum carrier amplitudes of 110 V and 90 V respectively:
a) Draw the waveform of AM wave thus created.
b) Write the expression of the resulting AM wave.
c) Find the total power radiated and efficiency.

## Short answer

Ac=100V, μ=0.1. Expression: s(t)=100[1+0.1cos(2π·325t)]cos(2π·750×10^3t) V. Total power Pt=5025W (1Ω load), efficiency η≈0.50%.

## Answer

Given: $f_c = 750$ kHz, $f_m = 325$ Hz, $A_{max} = 110$ V, $A_{min} = 90$ V. Assume load $R = 1\ \Omega$ (normalised).

$$
\begin{aligned}
A_c &= \frac{A_{max} + A_{min}}{2} = \frac{110 + 90}{2} = 100\ \text{V} \\
\mu &= \frac{A_{max} - A_{min}}{A_{max} + A_{min}} = \frac{20}{200} = 0.1
\end{aligned}
$$

#### a) Waveform

```text
 v(t)
 110 -  _   _   _   _   _   _   _   _      envelope
  90 - / \_/ \_/ \_/ \_/ \_/ \_/ \_/ \_    varies 90..110 V
       |||||||||||||||||||||||||||||||     carrier 750 kHz
   0 --+--------------------------------> t
       |||||||||||||||||||||||||||||||
 -90 - \_/ \_/ \_/ \_/ \_/ \_/ \_/ \_/
-110 -
       |<------ 1/325 s = 3.08 ms ---->|
```

The carrier swings between ±110 V at envelope peaks and ±90 V at envelope troughs; the envelope period is $1/325 = 3.08$ ms. Only 10% modulation, so the envelope ripple is shallow.

#### b) Expression

$$
s(t) = 100\,[1 + 0.1\cos(2\pi\,325\,t)]\cos(2\pi\times750\times10^3\,t)\ \text{V}
$$

#### c) Total power and efficiency

$$
\begin{aligned}
P_c &= \frac{A_c^2}{2R} = \frac{100^2}{2} = 5000\ \text{W} \\
P_t &= P_c\left(1 + \frac{\mu^2}{2}\right) = 5000(1 + 0.005) = 5025\ \text{W} \\
\eta &= \frac{\mu^2}{2 + \mu^2} = \frac{0.01}{2.01} = 0.4975\%
\end{aligned}
$$

**Answer:** $s(t) = 100[1 + 0.1\cos(2\pi325t)]\cos(2\pi750\times10^3t)$ V; $P_t = 5025$ W (for 1 Ω); $\eta \approx 0.50\%$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. With maximum and minimum amplitudes 110 V and 90 V, the unmodulated carrier amplitude Ac is:
   A. 20 V
   B. 110 V
   C. 200 V
   D. 100 V
   Answer: D. Ac = (Amax + Amin)/2 = (110 + 90)/2 = 100 V. (Key point: Carrier amplitude from extremes)

2. The modulation index is:
   A. 0.1
   B. 0.2
   C. 0.9
   D. 0.01
   Answer: A. mu = (Amax - Amin)/(Amax + Amin) = 20/200 = 0.1. (Key point: Modulation index)

3. For R = 1 ohm, Pc = 5000 W. The total power Pt = Pc(1 + mu^2/2) is:
   A. 5000 W
   B. 5025 W
   C. 5050 W
   D. 5500 W
   Answer: B. Pt = 5000(1 + 0.01/2) = 5000 x 1.005 = 5025 W. (Key point: Total power)

4. The efficiency mu^2/(2 + mu^2) for mu = 0.1 is about:
   A. 10%
   B. 5%
   C. 0.50%
   D. 33%
   Answer: C. 0.01/2.01 = 0.4975%, so the sidebands carry only about 0.5% of the power. (Key point: Efficiency)

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