---
title: "Modulated signal Z(t) = 50cos{98.6×10⁶×2πt + 15sin(2×10³×2πt)}"
subject: "Communication Systems"
chapter: 3
key: "semester_6/communication_systems:angle-mod-50cos-98-6mhz"
source: https://www.bctengineer.com/q/semester_6/communication_systems/angle-mod-50cos-98-6mhz
site: BCT Engineering
---

# Modulated signal Z(t) = 50cos{98.6×10⁶×2πt + 15sin(2×10³×2πt)}

Communication Systems · Chapter 3: Angle Modulation

- Asked in: 2067 Mangsir (CS I), 2×5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/angle-mod-50cos-98-6mhz
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-3

## Question

A modulated signal has the expression:
Z(t) = 50 cos{98.6 × 10⁶ × 2πt + 15 sin(2 × 10³ × 2πt)} volts
Determine: a) type of modulation, b) frequency deviation, c) frequency sensitivity of the modulator d) modulation index and e) power delivered to a 50 Ohms impedance transmitting antenna.

## Answer

**Given:** $Z(t) = 50\cos\{2\pi(98.6\times10^6)t + 15\sin(2\pi\times 2\times10^3\, t)\}$ V, $R = 50\ \Omega$.

Comparing with $A_c\cos[2\pi f_c t + \beta\sin 2\pi f_m t]$: $A_c = 50$ V, $f_c = 98.6$ MHz, $f_m = 2$ kHz, $\beta = 15$.

#### a) Type of modulation

The amplitude is constant and the phase varies sinusoidally, so it is **angle modulation**. If the message is $m(t) = A_m\cos 2\pi f_m t$, the phase term $15\sin 2\pi f_m t$ is proportional to its integral, so it is **frequency modulation (FM)**; the 98.6 MHz carrier also lies in the FM broadcast band. (If the message were a sine, the same wave would be PM; the expression alone cannot tell the two apart.)

#### b) Frequency deviation

$$
f_i(t) = \frac{1}{2\pi}\frac{d\theta}{dt} = f_c + \beta f_m\cos 2\pi f_m t
$$

$$
\Delta f = \beta f_m = 15 \times 2\ \text{kHz} = 30\ \text{kHz}
$$

#### c) Frequency sensitivity

$k_f = \Delta f / A_m$. The message amplitude is not given, so taking $A_m = 1$ V:

$$
k_f = \frac{30\ \text{kHz}}{1\ \text{V}} = 30\ \text{kHz/V}
$$

(For any other $A_m$, $k_f = 30/A_m$ kHz/V.)

#### d) Modulation index

$$
\beta = \frac{\Delta f}{f_m} = \frac{30}{2} = 15
$$

#### e) Power delivered to 50 Ω antenna

$$
P = \frac{A_c^2}{2R} = \frac{50^2}{2 \times 50} = 25\ \text{W}
$$

**Answer:** FM; $\Delta f = 30$ kHz; $k_f = 30$ kHz/V (for $A_m = 1$ V); $\beta = 15$; $P = 25$ W.

## Check yourself

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

1. The wave Z(t) = 50 cos{2π 98.6e6 t + 15 sin(2π 2e3 t)} has constant amplitude and varying phase. What type of modulation is it?
   A. AM
   B. Angle modulation, taken as FM since sin is the integral of a cosine message
   C. DSB-SC
   D. SSB
   Answer: B. Constant envelope means angle modulation; with a cosine message the sine phase term is its integral, so FM (it cannot be distinguished from PM by the expression alone). (Key point: Identify angle modulation)

2. What is the frequency deviation?
   A. 2 kHz
   B. 15 kHz
   C. 7.5 kHz
   D. 30 kHz
   Answer: D. delta f = beta fm = 15 x 2 kHz = 30 kHz. (Key point: Frequency deviation)

3. Why must an assumption be made to find kf, and what is its value if Am = 1 V?
   A. The message amplitude is not given; kf = delta f/Am = 30 kHz/V
   B. fm is not given; kf = 2 kHz/V
   C. Ac is unknown; kf = 50 kHz/V
   D. kf = beta = 15 kHz/V
   Answer: A. kf = delta f/Am, and Am is not given, so Am = 1 V gives 30 kHz/V. (Key point: Frequency sensitivity)

4. What power is delivered to the 50 Ω antenna?
   A. 50 W
   B. 25 W
   C. 12.5 W
   D. 50 x 15 W
   Answer: B. P = Ac^2/(2R) = 2500/100 = 25 W. (Key point: Power to antenna)

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