---
title: "Armstrong FM transmitter numerical (Δf1 = 24 Hz)"
subject: "Communication Systems"
chapter: 3
key: "semester_6/communication_systems:armstrong-numerical-24hz"
source: https://www.bctengineer.com/q/semester_6/communication_systems/armstrong-numerical-24hz
site: BCT Engineering
---

# Armstrong FM transmitter numerical (Δf1 = 24 Hz)

Communication Systems · Chapter 3: Angle Modulation

- Asked in: 2068 Bhadra (CS I), 8 marks
- Video: [Commercial FM Transmission Using Armstrong Method](https://www.youtube.com/watch?v=toZsQIG7O9c) — Dr. Krishnamurthy Ramanujam
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/armstrong-numerical-24hz
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-3

## Question

For the following Armstrong FM transmitter, compute maximum frequency deviation and the carrier frequency fc if f1 = 200 kHz, fLO = 10.8 MHz, Δf1 = 24 Hz, n1 = 65 and n2 = 50.
[Figure: m(t) → NBFM (output f1, Δf1) → frequency multiplier ×n1 → mixer with local oscillator fLO → frequency multiplier ×n2 → S_WBFM(t) with fc, Δfc]

## Answer

**Given:** $f_1 = 200$ kHz, $\Delta f_1 = 24$ Hz, $n_1 = 65$, $f_{LO} = 10.8$ MHz, $n_2 = 50$.

**Rules:** a multiplier multiplies both carrier and deviation; a mixer changes only the carrier (deviation unchanged). The mixer output is taken as the difference frequency (normal Armstrong design, to bring the carrier down before the second multiplier).

**Step 1: After the first multiplier ($\times 65$)**

$$
\begin{aligned}
f_2 &= n_1 f_1 = 65 \times 200\ \text{kHz} = 13\ \text{MHz} \\
\Delta f_2 &= n_1 \Delta f_1 = 65 \times 24 = 1560\ \text{Hz}
\end{aligned}
$$

**Step 2: After the mixer**

$$
f_3 = f_2 - f_{LO} = 13 - 10.8 = 2.2\ \text{MHz}, \qquad \Delta f_3 = 1560\ \text{Hz}
$$

**Step 3: After the second multiplier ($\times 50$)**

$$
\begin{aligned}
f_c &= n_2 f_3 = 50 \times 2.2\ \text{MHz} = 110\ \text{MHz} \\
\Delta f_c &= n_2 \Delta f_3 = 50 \times 1560 = 78\,000\ \text{Hz} = 78\ \text{kHz}
\end{aligned}
$$

Check: $\Delta f_c = n_1 n_2 \Delta f_1 = 3250 \times 24 = 78$ kHz.

| Stage | Carrier | Deviation |
|---|---|---|
| NBFM output | 200 kHz | 24 Hz |
| After ×65 | 13 MHz | 1.56 kHz |
| After mixer | 2.2 MHz | 1.56 kHz |
| After ×50 | 110 MHz | 78 kHz |

(If the sum frequency 23.8 MHz were selected, $f_c$ would be 1190 MHz with the same deviation, which is not a practical FM carrier.)

**Answer:** maximum frequency deviation $\Delta f_c = 78$ kHz, carrier frequency $f_c = 110$ MHz.

## Check yourself

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

1. What does a frequency multiplier do to the carrier and deviation?
   A. Multiplies only the carrier
   B. Multiplies both the carrier and the deviation
   C. Multiplies only the deviation
   D. Multiplies neither
   Answer: B. A multiplier of factor n multiplies both the carrier frequency and the frequency deviation by n. (Key point: Multiplier rule)

2. What does the mixer do to the FM signal in an Armstrong transmitter?
   A. Doubles the deviation
   B. Shifts the carrier frequency but leaves the deviation unchanged
   C. Removes the deviation
   D. Shifts only the deviation
   Answer: B. Mixing translates the carrier (difference frequency) while the deviation stays the same. (Key point: Mixer rule)

3. After the x65 multiplier the carrier is 13 MHz; with fLO = 10.8 MHz and difference frequency selected, what is the carrier before the x50 stage?
   A. 23.8 MHz
   B. 10.8 MHz
   C. 65 MHz
   D. 2.2 MHz
   Answer: D. f3 = 13 - 10.8 = 2.2 MHz. (Key point: Carrier after the mixer)

4. What are the final carrier frequency and deviation?
   A. fc = 110 MHz, delta f = 78 kHz
   B. fc = 110 MHz, delta f = 1.56 kHz
   C. fc = 1190 MHz, delta f = 78 kHz
   D. fc = 13 MHz, delta f = 24 Hz
   Answer: A. fc = 50 x 2.2 = 110 MHz and delta f = 65 x 50 x 24 = 78 kHz. (Key point: Final results)

## Related in Communication Systems

- [Armstrong FM modulator: find n1 and n2](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aarmstrong-n1-n2-numerical) — Communication Systems, chapter 3
- [FM numerical: 10 kHz baseband, 100 MHz carrier, 75 kHz deviation](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Afm-10khz-100mhz-75khz) — Communication Systems, chapter 3
- [Armstrong FM transmitter numerical](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aarmstrong-numerical-200khz-25hz) — Communication Systems, chapter 3
- [FM numerical: 102.4 MHz carrier, 75 kHz deviation](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Afm-102-4mhz-75khz) — Communication Systems, chapter 3
- [Generate FM (50 MHz, β = 1) from NBFM](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Anbfm-to-fm-50mhz) — Communication Systems, chapter 3
- [FM numerical: 5cos(18849.55t), 9 kHz/V, 25 V, 88.7 MHz (five parts)](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Afm-5cos18849-five-parts) — Communication Systems, chapter 3, asked 3×