---
title: "Generate FM (50 MHz, β = 1) from NBFM"
subject: "Communication Systems"
chapter: 3
key: "semester_6/communication_systems:nbfm-to-fm-50mhz"
source: https://www.bctengineer.com/q/semester_6/communication_systems/nbfm-to-fm-50mhz
site: BCT Engineering
---

# Generate FM (50 MHz, β = 1) from NBFM

Communication Systems · Chapter 3: Angle Modulation

- Asked in: 2072 Magh (CS I), 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/nbfm-to-fm-50mhz
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-3

## Question

Generate a FM signal with Fc = 50 MHz and β = 1 from a NBFM signal with Fc = 1 MHz, β = 0.1 and Fm = 10 kHz.

## Short answer

Multiply the NBFM (1 MHz, β=0.1) by n = β/β_1 =10 to get 10 MHz carrier, 10 kHz deviation. Then mix with a 40 MHz local oscillator and select the sum (50 MHz) to obtain fm with f_c=50 MHz, β=1.

## Answer

**Idea:** a frequency multiplier multiplies both the carrier and the deviation (so it multiplies $\beta$); a mixer shifts only the carrier and leaves the deviation unchanged.

**Given NBFM:** $f_{c1} = 1$ MHz, $\beta_1 = 0.1$, $f_m = 10$ kHz

$$
\Delta f_1 = \beta_1 f_m = 0.1 \times 10\ \text{kHz} = 1\ \text{kHz}
$$

**Required FM:** $f_c = 50$ MHz, $\beta = 1$

$$
\Delta f = \beta f_m = 1 \times 10\ \text{kHz} = 10\ \text{kHz}
$$

**Step 1: Multiplication factor.** The mixer does not change $\beta$, so the multiplier must give

$$
n = \frac{\beta}{\beta_1} = \frac{1}{0.1} = 10
$$

After a $\times 10$ multiplier:

$$
f_{c2} = 10 \times 1\ \text{MHz} = 10\ \text{MHz}, \qquad \Delta f_2 = 10 \times 1\ \text{kHz} = 10\ \text{kHz}
$$

**Step 2: Frequency translation.** The carrier is now 10 MHz but 50 MHz is needed. Mix with a local oscillator and select the sum frequency with a band-pass filter:

$$
f_{LO} + 10\ \text{MHz} = 50\ \text{MHz} \;\Rightarrow\; f_{LO} = 40\ \text{MHz}
$$

(Choosing the difference, $f_{LO} = 60$ MHz, also works.) The deviation stays 10 kHz.

```text
NBFM      +-----+ 10 MHz  +-------+  +-----+ FM out
1 MHz --->| x10 |-------->| Mixer |->| BPF |------->
df=1kHz   +-----+ df=10kHz+-------+  |50MHz| 50 MHz
b=0.1                         ^      +-----+ df=10kHz
                         LO 40 MHz           b = 1
```

**Check:** $\beta = \Delta f / f_m = 10\ \text{kHz}/10\ \text{kHz} = 1$, $f_c = 50$ MHz.

The order can also be reversed: mix the 1 MHz NBFM with a 4 MHz LO to get 5 MHz, then multiply by 10 to get 50 MHz with $\Delta f = 10$ kHz.

**Answer:** frequency multiplier $n = 10$ followed by a mixer with $f_{LO} = 40$ MHz (sum taken) gives $f_c = 50$ MHz, $\Delta f = 10$ kHz, $\beta = 1$.

## Check yourself

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

1. What multiplier is needed?
   A. n = 1/0.1 = 10
   B. n = 50
   C. n = 5
   D. n = 100
   Answer: A. Multiplier scales beta; mixer does not. (Key point: Multiplication factor)

2. What is the NBFM deviation?
   A. 10 kHz
   B. 0.1 x 10 kHz = 1 kHz
   C. 100 Hz
   D. 0.1 kHz x 10 = 10 kHz
   Answer: B. Delta f = beta fm. (Key point: Delta f)

3. What LO (sum taken) is needed after x10 gives 10 MHz?
   A. 50 MHz
   B. 60 MHz
   C. 40 MHz
   D. 10 MHz
   Answer: C. 40 + 10 = 50 MHz. (Key point: Mixer LO)

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- [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×
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- [Armstrong FM transmitter numerical (Δf1 = 24 Hz)](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aarmstrong-numerical-24hz) — Communication Systems, chapter 3

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