---
title: "DSB-FC generation using square law modulator"
subject: "Communication Systems"
chapter: 2
key: "semester_6/communication_systems:dsbfc-square-law-modulator"
source: https://www.bctengineer.com/q/semester_6/communication_systems/dsbfc-square-law-modulator
site: BCT Engineering
---

# DSB-FC generation using square law modulator

Communication Systems · Chapter 2: Amplitude Modulation

- Asked in: 2080 Chaitra (CS I), 6 marks
- Video: [Square Law Modulator and AM generation using Analog Multiplier | Generation of AM signal](https://www.youtube.com/watch?v=OIE7h6nn1aM) — ALL ABOUT ELECTRONICS
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/dsbfc-square-law-modulator
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-2

## Question

Explain the generation of DSB FC signal using square law modulator with appropriate diagram and spectrum of transmitted signal.

## Short answer

A square-law modulator adds m(t) to carrier, passes the sum through a nonlinear device with characteristic v₂ = a₁v₁ + a₂v₁². A bandpass filter centered at f_c extracts the DSB-FC output s(t) = a₁A_c[1 + (2a₂/a₁)m(t)]cosω_c t.

## Answer

A **square law modulator** generates DSB-FC (conventional AM) by passing the sum of carrier and message through a non-linear device (diode or transistor biased in its non-linear region) whose output contains a square term, and then filtering out the AM band.

#### Block diagram

```text
 m(t) -->(+)--> v1 --> Non-linear --> v2 --> BPF --> s(t)
          ^             device             (fc,2W)   (AM)
          |             (diode)
 c(t) ----+
 Ac cos wc t
```

#### Analysis

Input to the non-linear device:

$$
v_1(t) = m(t) + A_c\cos\omega_c t
$$

The device characteristic (square law approximation):

$$
v_2(t) = a_1v_1(t) + a_2v_1^2(t)
$$

Substituting:

$$
\begin{aligned}
v_2(t) = {} & a_1m(t) + a_1A_c\cos\omega_c t \\
& + a_2m^2(t) + 2a_2A_c\,m(t)\cos\omega_c t + a_2A_c^2\cos^2\omega_c t
\end{aligned}
$$

The terms and their frequencies:

| Term | Frequency range | Kept? |
|---|---|---|
| $a_1m(t)$ | 0 to $W$ | No |
| $a_2m^2(t)$ | 0 to $2W$ | No |
| $a_2A_c^2\cos^2\omega_ct$ | DC and $2f_c$ | No |
| $a_1A_c\cos\omega_ct$ | $f_c$ | Yes (carrier) |
| $2a_2A_cm(t)\cos\omega_ct$ | $f_c \pm W$ | Yes (sidebands) |

The BPF centred at $f_c$ with bandwidth $2W$ passes:

$$
s(t) = a_1A_c\left[1 + \frac{2a_2}{a_1}m(t)\right]\cos\omega_c t
$$

This is DSB-FC AM with amplitude sensitivity $k_a = 2a_2/a_1$. Condition for separation: $f_c > 3W$, so that the $m^2(t)$ band (up to $2W$) does not overlap the lower sideband ($f_c - W$).

#### Spectrum

```text
 M(f)                      S(f)
   /\                   /\    |    /\     /\    |    /\
  /  \                 /  \   |   /  \   /  \   |   /  \
-W 0  W        ...  -fc-W -fc -fc+W ... fc-W  fc  fc+W
                    (LSB carr USB)     (LSB carr USB)
```

The transmitted spectrum has the carrier impulse at $\pm f_c$ with the USB and LSB on either side, total bandwidth $2W$.

**Limitation:** distortion if higher-order terms are present and the device is not ideally square law; output is low-level, so it is used in low-power transmitters.

## Check yourself

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

1. Why is a non-linear device used in the square law modulator?
   A. It amplifies the message linearly
   B. Its square term produces the product m(t)cos(wc t), creating the sidebands
   C. It removes the carrier
   D. It filters out the message
   Answer: B. The a2 v1^2 term gives 2 a2 Ac m(t) cos wc t. (Key point: Non-linear device role)

2. What does the band-pass filter at fc with bandwidth 2W do?
   A. Keeps only the m(t) term
   B. Keeps only the 2fc term
   C. Keeps all terms for further processing
   D. Keeps the carrier and sidebands, rejects the baseband, m^2(t) and 2fc terms
   Answer: D. Only a1 Ac cos wc t and 2a2 Ac m(t) cos wc t lie around fc. (Key point: Role of BPF)

3. What is the amplitude sensitivity ka of the resulting AM wave?
   A. 2a2/a1
   B. a2/a1
   C. a1/2a2
   D. a1 a2
   Answer: A. s(t) = a1 Ac[1 + (2a2/a1) m(t)] cos wc t. (Key point: Amplitude sensitivity)

4. Why is fc > 3W required?
   A. So the carrier is above the audio range only
   B. So the 2fc term overlaps the sideband
   C. So the m^2(t) band (0 to 2W) does not overlap the lower sideband at fc - W
   D. So the BPF has zero bandwidth
   Answer: C. m^2(t) extends to 2W, and fc - W must exceed 2W. (Key point: Frequency condition)

## Related in Communication Systems

- [Square-law modulator for DSB-AM; characteristics of AM](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Asquare-law-dsbam-characteristics) — Communication Systems, chapter 2, asked 2×
- [Switching modulator for DSB-FC AM](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aswitching-modulator-dsbfc) — Communication Systems, chapter 2
- [AM modulation index; square law modulator](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aam-index-square-law-modulator) — Communication Systems, chapter 2
- [Define AM; switching modulator for DSB-FC](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Adefine-am-switching-modulator) — Communication Systems, chapter 2
- [Compare AM systems; one method of DSB-FC generation](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aam-types-compare-dsbfc-method) — Communication Systems, chapter 2
- [Square law approximation for DSB-AM](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Asquare-law-approximation-dsbam) — Communication Systems, chapter 2

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