---
title: "Switching modulator for DSB-FC AM"
subject: "Communication Systems"
chapter: 2
key: "semester_6/communication_systems:switching-modulator-dsbfc"
source: https://www.bctengineer.com/q/semester_6/communication_systems/switching-modulator-dsbfc
site: BCT Engineering
---

# Switching modulator for DSB-FC AM

Communication Systems · Chapter 2: Amplitude Modulation

- Asked in: 2071 Magh (CS I), 8 marks
- Video: [Generation of AM signal: Switching Modulator Explained](https://www.youtube.com/watch?v=M4P5QCEo9WM) — 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/switching-modulator-dsbfc
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-2

## Question

Explain the generation of DSB-FC AM using switching modulator with the help of diagrams and expressions.

## Short answer

The switching modulator uses a diode switching at the carrier rate with Ac>>|m(t)|. The output multiplies the input by a 50% duty-cycle pulse train; after bandpass filtering, s(t) = (Ac/2)[1 + (4/(πAc))m(t)]cosωct, a DSB-FC AM wave.

## Answer

A **switching modulator** generates DSB-FC AM by using a diode as an ON/OFF switch controlled by a large carrier. The switched output contains the AM wave, which is selected by a band-pass filter.

#### Circuit

```text
           +------+         D
 m(t) ---->|      |-- v1 --|>|---+--- v2 ---> BPF ----> s(t)
           | sum  |              |           (fc,2W)    AM
 Ac cos -->|      |             R_L
 wc t      +------+              |
                                ---
```

**Condition:** $A_c \gg |m(t)|$, so the diode state depends only on the carrier: ON when $c(t) > 0$, OFF when $c(t) < 0$.

#### Analysis

Input: $v_1(t) = A_c\cos\omega_ct + m(t)$.

The diode acts like a switch, so

$$
v_2(t) = v_1(t)\,g_p(t)
$$

where $g_p(t)$ is a periodic 0/1 pulse train at $f_c$ with 50% duty cycle:

$$
g_p(t) = \frac12 + \frac{2}{\pi}\cos\omega_ct - \frac{2}{3\pi}\cos3\omega_ct + \frac{2}{5\pi}\cos5\omega_ct - \dots
$$

Multiplying:

$$
\begin{aligned}
v_2(t) = {} & \frac{A_c}{2}\cos\omega_ct + \frac{2}{\pi}m(t)\cos\omega_ct \\
& + \frac{m(t)}{2} + \frac{A_c}{\pi} + \frac{A_c}{\pi}\cos2\omega_ct \\
& - \frac{2}{3\pi}m(t)\cos3\omega_ct + \dots
\end{aligned}
$$

| Term | Frequency | After BPF |
|---|---|---|
| $\frac{A_c}{2}\cos\omega_ct$ | $f_c$ | Kept (carrier) |
| $\frac{2}{\pi}m(t)\cos\omega_ct$ | $f_c \pm W$ | Kept (sidebands) |
| $m(t)/2$, $A_c/\pi$ | 0 to $W$ | Removed |
| $\cos2\omega_ct$, $3\omega_ct$, ... | $2f_c$, $3f_c$, ... | Removed |

BPF output:

$$
s(t) = \frac{A_c}{2}\left[1 + \frac{4}{\pi A_c}m(t)\right]\cos\omega_ct
$$

This is DSB-FC AM with amplitude sensitivity $k_a = \dfrac{4}{\pi A_c}$. Requirement: $f_c > 2W$ so the band around $f_c$ does not overlap the baseband or $2f_c$ terms.

#### Waveforms

```text
 v1(t):  carrier riding on m(t)
          /\    /\    /\    /\
         /  \  /  \  /  \  /  \
        /    \/    \/    \/    \
 v2(t):  only positive half-cycles
          /\    /\    /\    /\
         /  \  /  \  /  \  /  \
 ------ /    \/    \/    \/    \___ (clipped)
 s(t):  AM wave after BPF (envelope = m(t) shape)
```

**Advantages:** does not need an exact square-law device, works well at high power levels.

## Check yourself

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

1. In the switching modulator, what decides whether the diode is ON or OFF?
   A. The band-pass filter at the output
   B. The load resistance R_L
   C. The message m(t), because it is the largest signal
   D. The carrier, because Ac is much larger than |m(t)|
   Answer: D. With Ac >> |m(t)| the diode conducts when c(t) > 0 and is off when c(t) < 0, so it acts as a switch driven by the carrier. (Key point: Switching condition)

2. What is g_p(t) in v2(t) = v1(t) g_p(t)?
   A. A sinusoid at the message frequency
   B. A bipolar +1/-1 square wave
   C. A constant gain equal to Ac
   D. A periodic 0/1 pulse train at fc with 50% duty cycle
   Answer: D. The diode passes v1 only during positive carrier half-cycles, which is multiplication by a 0/1 pulse train. (Key point: Switching function)

3. Which terms of v2(t) are kept by the band-pass filter?
   A. Ac/pi cos 2wc t and m(t)cos 3wc t
   B. Only (2/pi) m(t) cos wc t
   C. (Ac/2)cos wc t and (2/pi) m(t) cos wc t
   D. m(t)/2 and Ac/pi
   Answer: C. The BPF at fc keeps the carrier and sideband terms near fc; baseband and 2fc, 3fc terms are removed. (Key point: Wanted terms)

4. What is the amplitude sensitivity of the generated DSB-FC wave?
   A. ka = 2/(pi Ac)
   B. ka = 4/(pi Ac)
   C. ka = 4 Ac/pi
   D. ka = 1/Ac
   Answer: B. s(t) = (Ac/2)[1 + (4/(pi Ac)) m(t)] cos wc t, so ka = 4/(pi Ac). (Key point: Result and sensitivity)

## Related in Communication Systems

- [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
- [DSB-FC generation using square law modulator](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Adsbfc-square-law-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
- [Single tone DSB-FC AM in time and frequency domain](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Asingle-tone-dsbfc-time-freq) — Communication Systems, chapter 2
- [SSB vs AM and DSB-SC; balanced modulator](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Assb-vs-am-balanced-modulator) — Communication Systems, chapter 2
- [Why AM wastes power and bandwidth; switching modulator](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aam-wasteful-switching-modulator) — Communication Systems, chapter 2