---
title: "Balanced modulator output is DSB-SC"
subject: "Communication Systems"
chapter: 2
key: "semester_6/communication_systems:balanced-modulator-dsbsc"
source: https://www.bctengineer.com/q/semester_6/communication_systems/balanced-modulator-dsbsc
site: BCT Engineering
---

# Balanced modulator output is DSB-SC

Communication Systems · Chapter 2: Amplitude Modulation

- Asked in: 2065 Kartik (CS I), 6+2 marks
- Video: [Balanced Modulator using AM Modulator for DSBSC Generation (Basics, Working, Circuit & Waveform)](https://www.youtube.com/watch?v=tPWAornTFzI) — Engineering Funda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/balanced-modulator-dsbsc
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-2

## Question

Show that the output of the balanced modulator is DSB-SC modulated wave. Draw DSB-SC modulated wave for sinusoidal modulating signal.

## Short answer

Subtracting outputs of two identical AM modulators driven by m(t) and -m(t) cancels the carrier, yielding DSB-SC: s(t) = 2ka Ac m(t) cos(omega_c t). The modulated wave's envelope shows phase reversals at message zero crossings.

## Answer

A **balanced modulator** uses two identical AM modulators driven with opposite-polarity messages and the same carrier. Their outputs are subtracted, so the carrier cancels and only the sidebands remain, giving DSB-SC.

#### Block diagram

```text
          +----------+  s1(t)
  m(t) -->|   AM     |-------+
          | modulator|       |  +
          +----------+       v
               ^           ( Σ )----> s(t) = DSB-SC
   Ac cos wct--+             ^  -
               v             |
          +----------+       |
 -m(t) -->|   AM     |-------+
          | modulator|  s2(t)
          +----------+
```

#### Proof

Both modulators are identical with amplitude sensitivity $k_a$:

$$
\begin{aligned}
s_1(t) &= A_c[1 + k_am(t)]\cos\omega_ct \\
s_2(t) &= A_c[1 - k_am(t)]\cos\omega_ct
\end{aligned}
$$

Subtracting:

$$
\begin{aligned}
s(t) &= s_1(t) - s_2(t) \\
&= A_c\cos\omega_ct + A_ck_am(t)\cos\omega_ct - A_c\cos\omega_ct + A_ck_am(t)\cos\omega_ct \\
&= 2k_aA_c\,m(t)\cos\omega_ct
\end{aligned}
$$

This is the product of the message and the carrier: a **DSB-SC** wave. The carrier term cancels exactly when the two modulators are perfectly matched (balanced). In practice each modulator can be a square-law or switching modulator; with square-law devices ($v_o = a_1v_i + a_2v_i^2$) the output is $4a_2A_cm(t)\cos\omega_ct$ after a band-pass filter.

In frequency domain: $S(f) = k_aA_c[M(f-f_c) + M(f+f_c)]$, i.e. only USB and LSB, no impulse at $f_c$.

#### DSB-SC wave for a sinusoidal message

For $m(t) = A_m\cos\omega_mt$: $s(t) = 2k_aA_cA_m\cos\omega_mt\cos\omega_ct$.

```text
 m(t):  __          __
       /  \        /  \
 -----/----\------/----\---
            \____/

 s(t): /\/\/\      /\/\/\
      /\/\/\/\    /\/\/\/\
 ----+--------X--+--------X--
      \/\/\/\/    \/\/\/\/
       \/\/\/      \/\/\/
            ^ phase reversal at
              each zero of m(t)
```

The envelope is $|m(t)|$, and the carrier phase reverses by 180° at each zero crossing of the message.

## Check yourself

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

1. How is a balanced modulator built from two AM modulators?
   A. Their outputs are added with the same message
   B. They are driven by m(t) and −m(t) with the same carrier, and their outputs are subtracted
   C. They use different carriers and the outputs are multiplied
   D. Both get m(t) and the output is differentiated
   Answer: B. Opposite-polarity messages with the same carrier, followed by subtraction, cancel the carrier and keep the sidebands. (Key point: Balanced modulator structure)

2. What is s1(t) − s2(t) with s1 = Ac[1+ka m]cos ωc t and s2 = Ac[1−ka m]cos ωc t?
   A. 2Ac cos ωc t
   B. 2 ka Ac m(t) cos ωc t
   C. ka Ac m(t)
   D. Ac[1 + ka m(t)] cos ωc t
   Answer: B. The Ac cos ωc t terms cancel and the message terms add, leaving 2 ka Ac m(t) cos ωc t, a DSB-SC wave. (Key point: Result of the subtraction)

3. When does the carrier cancel exactly?
   A. When the message frequency is zero
   B. When the modulators are perfectly matched
   C. When ka is large
   D. When the carrier is much larger than the message
   Answer: B. Exact cancellation needs identical (balanced) modulators. (Key point: Condition for carrier cancellation)

4. For a sinusoidal message, what happens to the DSB-SC carrier at each zero crossing of m(t)?
   A. Its frequency doubles
   B. Its amplitude becomes maximum
   C. Nothing; the phase is constant
   D. Its phase reverses by 180°
   Answer: D. The envelope is |m(t)| and the sign change of m(t) reverses carrier phase by 180°. (Key point: DSB-SC waveform)

## Related in Communication Systems

- [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
- [DSB-AM vs SSB; ring modulator for DSB-SC](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Adsbam-ssb-compare-ring-modulator) — Communication Systems, chapter 2
- [Ring modulator for DSB-SC](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aring-modulator-dsbsc) — Communication Systems, chapter 2
- [Ring modulator for DSB-SC; characteristics of DSB-SC](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aring-modulator-dsbsc-characteristics) — Communication Systems, chapter 2
- [FM expression after change of modulating signal](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Afm-new-waveform-500hz-2khz) — Communication Systems, chapter 3
- [Costas loop as synchronous receiver for DSB-SC](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Acostas-loop-dsbsc) — Communication Systems, chapter 2