---
title: "Why SSB preferred; AM power at 100% modulation"
subject: "Communication Systems"
chapter: 2
key: "semester_6/communication_systems:ssb-preferred-am-power-1000w"
source: https://www.bctengineer.com/q/semester_6/communication_systems/ssb-preferred-am-power-1000w
site: BCT Engineering
---

# Why SSB preferred; AM power at 100% modulation

Communication Systems · Chapter 2: Amplitude Modulation

- Asked in: 2078 Chaitra, 2+3 marks
- Video: [AMPLITUDE MODULATION  (SOLVED PROBLEMS)](https://www.youtube.com/watch?v=7f3Y2dvfnqI) — Prof. Barapate's Tutorials
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/ssb-preferred-am-power-1000w
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-2

## Question

Why SSB modulation scheme is preferred over DSB, DSB-SC modulation schemes? The total content of an AM signal is 1000 W. Determine the power being transmitted at carrier frequency and at each sideband when modulation percentage is 100%.

## Short answer

SSB is preferred because it uses only half the bandwidth (f_m instead of 2f_m) and transmits no wasted carrier or duplicate sideband, saving about 83% power. For 1000 W total AM power at 100% modulation, carrier power is 666.67 W, and each sideband carries 166.67 W.

## Answer

#### Why SSB is preferred

SSB (single sideband suppressed carrier) sends only one sideband. It is preferred over DSB-FC and DSB-SC because:

- **Half bandwidth:** SSB needs only $f_m$, while DSB-FC and DSB-SC need $2f_m$. Twice as many channels fit in a band.
- **Power saving:** The carrier carries no information. At $\mu = 1$, DSB-FC wastes 2/3 of the power in the carrier; DSB-SC still sends a redundant second sideband. SSB puts all power in one sideband (about 83% saving over DSB-FC at $\mu=1$).
- **Better S/N:** Narrower bandwidth means less noise power at the receiver.
- **Less selective fading:** Only one sideband, so phase differences between sidebands cannot cancel each other.

#### Numerical

Given $P_t = 1000$ W, $\mu = 1$ (100%).

$$
P_t = P_c\left(1 + \frac{\mu^2}{2}\right) \Rightarrow P_c = \frac{P_t}{1 + \mu^2/2}
$$

$$
\begin{aligned}
P_c &= \frac{1000}{1 + 0.5} = 666.67\ \text{W} \\
P_{SB,\,total} &= P_t - P_c = 1000 - 666.67 = 333.33\ \text{W} \\
P_{USB} = P_{LSB} &= \frac{\mu^2}{4}P_c = \frac{1}{4}(666.67) = 166.67\ \text{W}
\end{aligned}
$$

**Answer:** Carrier power = 666.67 W; power in each sideband = 166.67 W (USB = LSB).

## Check yourself

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

1. Why does SSB allow more channels in a band than DSB?
   A. It transmits a stronger carrier
   B. It needs only fm bandwidth instead of 2fm
   C. It uses wider guard bands
   D. It needs 4fm bandwidth per channel
   Answer: B. Sending one sideband halves the bandwidth, so twice as many channels fit. (Key point: SSB bandwidth advantage)

2. With Pt = 1000 W and mu = 1, which formula and value give carrier power?
   A. Pc = Pt/2 = 500 W
   B. Pc = Pt(1 + mu^2/2) = 1500 W
   C. Pc = Pt/3 = 333.33 W
   D. Pc = Pt/(1 + mu^2/2) = 666.67 W
   Answer: D. Pt = Pc(1 + 0.5) = 1.5Pc, so Pc = 1000/1.5 = 666.67 W. (Key point: Carrier power at 100%)

3. What is the power in each sideband at 100% modulation with Pt = 1000 W?
   A. 166.67 W
   B. 333.33 W
   C. 666.67 W
   D. 83.33 W
   Answer: A. Sideband total is 333.33 W, shared equally by USB and LSB, giving 166.67 W each (mu^2 Pc/4). (Key point: Sideband power)

## Related in Communication Systems

- [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
- [Compare DSB-AM, DSB-SC and SSB](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Acompare-dsbam-dsbsc-ssb) — Communication Systems, chapter 2
- [Compare DSB-AM, DSB-SC, SSB, VSB; SSB by phase discrimination](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Acompare-am-variants-ssb-phase) — 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
- [DSB-AM, DSB-SC and SSB spectra; synchronous detection](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Adsbam-dsbsc-ssb-spectrum-synchronous) — Communication Systems, chapter 2
- [Noise performance of DSB-AM, DSB-SC and SSB-SC](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Anoise-dsbam-dsbsc-ssbsc) — Communication Systems, chapter 9

## Related in other subjects

- [AM numerical: 20 sin2π500t on 60 sin2π10⁶t](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Aam-numeric-20sin500) — Data Communication, chapter 5