---
title: "Advantages of SSB over DSB-FC; SSB expression"
subject: "Communication Systems"
chapter: 2
key: "semester_6/communication_systems:ssb-advantages-derive"
source: https://www.bctengineer.com/q/semester_6/communication_systems/ssb-advantages-derive
site: BCT Engineering
---

# Advantages of SSB over DSB-FC; SSB expression

Communication Systems · Chapter 2: Amplitude Modulation

- Asked in: 2071 Magh (CS I), 2+4 marks
- Video: [Quick Revision of Amplitude Modulation classifications| SSB-SC| DSB-SC| VSB-SC|Analog Communication](https://www.youtube.com/watch?v=uJhflGLjUk8) — Easy Electronics
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/ssb-advantages-derive
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-2

## Question

What are the advantages of SSB-AM over DSBFC-AM? Derive the expression for SSB signal.

## Short answer

SSB needs only half bandwidth (W vs 2W), saves 83.3% power, has better noise performance, less selective fading, and all transmitted power carries information. The SSB signal is derived from DSB-SC by suppressing one sideband: s_SSB(t) = (Ac/2)[m(t)cosωct ∓ m̂(t)sinωct], where m̂(t) is the Hilbert transform (minus for USB, plus for LSB).

## Answer

#### Advantages of SSB over DSB-FC

- **Half bandwidth:** $W$ instead of $2W$, so twice the number of channels.
- **Power saving:** at $\mu = 1$, SSB needs only $0.25P_c$ against $1.5P_c$, a saving of 83.3%.
- **Better S/N:** narrower band admits less noise.
- **Less selective fading:** with one sideband and no carrier, phase shifts between components cannot cancel each other.
- **Higher efficiency:** all transmitted power carries information.

#### Derivation of SSB signal

Start with DSB-SC for a single tone $m(t) = A_m\cos\omega_mt$:

$$
s_{DSB}(t) = A_cA_m\cos\omega_mt\cos\omega_ct = \frac{A_cA_m}{2}[\cos(\omega_c+\omega_m)t + \cos(\omega_c-\omega_m)t]
$$

Keeping only the upper sideband:

$$
\begin{aligned}
s_{USB}(t) &= \frac{A_cA_m}{2}\cos(\omega_c + \omega_m)t \\
&= \frac{A_c}{2}[A_m\cos\omega_mt\cos\omega_ct - A_m\sin\omega_mt\sin\omega_ct]
\end{aligned}
$$

Similarly, the lower sideband:

$$
s_{LSB}(t) = \frac{A_c}{2}[A_m\cos\omega_mt\cos\omega_ct + A_m\sin\omega_mt\sin\omega_ct]
$$

Here $A_m\sin\omega_mt$ is the Hilbert transform of $A_m\cos\omega_mt$. Since any message is a sum of such tones, for a general $m(t)$:

$$
s_{SSB}(t) = \frac{A_c}{2}\left[m(t)\cos\omega_ct \mp \hat m(t)\sin\omega_ct\right]
$$

(minus sign: USB; plus sign: LSB), where $\hat m(t)$ is the Hilbert transform of $m(t)$.

**Frequency domain check (USB):** $\hat M(f) = -j\,\text{sgn}(f)M(f)$, so the in-phase and quadrature terms add for $|f| > f_c$ and cancel for $|f| < f_c$, leaving only the upper sideband.

**Power** (single tone, load $R$): $P_{SSB} = \dfrac{(A_cA_m/2)^2}{2R} = \dfrac{A_c^2A_m^2}{8R}$.

## Check yourself

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

1. Which is an advantage of SSB over DSB-FC?
   A. Simpler receiver
   B. Needs a carrier for detection
   C. Half bandwidth and large power saving
   D. Wider bandwidth for more fidelity
   Answer: C. SSB needs W instead of 2W and at mu = 1 saves about 83.3% of the power. (Key point: Advantages of SSB)

2. The SSB derivation starts from DSB-SC single tone and:
   A. Adds a carrier
   B. Squares the signal
   C. Differentiates it
   D. Keeps only one sideband, e.g. (AcAm/2) cos((wc + wm)t) for USB
   Answer: D. Removing the lower sideband from the DSB-SC product leaves the USB term. (Key point: Starting point of derivation)

3. The general SSB signal is:
   A. Ac m(t) cos(wc t)
   B. (Ac/2)[m(t) cos(wc t) minus/plus mhat(t) sin(wc t)], minus for USB
   C. (Ac/2)[m(t) + mhat(t)] cos(wc t)
   D. Ac cos(wc t) + m(t)
   Answer: B. Expanding cos(wc + wm)t gives the cos-cos minus sin-sin form, where sin wm t is the Hilbert transform of cos wm t. (Key point: General SSB expression)

4. Which sign choice is correct?
   A. Minus gives USB, plus gives LSB
   B. Plus gives USB, minus gives LSB
   C. Both signs give USB
   D. The sign does not matter
   Answer: A. cos(A + B) has a minus between the products (USB); cos(A - B) has a plus (LSB). (Key point: Sign for sidebands)

## Related in Communication Systems

- [Double-tone AM; DSB vs SSB](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Adouble-tone-am-dsb-vs-ssb) — Communication Systems, chapter 2
- [DSB-FC vs DSB-SC; phase shift method for SSB](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Adsbfc-dsbsc-ssb-phase-shift) — Communication Systems, chapter 2
- [Demodulation of DSB-FC AM](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Adsbfc-demodulation-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
- [Detection gains of DSB-FC, DSB-SC and SSB](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Adetection-gain-dsbfc-dsbsc-ssb) — Communication Systems, chapter 9
- [Switching modulator for DSB-FC AM](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aswitching-modulator-dsbfc) — Communication Systems, chapter 2