---
title: "SSB wave for a low pass message signal"
subject: "Communication Systems"
chapter: 2
key: "semester_6/communication_systems:ssb-lowpass-message"
source: https://www.bctengineer.com/q/semester_6/communication_systems/ssb-lowpass-message
site: BCT Engineering
---

# SSB wave for a low pass message signal

Communication Systems · Chapter 2: Amplitude Modulation

- Asked in: 2070 Magh (CS I), 6 marks
- Video: [Lecture 18: Generation of SSB-SC #amplitudemodulation #modulation #analogcommunication](https://www.youtube.com/watch?v=gVwk-xzJgak) — GyanStation
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/ssb-lowpass-message
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-2

## Question

Derive the expression for SSB wave modulated by a low pass signal m(t).

## Short answer

Using Hilbert transform m̂(t), SSB wave is s_USB=(A_c/2)[m(t)cosωct - m̂(t)sinωct], s_LSB=(A_c/2)[m(t)cosωct + m̂(t)sinωct], derived from DSB-SC by selecting one sideband with sign functions in frequency domain. Bandwidth equals message bandwidth W.

## Answer

An **SSB wave** keeps only one sideband (upper or lower) of a DSB-SC wave. For a general low-pass message $m(t)$ of bandwidth $W$, its time-domain expression uses the **Hilbert transform** $\hat m(t)$.

#### Step 1: DSB-SC spectrum

$$
s_{DSB}(t) = A_c m(t)\cos\omega_c t \;\Rightarrow\; S_{DSB}(f) = \frac{A_c}{2}[M(f-f_c) + M(f+f_c)]
$$

#### Step 2: Pick one sideband with sign functions

The USB contains $M(f-f_c)$ only for $f > f_c$ and $M(f+f_c)$ only for $f < -f_c$. Using $u(f) = \frac{1}{2}[1+\text{sgn}(f)]$:

$$
\begin{aligned}
S_{USB}(f) &= \frac{A_c}{2}\left[M(f-f_c)\,u(f-f_c) + M(f+f_c)\,u(-f-f_c)\right] \\
&= \frac{A_c}{4}\big[M(f-f_c) + M(f+f_c)\big] \\
&\quad + \frac{A_c}{4}\big[M(f-f_c)\text{sgn}(f-f_c) - M(f+f_c)\text{sgn}(f+f_c)\big]
\end{aligned}
$$

#### Step 3: Back to time domain

The Hilbert transform satisfies $\hat M(f) = -j\,\text{sgn}(f)M(f)$, so $M(f)\text{sgn}(f) = j\hat M(f)$. Then

- First bracket ↔ $\frac{A_c}{2} m(t)\cos\omega_c t$
- Second bracket $= \frac{jA_c}{4}[\hat M(f-f_c) - \hat M(f+f_c)]$ ↔ $-\frac{A_c}{2}\hat m(t)\sin\omega_c t$

(since $\hat m(t)\sin\omega_c t \leftrightarrow \frac{1}{2j}[\hat M(f-f_c) - \hat M(f+f_c)]$).

Hence

$$
s_{USB}(t) = \frac{A_c}{2}\left[m(t)\cos\omega_c t - \hat m(t)\sin\omega_c t\right]
$$

In the same way

$$
s_{LSB}(t) = \frac{A_c}{2}\left[m(t)\cos\omega_c t + \hat m(t)\sin\omega_c t\right]
$$

#### Check with a single tone

For $m(t) = A_m\cos\omega_m t$, $\hat m(t) = A_m\sin\omega_m t$:

$$
s_{USB}(t) = \frac{A_cA_m}{2}\cos(\omega_c+\omega_m)t
$$

which is only the upper side frequency, as expected.

#### Meaning

- The term $m(t)\cos\omega_c t$ is the **in-phase** component; $\hat m(t)\sin\omega_c t$ is the **quadrature** component.
- This equation is the basis of the **phase-shift (Hartley) method**: two balanced modulators, one fed with $m(t)$ and $\cos\omega_c t$, the other with $\hat m(t)$ (−90° shifted message) and $\sin\omega_c t$, and their outputs subtracted (USB) or added (LSB).
- Bandwidth of SSB = $W$, half that of DSB.

## Check yourself

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

1. How is one sideband selected from the DSB-SC spectrum in the derivation?
   A. Using the unit step u(f) = (1 + sgn f)/2
   B. Using a Fourier series
   C. Using a multiplication by cos 2wc t
   D. Using a Gaussian window
   Answer: A. The step function keeps M(f-fc) for f > fc and M(f+fc) for f < -fc. (Key point: Sideband selection)

2. What is the Hilbert transform in frequency domain?
   A. M^(f) = -j sgn(f) M(f)
   B. M^(f) = j M(f)
   C. M^(f) = sgn(f) M(f)
   D. M^(f) = M(f - fc)
   Answer: A. Hilbert transform shifts positive frequencies by -90 degrees and negative by +90. (Key point: Hilbert transform)

3. What is the USB wave for a general message?
   A. Ac m cos wc t
   B. (Ac/2)[m cos wc t - m^ sin wc t]
   C. (Ac/2)[m cos wc t + m^ sin wc t]
   D. (Ac/2)[m^ cos wc t - m sin wc t]
   Answer: B. The minus sign gives USB; plus gives LSB. (Key point: USB expression)

4. For m = Am cos wm t, what does s_USB reduce to?
   A. (AcAm/2) cos(wc + wm)t
   B. (AcAm/2) cos wc t
   C. AcAm cos wm t cos wc t
   D. (AcAm/2) cos(wc - wm)t
   Answer: A. Only the upper side frequency remains. (Key point: Single tone check)

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