SSB wave for a low pass message signal

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.