SSB in terms of Hilbert transform; SSB generation

A single-sideband (SSB) signal contains only one sideband (upper or lower) of a DSB-SC signal, so it needs only bandwidth $W$, half that of DSB.

Derivation of SSB in terms of Hilbert transform

Tools: the Hilbert transform $\hat m(t)$ of $m(t)$ has $\hat M(f)=-j\,\text{sgn}(f)M(f)$. The pre-envelope (analytic signal) is

$$m_+(t)=m(t)+j\hat m(t)\ \leftrightarrow\ M(f)+\text{sgn}(f)M(f)=2M(f)u(f)$$

So $m_+(t)$ contains only the positive-frequency part of $M(f)$, doubled. Similarly $m_-(t)=m(t)-j\hat m(t)\leftrightarrow 2M(f)u(-f)$ contains only negative frequencies.

Upper sideband: the USB spectrum is $M(f)$ shifted right by $f_c$ for its positive half and left by $f_c$ for its negative half:

$$S_{USB}(f)=\frac{A_c}{4}\left[2M(f-f_c)u(f-f_c)+2M(f+f_c)u(-f-f_c)\right]$$

Taking the inverse transform term by term:

$$\begin{aligned} s_{USB}(t)&=\frac{A_c}{4}\left[m_+(t)e^{j2\pi f_ct}+m_-(t)e^{-j2\pi f_ct}\right]\\ &=\frac{A_c}{4}\left[(m+j\hat m)e^{j\omega_ct}+(m-j\hat m)e^{-j\omega_ct}\right]\\ &=\frac{A_c}{4}\left[2m(t)\cos\omega_ct-2\hat m(t)\sin\omega_ct\right]\\ &=\frac{A_c}{2}\left[m(t)\cos\omega_ct-\hat m(t)\sin\omega_ct\right] \end{aligned}$$

Lower sideband: using $m_-$ at $+f_c$ and $m_+$ at $-f_c$, in the same way:

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

Combined: $s_{SSB}(t)=\frac{A_c}{2}\left[m(t)\cos\omega_ct\mp\hat m(t)\sin\omega_ct\right]$ (– for USB, + for LSB).

Check with a tone: $m=A_m\cos\omega_mt$, $\hat m=A_m\sin\omega_mt$. Then $s_{USB}=\frac{A_cA_m}{2}[\cos\omega_mt\cos\omega_ct-\sin\omega_mt\sin\omega_ct]=\frac{A_cA_m}{2}\cos(\omega_c+\omega_m)t$, a single line at $f_c+f_m$, as expected.

Generation: phase-shift (phase discrimination) method

This method implements the above equation directly.

           +->[Mod 1]<-- cos wct ---+
           |   m cos wct            |
 m(t)------+                   Carrier osc
           |                        |
           +->[-90 deg]->[Mod 2]<-[-90 deg]
              m^(t)       m^ sin wct
                              |
     [Mod1] --(+)-- [Mod2]: sum (-/+) -> SSB

1. Balanced modulator 1 multiplies $m(t)$ by $\cos\omega_ct$, giving DSB-SC $m\cos\omega_ct$.
2. A wideband $-90°$ phase shifter (Hilbert transformer) produces $\hat m(t)$; a second $-90°$ shifter gives $\sin\omega_ct$ from the carrier. Balanced modulator 2 gives $\hat m\sin\omega_ct$.
3. Subtracting the outputs gives USB; adding gives LSB, because one sideband adds in phase and the other cancels.

It needs no sharp sideband filter, so it works at any carrier frequency and with message spectra extending to low frequencies; its difficulty is building an exact $90°$ shifter over the whole audio band.