USB-SSB expression for a random band limited message

SSB-SC transmits only one sideband of a DSB-SC wave. For a random band-limited message $m(t)$ (bandwidth $W$), the upper sideband (USB) signal is expressed using the Hilbert transform $\hat m(t)$ of the message.

1. Starting point: DSB-SC

$$s_{DSB}(t) = A_cm(t)\cos\omega_ct \;\Leftrightarrow\; S_{DSB}(f) = \frac{A_c}{2}[M(f-f_c) + M(f+f_c)]$$

      M(f)                    S_DSB(f)
       /\               USB LSB      LSB USB
      /  \           __/|\__        __/|\__
 ----/----\----   --/---|---\-----/---|---\--> f
    -W  0  W        -fc          fc

2. Selecting the USB

The USB is the part of $S_{DSB}(f)$ with $|f| > f_c$. Using the unit step in frequency, $u(f) = \frac{1}{2}[1 + \text{sgn}(f)]$:

$$S_{USB}(f) = \frac{A_c}{2}M(f-f_c)u(f-f_c) + \frac{A_c}{2}M(f+f_c)u(-(f+f_c))$$

Substituting $u$:

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

3. Hilbert transform

The Hilbert transform shifts every frequency component of $m(t)$ by −90°:

$$\hat m(t) = m(t) * \frac{1}{\pi t} \;\Leftrightarrow\; \hat M(f) = -j\,\text{sgn}(f)\,M(f)$$

So $M(f)\,\text{sgn}(f) = j\hat M(f)$, and

$$M(f-f_c)\text{sgn}(f-f_c) - M(f+f_c)\text{sgn}(f+f_c) = j[\hat M(f-f_c) - \hat M(f+f_c)]$$

4. Inverse Fourier transform

Using the modulation theorem:

$$\begin{aligned} m(t)\cos\omega_ct &\Leftrightarrow \frac{1}{2}[M(f-f_c) + M(f+f_c)] \\ \hat m(t)\sin\omega_ct &\Leftrightarrow \frac{1}{2j}[\hat M(f-f_c) - \hat M(f+f_c)] \end{aligned}$$

So the first bracket of $S_{USB}(f)$ gives $\frac{A_c}{2}m(t)\cos\omega_ct$, and the second gives

$$\frac{jA_c}{4}[\hat M(f-f_c) - \hat M(f+f_c)] = \frac{jA_c}{4}\cdot 2j\,\mathcal{F}\{\hat m(t)\sin\omega_ct\} = -\frac{A_c}{2}\mathcal{F}\{\hat m(t)\sin\omega_ct\}$$

Therefore

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

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

5. Verification with a single tone

Let $m(t) = A_m\cos\omega_mt$, so $\hat m(t) = A_m\sin\omega_mt$:

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

Only the upper side frequency is present, which confirms the result.

6. Interpretation

• $m(t)\cos\omega_ct$ is the in-phase part; $\hat m(t)\sin\omega_ct$ is the quadrature part that cancels the lower sideband.
• The bandwidth is $W$, half of DSB, and the power is that of one sideband only.
• The envelope is $\frac{A_c}{2}\sqrt{m^2 + \hat m^2}$, not $m(t)$, so coherent detection is required.
• The equation directly gives the phase-shift method of SSB generation:

 m(t) -+---------->[ X ]--+
       |            ^cos  |       +
       |            |     +-->(+/-)--> s_SSB
       |   [ Osc ]--+     |       -
       |      |-90 deg    |
       +-[-90 deg]->[ X ]-+
          (Hilbert)  ^sin

(subtract for USB, add for LSB).