Switching modulator for DSB-FC AM

A switching modulator generates DSB-FC AM by using a diode as an ON/OFF switch controlled by a large carrier. The switched output contains the AM wave, which is selected by a band-pass filter.

Circuit

           +------+         D
 m(t) ---->|      |-- v1 --|>|---+--- v2 ---> BPF ----> s(t)
           | sum  |              |           (fc,2W)    AM
 Ac cos -->|      |             R_L
 wc t      +------+              |
                                ---

Condition: $A_c \gg |m(t)|$, so the diode state depends only on the carrier: ON when $c(t) > 0$, OFF when $c(t) < 0$.

Analysis

Input: $v_1(t) = A_c\cos\omega_ct + m(t)$.

The diode acts like a switch, so

$$v_2(t) = v_1(t)\,g_p(t)$$

where $g_p(t)$ is a periodic 0/1 pulse train at $f_c$ with 50% duty cycle:

$$g_p(t) = \frac12 + \frac{2}{\pi}\cos\omega_ct - \frac{2}{3\pi}\cos3\omega_ct + \frac{2}{5\pi}\cos5\omega_ct - \dots$$

Multiplying:

$$\begin{aligned} v_2(t) = {} & \frac{A_c}{2}\cos\omega_ct + \frac{2}{\pi}m(t)\cos\omega_ct \\ & + \frac{m(t)}{2} + \frac{A_c}{\pi} + \frac{A_c}{\pi}\cos2\omega_ct \\ & - \frac{2}{3\pi}m(t)\cos3\omega_ct + \dots \end{aligned}$$

BPF output:

$$s(t) = \frac{A_c}{2}\left[1 + \frac{4}{\pi A_c}m(t)\right]\cos\omega_ct$$

This is DSB-FC AM with amplitude sensitivity $k_a = \dfrac{4}{\pi A_c}$. Requirement: $f_c > 2W$ so the band around $f_c$ does not overlap the baseband or $2f_c$ terms.

Waveforms

 v1(t):  carrier riding on m(t)
          /\    /\    /\    /\
         /  \  /  \  /  \  /  \
        /    \/    \/    \/    \
 v2(t):  only positive half-cycles
          /\    /\    /\    /\
         /  \  /  \  /  \  /  \
 ------ /    \/    \/    \/    \___ (clipped)
 s(t):  AM wave after BPF (envelope = m(t) shape)

Advantages: does not need an exact square-law device, works well at high power levels.