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Define AM; switching modulator for DSB-FC

Short answer

Amplitude modulation varies carrier amplitude in proportion to message. In a switching modulator, a diode switched by a large carrier multiplies the sum of message and carrier with a square wave; after bandpass filter, output is s(t) = (Ac/2)[1 + (4/(pi Ac)) m(t)] cos(omega_c t), a DSB-FC wave.

Define Amplitude Modulation. With block diagram and necessary derivations, show that switching modulator can be used to generate Double Sideband Full Carrier AM signal.

Amplitude modulation

Amplitude modulation is the process in which the amplitude of a high-frequency carrier is varied in proportion to the instantaneous value of the message, keeping frequency and phase constant:

s(t)=Ac[1+kam(t)]cos⁡2πfcts(t) = A_c[1 + k_am(t)]\cos 2\pi f_ct

Switching modulator

A switching modulator uses a diode as an ideal switch driven by a large carrier. The carrier and the message are added and applied to the diode with a resistive load, followed by a band-pass filter.

          D
 m(t)-->(+)--|>|----+-------> [ BPF  ] ---> s(t)
         ^          |         [ at fc]
 Accos   |          > RL      [BW=2W ]
 (wct)---+          >
                    |
                   GND

Assumptions: Ac≫∣m(t)∣A_c \gg |m(t)|, so the carrier alone decides when the diode is on; the diode is ideal (zero resistance when on, open when off).

Input

v1(t)=Accos⁡ωct+m(t)v_1(t) = A_c\cos\omega_ct + m(t)

Diode as a switch: the diode conducts in the positive half cycles of the carrier and is off in the negative ones:

v2(t)={v1(t),c(t)>00,c(t)<0v_2(t) = \begin{cases} v_1(t), & c(t) > 0 \\ 0, & c(t) < 0 \end{cases}

So v2(t)=v1(t) gp(t)v_2(t) = v_1(t)\,g_p(t), where gp(t)g_p(t) is a periodic pulse train of 50% duty cycle at fcf_c. Its Fourier series is

gp(t)=12+2π∑n=1∞(−1)n−12n−1cos⁡[(2n−1)ωct]=12+2πcos⁡ωct−23πcos⁡3ωct+…g_p(t) = \frac{1}{2} + \frac{2}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{2n-1}\cos[(2n-1)\omega_ct] = \frac{1}{2} + \frac{2}{\pi}\cos\omega_ct - \frac{2}{3\pi}\cos3\omega_ct + \dots

Diode output

v2(t)=[Accos⁡ωct+m(t)][12+2πcos⁡ωct−… ]=Ac2cos⁡ωct+2πm(t)cos⁡ωct+12m(t)+2Acπcos⁡2ωct+terms near 3fc,5fc,…\begin{aligned} v_2(t) &= [A_c\cos\omega_ct + m(t)]\left[\frac{1}{2} + \frac{2}{\pi}\cos\omega_ct - \dots\right] \\ &= \frac{A_c}{2}\cos\omega_ct + \frac{2}{\pi}m(t)\cos\omega_ct \\ &\quad + \frac{1}{2}m(t) + \frac{2A_c}{\pi}\cos^2\omega_ct + \text{terms near } 3f_c, 5f_c, \dots \end{aligned}

The term 2Acπcos⁡2ωct=Acπ(1+cos⁡2ωct)\frac{2A_c}{\pi}\cos^2\omega_ct = \frac{A_c}{\pi}(1 + \cos2\omega_ct) gives DC and 2fc2f_c components.

Spectrum of v2(t)v_2(t): baseband m(t)m(t) and DC (0 to WW), the wanted AM around fcf_c, components at 2fc2f_c, 3fc3f_c and so on.

Band-pass filter: centred at fcf_c with bandwidth 2W2W, it removes everything except the first two terms:

s(t)=Ac2cos⁡ωct+2πm(t)cos⁡ωct=Ac2[1+4πAcm(t)]cos⁡ωcts(t) = \frac{A_c}{2}\cos\omega_ct + \frac{2}{\pi}m(t)\cos\omega_ct = \frac{A_c}{2}\left[1 + \frac{4}{\pi A_c}m(t)\right]\cos\omega_ct

This is a DSB-FC AM wave with amplitude sensitivity

ka=4πAck_a = \frac{4}{\pi A_c}

Conditions: fc>2Wf_c > 2W so that the AM band (fc±Wf_c \pm W) does not overlap the baseband or the 2fc2f_c term, and ka∣m(t)∣≤1k_a|m(t)| \le 1 to avoid over-modulation.

✓ Answer checked by Claude Opus 5.5 (high effort)

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