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Why AM wastes power and bandwidth; switching modulator

Short answer

In DSB‑FC AM the carrier carries no information yet uses most of the power, and both sidebands carry identical information so half the bandwidth is wasted. A switching modulator uses a diode switch driven by a large carrier; the switched output passes through a band‑pass filter to yield s(t) = (Ac/2)[1 + (4/(πAc))m(t)]cos ωct.

Why is conventional AM wasteful of power and bandwidth? Explain the method of conventional AM generation by using switching modulator.

Why conventional AM wastes power and bandwidth

Power: In DSB-FC AM, Pt=Pc(1+μ2/2)P_t = P_c(1 + \mu^2/2). The carrier carries no information, yet it takes most of the power. Even at μ=1\mu = 1, the carrier uses 11.5=66.7%\frac{1}{1.5} = 66.7\% of the total and the sidebands only 33.3%. In practice μ≈0.3\mu \approx 0.3 on average, giving efficiency 0.092.09≈4.3%\frac{0.09}{2.09} \approx 4.3\% only.

Bandwidth: Both sidebands carry the same information (each is a mirror image of the other about fcf_c). Sending both needs 2W2W bandwidth, while WW (one sideband) is enough. So half the bandwidth is wasted.

Switching modulator

A diode used as a switch, driven by a large carrier, generates DSB-FC AM.

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

Assumption: carrier amplitude Ac≫∣m(t)∣A_c \gg |m(t)|, so the diode is ON during the positive half-cycles of the carrier and OFF during the negative half-cycles, regardless of m(t)m(t).

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

Output:

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

where gp(t)g_p(t) is a unit-amplitude (0/1) pulse train at fcf_c with 50% duty:

gp(t)=12+2π∑n=1∞(−1)n−12n−1cos⁡[(2n−1)ωct]g_p(t) = \frac12 + \frac{2}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{2n-1}\cos[(2n-1)\omega_ct] v2(t)=[Accos⁡ωct+m(t)][12+2πcos⁡ωct−23πcos⁡3ωct+… ]=Ac2cos⁡ωct+2πm(t)cos⁡ωct+m(t)2+Acπ+Acπcos⁡2ωct+…\begin{aligned} v_2(t) &= [A_c\cos\omega_ct + m(t)]\left[\frac12 + \frac{2}{\pi}\cos\omega_ct - \frac{2}{3\pi}\cos3\omega_ct + \dots\right] \\ &= \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 + \dots \end{aligned}

A BPF at fcf_c with bandwidth 2W2W keeps only the first two terms:

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

This is DSB-FC AM with ka=4πAck_a = \dfrac{4}{\pi A_c}.

 v1:  carrier + m(t)      v2: positive half-cycles only
   /\  /\  /\               /\  /\  /\
  /  \/  \/  \/            /  \_/  \_/  \_
                         (then BPF -> AM wave)

Advantage: works at high power levels with good efficiency, does not depend on an exact square-law characteristic.

✓ Answer checked by Claude Opus 5.5 (high effort)

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