--- title: "Why AM wastes power and bandwidth; switching modulator" subject: "Communication Systems" chapter: 2 key: "semester_6/communication_systems:am-wasteful-switching-modulator" source: https://www.bctengineer.com/q/semester_6/communication_systems/am-wasteful-switching-modulator site: BCT Engineering --- # Why AM wastes power and bandwidth; switching modulator Communication Systems · Chapter 2: Amplitude Modulation - Asked in: 2073 Bhadra (CS I), 4+6 marks - Video: [Generation of AM signal: Switching Modulator Explained](https://www.youtube.com/watch?v=M4P5QCEo9WM) — ALL ABOUT ELECTRONICS - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/am-wasteful-switching-modulator - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-2 ## Question Why is conventional AM wasteful of power and bandwidth? Explain the method of conventional AM generation by using 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. ## Answer #### Why conventional AM wastes power and bandwidth **Power:** In DSB-FC AM, $P_t = P_c(1 + \mu^2/2)$. The carrier carries **no information**, yet it takes most of the power. Even at $\mu = 1$, the carrier uses $\frac{1}{1.5} = 66.7\%$ of the total and the sidebands only 33.3%. In practice $\mu \approx 0.3$ on average, giving efficiency $\frac{0.09}{2.09} \approx 4.3\%$ only. **Bandwidth:** Both sidebands carry the same information (each is a mirror image of the other about $f_c$). Sending both needs $2W$ bandwidth, while $W$ (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. ```text m(t) --->(+)--- v1 ----|>|----+------> v2 --> BPF --> s(t) ^ D | (fc, 2W) | R_L Ac cos wc t | (Ac >> |m(t)|) --- ``` **Assumption:** carrier amplitude $A_c \gg |m(t)|$, so the diode is ON during the positive half-cycles of the carrier and OFF during the negative half-cycles, regardless of $m(t)$. Input: $v_1(t) = A_c\cos\omega_ct + m(t)$. Output: $$ v_2(t) = \begin{cases} v_1(t), & c(t) > 0 \\ 0, & c(t) < 0 \end{cases} \;=\; v_1(t)\,g_p(t) $$ where $g_p(t)$ is a unit-amplitude (0/1) pulse train at $f_c$ with 50% duty: $$ g_p(t) = \frac12 + \frac{2}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{2n-1}\cos[(2n-1)\omega_ct] $$ $$ \begin{aligned} v_2(t) &= [A_c\cos\omega_ct + m(t)]\left[\frac12 + \frac{2}{\pi}\cos\omega_ct - \frac{2}{3\pi}\cos3\omega_ct + \dots\right] \\ &= \frac{A_c}{2}\cos\omega_ct + \frac{2}{\pi}m(t)\cos\omega_ct + \frac{m(t)}{2} + \frac{A_c}{\pi} + \frac{A_c}{\pi}\cos2\omega_ct + \dots \end{aligned} $$ A BPF at $f_c$ with bandwidth $2W$ keeps only the first two terms: $$ s(t) = \frac{A_c}{2}\left[1 + \frac{4}{\pi A_c}m(t)\right]\cos\omega_ct $$ This is DSB-FC AM with $k_a = \dfrac{4}{\pi A_c}$. ```text 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is conventional AM wasteful of power? A. The sidebands are suppressed B. The carrier carries no information yet takes most of the power C. The message is transmitted twice D. The modulator is a class A amplifier Answer: B. Even at mu = 1 the carrier takes 66.7% of total power and the sidebands, which carry the information, only 33.3%. (Key point: Power waste) 2. Why is bandwidth wasted in conventional AM? A. The carrier is wideband B. Harmonics are generated C. Both sidebands carry the same information, so 2W is used when W suffices D. The filter is too wide Answer: C. The sidebands are mirror images about fc, so one would be enough. (Key point: Bandwidth waste) 3. In the diode switching modulator, the key assumption is: A. Ac << |m(t)| B. The diode has a square-law characteristic C. m(t) has no DC D. Ac >> |m(t)|, so the diode is on during positive carrier half cycles regardless of m(t) Answer: D. A large carrier makes the carrier alone control the switching, so v2 = v1 times a 0/1 pulse train. (Key point: Switching modulator condition) 4. After the BPF, the output of the switching modulator is: A. (Ac/2)[1 + (4/(pi Ac)) m(t)] cos(wc t), a DSB-FC AM wave B. (4/pi) m(t) cos(wc t), DSB-SC C. m(t)/2 only D. Ac/pi only Answer: A. The terms (Ac/2) cos wc t and (2/pi) m(t) cos wc t survive, giving AM with ka = 4/(pi Ac). (Key point: Result of filtering) ## Related in Communication Systems - [Switching modulator for DSB-FC AM](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aswitching-modulator-dsbfc) — Communication Systems, chapter 2 - [Define AM; switching modulator for DSB-FC](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Adefine-am-switching-modulator) — Communication Systems, chapter 2 - [SSB vs AM and DSB-SC; balanced modulator](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Assb-vs-am-balanced-modulator) — Communication Systems, chapter 2 - [DSB-AM vs SSB; ring modulator for DSB-SC](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Adsbam-ssb-compare-ring-modulator) — Communication Systems, chapter 2 - [SSB vs full carrier AM; ring modulator](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Assb-vs-am-ring-modulator) — Communication Systems, chapter 2 - [Multi-tone AM: spectrum, bandwidth, power efficiency](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aam-50-0-3-0-2-spectrum) — Communication Systems, chapter 2 ## Related in other subjects - [AM wave power and bandwidth numerical](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Aam-numeric-40-carrier-power) — Data Communication, chapter 5 - [Conventional vs renewable energy for generation](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aconventional-vs-renewable-generation) — Power System Analysis I, chapter 1, asked 2× - [Conventional exchange functions; PCM switching advantages](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Aconventional-exchange-pcm-advantages) — Telecommunications, chapter 4 - [Series RLC resonance, half power, bandwidth](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Aseries-resonance-half-power-bandwidth) — Electrical Circuit Theory, chapter 1, asked 3×