--- title: "Square law approximation for DSB-AM" subject: "Communication Systems" chapter: 2 key: "semester_6/communication_systems:square-law-approximation-dsbam" source: https://www.bctengineer.com/q/semester_6/communication_systems/square-law-approximation-dsbam site: BCT Engineering --- # Square law approximation for DSB-AM Communication Systems · Chapter 2: Amplitude Modulation - Asked in: 2071 Magh (old course), 2+6 marks - Video: [Square Law Modulator and AM generation using Analog Multiplier | Generation of AM signal](https://www.youtube.com/watch?v=OIE7h6nn1aM) — 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/square-law-approximation-dsbam - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-2 ## Question What do you mean by square law approximation? How can you use it for the modulation of DSB-AM? ## Short answer Square law approximation models nonlinear device output as i = a1v + a2v² when higher terms are negligible. For DSB-AM, the message plus carrier are applied; the v² term yields 2a2Ac m(t)cosωct, and a bandpass filter extracts s(t)=a1Ac[1+(2a2/a1)m(t)]cosωct. ## Answer #### Square law approximation Any non-linear device (diode, transistor) has an output–input characteristic that can be expanded as a power series: $$ i = a_0 + a_1v + a_2v^2 + a_3v^3 + \dots $$ When the input is small and the device is biased in its non-linear region, terms above the second order are negligible. Then $$ i \approx a_1v + a_2v^2 $$ This is the **square law approximation**. The $v^2$ term multiplies the input components with each other, so it can produce product terms such as $m(t)\cos\omega_ct$ needed for modulation (and for detection). #### Square law modulator for DSB-AM ```text m(t) -->(+)--> v1 --> Non-linear --> v2 --> BPF --> s(t) ^ device (diode) (fc,2W) DSB-AM | v2 = a1 v1 + a2 v1^2 Ac cos wc t ``` Input: $$ v_1(t) = m(t) + A_c\cos\omega_ct $$ Output: $$ \begin{aligned} v_2(t) &= a_1[m(t) + A_c\cos\omega_ct] + a_2[m(t) + A_c\cos\omega_ct]^2 \\ &= a_1m(t) + a_1A_c\cos\omega_ct + a_2m^2(t) \\ &\quad + 2a_2A_cm(t)\cos\omega_ct + \frac{a_2A_c^2}{2}(1 + \cos2\omega_ct) \end{aligned} $$ | Term | Spectrum location | |---|---| | $a_1m(t)$ | 0 to $W$ | | $a_2m^2(t)$ | 0 to $2W$ | | $\frac{a_2A_c^2}{2}$, $\frac{a_2A_c^2}{2}\cos2\omega_ct$ | DC, $2f_c$ | | $a_1A_c\cos\omega_ct$ | $f_c$ (carrier) | | $2a_2A_cm(t)\cos\omega_ct$ | $f_c - W$ to $f_c + W$ (sidebands) | A BPF centred at $f_c$ with bandwidth $2W$ passes: $$ s(t) = a_1A_c\left[1 + \frac{2a_2}{a_1}m(t)\right]\cos\omega_ct $$ which is DSB-AM with amplitude sensitivity $k_a = 2a_2/a_1$. ```text |V2(f)| m, m^2 carrier+sidebands 2fc /\____ /\ | /\ | 0 2W fc-W fc fc+W 2fc [ BPF passes ] ``` **Conditions:** $f_c > 3W$ (so the $m^2$ band up to $2W$ does not overlap $f_c - W$); $|k_am(t)| \le 1$ to avoid over-modulation; small signals so higher-order terms are negligible, otherwise distortion appears. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the square law approximation? A. For small inputs in the non-linear region, output is approximated as a1 v + a2 v^2 B. The output is a1 v only C. Terms up to v^3 are kept D. The output is exactly proportional to v^2 Answer: A. Higher-order terms are neglected, leaving i = a1 v + a2 v^2. (Key point: Square law approximation) 2. Why is the v^2 term needed for modulation? A. It produces the product term m(t) cos wc t B. It generates the DC bias C. It amplifies the carrier only D. It removes the message Answer: A. Squaring m(t) + Ac cos wc t gives the cross term 2 a2 Ac m(t) cos wc t, which forms the sidebands. (Key point: Role of v^2) 3. For the square law modulator, what is ka of the BPF output? A. a2 / a1 B. 2 a1 / a2 C. 2 a2 / a1 D. a1 a2 Answer: C. s(t) = a1 Ac [1 + (2a2/a1) m(t)] cos wc t. (Key point: Amplitude sensitivity) 4. Which condition is required so the m^2(t) band (0 to 2W) does not overlap the sideband region? A. fc > W/2 B. fc < W C. fc = 2W exactly D. fc > 3W Answer: D. The lower sideband begins at fc - W, which must exceed 2W, giving fc > 3W. 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