--- title: "Square law detector limitation; envelope detector" subject: "Communication Systems" chapter: 2 key: "semester_6/communication_systems:square-law-detector-limitation-envelope" source: https://www.bctengineer.com/q/semester_6/communication_systems/square-law-detector-limitation-envelope site: BCT Engineering --- # Square law detector limitation; envelope detector Communication Systems · Chapter 2: Amplitude Modulation - Asked in: 2070 Magh (CS I), 2+6 marks - Video: [Square Law Detector (Basics, Derivation, Working, Block Diagram & Circuit) Explained | AM detection](https://www.youtube.com/watch?v=MZmMCj0h_FA) — Engineering Funda - 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-detector-limitation-envelope - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-2 ## Question What is the limitation of square law detector for DSB-AM detection? Explain the operation of envelope detector with required diagrams and conditions. ## Short answer Square-law detector produces m²(t) distortion, requiring low modulation index; cannot recover DSB-SC. Envelope detector: diode rapidly charges capacitor to carrier peak, then capacitor slowly discharges through R_L, tracing envelope. Required: 1/f_c << R_LC << 1/W, μ≤1 to prevent ripple and diagonal clipping. ## Answer #### Limitation of the square-law detector A square-law detector passes the AM wave through a non-linear device $v_2 = a_1v_1 + a_2v_1^2$ and then a low-pass filter. For $v_1 = A_c[1+k_am(t)]\cos\omega_ct$ the useful low-frequency output is $$ v_o(t) = \frac{a_2A_c^2}{2}\left[2k_am(t) + k_a^2m^2(t)\right] $$ The $k_a^2 m^2(t)$ term is **distortion** (second harmonic and intermodulation of the message). The ratio of distortion to wanted signal is about $k_a|m(t)|/2$, so it is small only when $\mu$ is small (lightly modulated, weak signals). It therefore needs low modulation depth, works only at low signal levels, and has low efficiency. For DSB-SC (no carrier) it gives only $m^2(t)$, so the message cannot be recovered at all. #### Envelope detector An envelope detector is a diode–RC circuit whose output follows the envelope $A_c[1+\mu m(t)]$ of a standard AM wave. ```text D o----->|-----+--------+-----o AM | | input C === > RL v_o | > o------------+--------+-----o ``` **Operation** 1. In the positive half of each carrier cycle the diode is forward biased and the capacitor charges rapidly, through the small source resistance $R_s$, up to the peak value. 2. When the input drops below the capacitor voltage the diode turns off, and $C$ discharges slowly through $R_L$. 3. At the next peak the diode conducts again. The capacitor voltage thus traces the peaks, i.e. the envelope, with a small ripple at $f_c$. 4. An RC low-pass filter removes the ripple, and a blocking capacitor removes the DC level, leaving $m(t)$. ```text Input AM Output /\ /\/\ /\ _/\_ _ /\/\/\ \/\/\/ -> / \_/ \/\/\/ /\/\/\ \/ \/\/ \/ ``` **Conditions for correct working** 1. **Fast charging:** $R_sC \ll 1/f_c$, so the capacitor reaches the peak in each cycle. 2. **Slow discharge between carrier peaks:** $R_LC \gg 1/f_c$, to keep ripple small. 3. **Fast enough to follow the message:** $R_LC \ll 1/W$. If $R_LC$ is too large, the output cannot follow a falling envelope (**diagonal clipping**). For a single tone the limit is $R_LC \le \dfrac{\sqrt{1-\mu^2}}{\mu\,\omega_m}$. 4. **No over-modulation:** $\mu \le 1$, otherwise the envelope crosses zero and no longer equals $m(t)$. 5. $f_c \gg W$, so the carrier ripple can be separated from the message. Combined: $\dfrac{1}{f_c} \ll R_LC \ll \dfrac{1}{W}$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the distortion in the square-law detector output? A. The 2 ka m(t) term B. The k_a^2 m^2(t) term C. The 2 wc term D. The carrier term Answer: B. Output is proportional to 2ka m(t) + ka^2 m^2(t); the second term is distortion. (Key point: Distortion term) 2. Why does a square-law detector fail for DSB-SC? A. It needs a high carrier B. It gives 2 m(t) with a DC offset C. It gives only m^2(t), so the message cannot be recovered D. It requires a local oscillator Answer: C. Without a carrier there is no linear m(t) term. (Key point: DSB-SC) 3. Which condition ensures the capacitor reaches the peak each cycle? A. mu > 1 B. fc << W C. Rs C << 1/fc D. R_L C >> 1/W Answer: C. A small charging time constant lets C follow the peaks. (Key point: Fast charging) 4. For a single tone, the limit to avoid diagonal clipping is A. R_L C <= mu wm B. R_L C >= mu/wm C. R_L C <= 1/fc D. R_L C <= sqrt(1 - mu^2)/(mu wm) Answer: D. Higher mu or wm requires a smaller time constant. 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