📡 AM Modulation & Power Simulator Carrier • Message • AM Waveform • Modulation Index • Power Teaching objective: This simulation demonstrates an important concept: Carrier frequency and message frequency determine the frequency location of the AM signal, but they do not by themselves determine the transmitted power. The transmitted power depends mainly on the carrier power and modulation index. 🎛️ Signal Controls Carrier Frequency f c 100 kHz Carrier frequency changes the number of carrier cycles. Message Frequency f m 5 kHz Message frequency controls the envelope variation. Carrier Power P C Reference unmodulated carrier power. Modulation Index μ 1.00 μ = 1 → 100% modulation VSB Vestige k 50% 0% → SSB-like, 100% → DSB-like 📈 AM Waveform Visualization Carrier Message AM envelope AM signal ✓ μ = 1.00 → 100% modulation....

  The Mathematics Behind the RSA Algorithm RSA is one of the classic public-key cryptographic algorithms. Its mathematics comes primarily from number theory , especially prime numbers, modular arithmetic, Euler's theorem, and modular inverses. 1. The Big Picture RSA has two keys: Public key — anyone can know it. Private key — must be kept secret. Message   →   Encryption using public key   →   Ciphertext Ciphertext   →   Decryption using private key   →   Original Message The fascinating part is that encryption and decryption are based on exponentiation modulo a large number. Encryption: c = m e mod n Decryption: m = c d mod n 2. First: What Does "mod" Mean? The word mod means "remainder after division." Example 1 17 divided by 5 gives: 17 = 5 × 3 + 2 Therefore: 17 mod 5...

Home / UGC NET PYQ / June 2026 Solved UGC NET Electronic Science June 2026 Question Paper with Answer Key and Full Explanations 📥 Download Question Paper (PDF) 2025 2024 2023 2022 2021 2020 Explanations 1. Answer: Option (3) Total Students = 900 Urban Students = 7/10*400 + 4/5*500 = 680 680/900 * 100 = 75.56% 2. Answer: Option (1) Urban Students = 7/10*400 + 4/5*500 + 4/7*700 + 1/2 * 600 = 1380 3. Answer: Option (3) Male = 700*4/7 = 400 Female = 700*3/7=300 400-300=100 100/400*100 = 25% 4. Answer: Option (2) 400*3/10*40/100 + 500*1/5*60/100+700*3/7*50/100+600*1/3*30/100 = 318 5. Answer: Option (2) Total Female Students in B,C, and D colleges = 600 Female Post-graduates  = 100*60/100 + 300*50/100 + 200*30/100 = 270 270/600 *100 = 45% 6. Answer: Option (3) 7. Answer: Option (2) 8. Answer: Option (4) 9. Answer: Option (4) 10. Answer: Option (2) 11. Answer: Option (1) 12. Answer...

  Given, 𝑧^2= 2𝑥^2− 𝑦^2 F(x, y, z) = 2𝑥^2− 𝑦^2 - 𝑧^2 normal vector, ∇F = (4x, -2y, -2z) Evaluating at P = (1,0,√2) ∇F(1,0,√2) = (4, 0, 2√2) Normalizing the vector we get √2 i / √3 - 1 k /√3 Answer D

  Given,  𝑤⃗⃗ ̇ =𝐴𝑤⃗⃗ ,  𝑤⃗⃗ = e^t *  ux + e^-2t *  uy  𝑤⃗⃗ ̇ = e^t * ux - 2* e^-2t * uy with 𝑤⃗⃗ (𝑡=0)= ux + uy [ 1 ]   1 A.[e^t     ]     =   [  e^t     ]      e^-2t               -2e^-2t Or, A = [1   0  ]               0   -2 Answer A

  GATE Electronics and Communication (EC) Questions Paper With Answer Key Download Pdf [2026] Download Question Paper                See Answers   2026 | 2025 | 2024 | 2023 | 2022 | 2021 | 2020 GATE - EC 2026 Answers with Explanations  Q.1 Answer B Q.2 1st = 100 4th = 76 So the four top students are somewhere between 100 and 76. That's only a 24-mark range: 100 - 76 = 24 < 25 Hence those four students automatically satisfy S3. Answer A Q.3 Answer D Q.4 (log 𝑝^1/𝑛 𝑦)(log 𝑦^1/𝑛 𝑝)=16 Or, n.log p y * n.log y P = 16 Or, n^2 = 16 Or, n = 4 Answer B Q.5 Answer B Q.6 Answer A Q.7 Answer C Q.8 P k = 4*k P 1 +P 2 +P 3 +...+P 10 = 4 + 8 + 12 + ... +40 =4(1+2+3+...+10) =4*10*(10+1)/2 =220 Answer C Q.9 Answer C Q.10 Answer C Q.11 Answer A  See solution Q.12 Answer D  See solution Q.13 x(t) = u(t-2) * tu(t) L{x(t)} = (e^-2s) / s  . 1/s^2 = (e^-2s)/s^3 Answer D Q.14 Answer C Q.15 x1[n] = cos(2* π*10*n/...

  For T= 0,  Delay = (2+1) = 3ns For T= 1 Delay = (2+1) + (2+1) = 6ns For worst case it is 6 ns Answer (c) 6 ns

Flat vs Frequency-Selective Fading Simulator Flat vs Frequency-Selective Fading Interactive Multipath Channel Simulator Change the channel mode, delay spread, signal bandwidth, Doppler frequency, and SNR. Observe how the channel response changes. Parameters Channel Mode Flat Fading Frequency-Selective Fading Signal Type Pure Sine Wave Wideband Multitone Signal Center Frequency 1.00 MHz Signal Bandwidth 1.00 MHz Maximum Doppler 30 Hz SNR 25 dB Reflected Path Power -3 dB Maximum Path Delay 2.0 μs Number of Paths 4 Run Simulation Channel -- Signal BW -- Coherence BW -- RMS Delay Spread -- 1. Transmitted Signal 2. Channel Impulse Response 3. Channel Frequency Response 4. Received Signal 5. Signal Spectrum 6. Received Envelope 7. Doppler ...