If V r e f V_{ref} is the output reference value and V f s V_{fs} is the ideal full-scale output voltage when all digital inputs are 1. Which of the following equations is correct? A. V r e f = V f s ( 1 − 1 2 n ) V_{ref} = V_{fs}\left(1-\frac{1}{2^n}\right) B. V f s = V r e f ( 1 2 n − 1 ) V_{fs} = V_{ref}\left(\frac{1}{2^n}-1\right) C. V f s = V r e f ( 1 − 1 2 n ) V_{fs} = V_{ref}\left(1-\frac{1}{2^n}\right) D. V r e f = V f s ( 1 2 n − 1 ) V_{ref} = V_{fs}\left(\frac{1}{2^n}-1\right) The correct answer is C : V f s = V r e f ( 1 − 1 2 n ) \boxed{V_{fs}=V_{ref}\left(1-\frac{1}{2^n}\right)} Why? For an n-bit DAC , the largest digital input is: 111 … 111 = 2 n − 1 111\ldots111 = 2^n-1 The DAC output is: V o = V r e f Digital input 2 n V_o=V_{ref}\frac{\text{Digital input}}{2^n} At the maximum input: V f s = V r e f 2 n − 1 2 n V_{fs}=V_{ref}\frac{2^n-1}{2^n} Separate the fraction: V f s = V r e f ( 2 n 2 n − 1 2 n ) V_{fs}=V_{ref}\left(\frac{2^n}{2^n}-\frac{1}{2^n}\r...

  Q. The final code after encoding data bits 1101 into 7-bit even parity Hamming Code is: A. 1110101 B. 1011101 C. 1010101 D. 0110101 Hamming (7,4) – Step by Step Data bits: D = [1,1,0,1] Step 1: Parity-Check Matrix H H = [1 0 1 0 1 0 1 0 1 1 0 0 1 1 0 0 0 1 1 1 1] Parity bits are at positions 1,2,4; data bits at positions 3,5,6,7. Step 2: Extract P from H Using standard form H = [P^T | I], we get: P^T (columns 3,5,6,7) = [1 1 0 1 1 0 1 1 0 1 1 1] Transpose to get P : P = [1 1 0 1 0 1 0 1 1 1 1 1] Step 3: Generator Matrix G = [I | P] G = [1 0 0 0 | 1 1 0 0 1 0 0 | 1 0 1 0 0 1 0 | 0 1 1 0 0 0 1 | 1 1 1] Step 4: Encode Data → Codeword Multiply D × G (mod 2): D = [1 1 0 1] C = D × G = [1 1 0 1 1 0 0] In the above the sequence is D1 D2 D3 D4 P1 P2 P3   (But Transmitted Codeword should be = [P1 P2 D1 P3 D2 D3 D4] or 1010101 ) Answer: C. 1010101 Another Approach Data bits: 1101 1101 For a 7-bit Hamming code , parity bi...

  Q. The A/O gates in which an additional variable or a combination of variables can be included in the logic operation are called: Options: A. AOI Gates B. Expandable Gates C. Variable Gates D. Scalable Gates The correct answer is: B. Expandable Gates \boxed{\text{B. Expandable Gates}} Why? Expandable gates allow additional variables/inputs to be incorporated into the logic operation. Final answer C. 1010101   Browse All Solved Papers (2012 - 2025) → UGC-NET : Electronics Science Study Material (Subject: 088) → Further Reading: GATE EC Previous Year Papers with Solutions

Home / UGC NET PYQ / June 2020 Solved UGC NET Electronic Science June 2020 Question Paper with Answer Key and Full Explanations 📥 Download Question Paper (PDF) 2025 2024 2023 2022 2021 2020 Explanations 1.  Answer: Option (2) For Varactor Diode, m = ln(C2/C1) / ln{(1+V2/phi) / (1+V1/phi)} 2.  Answer: Option (4) 3.  Answer: Option (2) 4.  Answer: Option (2) 5.  Answer: Option (4) 6.  Answer: Option (4)  L= (5*10^-3) / P Or, P = (5*10^-3) / (5*10^7) = 10^-10 7.  Answer: Option (4) 8.  Answer: Option (3) 9.  Answer: Option (3) 10.  Answer: Option (1) 11.  Answer: Option (2) 12.  Answer: Option (2) 13.  Answer: Option (1) Slew Rate Slew rate tells us how fast the output voltage of an amplifier/op-amp can change . It is defined as: S R = max ⁡ ∣ d...

Phase Demodulation Instructions for Phase Modulation (PM) Step 1: Click on 'Generate Message' button to generate input message signal Step 2: Then click on 'Generate Carrier' button to generate carrier signal. The carrier frequency has to be more than the message frequency and You can change frequencies using sliders Step 3: Click on 'Generate Phase Modulated Signal' button to generate Phase Modulated Signal Step 4: Click on the 'Show Frequency Spectrums of PM' button to see spectrums of the PM signal Here, β represents the phase modulation index, given by β=kp*Am​, where Am​ is the amplitude of the message signal (assumed to be fixed), and kp​ is the phase sensitivity of the modulator 5 Hz Step 1: Generate Message ...

For an electricity bill, we are charged based on the amount of electrical energy consumed. 1 unit of electricity = 1 kWh (kilowatt-hour) For a constant electrical load, the energy consumed can be calculated as: Energy (kWh) = V × I × h / 1000 Where: V = Voltage in volts (V) I = Current in amperes (A) h = Operating time in hours (h) Calculation For a resistive appliance: I = V/R P = VI = V²/R At 230 V : Appliance Resistance Current Power A 100 Ω 230/100 = 2.3 A 230²/100 = 529 W B 300 Ω 230/300 = 0.767 A 230²/300 = 176.3 W So the 100 Ω appliance consumes about 3× as much power as the 300 Ω appliance. If both run for 10 hours : 100 Ω: 0.529 kW × 10 h = 5.29 kWh 300 Ω: 0.1763 kW × 10 h = 1.763 kWh Therefore, assuming the electricity tariff is the same, the 100 Ω appliance costs about 3× more to operate. ⚡ Smart Household Electricity Bill Calculator Estimate your monthly electricity bill from units consumed, calculate slab-wise charges, model household appliances, ...