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UGC NET Electronic Science Previous Year Question Papers with Solutions

Download Papers and Solutions Exam Pattern Preparation Tips FAQs More Home / Engineering & Other Exams / UGC NET 2026 PYQ ๐Ÿ“Š Exam Highlights: Electronic Science (88) Feature Details Junior Research Fellowship (JRF) ₹37,000 + HRA per month Eligibility M.Sc/M.Tech in Electronics (55%) Validity of Certificate JRF (3 Years) | Lectureship (Lifetime) ๐Ÿ“ฅ Download UGC NET Electronics PDFs Complete collection of previous year question papers, answer keys and explanations for Subject Code 88. Start Downloading ๐Ÿ“‚ View All Question Papers June 2026 - Question Paper Download PDF June 202...
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The output voltage V0 (in Volt) for the network given in the Figure is ___

  Find the Output Voltage V o Solve the given circuit using Kirchhoff's Current Law (KCL), Ohm's law, and the supernode method. 1. Given Circuit Parameters Current source 1 mA Resistors 1 kฮฉ Each resistor Voltage source 6 V Required V o 2. Understand the Circuit The circuit contains a 1 mA current source, four 1 kฮฉ resistors, and a 6 V source connected between the two midpoint nodes. The output voltage is measured between the top and bottom rails. Reference node Take the bottom rail as ground: 0 V . Let the top rail voltage be V T , and...
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GMSK BER vs SNR (with Simulation)

  GMSK Performance: Why the BER vs. SNR Curve Matters If you’ve ever wondered why GMSK (Gaussian Minimum Shift Keying) became the backbone of the global GSM revolution, the answer lies in its unique balance of spectral efficiency and power performance. But there is no free lunch in engineering. To get that sleek, narrow spectrum, you have to pay a price in Bit Error Rate (BER) . In this guide, we break down the theoretical BER vs. SNR performance of GMSK, the "magic" of the BT product, and how to optimize your link budget for real-world conditions. What is GMSK and Why Should You Care? GMSK is a derivative of MSK (Minimum Shift Keying) that uses a Gaussian pre-modulation filter. This filter "smooths" the phase transitions, significantly reducing sideband power. The result? A constant envelope signal that allows you to use high-efficiency Class-C amplifiers without distorting the signal. This is the secr...
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Parametric Down-Conversion of Photons

  Nonlinear Optics One incoming photon is converted into two lower-energy photons, typically in a nonlinear optical medium. Energy is conserved: E in = E 1 + E 2 E in ​ = E 1 ​ + E 2 ​ . Parametric Down-Conversion A nonlinear optical process in which one pump photon is converted into two lower-energy photons, known as the signal and idler photons. 1 Energy Conservation The energy of the pump photon equals the sum of the energies of the signal and idler photons. \[ \boxed{\hbar\omega_p=\hbar\omega_s+\hbar\omega_i} \] \[ \boxed{\omega_p=\omega_s+\omega_i} \] ฯ‰โ‚š Pump angular frequency ...
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SRAM, DRAM, EPROM, MRAM, and DDRAM Explained

Computer Fundamentals Types of Memory and Their Technologies Understand the fundamental relationship between different types of computer memory and the technologies used to implement them. Fundamental Matching Memory type, technology, and key characteristics Memory Type Technology Key Characteristic Cache Memory SRAM Very fast; stores frequently accessed data Main Memory DRAM Holds programs and data currently in use Au...
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The cutoff frequency (in GHz) for the dominant TE10 mode of an air-filled rectangular waveguide of ...

  Q. The cutoff frequency (in GHz) for the dominant TE10 mode of an air-filled rectangular waveguide of inner dimension 0.28 inch×0.14 inch is __. (rounded off to two decimal places). Given Information Waveguide dimensions = 0.28 inch × 0.14 inch Wider dimension (a) = 0.28 inch Smaller dimension (b) = 0.14 inch Medium = Air Dominant mode = TE 10 Speed of light (c) = 3 × 10 8 m/s Step 1: Cutoff Frequency Formula The cutoff frequency of a rectangular waveguide is: f c = (c / 2) √[(m/a) 2 + (n/b) 2 ] For the dominant TE 10 mode: m = 1, n = 0 f c = c / 2a Step 2: Convert Inch into Meter We know that 1 inch = 0.0254 meter. a = 0.28 × 0.0254 a = 0.007112 m ...
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The electric field of a monochromatic plane wave travelling in a lossless isotropic and homogenous medium is given by E(z, t) = ...

  Q. The electric field of a monochromatic plane wave travelling in a lossless isotropic and homogenous medium is given by E(z, t) = E₀ [x̂ cos(ฯ‰t − kz) + ลท sin(ฯ‰t − kz)] in a right-handed orthogonal co-ordinate system. Which of the following is the correct polarization of the electromagnetic wave? (A) Right-handed circularly polarized Correct answer under the IEEE convention. (B) Left-handed circularly polarized (C) Linearly polarized (D) Linearly polarized at −45° to x̂ Solution 1. Identify the electric-field components Let ฯˆ = ฯ‰t − kz. The components along the x and y directions are: Eโ‚“ = E₀ cos ฯˆ Eแตง = E₀ sin ฯˆ Both components have equal amplitudes, the same frequency, and a phase difference of 90°. These are the conditions for circular polarization. ...
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A wireless digital transmission scheme is using 16-QAM over an additive white Gaussian noise channel ...

  Given Information Modulation scheme: 16-QAM Bit rate: R b = 4 × 10 6 bits/sec Channel: Additive White Gaussian Noise (AWGN) Receiver: Maximum-Likelihood Receiver Step 1: Calculate Bits per Symbol For M-ary QAM, the number of bits transmitted per symbol is: Bits per symbol = log 2 (M) = log 2 (16) = 4 bits/symbol Step 2: Calculate Symbol Rate Symbol rate is calculated by dividing the bit rate by the number of bits per symbol. R s = R b / log 2 (M) = (4 × 10 6 ) / 4 = 10 6 symbols/sec Step 3: Calculate Minimum Bandwidth For ideal Nyquist transmission with zero roll-off factor: B min = R s = 10 6 Hz...
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