Difference between DCO-OFDM and Regular OFDM 1. Regular OFDM (Orthogonal Frequency Division Multiplexing) In standard OFDM (used in RF communication): The signal is complex-valued (has real and imaginary parts). Subcarriers carry data in QAM/PSK modulation. Signal can take both positive and negative values . Usually transmitted over RF channels . No need for DC bias or clipping because the channel can handle bipolar signals. 2. DCO-OFDM (Direct Current-biased Optical OFDM) DCO-OFDM is a variant of OFDM designed for optical communication , e.g., VLC or optical fiber . Key differences from regular OFDM: Unipolar requirement Optical transmitters (LEDs, lasers) can only emit positive light intensity . OFDM signals ar...

Galois Fields: GF(2) and GF(2 m ) 1. What is a Galois Field (GF)? A Galois Field (GF) is a finite set of elements in which the four basic arithmetic operations—addition, subtraction, multiplication, and division (except by zero)— are all well defined and closed. GF(q) ⇒ a field with exactly q elements 2. The Simplest Field: GF(2) GF(2) is the smallest possible finite field and forms the foundation of all digital systems. GF(2) = {0, 1} Addition in GF(2) Addition is performed modulo 2 (XOR operation): + 0 1 0 0 1 1 1 0 Multiplication in GF(2) × 0 1 0 0 0 1 0 1 GF(2) is used in binary logic, XOR operations, and simple error-control codes. 3. Meaning of GF(2 m ) GF(2 m ) is a finite field containing exactly 2 m elements . Each element represents an m-bit symbol . Field Number of Elements GF(2) 2 GF(2²) 4 GF(2³) 8 GF(2⁸) 256 Important: GF(2 m ) is not integer arithmetic modulo 2 m . It is polynomial-based arithmetic. 4....

Reed–Solomon Coding and Decoding 1. Input Bitstream to Symbols Given input bitstream: 101011000… Choose symbol size: m = 3 ⇒ symbols in GF(2³) Grouping bits: 101 | 011 | 000 Binary to decimal symbols: [5, 3, 0] 2. Finite Field Construction GF(2³) Primitive polynomial: p(x) = x³ + x + 1 Element Polynomial Binary Decimal α⁰ 1 001 1 α¹ α 010 2 α² α² 100 4 α³ α + 1 011 3 α⁴ α² + α 110 6 α⁵ α² + α + 1 111 7 α⁶ α² + 1 101 5 3. Message Polynomial Choose RS(7,3): n = 7, k = 3 Message symbols: [5, 3, 0] Message polynomial: m(x) = 5 + 3x + 0x² 4. Generator Polynomial Number of parity symbols: n − k = 4 Generator polynomial: g(x) = (x − α)(x − α²)(x − α³)(x − α⁴) Expanded form: g(x) = x⁴ + 6x³ + x² + 6x + 1 5. RS Encoding (Polynomial Division) Multiply message by x⁴: x⁴m(x) = 5x⁴ + 3x⁵ Divide by generator polynomial: r(x) = 6 + 4x + 2x² + 5x³ Codeword polynomial: c(x) = x⁴m(x) + r(x) Final codeword symbols: [...

Quantum Computing Fundamentals Understanding qubits, basis, Hadamard transform, and quantum probability 1. Qubit (quantum bit) A qubit is a unit vector in a 2-dimensional complex Hilbert space: \[ |\psi\rangle = \alpha |0\rangle + \beta |1\rangle \] where \[ \alpha, \beta \in \mathbb{C}, \quad |\alpha|^2 + |\beta|^2 = 1 \] \(|0\rangle, |1\rangle\) are basis states \(|\alpha|^2\) = probability of measuring 0 \(|\beta|^2\) = probability of measuring 1 2. Basis (measurement basis) A basis is a set of orthonormal vectors used to describe or measure a qubit. (a) Computational (Z) basis \[ |0\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix...

How to Calculate Thevenin Equivalent Resistance To calculate the Thevenin equivalent resistance ( R th ), follow these clear steps. Method 1: Turning Off Independent Sources (Most Common) Step 1: Identify the Output Terminals Choose the two terminals where you want the Thevenin equivalent. Step 2: Turn Off All Independent Sources Voltage sources → Replace with a short circuit Current sources → Replace with an open circuit Note: Do not turn off dependent sources. Step 3: Simplify the Circuit Combine resistors in series and parallel Use circuit-reduction techniques The remaining resistance between the terminals is R th . Method 2: Using Open-Circuit Voltage and Short-Circuit Current Open-circuit voltage ( V oc ) Short-circuit current ( I sc ) R th = V oc / I sc Method 3: Using a Test Source (For Dependent Sou...

Quantum vs Classical Communication The difference between having only 10–12 qubits in quantum computers and kbps–Gbps speeds in conventional communication can be confusing. This difference exists because quantum and classical systems work in fundamentally different ways. 1. Qubits and Quantum Superposition Qubits are the basic units of quantum computing. Unlike classical bits (which are either 0 or 1), a qubit can exist in a superposition of both 0 and 1 at the same time. This means that even a small number of qubits can represent a very large number of possible states. The computing power grows exponentially as more qubits are added. Qubit (quantum bit) A qubit is a unit vector in a 2-dimensional complex Hilbert space: \[ |\psi\rangle = \alpha |0\rangle + \beta |1\rangle \] where ...

DCO-OFDM: Mathematical Model and MATLAB Implementation MATLAB Code: DCO-OFDM Modulation & Demodulation clc; clear all; close all; %% DCO-OFDM System Parameters N = 1024; % FFT size M = 4; % QPSK k = log2(M); % Bits per symbol Ncp = 256; % Cyclic Prefix num_symbols = 100; % OFDM symbols SNRdB = 0:2:30; % NEW: SNR range BER = zeros(size(SNRdB)); %% Transmitter % 1. Data Generation and QAM Modulation data_bits = randi([0 M-1], num_symbols, N/2-1); data_qam = qammod(data_bits, M, 'UnitAveragePower', true); % 2. Serial to Parallel Conversion datamat = zeros(N, num_symbols); datamat(2:N/2, :) = data_qam.'; % Data subcarriers % 3. Hermitian Symmetry datamat(1,:) = 0; % DC datamat(N/2+1,:) = 0; % Nyquist datamat(N/2+2:N,:) = flipud(conj(datamat(2:N/2,:))); % 4. IFFT signal_ifft = ifft(datamat, N, 1); % 5. Add Cyclic Prefix signal_ifft_parallel = signal_ifft.'; ...

  UGC-NET Paper 1 with Answer Key Download Pdf [2023] Download Question Paper                   2025 | 2024 | 2023 | 2022 | 2021 | 2020 UGC-NET Paper  1 - 2023 Answers with Explanations Q.1 Answer: B  Q.2 Answer: A  (40.(5/8)% decrease)   Q.3 Answer: D  (196)   Q.4 Answer: A   Q.5 Answer:  B  (150)       Q.6 Answer: B   Q.7 Answer: D     Q.8 Answer: B  (Massive Open Online Courses)   Q.9 Answer: C     Q.10 Answer: C Total = A.2^(t/2) or, 500 = 2^(t/2) or, t = 18     Q.11 Answer: B     Q.12 Answer: A   Q.13 Answer: A     Q.14 Answer: B       Q.15 Answer: C       Q.16 Answer: C       Q.17 Answer: A     Q.18 Answer: B   Q.19 Answer: B     Q.20 Answer: A     Q.21 Answer: B     Q.22 Answer: A, B ...