31) A voice signal band limited to 3.4 kHz is sampled at 8 kHz and pulse code modulated using 64 quantization levels. Ten such signals are time division multiplexed using an 5-bit synchronizing word. The minimum channel band width will be (1) 64 kHz (2) 128 kHz (3) 320 kHz (4) 520 kHz Answer: Option D Solution Step 1: Calculate bits per sample (n) The number of quantization levels (L) is 64. Formula: n = log 2 (L) n = log 2 (64) = 6 bits per sample Step 2: Calculate total bits per TDM frame A frame contains one sample from each of the 10 signals plus the sync bits. Total bits = (Number of signals × bits per sample) + synchronization bits Total bits = (10 × 6) + 5 = 65 bits per frame Step 3: Calculate the bit rate (R b ) The sampling rate is given as 8 kHz (8,000 samples per second). R b = Sampling Rate × Bits per frame R b = ...

  MATAB Code %% ========================================================== %  WSS / Noise Demonstration Simulator %  Sinusoid + Multiple Noise Types %  Distribution + Autocorrelation + PSD %  LTI System Demonstration %% ========================================================== clear; close all; clc; %% Parameters Fs = 1000;              % Sampling Frequency N  = 5000;              % Number of Samples t  = (0:N-1)/Fs; A  = 1;                 % Sinusoid Amplitude f0 = 20;                % Sinusoid Frequency signal = A*sin(2*pi*f0*t); noiseVariance = 0.25; noiseStd = sqrt(noiseVariance); %% ========================================================== % Generate Noise Types %% ========================================================== % Gaussian Noise gaussianNoise = noiseStd*randn(1,N); % Uniform No...

Wide Sense Stationary Simulator Number of Samples: Mean: Variance: Channel Impulse Response: Choose Input Type: WSS (Gaussian) Noisy Sinusoid Uniform Noise Laplace Noise Binary Noise Pink Noise Amplitude: Frequency (Hz): Upload CSV and See Output Channel Impulse Response Channel Impulse Response Output About the Wide Sense Stationary Simulator This simulator provides an interactive environment for studying the behavior of Wide Sense Stationary (WSS) random processes and their response when passed through a Linear Time-Invariant (LTI) system . Users can gen...

  UGC-NET Electronic Science Question Paper With Answer Key Download Pdf [Sept. 2022] Download Question Paper               See Answers   2025 | 2024 | 2023 | 2022 | 2021 | 2020 UGC-NET Electronic Science  Sept. 2022 Answers with Explanations 1. (C) 2. (A) 3. (B) 4. (B) 5. (B) 6. (A) 7. (B) 8. (A) 9. (C) 10. (B) 11. A, B, C. D 12. (B) 13. (D) 14. (A) 15. B Solution min slew rate = 2*pi*f*Vpeak    (Unit V/s) = 2*pi* 20*10^3*10 = 12.56*10^5 V/s = 1.25 V/microsec 16. (B)   17) Consider the expression Y = P ⊕ Q ⊕ R where P, Q, R are the input variables and Y is the output variable. Y will be logic 0 if: an odd number of input variables are 1 an even number of input variables are 1 an odd number of input variables are 0 an even number of input variables are 0 an odd number of input variables are between 0 and 1 ...

26. A sphere of radius r₁ = 30 cm has a charge density variation ρ(r) = ρ₀ (r / r₁) where ρ₀ = 200 pC/m³ . Find the total charge on the sphere. Options: 17 pC 8.5 pC 34 pC 200 pC Answer: Option A Solution The charge element is: dQ = ρ dV For a spherical shell: dV = 4πr² dr Substitute ρ(r): dQ = ρ₀ (r / r₁) × 4πr² dr Total charge: Q = ∫₀ʳ¹ ρ₀ (r / r₁) 4πr² dr Q = (4πρ₀ / r₁) ∫₀ʳ¹ r³ dr Q = (4πρ₀ / r₁) [r⁴ / 4]₀ʳ¹ Q = πρ₀ r₁³ Now substitute values: r₁ = 0.3 m Q = π × 200 × (0.3)³ pC Q = π × 200 × 0.027 pC Q = 5.4π pC Q ≈ 17 pC Final Answer: 17 pC ...

  17) Consider the expression Y = P ⊕ Q ⊕ R where P, Q, R are the input variables and Y is the output variable. Y will be logic 0 if: an odd number of input variables are 1 an even number of input variables are 1 an odd number of input variables are 0 an even number of input variables are 0 an odd number of input variables are between 0 and 1 Choose the correct answer from the options given below: (A), (C) only (E), (D) only (A), (D) only (B) only Answer: Option D  Previous yr Question papers with Full Explanations → UGC-NET : Electronics Science Study Materials →

The Derivation of 2 or 4-Fold Transforms In digital signal processing (DSP), the Discrete Fourier Transform (DFT) is more than just a tool for frequency analysis; it possesses elegant mathematical symmetries. One of the most fascinating is the Duality Property . But what happens when you nest the DFT operator four times? Let’s derive the proof step-by-step. 1. The Fundamental Definitions To begin, we define an N -point sequence x[n] and its DFT X[k] . Let the symbol Å represent the DFT operation. DFT: X[k] = F{x[n]} = ∑ n=0 N-1 x[n] e -j(2π/N)nk IDFT: x[n] = (1/N) ∑ k=0 N-1 X[k] e j(2π/N)nk 2. Applying DFT Twice (The Duality Core) The Challenge: What is the result of applying the DFT to the frequency sequence X[k] ? Let g[m] = F{X[k]} . Following the DFT definition: g[m] = ∑ k=0 N-1 X[k] e -j(2π/N)km If we look closely at the IDFT formula and multiply both sides by N , we get:...

  Q.59 The relationship between any N -length sequence x[n] and its corresponding N -point discrete Fourier transform X[k] is defined as X[k] = F{x[n]}. Another sequence y[n] is formed as below y[n] = F{F{F{F{x[n]}}}}. For the sequence x[n] = {1, 2, 1, 3} , the value of y[0] is ___________ Answer: 16 Solution This problem relies on the Duality Property of the Discrete Fourier Transform (DFT). 1. The Property: Applying the DFT twice to a sequence x[n] of length N results in a time-reversed and scaled version of itself: F{F{x[n]}} = N ⋅ x[(-n) N ] 2. Deriving y[n]: The problem asks for the result of four consecutive DFT operations: y[n] = F 4 {x[n]} = F 2 { Å 2 {x[n]} } = N ⋅ [ N ⋅ x( -(-n) N ) ] = N 2 ⋅ x[n] 3. Final Calculation: • Length of sequence x[n] = {1, 2, 1, 3} is N = 4 • The initial value is x[0] = 1 ...