Question For the given circuit, determine the correct descending order of time constants. Given: (a) V supply = 250 V, C = 3 μF (b) V supply = 150 V, C = 1 μF (c) V supply = 100 V, C = 2 μF (d) V supply = 250 V, C = 6 μF (e) V supply = 80 V, C = 0.5 μF Options: A. (a), (c), (b), (e), (d) B. (c), (e), (a), (b), (d) C. (b), (d), (a), (c), (e) D. (d), (a), (c), (b), (e) Solution Step 1: Formula Time constant is given by: τ = R th × C Step 2: Find Thevenin Resistance Deactivate the voltage source (replace with short circuit). Compute equivalent resistance: 30kΩ || 70kΩ = (30 × 70) / (30 + 70) = 21kΩ 9kΩ + 21kΩ = 30kΩ 20kΩ || 30kΩ = (20 × 30) / (20 + 30) ...

  MMF Wave, Pole Pitch & Rotating Magnetic Field MMF Wave, Pole Pitch & Rotating Magnetic Field 1. Pole Pitch Pole pitch is the distance between two adjacent poles (N to S). 1 pole pitch = 180° electrical 1 MMF cycle = 2 pole pitches 2. MMF Wave The MMF wave represents how magnetomotive force is distributed in space. Half cycle = 1 pole pitch Full cycle = 2 pole pitches Total cycles = P / 2 3. Rotating Magnetic Field (RMF) In a 3-phase system: Three currents are 120° apart Each produces an MMF wave Combined effect → Rotating Magnetic Field The wave rotates but keeps the same shape. 4. Synchronous Speed Ns = (120 × f) / P Ns = speed in RPM f = frequency (Hz) P = number of poles 5. Concept Behind Formula Frequency gives cycles per...

  Dual Converter Problem with Solution Circuit Diagram: Dual converter system with two converters (Converter 1 and Converter 2) connected back-to-back. Each converter consists of thyristors (T1, T2, T3, T4) with an inductive load in between. Inductors of value L/2 are present on both sides of the load. The above circuit is operated at 120 V, 60 Hz supply and the load resistance R = 20 Ω , L = 40 mH , delay angles are α₁ = 60° , α₂ = 120° . The value of: Peak circulating current = 11.250 A Peak circulating current = 6.250 A Peak current of converter 1 = 8.485 A Peak current of converter 1 = 19.735 A Choose the most appropriate answer from the options given below: A. (a) and (d) only B. (b) and (c) only C. (b) and (d) only D. (a) and (c) only Solution Step 1: Calculate peak supply voltage Vm = √2 × 120 = 169.7 V S...

Real-time Signal Cleaner Simulator Advanced Filter Simulator: Signal Cleaner Instructions: Adjust the sliders below to change the amplitudes of the signals. The simulator automatically applies the filters: 20 Hz → Desired signal 50 Hz → Power-line hum (removed by Notch) 150 Hz → High-frequency noise (blocked by Low-pass) Optional controls for advanced users: Quality Factor (Q) → Controls how deep/narrow the 50 Hz notch is Low-pass cutoff frequency → Filter out high-frequency noise above this value 20 Hz Signal Amplitude 50 Hz Hum Amplitude 150 Hz Noise Amplitude Notch Quality Factor (Q) Low-pass Cutoff Frequency (Hz)

  MATLAB Code %% Advanced Filter Visualization: The "Signal Cleaner" clc; clear; close all ; % --- 1. SETUP PARAMETERS --- fs = 1000; % Sampling frequency (1kHz) T = 1.5; % Duration in seconds t = (0:1/fs:T-1/fs)'; % Time vector % --- 2. CREATE A "MESSY" INPUT SIGNAL --- % Desired Signal: 20 Hz (Clean sine wave) s1 = 1.0 * sin(2*pi*20*t); % Interference 1: 50 Hz (Strong Power-line Hum) - TARGET FOR NOTCH n1 = 0.8 * sin(2*pi*50*t); % Interference 2: 150 Hz (High-frequency Noise) - TARGET FOR LOW-PASS n2 = 0.6 * sin(2*pi*150*t); % The combined "Dirty" signal x = s1 + n1 + n2; % --- 3. DESIGN ANALOG FILTERS (using coefficients) --- % A. Notch Filter @ 50 Hz (Deep cut at exactly 50Hz) f0 = 50; w0 = 2*pi*f0; Q = 10; % Quality factor (higher = narrower notch) b_notch = [1 0 w0^2]; a_notch = [1 w0/Q w0^2]; % B. Low-Pass Filter @ 80 Hz (Blocks 150Hz) fc = 80; wc = 2*pi*fc; [b_lp, a_lp] = butter(2, wc, 's' ); % 2nd order Bu...

The Bandpass Filter: Isolating the Signal Unlike the Notch filter which rejects a frequency, the Bandpass Filter (BPF) acts as a frequency "gatekeeper." It is defined by its center frequency and its bandwidth. Analog Transfer Function \[ H(s) = \frac{B \cdot s}{s^2 + B \cdot s + \omega_0^2} \] Center (\(\omega_0\)) The peak frequency of the filter. Bandwidth (\(B\)) The width of the passband. Quality (\(Q\)) \( Q = \frac{\omega_0}{B} \) (Selectivity of the filter). MATLAB Implementation: % Define Bandpass: Center 40Hz, Bandwidth 10Hz f0 = 40; w0 = 2*pi*f0; B = 2*pi*10; b_bp = [B 0]; % Numerator: B*s a_bp = [1 B w0^2]; % Denominator: s^2 + B*s + w0^2 % View Response freqs(b_bp, a_bp); Designing a 100Hz – 200Hz Bandpass Filter Suppose we want to isolate a signal that lives between 100 H...

Pole-Zero Frequency Manipulation: 40 Hz, 50 Hz, 60 Hz This example demonstrates how to cancel, amplify, and attenuate specific frequencies using pole-zero placement in analog and digital filters. We target sinusoids at 40 Hz, 50 Hz, and 60 Hz. 1. Analog Example (s-domain) We have a signal: x(t) = sin(2π·40 t) + sin(2π·50 t) + sin(2π·60 t) Step 1: Convert frequencies to angular frequency 40 Hz → ω₁ = 2π·40 ≈ 251.33 rad/s 50 Hz → ω₂ = 2π·50 ≈ 314.16 rad/s 60 Hz → ω₃ = 2π·60 ≈ 376.99 rad/s Step 2: Construct filters Notch filter at 50 Hz (cancel 50 Hz): H notch (s) = (s² + ω₂²) / (s² + 2ζω₂s + ω₂²) Zero at ±ω₂ cancels 50 Hz; poles define notch width via damping ζ. Resonator at 40 Hz (amplify 40 Hz): H res (s) = 1 / (s² + 2ζ₁ω₁ s + ω₁²) Poles near ±ω₁ amplify 40 Hz; damping ζ₁ controls resonance sharpness. Low-pass effect for 60 Hz ...

All-Pole Filter What it Does An all-pole filter shapes the frequency response using only poles and no zeros. It can act as a low-pass, high-pass, or resonator depending on the pole locations. All-pole filters are commonly used to model resonances in physical systems or speech formants. Mathematical Model (Continuous-Time) H(s) = G / ((s - p₁)(s - p₂) ... (s - pₙ)) Where: G = gain constant pᵢ = poles of the system n = filter order All poles → numerator is just a constant → no frequency is completely canceled Frequency Response Magnitude depends entirely on pole locations Poles closer to the imaginary axis → sharper resonance Poles closer to the origin → smoother response Digital (Discrete-Time) Version H(z) = G / (1 - a₁ z⁻¹ - a₂ z⁻² - ... - aₙ z⁻ⁿ) aᵢ = filter coefficients; poles ar...