Lumped vs Distributed Circuits Lumped Circuit vs Distributed Circuit The difference between lumped and distributed circuits comes down to how voltage, current, and physical size relate to the signal wavelength. Lumped Circuit Concept: Components are assumed to be concentrated at discrete points. Condition: \[ \text{Size of circuit} \ll \lambda \] Where \( \lambda \) is the wavelength of the signal. Assumptions: Voltage and current are uniform across components No propagation delay Governing Laws: Ohm's Law: \( V = IR \) Kirchhoff’s Laws (KVL and KCL) Examples: DC circuits Low-frequency AC circuits Basic electronic circuits Distributed Circuit Concept: Electrical parameters are distributed continuously along the conductor. Condition: \[ \text{Size of circuit} \approx \lambda \ \text{or larger} \] Characteristics: Voltage and current vary with ...

📘 Instructions ⚙️ Simulator 📡 Demodulate 📈 Graphs OFDM Simulation Tool Interactive Virtual Lab: Configure parameters and visualize BPSK mapping & IFFT transformations. Start Simulator Now Instructions for OFDM Modulation with BPSK Follow these steps to complete the modulation process: Note: Use the input fields to enter symbols, subcarriers, CP length, frequency, and Baud Rate. Step 1: Generate Message button for input bitstream. Step 2: 'Make OFDM Symbol' to map bits. Step 3: 'Generate Subcarr...

  FM vs AM Power Comparison Does FM Need More Power Than AM? FM generally requires more transmitted power in practical communication systems, but the reason is not straightforward. In theory, both AM and FM can transmit information without inherently requiring more average power. However, FM is designed for better noise immunity, which changes bandwidth and system requirements. 1. AM Power (DSB-LC Case) Standard AM signal: s(t) = Ac (1 + m cos ωm t) cos ωc t Where: Ac = carrier amplitude m = modulation index (0 to 1) Total AM Power: PAM = Pc (1 + m²/2) Where Pc is the carrier power. A major portion of AM power is wasted in the carrier. At maximum modulation (m = 1): PAM(max) = 1.5 Pc Only about 33% of the power is actually useful for information transfer. 2. FM Power FM signal equation: s(t) = Ac cos(ωc t + β sin ωm t) Key point...

  Pinch-Off Voltage Pinch-Off Voltage (Semiconductor Physics) Concept: Pinch-off occurs when the depletion region width becomes equal to the channel thickness, fully depleting the channel. Step 1: Depletion Width Formula For an abrupt junction: W = sqrt((2 × εs × V) / (q × N)) W = depletion width εs = semiconductor permittivity V = applied voltage (pinch-off voltage) q = 1.6 × 10⁻¹⁹ C N = doping concentration Step 2: Pinch-Off Condition At pinch-off: W = a where a = channel thickness Step 3: Derivation a² = (2 × εs × Vp) / (q × N) Final Formula Vp = (q × N × a²) / (2 × εs) Step 4: Permittivity εs = εr × ε0 Where: ε0 = 8.85 × 10⁻¹² F/m εr = dielectric constant Important Observations Higher doping (N ↑) → higher pinch-off voltage Thicker channel (a ↑) → Vp increases sharply (a² dependence) Higher dielectric constant → lowers Vp Vp = (1.6×10⁻¹⁹ × N × a²) / (2 × εr × ...

  Fan Regulator Math (Phase Control) Modern Fan Regulator (Phase Control) Modern fan regulators control power instead of just reducing voltage. They use a technique called phase control , which cuts parts of the AC waveform. Step 1: AC Voltage Equation v(t) = Vm sin(ωt) Vm = peak voltage ω = angular frequency Step 2: Cutting the Wave A TRIAC turns ON at angle α (firing angle) . 0 to α → No voltage α to π → Voltage passes Step 3: RMS Voltage The fan speed depends on RMS voltage: Vrms = Vm √[ (1 / 2π) ∫(α to π) sin²(θ) dθ ] After solving: Vrms = Vm √[ (1 / 2π) ( (π − α)/2 + sin(2α)/4 ) ] What This Means α = 0° → Full speed α = 90° → Medium speed α → 180° → Very low speed Step 4: Power Delivered P ∝ Vrms² ...

  Sampling and Z-Transform Example Sampling and Z-Transform (Step-by-Step) Step 1: Given Signal Continuous-time signal: x(t) = e -t Step 2: Sampling Sampling frequency: f s = 10 Hz Sampling period: T = 1 / f s = 0.1 sec Discrete signal: x[n] = x(nT) = e -0.1n Step 3: Z-Transform Definition: X(z) = Σ x[n] z -n Substitute x[n]: X(z) = Σ (e -0.1 ) n z -n This is a geometric series: X(z) = 1 / (1 - e -0.1 z -1 ) Multiply numerator and denominator by z: X(z) = z / (z - e -0.1 ) Step 4: Numerical Value Compute: e -0.1 ≈ 0.9048 Final answer: X(z) = z / (z - 0.9048) Step 5: Special Case If: x[n] = (0.5) n Then: X(z) = z / (z - 0.5) This corresponds to a different continuous signal: x(t) = e -6.93t Conclusion Sampling converts continuous signal to discrete signal. ...

  Von Neumann Bottleneck Von Neumann Bottleneck Introduction The von Neumann bottleneck is a fundamental limitation in traditional computer architectures where the CPU (processing unit) and memory are physically separate. Due to this separation, data must continuously move between memory and processor, creating a bottleneck that limits system performance. Why It Happens Single shared bus for data and instructions Limited data transfer bandwidth Separation of memory and processing units Problems Caused High latency Increased power consumption Limited scalability Inefficiency in data-intensive tasks Impact on AI/ML Modern AI systems require large-sca...

  Impedance Matching in GPR Antennas Impedance Matching in Ground-Penetrating Radar (GPR) Antennas 1. What “Matched Impedance” Means In a GPR system, impedance matching refers to making the antenna impedance ( Z a ) as close as possible to the soil impedance ( Z s ). Z a ≈ Z s This minimizes reflections at the boundary between the antenna and the ground. 2. Soil (Medium) Impedance The impedance of soil is given by: Z s = √(jωμ / (σ + jωε)) Where: ω = 2πf → angular frequency μ → permeability ε → permittivity σ → conductivity j → imaginary unit 3. Low-Loss Soil Approximation If conductivity is very low: Z s ≈ √(μ / ε) = Z₀ / √ε r Where Z₀ ≈ 377 Ω (free-space impedance). Example: For dry soil (ε r ≈ 4) Z s = 377 / √4 = 188.5 Ω ...