Quantum Key Distribution (QKD): An Intuitive Explanation This explanation rebuilds the ideas behind Quantum Key Distribution (QKD) from the ground up, focusing on intuition first and math second. Big Picture Quantum Key Distribution is a way for two parties (Alice and Bob) to: Create a shared random secret key Detect if anyone (Eve) tried to eavesdrop The core idea: measuring a quantum system changes it . Everything else in QKD follows from this rule. 1. What Are Quantum “States”? A photon can be polarized in different directions: Horizontal (↔) Vertical (↕) Diagonal (↗) Other diagonal (↘) We label these using symbols: |0⟩ = horizontal |1⟩ = vertical These form one basis (a way of asking questions). ...

Lossless Transmission Line Calculator Assumptions: R = 0, G = 0 (Lossless line) Input Parameters Inductance per unit length (L) [H/m] Capacitance per unit length (C) [F/m] Frequency (f) [Hz] Load Impedance Z L [Ω] Incident Voltage V⁺ [V] Line Length λ/4 (odd multiple) λ/2 (any multiple) Calculate Results Results will appear here.

Instructions for Fast Fourier Transform (FFT) Simulator Step 1: Click on "Generate Input Signal" to generate the signal. Step 2: Enter the input frequency value in Hz in the parameters section. Step 3: Select the base signal from the dropdown menu. Step 4: Choose an operation such as addition, multiplication, or convolution from the dropdown menu. Step 5: Click the "Simulate" button to run the simulation. Step 6: Reset the simulator by clicking the "Reset Simulator" button. Generate Input Signal Parameters Input Signal Frequency (Hz) Pulse Width (s) Magnitude Plot Phase Plot Base Signal Select an input Sine Wave Cosine...

Dynamic Uniform Mid-Rise Quantizer Signal (comma separated): Vmin: Vmax: Levels (L): Quantize

Uniform Mid-Rise Quantizer (8 Levels, Range [-8, +8]) Your function implements a uniform mid-rise quantizer over: \[ V_{\min} = -8, \quad V_{\max} = 8 \] with: \[ L = 8 \text{ total quantization levels} \] Step Size (Resolution) \[ \Delta = \frac{V_{\max} - V_{\min}}{L} \] Substitute values: \[ \Delta = \frac{16}{8} = 2 \] Each quantization interval has width 2 . Decision Boundaries (Partitions) \[ P_i = V_{\min} + i\Delta \] For \( L = 8 \): \[ [-8, -6, -4, -2, 0, 2, 4, 6, 8] \] Representation Levels (Midpoints) \[ C_i = V_{\min} + (i + 0.5)\Delta \] The 8 quantized output levels: \[ [-7, -5, -3, -1, 1, 3, 5, 7] \] 4 negative levels 4 positive levels No level exactly at 0 → mid-rise quantizer Quantization Rule Step 1 — Compute interval index \[ k = \left\lfloor \frac{x - V_{\min}}{\Delta} \right\rfloor \] Step 2 — Compute quantized value \[ Q(x) = V_{\min} + (k + 0.5)\Delta \] Step 3 — Clip to valid range \[ Q(x)...

Derivation of Frequency Separation for Orthogonality Step 1: Define BFSK Signals s₁(t) = √(2E b /T) cos(2πf₁t) s₂(t) = √(2E b /T) cos(2πf₂t) Defined over: 0 ≤ t ≤ T For orthogonality: ∫₀ᵀ s₁(t)s₂(t) dt = 0 Step 2: Remove Constants ∫₀ᵀ cos(2πf₁t) cos(2πf₂t) dt = 0 Step 3: Use Trigonometric Identity cos A cos B = ½ [ cos(A − B) + cos(A + B) ] Applying identity: ½ ∫₀ᵀ [ cos(2π(f₁ − f₂)t) + cos(2π(f₁ + f₂)t) ] dt Step 4: Focus on Frequency Difference The second term integrates to zero for high carrier frequencies. ∫₀ᵀ cos(2πΔf t) dt ...

Q In a digital communication system employing Frequency Shift Keying (FSK), the 0 and 1 bits are represented by sine waves of 10 kHz and 25 kHz respectively. These waveforms will be orthogonal for a bit interval of: (a) 45 µsec    (b) 200 µsec    (c) 50 µsec    (d) 250 µsec Solution Given: Binary FSK system with frequencies \( f_1 = 10 \text{ kHz} \) \( f_2 = 25 \text{ kHz} \) For orthogonality in BFSK: \[ |f_2 - f_1| = \frac{1}{2T} \] Frequency separation: \[ \Delta f = 25 - 10 = 15 \text{ kHz} \] So, \[ 15 \times 10^3 = \frac{1}{2T} \] \[ T = \frac{1}{2 \times 15 \times 10^3} \] \[ T = \frac{1}{30 \times 10^3} \] \[ T = 33.3 \,\mu s \] Among the given options, the closest value is: \[ \boxed{45 \,\mu s} \] Correct Answer: (a) 45 µsec

Jordan Decomposition The goal of a Jordan decomposition is to diagonalize a given square matrix. If there is an invertible n×n matrix C and a diagonal matrix D such that A=CDC-1, then an n×n matrix A is diagonalizable. Procedure- Choose a square matrix ( m X m) (e.g., 3 X 3, 4 X 4, 5 X 5, etc.,) Otherwise-Pop up error – select number of rows and Columns should be same (or matrix dimension mismatched) For a given matrix, A = $\begin{bmatrix} 2 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 2 \end{bmatrix}$ The aim of Jordan decomposition is to diagonalize a given square matrix A, if A=PDP -1 is possible, where P is an invertible matrix and D is diagonal matrix. We'll go into the specifics of how matrix P and matrix D are formed later. Matrix P and D are derived from matrix A. Firstly, we’ll find the eigen values of the matrix A | A – λ*I | = 0 (I = identity matrix) Or, $\begin{bmatrix} 2 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 2 \end{bma...