Distance and Angle in Lambertian Optical Link Let’s break this step by step and understand the math behind these lines. This is from optical wireless communication , specifically calculating distances and angles for a Lambertian LED-to-photodetector link. 1. Distance from the LED to a point on the receiver plane The MATLAB line: D1 = sqrt((XR - XT(1,1)).^2 + (YR - YT(1,1)).^2 + h^2); What it means mathematically: (XT(1,1), YT(1,1)) → x, y coordinates of LED #1 on the transmitter plane. (XR, YR) → x, y coordinates of points on the receiver plane . h → vertical distance between the LED plane and the receiver plane (height). We are computing the 3D Euclidean distance from the LED to each point on the receiver plane: \[ D_1 = \sqrt{ (X_R - X_T)^2 + (Y_R - Y_T)^2 + h^2 } \] This is just the standard 3D distance formula: \[ \text{distance} = \sqrt{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2} \] \(\Delta x = X_R - X_T\) \(\Delta y = Y_R - Y_T\) ...

Energy of a Time-Scaled Signal If a signal f(t) has energy E , then the energy of f(2t) will be: E / 2 Explanation The energy of a continuous-time signal is defined as: E = ∫ |f(t)|² dt (from -∞ to ∞) Now consider the time-scaled signal: g(t) = f(2t) Energy of g(t): Eg = ∫ |f(2t)|² dt Let: x = 2t dt = dx / 2 Substituting: Eg = ∫ |f(x)|² (dx / 2) Eg = (1/2) ∫ |f(x)|² dx Eg = E / 2 General Rule If a signal is time-scaled as f(at) , then its energy becomes: Energy = E / |a| For a = 2: Energy = E / 2 Final Answer Energy of f(2t) = E / 2 Time compression (like f(2t)) reduces the signal energy by the scaling factor.

Directivity Comparison Between AVS Arrays and Hydrophone Arrays Underwater acoustic sensing plays an important role in applications such as marine exploration, underwater communication, sonar systems, and target tracking. To detect acoustic signals in water and determine their direction of arrival, engineers commonly use sensor arrays. Two widely used sensing technologies are hydrophone arrays and Acoustic Vector Sensor (AVS) arrays . Although both can detect underwater sound, their ability to determine the direction of incoming signals differs significantly. In this article, we explore how these two sensing approaches compare in terms of directivity , which is a key parameter describing how well a sensor focuses on signals coming from a particular direction while suppressing signals from other directions. Understanding Directivity The performance of an acoustic sensor or array is often evaluated using a metric called directivity . In simple term...

Binary Cross-Entropy Loss (BCELoss) Binary Cross-Entropy Loss (BCELoss) is a loss function commonly used in binary classification problems. It measures how well a model’s predicted probabilities match the true binary labels (0 or 1). Why BCELoss is Important BCELoss is widely used in tasks such as: Binary classification Logistic regression Discriminator training in Generative Adversarial Networks (GANs) In GANs, the discriminator decides whether an image is real or fake , making BCELoss a natural choice. Mathematical Definition The Binary Cross-Entropy Loss is defined as: L(y, ŷ) = − [ y · log(ŷ) + (1 − y) · log(1 − ŷ) ] Where: y is the true label (0 or 1) ŷ is the...

1. What is Categorical Cross-Entropy (CCE)? Categorical Cross-Entropy is used when the model predicts more than two classes (multi-class classification). Examples: Image classification (cat, dog, horse) Digit recognition (0–9) News category classification Formula: Loss = - Σ y i log(p i ) Where: y i = actual class (one-hot encoded) p i = predicted probability for class i Example: Cat Dog Horse 0 1 0 Model prediction: Cat Dog Horse 0.1 0.8 0.1 Loss becomes small because the correct class probability is high. Usually used with Softmax activation . 2. What is Binary Cross-Entropy (BCE)? Binary Cross-Entropy is a loss function used when the model predicts two classes only (binary classification). Examples: Spam vs Not Spam Fraud vs Not Fraud Disease vs No Disease Formula: Loss = -[y log(p) + (1-y) log(1-p)] Where: y = actual label (0 or 1) p = predicted probability Explanation: It measures how far the pr...

Understanding Hydrophones and Vector Sensors in Underwater Acoustic Systems Right–Left Ambiguity and Hydrophones in Underwater Acoustic Systems Underwater communication and sensing rely heavily on specialized equipment designed to work efficiently in aquatic environments. One such device is the hydrophone , a sensor that functions similarly to a microphone but is specifically built to detect sound waves in water. What is a Hydrophone? A hydrophone is an underwater acoustic sensor that captures sound signals traveling through water. Most hydrophones operate using a piezoelectric transducer . This component converts pressure variations created by sound waves into electrical signals that can be analyzed and processed. While some piezoelectric devices can also transmit sound, many hydrophones are optimized primarily for receiving acoustic signals. Attempting to use certain receivers as transmitters can even damage them. Sound behaves very differently underwater. It trav...

Final Value of \(F(s)=\frac{1}{s(s-1)}\) We use the Final Value Theorem : \[ \lim_{t \to \infty} f(t) = \lim_{s \to 0} sF(s) \] Step 1: Given \[ F(s) = \frac{1}{s(s-1)} \] Step 2: Apply Final Value Theorem \[ \lim_{t\to\infty} f(t) = \lim_{s\to0} sF(s) \] Substitute \(F(s)\): \[ sF(s) = s\left(\frac{1}{s(s-1)}\right) = \frac{1}{s-1} \] Step 3: Take \(s \to 0\) \[ \lim_{s\to0} \frac{1}{s-1} = \frac{1}{0-1} = -1 \] Important Note: The Final Value Theorem is not valid here because \(F(s)\) has a pole at \(s=1\) (right half plane), meaning the system is unstable . Verification via Inverse Laplace Partial fraction expansion: \[ F(s) = \frac{1}{s(s-1)} = \frac{-1}{s} + \frac{1}{s-1} \] Inverse Laplace transform: \[ f(t) = -1 + e^{t} \] As \(t \to \infty\): \[ e^{t} \to \infty \] So the function diverges and does not have a finite final value. Conclusion Final Value Theorem (applied blindly) gi...

Impulse Response of an LTI System Consider the differential equation of a second-order system: \[ \frac{d^2 y(t)}{dt^2} - \frac{dy(t)}{dt} - 6y(t) = x(t) \] We want to find the impulse response \(h(t)\) . For an LTI system, the impulse response is the output when the input is a unit impulse \( \delta(t) \). So we set: \[ x(t)=\delta(t) \] Step 1: Take Laplace Transform Applying Laplace transform with zero initial conditions: \[ s^2Y(s) - sY(s) - 6Y(s) = X(s) \] Factor \(Y(s)\): \[ Y(s)(s^2 - s - 6) = X(s) \] Step 2: Find the Transfer Function \[ H(s)=\frac{Y(s)}{X(s)} \] \[ H(s)=\frac{1}{s^2 - s - 6} \] Factor the denominator: \[ s^2 - s - 6 = (s-3)(s+2) \] \[ H(s)=\frac{1}{(s-3)(s+2)} \] Step 3: Partial Fraction Expansion \[ \frac{1}{(s-3)(s+2)}=\frac{A}{s-3}+\frac{B}{s+2} \] Multiply both sides: \[ 1=A(s+2)+B(s-3) \] Solving gives: \(A=\frac{1}{5}\) \(B=-\frac{1}{5}\) T...