A bar of silicon is doped with boron concentration of $10^{16}\,\text{cm}^{-3}$ and assumed to be fully ionized. It is exposed to light such that electron-hole pairs are generated throughout the volume of the bar at the rate of $10^{20}\,\text{cm}^{-3}\text{s}^{-1}$. If the recombination lifetime is $100\,\mu s$, intrinsic carrier concentration of silicon is $10^{10}\,\text{cm}^{-3}$, and assuming 100% ionization of boron, then the approximate product of steady-state electron and hole concentrations due to this light exposure is (A) $10^{20}\,\text{cm}^{-6}$ (B) $2\times10^{20}\,\text{cm}^{-6}$ (C) $10^{32}\,\text{cm}^{-6}$ (D) $2\times10^{32}\,\text{cm}^{-6}$ Step-by-Step Solution Step 1 : Identify the Type of Semiconductor The silicon is doped with boron. Boron is an acceptor impurity , therefore the semiconductor is p-type . Since boron is fully ionized, \[ p_0=N_A=10^{16}\,\text{cm}^{-3} \] This is the equilibrium hole ...

Let the linear convolution of two discrete-time finite-length sequences $x[n]$ and $h[n]$, each of length 16, be denoted by $y[n]$. Let $z[n]$ denote the 16-point inverse discrete Fourier transform (IDFT) of the product of the 16-point discrete Fourier transforms (DFTs) of $x[n]$ and $h[n]$. The value of $k$ for which $z[k]=y[k]$ is (A) 0 (B) 1 (C) 15 (D) 16 Solution Step 1: Length of Linear Convolution Both sequences have length 16. Length of linear convolution: $ 16+16-1=31 $ Hence the linear convolution contains the samples $ y[0],y[1],\ldots,y[30] $ Notice that there is no sample $y[31]$. Step 2: What does the IDFT of the Product of DFTs Give? A standard DFT property states: $ \text{IDFT}\{X[k]H[k]\} = \text{Circular Convolution} $ Therefore, $ z[n] = \text{16-point circular convolution of }x[n]\text{ and }h[n]. $ So, $y[n]$ = Linear Convolution $z[n]$ = 16-point Circular Convolution Step 3: Folding (Wrap-A...

Covariance vs Correlation: Difference, Formula, Examples and Comparison Introduction Covariance and correlation are two important statistical concepts used to measure the relationship between two variables. They help answer questions such as: Does advertising increase sales? Does study time affect exam marks? Are height and weight related? Although both measure relationships between variables, covariance and correlation have important differences in interpretation and usage. What is Covariance? Covariance measures the direction in which two variables change together. $$ Cov(X,Y)= \frac{\sum(X_i-\bar X)(Y_i-\bar Y)} {n} $$ Interpretation of Covariance Positive Covariance When: $$ Cov(X,Y)>0 $$ Both variables increase or decrease together. Negative Covariance $$ Cov(X,Y) One variable increases while the other decreases. Zero Covariance $$ Cov(X,Y)=0 $$ There is no linear relationship between v...

Question Find the integrating factor of the differential equation $$ \frac{dy}{dx} + \frac{x}{1-x^2}y = x\sqrt{y} $$ Solution The given differential equation is $$ \frac{dy}{dx} + \frac{x}{1-x^2}y = x\sqrt{y} $$ Since the equation contains the term $\sqrt{y}$, it is not linear. Let $$ v=\sqrt{y} $$ Then, $$ y=v^2 $$ Differentiating, $$ \frac{dy}{dx} = 2v\frac{dv}{dx} $$ Substitute these values into the given equation: $$ 2v\frac{dv}{dx} + \frac{x}{1-x^2}v^2 = xv $$ Divide both sides by $2v$: $$ \frac{dv}{dx} + \frac{x}{2(1-x^2)}v = \frac{x}{2} $$ This is now a linear differential equation of the form $$ \frac{dv}{dx}+P(x)v=Q(x) $$ where $$ P(x)= \frac{x}{2(1-x^2)}. $$ The integrating factor is $$ \text{I.F.} = e^{\int P(x)\,dx} = e^{\int \frac{x}{2(1-x^2)}dx} $$ Let $$ u=1-x^2 $$ Then, $$ du=-2x\,dx $$ Therefore, $$ \int \frac{x}{2(1-x^2)}dx = -\frac14 \int \frac{du}{u} $$ Hence, $$ = -\frac14\ln|1-x^2| $$ ...

  Line Integral Solution Solution $ \mathbf{F}(x,y)=-x\hat{i}+y\hat{j} $ $$ \int_C \mathbf{F}\cdot d\mathbf{r} $$ $$ =\int_C(-x\,dx+y\,dy) $$ Parameterization: $$ x=\cos t,\qquad y=\sin t $$ $$ dx=-\sin t\,dt,\qquad dy=\cos t\,dt $$ $$ \int_{0}^{\pi/4} \left[ (-\cos t)(-\sin t) + (\sin t)(\cos t) \right]dt $$ $$ = \int_{0}^{\pi/4} 2\sin t\cos t\,dt $$ where $$ 0\le t\le\frac{\pi}{4} $$ $$ 2\sin t\cos t=\sin2t $$ $$ =\int_{0}^{\pi/4}\sin2t\,dt $$ $$ =\left[-\frac12\cos2t\right]_{0}^{\pi/4} $$ $$ =-\frac12\cos\frac{\pi}{2} +\frac12\cos0 $$ $$ =0+\frac12 =\frac12 $$ $$ \boxed{\int_C \mathbf{F}\cdot d\mathbf{r}=\frac12} $$ Option A GATE EC Previous Year Papers with Solutions → Electronics and Communication Study Material →

Download Question Paper                   2025 | 2024 | 2023 | 2022 | 2021 | 2020 GATE - EC 2025 Answers with Explanations  Q.1 Current population 11,02,500 increasing 5% per annum Let assume 2 years age, the population were, x then, x*(1+5/100)^2 = 11,02,500 Or, x = 1000000 Option C Q.2 p/q + q/p = 3 Or, (p/q)^2 + (q/p)^2 = (p/q + q/p)^2 - 2 Or, (p/q)^2 + (q/p)^2 = 9-2=7 Option B Q.3 Option C Q.4 Option C Q.5 Option B Q.6 Option C Q.7 Option D Q.8 Total Volume = 1/3*pi*r^2*l = 1/3*pi*1*1 = pi/3 Option A Q.9 Option C Q.10 The area of the triangle is  1/2*3x*3x = 9/2*x^2 The area of the regular convex hexagon is  9/2*x^2 - 3*1/2*x*x = 6/2*x^2 Or, area of hexagon: triangle = 2:3 Option D Electronics and Communication (EC) Q.1 $ \mathbf{F}(x,y)=-x\hat{i}+y\hat{j} $ $$ \int_C \mathbf{F}\cdot d\mathbf{r} $$ $$ =\int_C(-x\,dx+y\,dy) $$ Parameterization: $$ x=\cos t,\qquad y=\sin t $$ $$ dx=-\sin t\,dt,\qqua...

  Understanding OFDM Overhead: Why Real-World 5G Speeds Differ from Theory Why Internet Isn't as Fast as the Theory: Understanding Overhead Have you ever wondered why a 100 Mbps wireless connection often delivers only 70 or 80 Mbps in a speed test? In the world of OFDM (Orthogonal Frequency Division Multiplexing) —the technology powering 5G, LTE, and Wi-Fi 6—the difference between theoretical limits and real-world performance is caused by Overhead . What is Overhead in OFDM? In telecommunications, overhead is any data transmitted that isn’t the actual user payload. It is the "management tax" required to keep a wireless connection stable, synchronized, and error-free. The 4 Main Types of Practical Overhead 1. Frequency Guard Bands Wireless standards cannot use 100% of their allocated frequency. In a 20 MHz channel, roughly 10% is left empty at the edges. This acts as a buffer to prevent ...

  1. Noise Remover (Dead-Zone Clipper - Image 1) Vi Vo V1 V2 Signal Amp (Vp): Noise Level: Deadzone V1: Deadzone V2: --- Noisy Input    — Cleaned Output (Series) 2. Dual Parallel Clipper (Shunt - Image 2) Vi Vo ...