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Consider a real baseband signal: x(t) = e-2t, t ≥ 0 If 99% of the energy of x(t) lies within B Hz, which of the ...

  Q. Consider a real baseband signal: x(t) = e -2t ,   t ≥ 0 If 99% of the energy of x(t) lies within B Hz, which of the following options is TRUE for the value of B? (A)   B > 1 kHz (B) 63/π Hz < B < 64/π Hz Correct (C)   126/π Hz < B < 128/π Hz (D)   B < 1 Hz Step-by-Step Solution Step 1: Find the Fourier Transform X(f) = ∫₀∞ e -2t e -j2πft dt X(f) = ∫₀∞ e -(2+j2πf)t dt X(f) = 1 / (2 + j2πf) |X(f)|² = 1 / (4 + 4π²f²) Step 2: Calculate Total Energy E = ∫₀∞ |x(t)|² dt E = ∫₀∞ e -4t dt E = 1/4 ...

A QPSK modulated signal from an additive white Gaussian noise (AWGN) channel is received with an Eb/No = 8.4 dB

Q. A QPSK modulated signal from an additive white Gaussian noise (AWGN) channel is received with an E b /N 0 = 8.4 dB at the input of a coherent QPSK demodulator. A maximum-likelihood reception method is used in the demodulator. Assume the complementary error function: erfc(u) ≈ 1 u√π exp(−u²) Which is the nearest bit error rate (BER) at the output of the demodulator? (A) 10 −3 (B) 10 −4 Correct (C) 10 −5 (D) 10 −6 Step-by-Step Solution Q(x) = 1/2*erfc(x/√2) Step 1: Convert SNR from dB to linear scale Given: E b /N 0 = 8.4 dB E b /N 0 = 10 8.4/10 = 10 0.84 = 6.918 ...

Consider a real signal x(t), where −∞ \< t \< ∞, such that x(t) = 0 for t \< 0, x(t) = 2 for 0 ≤ t \< 1, and x(t) = 0 for t ≥ 1...

  Consider a real signal \(x(t)\), such that \(x(t)=\begin{cases} 0, & t Let the energy of the signal be defined as: \[ E[x(t)]=\int_{-\infty}^{\infty}|x(t)|^2dt \] Which of the following options correctly represents the ratio? \[ \frac{E[x(t)]}{E[3x(-3t+5)]} \] (A)   3 (B)   1 (C)   1/3 (D)   1/9 Step-by-Step Solution Step 1: Calculate the energy of x(t) The energy of a signal is given by: \[ E[x(t)]=\int_{-\infty}^{\infty}|x(t)|^2dt \] \[ =\int_0^1 2^2dt \] \[ =4[t]_0^1 \] \[ \boxed{E[x(t)]=4} \] Step 2: Calculate the energy of 3x(-3t+5) Using the energy t...

Decimal to Hexadecimal Conversion

  Computer Organization | Number System | GATE Preparation How to Convert 262144 Decimal to 40000 Hexadecimal? Learn how to convert decimal numbers into hexadecimal numbers using the repeated division-by-16 method. This tutorial explains the conversion of 262144 decimal into 40000 hexadecimal with step-by-step calculations, a division table, verification and shortcut techniques for GATE Computer Science examinations. Number System Conversion Question: Convert Decimal to Hexadecimal Convert the decimal number 262144 into its equivalent hexadecimal number. Given Decimal Number 262144 10 1. Understand the Hexadecimal Number System The hexadeci...

Why Do We Use RMS Voltage?

  Electrical Engineering Why Do We Use RMS Voltage? We use RMS voltage because it gives the correct average power dissipated in a resistor supplied by an AC source. Given Input Voltage v(t) = 12 sin(ωt) V Peak voltage = 12 V 1 Why Not Peak Voltage? Peak voltage is the maximum instantaneous voltage. An AC voltage continuously changes with time, so it does not remain at its peak value. Incorrect average-power calculation P = V peak 2 / R P = 12 2 / 1000 P = 144 mW This is the maximum instantaneous power, not the average power over a complete cycle. 2 Why Not Average Voltage? ...

Efficiency of a solar cell

  The practical efficiency of a solar cell depends on its technology, material, temperature, and manufacturing quality. For commonly used commercial solar panels, efficiency is generally around 18%–23% . 1. Typical efficiency of different solar cell technologies Solar cell technology Typical practical efficiency Monocrystalline silicon 20–23% Polycrystalline silicon 15–18% Thin-film (CdTe) 15–19% Amorphous silicon 6–10% Laboratory silicon cells Can exceed 26% These are approximate ranges; actual module efficiency varies by manufacturer and model. 2. What is the efficiency of a practical silicon solar cell? For practical commercial crystalline silicon solar cells, efficiency is generally around 22%–26% , while finished commercial modules typically achieve around 20%–23% . Department of Energy +1 3. Why is practical efficiency lower than theoretical efficiency? Some energy is lost due to: Reflection of sunlight from the cell surface. Recombination of electrons and holes. Series resi...


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