/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 In communication theory, wavefor... [FREE SOLUTION] | 91Ó°ÊÓ

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In communication theory, waveforms of the form $$ A(t)=x(t) \cos (\omega t)-y(t) \sin (\omega t) $$ appear quite frequently. At a fixed time instant, \(t=t_1, X=X\left(t_1\right)\), and \(Y=\) \(Y\left(t_1\right)\) are known to be independent Gaussian random variables, specifically, \(N\left(0, \sigma^2\right)\). Show that the distribution function of the envelope \(Z=\sqrt{X^2+Y^2}\) is given by $$ F_Z(z)= \begin{cases}1-e^{-z^2 / 2 \sigma^2}, & z>0 \\ 0, & \text { otherwise. }\end{cases} $$ This distribution is called the Rayleigh distribution. Compute and plot its pdf.

Short Answer

Expert verified
The distribution function of the envelope is given by \[ F_Z(z) = 1 - e^{-z^2 / 2 \sigma^2} \] for \( z > 0 \). The PDF is \[ f_Z(z) = \frac{z}{\sigma^2} e^{-z^2 / 2 \sigma^2} \].

Step by step solution

01

Define the Envelope

Given the waveform at a fixed time, we have the envelope defined by \[ Z = \sqrt{X^2 + Y^2} \]
02

Joint Distribution of X and Y

Since X and Y are independent Gaussian random variables with mean 0 and variance \( \sigma^2 \) each, their joint distribution is \[ f_{X,Y}(x,y) = \frac{1}{2 \pi \sigma^2} e^{-\frac{x^2 + y^2}{2 \sigma^2}} \]
03

Transform to Polar Coordinates

Transform the coordinates from Cartesian (X, Y) to polar coordinates (Z, \theta). Use the transformation: \[ X = Z \cos(\theta) \] and \[ Y = Z \sin(\theta) \]The Jacobian of this transformation is Z, so \[ dX dY = Z dZ d\theta \].
04

Joint Distribution in Polar Coordinates

Substitute the polar coordinates into the joint distribution. The joint density function in polar coordinates is \[ f_{Z,\theta}(z,\theta) = \frac{1}{2 \pi \sigma^2} e^{-\frac{z^2}{2 \sigma^2}} \cdot z \]This simplifies to \[ f_Z(z) f_{\theta}(\theta) = \frac{z}{2 \pi \sigma^2} e^{- \frac{z^2}{2 \sigma^2}} \]
05

Marginal Distribution of Z

To find the marginal distribution of Z, integrate out \(\theta\) over its entire range \[ f_Z(z) = \int_0^{2\pi} \frac{z}{2 \pi \sigma^2} e^{-\frac{z^2}{2 \sigma^2}} d\theta \] Since \(\theta\) is uniformly distributed over \([0, 2\pi] \), this simply integrates to\ 2\pi: \[ f_Z(z) = \frac{z}{\sigma^2} e^{-\frac{z^2}{2 \sigma^2}} \]
06

Distribution Function of Z

Find the CDF by integrating \( f_Z(z)\) with respect to z: \[ F_Z(z) = \int_0^z \frac{t}{\sigma^2} e^{-\frac{t^2}{2 \sigma^2}} dt \]Using the substitution \( u = \frac{t^2}{2 \sigma^2} \), \( du = \frac{t}{\sigma^2} dt \), we get \[ F_Z(z) = 1 - e^{-\frac{z^2}{2 \sigma^2}} \] for \(z > 0\). For \(z \leq 0\), \(F_Z(z) = 0\).
07

Compute and Plot the PDF

The PDF \( f_Z(z) = \frac{z}{\sigma^2} e^{-\frac{z^2}{2 \sigma^2}}\) can be plotted using software like MATLAB, Python, or any graphing tool. It is characterized by a peak at \( \sigma \) and tailing off exponentially for large z.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gaussian random variables
Gaussian random variables, also known as normally distributed variables, are fundamental in statistics and communication theory. A Gaussian random variable has a bell-shaped probability distribution characterized by its mean (µ) and variance (σ²). In the context of communication theory, such variables are often encountered as they model noise and signal variations effectively. The Gaussian distribution's probability density function (pdf) is given by:
\[ f_X(x) = \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{(x-\mu)^2}{2 \sigma^2}} \]
In our given problem, X and Y are independent Gaussian random variables with a mean of 0 and variance σ², symbolized as N(0, σ²). This implies their joint distribution can be compactly written as: \[ f_{X,Y}(x,y) = \frac{1}{2 \pi \sigma^2} e^{-\frac{x^2 + y^2}{2 \sigma^2}} \]
joint distribution
When dealing with two or more random variables, the joint distribution describes the probability of different outcomes that each variable can take simultaneously.
For our problem, the joint distribution of X and Y is given by:
\[ f_{X,Y}(x,y) = \frac{1}{2 \pi \sigma^2} e^{-\frac{x^2 + y^2}{2 \sigma^2}} \]
This function tells us the likelihood of X taking a specific value x and Y taking a value y concurrently. Because X and Y are independent, their joint pdf is simply the product of their individual pdfs. Additionally, the joint distribution is important to derive the properties of the envelope Z = √(X² + Y²).
CDF transformation
The Cumulative Distribution Function (CDF) transformation is a powerful tool for understanding the distribution of a random variable derived from other variables. The CDF of a random variable Z, denoted F_Z(z), is the probability that Z takes values less than or equal to z:
\[ F_Z(z) = P(Z \le z) \]
In our problem, we start with the joint distribution of X and Y and aim to find the CDF of Z. We transform from Cartesian to polar coordinates and integrate the probability density function (pdf) to get the CDF. By integrating the pdf from 0 to z, we obtain the CDF:
\[ F_Z(z) = \int_0^z \frac{t}{\sigma^2} e^{-\frac{t^2}{2 \sigma^2}} dt \]
Using appropriate substitutions, we derive:
\[ F_Z(z) = 1 - e^{-\frac{z^2}{2 \sigma^2}} , \quad z > 0 \]
polar coordinates
In many mathematical and engineering problems, polar coordinates simplify transformations and integrations. Instead of using Cartesian coordinates (x, y), we use (r, θ) in polar coordinates, where r is the radius (distance from the origin) and θ is the angle.
In this problem, transforming to polar coordinates is crucial. We transform the Gaussian variables X and Y into polar coordinates Z and θ using:
\[ X = Z \cos(\theta) \] \[ Y = Z \sin(\theta) \]
The Jacobian determinant for this coordinate transformation is Z, which modifies the integration measure to dX dY = Z dZ dθ. Applying this transformation simplifies our integration of the joint distribution significantly, as seen in transforming the pdf f_{X,Y}(x,y).
probability density function
A probability density function (pdf) gives the likelihood of a random variable to take on specific values. It is integral in understanding distributions. For our variable Z, derived from X and Y, we find its pdf f_Z(z) by transforming the joint pdf in polar coordinates and integrating out irrelevant variables.
The pdf for Z is given by:
\[ f_Z(z) = \frac{z}{\sigma^2} e^{-\frac{z^2}{2 \sigma^2}} , \quad z > 0 \]
This pdf describes the Rayleigh distribution, common in communication theory for signal amplitude modeling. The function peaks at z = σ and decreases exponentially for larger z values, indicating less likelihood of very high amplitudes.

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Most popular questions from this chapter

Errors occur in data transmission over a binary communication channel due to Gaussian white noise. The probability of an error \(P_e\), can be shown to be $$ P_c=\frac{1}{2}-\frac{1}{\sqrt{\pi}} \int_0^u e^{-y^2} d y=\frac{1}{2}[1-\operatorname{erf}(u)], $$ where \(z=u^2\) is a measure of the ratio of the signal power to noise power. The variable \(u\) is usually specified in terms of \(10 \log _{10} z\) in decibel units (dB). Plot \(P_e\) as a function of \(10 \log _{10} z\).

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