CALCU-Problem 15 marksFor each of the following, d

Problem 1 5 marks For each of the following, determine whether or not the random process is Mean ergodic. (a) A Bernoulli process with

\operatorname{Pr}\{x(n)=1\}=0.3

and

\operatorname{Pr}\{x(n)=-1\}=0.7

. (b)

y(n)=x(n)+x(n-1)

, where

x(n)

is the Bernoulli process defined in (a). (c)

y(n)=x(n)+5

, where

x(n)

is the Bernoulli process defined in (a).

Generated Graph

Solution by Steps

step 1

To determine if a random process is mean ergodic, we need to check if the time average converges to the ensemble average as time goes to infinity

step 2

For a Bernoulli process, the ensemble average is calculated by taking the expected value of the random variable

step 3

For part (a), the expected value E[x(n)] is

0.3 \cdot 1 + 0.7 \cdot (-1) = -0.4

step 4

Since the Bernoulli process is a stationary process and the expected value is constant, the time average will converge to the ensemble average. Therefore, the process is mean ergodic

step 5

For part (b), we need to find the expected value of

y(n) = x(n) + x(n-1)

step 6

Since

x(n)

and

x(n-1)

are identically distributed and independent,

E[y(n)] = E[x(n)] + E[x(n-1)] = 2 \cdot E[x(n)] = 2 \cdot (-0.4) = -0.8

step 7

The process

y(n)

is also stationary, and its time average will converge to the ensemble average, making it mean ergodic

step 8

For part (c), the process

y(n) = x(n) + 5

has an expected value of

E[y(n)] = E[x(n)] + 5 = -0.4 + 5 = 4.6

step 9

Adding a constant to a mean ergodic process does not affect its ergodicity, so

y(n)

is mean ergodic as well

Answer

(a) Mean ergodic

(b) Mean ergodic

(c) Mean ergodic

Key Concept

Mean Ergodicity in Random Processes

Explanation

A random process is mean ergodic if the time average of the process converges to the ensemble average as time goes to infinity. For stationary processes like the Bernoulli process, this condition is satisfied, and adding a constant to a mean ergodic process does not change its ergodicity.

Problem 2 2 marks A discrete-time random process

x(n)

is generated as follows:

x(n)=\sum_{k=1}^{p} a(k) x(n-k)+w(n)

where

w(n)

is white noise process with variance 0.3 . Another process,

z(n)

, is formed by adding noise to

x(n)

,

z(n)=x(n)+v(n)

where

v(n)

is white noise process with variance 0.5 that is uncorrelated with

w(n)

. (a) Find the power spectrum of

x(n)

. (b) Find the power spectrum of

z(n)

.

Solution by Steps

step 1

To find the power spectrum of

x(n)

, we need to consider the effect of the white noise process

w(n)

on

x(n)

step 2

The power spectrum of

x(n)

is the Fourier transform of its autocorrelation function. Since

w(n)

is white noise with variance 0.3, its power spectrum is constant and equal to its variance across all frequencies

step 3

Therefore, the power spectrum of

x(n)

, denoted as

P_x(f)

, is 0.3

Answer

P_x(f) = 0.3

Key Concept

Power Spectrum of a White Noise Process

Explanation

The power spectrum of a white noise process is constant across all frequencies and equal to its variance.

Solution by Steps

step 1

To find the power spectrum of

z(n)

, we need to consider the power spectra of both

x(n)

and

v(n)

step 2

Since

v(n)

is also a white noise process with variance 0.5 and is uncorrelated with

w(n)

, its power spectrum is also constant across all frequencies and equal to its variance

step 3

The power spectrum of

z(n)

, denoted as

P_z(f)

, is the sum of the power spectra of

x(n)

and

v(n)

because they are uncorrelated

step 4

Therefore,

P_z(f) = P_x(f) + P_v(f) = 0.3 + 0.5 = 0.8

Answer

P_z(f) = 0.8

Key Concept

Power Spectrum of the Sum of Two Uncorrelated Processes

Explanation

The power spectrum of the sum of two uncorrelated processes is the sum of their individual power spectra.

Problem 3 3 marks Consider an moving average (MA) process that is generated by the difference equation:

\begin{array}{c} x(n)=\sum_{k=0}^{q} b(k) w(n-k) \\ 1 \end{array}

where

w(n)

is a zero mean white noise random process with variance

\sigma_{w}^{2}

. Another process,

z(n)

, is formed by subtracting noise to

x(n)

,

z(n)=x(n)-v(n),

where

v(n)

is white noise process with variance

\sigma_{v}^{2}

that is uncorrelated with

w(n)

. (a) Find the power spectrum of

x(n)

. (b) Find the power spectrum of

z(n)

.

Solution by Steps

step 1

To find the power spectrum of

x(n)

, we need to calculate the Fourier transform of the autocorrelation function of

x(n)

step 2

The autocorrelation function

R_x(k)

of

x(n)

is the convolution of

b(k)

with itself, scaled by the variance

\sigma_w^2

of the white noise

w(n)

step 3

The power spectrum

P_x(f)

is the Fourier transform of

R_x(k)

. Since

w(n)

is white noise,

P_x(f)

will be

|B(f)|^2 \sigma_w^2

, where

B(f)

is the Fourier transform of

b(k)

Answer

P_x(f) = |B(f)|^2 \sigma_w^2

Key Concept

Power Spectrum of a MA Process

Explanation

The power spectrum of a moving average process is the magnitude squared of the Fourier transform of its coefficients, scaled by the variance of the white noise input.

Solution by Steps

step 1

To find the power spectrum of

z(n)

, we need to consider the power spectra of both

x(n)

and

v(n)

since

z(n) = x(n) - v(n)

step 2

The power spectrum of

v(n)

is

\sigma_v^2

since

v(n)

is white noise

step 3

The power spectrum of

z(n)

is the sum of the power spectra of

x(n)

and

v(n)

because they are uncorrelated

step 4

Therefore,

P_z(f) = P_x(f) + \sigma_v^2

Answer

P_z(f) = |B(f)|^2 \sigma_w^2 + \sigma_v^2

Key Concept

Power Spectrum of a Process with Added White Noise

Explanation

The power spectrum of a process with added white noise is the sum of the power spectra of the original process and the added white noise, provided they are uncorrelated.

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