Communications Test 8

Concept:

Properties of the autocorrelation of an ergodic process include:

1. RX(τ) is an even function, i.e.

\({R_X}\left( \tau \right) = {R_X}\left( { - \tau } \right)\)

2. If X(t) has no periodic component, RX(τ) will also have no periodic component.

3. RX(τ) is independent of absolute time.

4. RX(0) = X̅2

Analysis:

Given, the average power of the random process:

X̅² = 24

Also, given that X(t) has no periodic component.

Option (1): We have a periodic component and R_X(0) = 34, so, it is not a valid autocorrelation function.

Option (2): R_X(τ) is not even and it also depends on absolute time t. So, it is not a valid autocorrelation function.

Option (3): R_X(τ) is not even, and also we have:

\({R_X}\left( 0 \right) = \frac{1}{1} = 1\)

So, the autocorrelation is not valid.

Option (4): R_X(τ) = 24δ (2π τ²)

It is an even function and having no periodic component.

The autocorrelation function (ACF) tells how the correlation between any two values of the signal changes as their separation changes. It is denoted as R_XX (τ), with the Fourier Transform given as:

\({R_{XX}}\left( \tau \right)\mathop \leftrightarrow \limits^{F.T.} {S_{XX}}\left( \omega \right)\)

S_XX(ω) = Energy Spectral Density calculated as:

\({S_{XX}}\left( \omega \right) = \;\mathop \smallint \nolimits_\infty ^\infty {R_{XX}}\left( \tau \right){e^{ - j\omega \tau }}\;d\tau \) 

Option (1) and (4) both are therefore correct i.e. the Autocorrelation function and the energy spectral density forms Fourier Transform

pairs.

For a real-valued function, the autocorrelation function will be real and even, i.e.

R_xx(τ) = R_XX(-τ)

So, Option (2) which states that the autocorrelation function of a real-valued energy signal is valued odd function is wrong.

​\({R_{XX}}\left( \tau \right) = \left( {\frac{1}{{2\pi }}} \right)\mathop \smallint \nolimits_{ - \infty }^\infty {S_{XX}}\left( \omega \right)\;{e^{ + j\omega \tau }}\;d\omega \)

At Origin, i.e when τ = 0,

\({R_{XX}}\left( 0 \right) = \frac{1}{{2\pi }}\mathop \smallint \nolimits_{ - \infty }^\infty {S_{XX}}\left( \omega \right)d\omega\) , which gives the total power of a power signal.

So, Option (3), which states that the value of the autocorrelation function of a power signal at the origin is equal to the average power

of the signal, is correct.

The auto correlation function is the inverse Fourier power spectrum. Now, inverse Fourier of  \(\frac{{\sin {\rm{f}}}}{{\rm{f}}}\) is a rectangular pulse, and inverse Fourier of \({\left( {\frac{{\sin {\rm{f}}}}{{\rm{f}}}} \right)^2}\) is a triangular pulse.

Given the random process Z(t) = (Y + 3Xt)t

For the random variable, Y

Y̅ = 2, Y̅² = 6

We have just obtained the mean of X as:

X̅ = 0

Now, the mean square value of the random variable X is

\({\bar X^2} = \mathop \smallint \limits_{ - \infty }^\infty {x^2}{f_X}\left( x \right)dx\)

\( = \mathop \smallint \limits_{ - 1}^1 {x^2}\left( {\frac{1}{2}} \right)dx = \left[ {\frac{{{x^3}}}{6}} \right]_{ - 1}^1 = \frac{1}{3}\)

So, the autocorrelation of random process v(t) is obtained as:

\({R_Z}\left( {{t_1},\;{t_2}} \right) = E\left[ {Z\left( {{t_1}} \right)Z\left( {{t_2}} \right)} \right]\)

\(= \overline {\left[ {\left( {Y + 3X{t_1}} \right)} \right]\left[ {\left( {Y + 3X{t_2}} \right){t_2}} \right]} \)

\(= \overline {{Y^2}{t_1}{t_2} + 9{X^2}t_1^2t_2^2 + 3XY{t_1}{t_2}\left( {{t_1} + {t_2}} \right)} \)

\(= {\bar Y^2}{t_1}{t_2} + 9\;{\bar X^2}\;t_1^2t_2^2 + 3\;\overline {XY} \;{t_1}{t_2}\left( {{t_1} + {t_2}} \right)\)

\(= 6{t_1}{t_2} + \frac{9}{3}t_1^2t_2^2 + 3\bar X\bar Y\;{t_1}{t_2}\left( {{t_1} + {t_2}} \right)\)

Since X and Y are independent:

\(\overline {XY} = \bar X\;\bar Y = 0\)

\(= 6{t_1}{t_2} + 3t_1^2t_2^2 + 0\) 

\(= 3{t_1}{t_2}\left( {2 + {t_1}{t_2}} \right)\)

For DSB AM

\({\left( {\frac{S}{N}} \right)_i} = {\left( {\frac{S}{N}} \right)_o}\)

\({\left( {\frac{S}{N}} \right)_i} = 50dB = {10^5}watts\)

N_i = N_o W 

= (10^-12 × 2) (10 × 10³)

⇒ 2 × 10^-8

Si = 10⁵ × N_i

= 10⁵ × 2 × 10^-8

= 2× 10^-3

(S_i)dB = (P_t)dB - (P_L)dB

\({S_i} = \frac{{{P_t}}}{{{P_L}}}\)

P_t = S_i × P_L

P_L = 80 dB = 10⁸ W

P_t = 2 × 10^-3 × 10⁸ = 2 × 10⁵

Concept:

\(ACF\;\begin{array}{*{20}{c}} {FT} \\ \rightleftharpoons \\ {IFT} \end{array}PSD\)

PSD_o = PSD_i |H(f)|²

Calculations:

h(t) = e^-t u(t)

\(H\left( f \right) = \frac{1}{{1 + {j{2\pi }}f}}\)

S_y (f) = S_x(f) |H(f)|²

\(= \frac{{No/\;2}}{{1 + {{\left( {2\pi f} \right)}^2}}}\)

To find Auto correlation of output

R_y(z) = IFT {S_y (b)}

\(IFT\left\{ {\left( {\frac{{No}}{4}} \right)\frac{{2\left( 1 \right)}}{{1 + {\omega ^2}}}} \right\}\)

\(\Rightarrow \frac{{No}}{4}IFT\;\left\{ {\frac{2}{{1 + {\omega ^2}}}} \right\}\)

\({R_y}\left( z \right) = \frac{{No}}{4}{e^{ - \left| \tau \right|}}\)

\(E\left[ {{y^2}\left( t \right)} \right] = Total\;power = \;{R_y}\left( {\tau = 0} \right)\)

\(= \frac{{No}}{4}\)