Random Processes MCQ Quiz in मल्याळम - Objective Question with Answer for Random Processes - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Analysis:

The power spectral density (PSD) of a WSS process is defined as the Fourier transform of its Autocorrelation function.

It is given as:

\({S_X}\left( f \right) = \mathop \smallint \nolimits_{ - \infty }^\infty {R_x}\left( \tau \right){e^{ - j2\pi f\tau }} \cdot d\tau \)

Note:

The Autocorrelation can be obtained from the PSD through inverse transform as:

\({R_x}\left( \tau \right) = \mathop \smallint \nolimits_{ - \infty }^\infty {S_X}\left( f \right){e^{j2\pi f\tau }}df\)

S_X (f) is real and S_X (f) ≥ 0

S_X (f) = S_X (-f)

R_X (0) = ∫ S_X (f) ⋅ df = Power of x(t)

Concept:

R_X = R_XX (0)

Calculation:

\(R_{XX}(\tau)=5+4e^{-4\tau^2}\)

\(R_{XX}(0)=5+4e^0=9 W\)

\(S_{XX}(ω)=5(2\pi\delta(ω))+4√{\dfrac{\pi}{2}}e^{\frac{-ω^2}{4(4)}}\)

Power within (-√2 ≤ ω ≤ √2) is:-

\(\dfrac{1}{2\pi}\displaystyle\int^{\sqrt{2}}_{-\sqrt{2}}\delta_{XX}(ω)\ dω=\dfrac{1}{2\pi}5(2\pi)+\dfrac{4}{2\pi}\displaystyle\int^{\sqrt{2}}_{-\sqrt{2}}\sqrt{\dfrac{\pi}{4}}e^{\frac{-ω^2}{4\times4}}\)

Let, \(\dfrac{ω}{√{2}}=x\)

∴ √ω = 2√2 dx

\(=5+\dfrac{1}{2\pi}\displaystyle\int^{\frac{1}{2}}_{-\frac{1}{2}}\sqrt{\dfrac{\pi}{4}}e^{\frac{-x^2}{2}}(2\sqrt{2})\ dx\)

\(=5+\dfrac{\sqrt{2\pi}}{2\pi}\displaystyle\int^{\frac{1}{2}}_{-\frac{1}{2}}e^{\frac{-x^2}{2}}\ dx\)

\(=5+\dfrac{1}{2\pi}\displaystyle\int^{\frac{1}{2}}_{-\frac{1}{2}}e^{\frac{-x^2}{2}}\ dx\)

\(5+\left[1-2Q\left(\dfrac{1}{2}\right)\right]\)

= 5 + [1 - 0.6]

= 5 + 0.4 = 5.4 W

Fraction of power within, -√2 ≤ ω ≤ √2 is \(\dfrac{5.4}{9}=0.6\)

Concept: 

The power spectral density (PSD) of a WSS process is defined as the Fourier transform of its Autocorrelation function. It is given as:

\({S_X}\left( f \right) = \mathop \smallint \nolimits_{ - \infty }^\infty {R_x}\left( \tau \right){e^{ - j2\pi f\tau }} \cdot d\tau \)

The average power of a stationary random process is given as:

\(E\left[ {{X^2}} \right] = \mathop {\lim }\limits_{\tau \to 0} {R_{xx}}\left( \tau \right)\)

Calculation:

\(E\left[ {{X^2}\left( t \right)} \right] = \frac{1}{{2\pi }}\mathop \smallint \nolimits_{ - \infty }^\infty {S_X}\left( \omega \right)d\omega \)

\(= \frac{1}{{2\pi }}2\mathop \smallint \nolimits_0^\infty {S_X}\left( \omega \right)d\omega \)

\(= \frac{1}{\pi }\left[ {200\pi + 30\pi \times {{10}^2}} \right]\)

Concept:

Power spectral density of a wide sense stationary process is a positive real function.

It is related to Auto co-relation function Rx(τ) by:

For wide – sense stationary random process, the power of a signal is given by R_x(0).

Application:

R_x(0) = 2

Average power of signal = R_x(0) = 2

White noise is that signal whose frequency spectrum is uniform i.e. it has flat spectral density.

The power spectral density (PSD) of white noise is uniform throughout the frequency spectrum as shown:

The spectral density of white noise is Uniform and the autocorrelation function of White noise is the Delta function.

Derivation:

The power spectral density is basically the Fourier transform of the autocorrelation function of the power signal, i.e.

\({S_x}\left( f \right) = F.T.\left\{ {{R_x}\left( \tau \right)} \right\}\)

Also, the inverse Fourier transform of a constant function is a unit impulse.

Now, the auto-correlation is the inverse Fourier transform (IFT) of power spectral density function.

\(\rm{R_x\left(\tau\right)\mathop \to \limits^{IFT}S_X\left(f\right)}\)

The inverse Fourier transform of the power spectrum of white noise will be an impulse as shown:

\(\rm{R_x\left(\tau\right)=\frac{\eta}{2}\delta\left(\tau\right)}\)

Concept:

The mean of random process X(t), also called as the expectation of the random process is represented by E[X(t)].

Properties:

1) \(E\left[ {AX\left( t \right) + B} \right] = AE\left[ {X\left( t \right)} \right] + B\)

2) \(\;E\left[ {{{\left( {AX\left( t \right) + BY\left( t \right)} \right)}^2}} \right] = {A^2}\;E\left[ {{X^2}\left( t \right)} \right] + {B^2}\;E\left[ {{Y^2}\left( t \right)} \right] + 2A.B\;E\left[ {X\left( t \right)Y\left( t \right)} \right]\)

3) E[X²(t)] of a random process X(t) gives the power in X(t) and is equal to the value of autocorrelation function at τ = 0, i.e.

\(E\left[ {{X^2}\left( t \right)} \right] = {R_{XX}}\left( 0 \right)\);

R_XX(τ) is the autocorrelation function of X(t)

4) If the two processes are orthogonal, then their cross-correlation is zero, i.e. for orthogonal processes:

\({R_{XY}}\left( \tau \right) = E\left[ {X\left( t \right)Y\left( t \right)} \right] = 0\)

Calculation:

Given: E[X(t)] = 0, E[Y(t)] = 0, and E[X(t)Y(t)] = 0 (mutually orthogonal)

\(E\left[ {Z\left( t \right)} \right] = E\left[ {2X\left( t \right) + 3Y\left( t \right)} \right]\)

\( = 2E\left[ {X\left( t \right)} \right] + 3E\left[ {Y\left( t \right)} \right]\)

Putting on the respective values, we get:

\(E\left[ {Z\left( t \right)} \right] = 2 \times 0 + 3 \times 0\)​

\(E\left[ {Z\left( t \right)} \right] = 0\)

E[X²(t)] = R_XX(0) = 5e^-2×0 = 5

E[Y²(t)] = R_YY(0) = 3 (0) + 7 = 7

\(E\left[ {{Z^2}\left( t \right)} \right] = E\left[ {{{\left( {2X\left( t \right) + 3Y\left( t \right)} \right)}^2}} \right]\)

\(= E\left[ {4{X^2}\left( t \right) + 9{Y^2}\left( t \right) + 12X\left( t \right)Y\left( t \right)} \right]\)

\( = 4E\left[ {{X^2}\left( t \right)} \right] + 9E\left[ {{Y^2}\left( t \right)} \right] + 12E\left[ {X\left( t \right)Y\left( t \right)} \right]\)

\(= 4 \times 5 + 9 \times 7 = 83\)

∴ The Power in Z(t), E[Z²(t)] = 83

Concept:

Autocorrelation function [R (τ)] of the power signal is given as:

\(R\left( \tau \right) = \mathop {{\rm{lt}}}\limits_{t \to \infty } \frac{ \bot }{T}\mathop \smallint \nolimits_{ - T/2}^{T/2} x\left( t \right)x\left( {t - \tau } \right)dt\)

Calculation:

Given:

x(t) = A cos (ω₀t + ϕ)

\(ACF:R\left( \tau \right) = \mathop {{\rm{lt}}}\limits_{t \to \infty } \frac{ \bot }{T}\mathop \smallint \nolimits_{ - T/2}^{T/2} x\left( t \right) \cdot x\left( {t - \tau } \right)dt\)

\(\mathop {{\rm{Lt}}}\limits_{t \to \infty } \frac{ \bot }{T}\mathop \smallint \nolimits_{ - T/2}^{T/2} \left[ {A\cos \left( {{\omega _0}t + \phi } \right) \cdot A\cos \left( {{\omega _0}\left( {t - \tau } \right) + \phi } \right)} \right] \cdot dt\)

\(\mathop {{\rm{Lt}}}\limits_{T \to \infty } \frac{{{A^2}}}{{2T}}\mathop \smallint \nolimits_{ - T/2}^{T/2} \left[ {\cos \left( {2{\omega _0}t - {\omega _0}\tau + 2\phi } \right) + \cos {\omega _0}\tau } \right]dt\)

\(\mathop {{\rm{Lt}}}\limits_{T \to \infty } \frac{{{A^2}}}{{2T}}\cos {\omega _0}\tau \mathop \smallint \nolimits_{ - T/2}^{T/2} \cdot dt\)        

\(\therefore \mathop \smallint \nolimits_{ - T/2}^{T/2} \cos \left( {2{\omega _0}t - {\omega _0}\tau + 2\phi } \right) = 0\)

\(\mathop {{\rm{Lt}}}\limits_{T \to \infty } \frac{{{A^2}}}{{2T}}\cos {\omega _0}\tau .T\)

\(R\left( \tau \right) = \frac{{{A^2}}}{2}\cos {\omega _0}\tau \)

White noise is that signal whose frequency spectrum is uniform i.e. it has flat spectral density.

The power spectral density (PSD) of white noise is uniform throughout the frequency spectrum as shown:

Solution:

Quantization noise power for positive values of x,

(QNP)_P = E[(x-α)²]

\(\begin{aligned} &=\left.\frac{(x-α)^{3}}{3} \cdot \frac{1}{2}\right|_{0} ^{1} \\ \end{aligned}\)

\(\begin{aligned} &=\frac{1}{6}\left[(1-α)^{3}-(-α)^{3}\right] \\ \end{aligned}\)

\(\begin{aligned} &=\frac{1}{6}\left[(1-α)^{3}+α^{3}\right] \\ \end{aligned}\)

\(\begin{aligned} &=\frac{1}{6}\left[1-α^{3}+3 α^{2}-3 α+α^{3}\right] \\ \end{aligned}\)

\(\begin{aligned} &=\frac{1}{6}\left[1+3 α^{2}-3 α\right] \\ \end{aligned}\)

Now , taking differentiation on both side with respect to α .

\(\frac{d}{d α} (Q N P)_P=0+6 α-3=0\)

α = 1/2

α* = 1/2 =0.5

Quantization noise power for negative values of x,

(QNP)_N = E[(x-β)2]

\(\begin{aligned} &=\int_{-2}^{0}(x-β)^{2}\left(\frac{1}{4} x+\frac{1}{2}\right) d x \\ \end{aligned}\)

\(\begin{aligned} &=\frac{1}{4} \int_{-2}^{0}(x-β)^{2} x d x+\frac{1}{2} \int_{-2}^{0}(x-β)^{2} d x \\ \end{aligned}\)

\(\begin{aligned} &=\frac{1}{4}\left[\int_{-2}^{0}\left(x^{2}+β^{2}-2 β x\right) x d x\right] +\frac{1}{2}\left[\int_{-2}^{0}\left(x^{2}+β^{2}-2 β x\right) d x\right] \\ \end{aligned}\)

\(\begin{aligned} &=\frac{1}{4}\left[\frac{\left(x^{4}\right)_{-2}^{0}}{4}+\frac{β^{2}}{2}\left(x^{2}\right)_{-2}^{0}-\frac{2}{3} β\left(x^{3}\right)_{-2}^{0}\right] +\frac{1}{2}\left[\frac{8}{3}+2 β^{2}+4 β\right] \end{aligned}\)

\(=\frac{1}{4}\left[-4-2 β^{2}-\frac{16}{3} β\right]+\frac{1}{2}\left[\frac{8}{3}+2 β^{2}+4 β\right]\)

Now , taking differentiation on both side with respect to β .

\(\frac{d}{d β} (Q N P)_N=-4 β-\frac{16}{3}+2 β+4=0 ~\)

2β = -4/3

β = -2/3

β* = -2/3 = - 0.667

So,

α* -β* = 0.5 - (- 0.667)

α* -β* = 1.167

Concept:

Autocorrelation function and the energy spectral density form Fourier Transform pairs, i.e.

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

SXX(ω) = Energy Spectral Density calculated as:

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

Analysis:

Fourier transform of Autocorrelation function ↔ Power spectral density

\(\frac{\eta }{2}\int\limits_{ - \infty }^{ + \infty } {{e^{j\omega \tau }}df ↔ \frac{\eta }{2}} \)

\(\int\limits_{ - \infty }^{ + \infty } {{e^{j\omega \tau }}df ↔ 1} \)

Hence it's a delta function.