Special Classes of Random Processes MCQ Quiz in தமிழ் - Objective Question with Answer for Special Classes of Random Processes - இலவச PDF ஐப் பதிவிறக்கவும்

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:

Additional Information

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.

The power spectral density of white noise is given by:

\(\rm{S_{X}\left(f\right)=\frac{\eta}{2}}\) for all frequency 'f', i.e.

Now auto-correlation is 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:

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

E[y(t)] = E[X²(t)] = R_XX(0)

= e^-(0)

= 1

∴ mean of y(t) = 1

hence option (2) is correct.

Concept: (1) For WSS Process  E[X²] = R_xx(0) = E²[x] + σ²

(ii) In a Random process if time become fixed then R.Process become Random variable.

Calculation: X(4) is a random variable

So we need to find P[X(4) ≤ 3], now as μ_x = 2, \(\sigma _x^2 = E\left[ {{X^2}} \right] - \mu _x^2\)

\(\sigma _x^2 = {R_{xx}}\left( 0 \right) - \;\mu _x^2\) 

\(\sigma _x^2 = 5 - 4 = 1\) 

σ_x = 1

P [1 < X (4) ≤ 3] = 0.6828

P [2 ≤ X (4) ≤ 3] = 0.3414

P [ X (4) ≤ 2) = 0.50 due to symmetry about mean.

∴ P [ X(4) ≤ 3 ] = 0.5 + 0.3414

P [X(4) ≤  3] = 0.8414

Hence the probability that X(4) ≤ 3 is 0.8414.

The rms noise voltage generated

\(\begin{array}{l} {V_n} = \sqrt {4kTBR} \\ = \sqrt {4 \times 1.38 \times {{10}^{ - 23}} \times 300 \times 2 \times {{10}^7} \times {{10}^3}} \end{array}\)

P_T = 1 mV

B = 1 GHz

Loss = 20 dB = 100

Power received \({P_v} = 1 \times {10^{ - 3}} \times {10^{ - 2}}\;W\)

= 10^-5 W

Noise at receiver = 10^-16 × 10⁹

= 10^-7 W

\(\therefore {\rm{SNR\;at\;receiver}} = \frac{{{{10}^{ - 5}}}}{{{{10}^{ - 7}}}}\)

We know that \({T_{eq}} = {T_{e1}} + \frac{{{T_{e2}}}}{{{G_1}}} + \frac{{{T_{e3}}}}{{{G_1}{G_2}}} + \ldots \)

Given,

T_e1 = 5 k

G₂ = 5 dB = 3.16

G₃ = 12 dB = 15.85

G₁ = 20 dB = 100

F₂ = 2.51

F₃ = 8 dB = 6.31

Since T_e = T_o(F – 1)

\({T_{eq}} = 5 + \frac{{300\left( {2.51 - 1} \right)}}{{100}} + \frac{{300\left( {6.31 - 1} \right)}}{{100 \times 3.16}}\)

= 5 + 4.53 + 5.04

The narrow band representation of noise is

\(n\left( t \right) = {n_c}\left( t \right)\cos \left( {{\omega _c}t} \right) + {n_s}\left( t \right)\sin \left( {{\omega _c}t} \right)\)

Its envelope is \( = \sqrt {({n_c}{{\left( t \right)}^2} + {n_s}{{\left( t \right)}^2}\;} \) 

Given signal strength S_t = 10^-3 W

Given cable loss factor = 40 dB = 10⁴ i.e signals strength is attenuated by 40 dB

Noise one sided P.S.D at receiver is η = 10^-20 Watt/Hz

Since bandwidth B = 100 MHz

The noise power at receiver is N = 10^-20 × 10⁸ Watt

Signal strength at receiver S = 10^-3 × loss factor

= 10^-3 × 10^-4

= 10^-7 watt

∴ SNR at receiver \(= \frac{{{{10}^{ - 7}}}}{{{{10}^{ - 12}}}}\)

= 10⁵

Let \(\rm Z\) be a standard normal random variable. Thus, its pdf is given by \(\rm N\left(0;1\right)\). Using the relation that two Gaussian random variables random variables related by, \(\rm Y=aW+b\) such that \(\rm W=N\left(\mu;\sigma^2\right)\), then \(\rm \rm Y=N\left(a\mu+b;a^2\sigma^2\right)\). Using this can write \(\rm X\) as \(\rm \rm X=\sigma Z\).Thus, we have

\(\rm E\left[\left|X\right|\right]=E\left[\left|\sigma Z\right|\right]=\sigma E\left[\left|Z\right|\right]\)  \(\rm \rm \left(\because\sigma\text{ being positive } \left|\sigma Z\right|=\sigma\left|Z\right|\right)\)

Recall that the pdf \(\rm N\left(\mu, \sigma^2 \right)=\frac{1}{\sqrt{2\pi}\sigma}\exp{\frac{\left(x-\mu\right)^2}{2\sigma^2}}\) and that 

\(\rm E[X]=\int_{-\infty}^{\infty}xf_X(x)dx\)

Using this, we have \(\rm E\left[\left|Z\right|\right]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{\left|z\right|\exp\left({-\frac{z^2}{2}}\right)dz}\)

Since, the function above is even, we may rewrite it as

\(\Rightarrow \rm E\left[\left|Z\right|\right]=\frac{2}{\sqrt{2\pi}}\int_{0}^{\infty}{z\exp\left({-\frac{z^2}{2}}\right)dz}\)

Putting

 \(\rm \frac{z^2}{2}=p\Rightarrow zdz=dp\\ \Rightarrow dz=\frac{dp}{z}\)

Thus, \(\rm E\left[\left|Z\right|\right]=\sqrt\frac{2}{{\pi}}\int_{0}^{\infty}{\exp\left({-p}\right)dp}\)

\(\Rightarrow \rm E\left[\left|Z\right|\right]=\sqrt\frac{2}{{\pi}}\left.\left[-e^{-p}\right]\right|_{0}^{\infty}\)

\(\rm \Rightarrow E\left[\left|Z\right|\right]=\sqrt{\frac{2}{\pi}}\)

Now, \(\rm E\left[\left|X\right|\right]=\sigma E\left[\left|Z\right|\right]=\sigma\sqrt{\frac{2}{\pi}}\)

Alternatively, we can directly calculate \(\rm E\left[\left|X\right|\right]\) by using the substitution 

\(\rm \frac{z^2}{2\sigma^2}=p\Rightarrow \frac{z}{\sigma^2}dz=dp\\ \Rightarrow dz=\sigma^2\frac{dp}{z}\)

while integration.