Communications Test 1

Concept:

The total transmitted power for an AM system is given by:

\({P_t} = {P_c}\left( {1 + \frac{{{μ^2}}}{2}} \right)\)

Pt = It2 R (Transmitted Power)

PC = IC2 R (Carrier Power)

It = RMS value of antenna current after modulation

IC = RMS value of antenna current before modulation

R = Antenna Resistance

∴ The antenna current of modulated and un-modulated signal is related as:

\(I_t^2R=I_C^2R​​​​\left( {1 + \frac{{{μ^2}}}{2}} \right)\)

\({I_t} = {I_C}\sqrt {1 + \frac{{{μ ^2}}}{2}} \)  ----(1)

Calculation:

Given:

It = (1.2)Ic

From equation (1),

\(1.2 I_c = {I_C}\sqrt {1 + \frac{{{μ ^2}}}{2}} \)

\(1.2 = \sqrt {1 + \frac{{{μ ^2}}}{2}} \)

\(\frac{μ^2}{2}=(1.2)^2-1\)

μ2 = 0.88

μ = 0.938

Concept:

The trick in this question is:

QAM is a baseband signal hence bandwidth to be calculated is without considering Encoding. (ignore n =8 )

The bandwidth of M-ary QAM

\(W = \frac{{{R_b}}}{{2{{\log }_2}M\;}}\left( {1 + \alpha } \right)\)

Calculation:

\(W = \frac{{8 \times {{10}^3}}}{{2{{\log }_2}8}}\left( {1 + 0.2} \right)\)

\(W = \frac{{8 \times {{10}^3}}}{{2 \times 3}}\left( {1.2} \right)\)

W = 1600 Hz

Concept:

Power spectral density function (PSD) is the Fourier transform of the Autocorrelation function of the WSS process.

\(h\left( t \right)\mathop \to \limits^{FT} H\left( f \right)\) 

\(h\left( {at} \right)\mathop \to \limits^{FT} \frac{1}{a}H\left( {\frac{f}{a}} \right)\) 

The shifting in the time domain does not change the PSD.

Calculation:

Let the autocorrelation function be denoted as R(τ).

Therefore,

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

Then,

\(Y\left( t \right) = X\left( {2t - 1} \right)\mathop \to \limits^{Autocorrelation} {R_y}\left( \tau \right) = {R_x}\left( {2\tau } \right)\) 

\({R_x}\left( {2\tau } \right)\mathop \to \limits^{FT} \frac{1}{2}{S_x}\left( {\frac{f}{2}} \right)\)  

\({R_y}\left( \tau \right)\mathop \to \limits^{FT} {S_y}\left( f \right) = \frac{1}{2}{S_x}\left( {\frac{f}{2}} \right)\)  

For on – off signal

S₁(t) = A

S₂(t) = O

\({E_d} = \mathop \smallint \limits_o^{{T_b}} {A^2}dt = {A^2}{T_b}\)

\({P_e} = Q\left( {\sqrt {\frac{{{A^2}{T_b}}}{{2{N_o}}}} } \right)\)

For ASK

\({P_e} = Q\left( {\sqrt {\frac{{A_c^2{T_b}}}{{4{N_o}}}} } \right)\)

\(K = \frac{{{A^2}{T_b}}}{{2{N_o}}}\)

\(zk = \frac{{{A^2}{T_b}}}{{4{N_o}}}\)

Concept:

For a PCM system, the data rate is given by:

R_b = nf_s; where ‘n’ is the number of bits/sample.

And f_s is the sampling frequency.

Calculation:

Given, f_m = 4.5 MHz

Number of Quantization levels (q) = 1024

Which requires 10 bits to encode.

So, n = 10

Also, f_s = Sampling frequency which is 10% more than 2f_m

i.e. 1.1 × 2f_m

= 2.2 f_m

= 2.2 × 4.5 MHz

= 9.9 MHz

So, R_b (bits/second) = nf_s = 10 × 9.9 MHz

Concept:

Variance (x) = E[x²] - (E [x])²

Calculation:

\(E\left[ X \right] = \mathop \smallint \nolimits_{ - \infty }^{ + \infty } xf_{x}\left( x \right)dx\)

\(= \mathop \smallint \nolimits_1^\infty \frac{3}{{{x^3}}}dx\)

\(= \left[ {\frac{{ - 3}}{2}{x^{ - 2}}} \right]_1^\infty = \frac{3}{2}\)

\(E\left[ {{X^2}} \right] = \mathop \smallint \nolimits_{ - \infty }^{ + \infty } {x^2}f_{x}\left( x \right)dx\)

\(= \mathop \smallint \nolimits_1^\infty \frac{3}{{{x^2}}}dx\)

\(= \left[ { - 3{x^{ - 1}}} \right]_1^\infty \)

= 3
Var(X) = E[X²] - (E [X])²

\(= 3 - \frac{9}{4}\)

P(x₁ x₂ x₁ x₃) = (0.4) (0.3) (0.4) (0.2)

= 0.0096

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:

The general form of an amplitude modulation signal is given by:

x(t) = [Ac + m(t)] cos (ωct)  

The total transmitted power for an AM system is given by:

\({P_t} = {P_c}\left( {1 + \frac{{{μ^2}}}{2}} \right)\)   ---(1)

Pc = Carrier Power

μ = Modulation Index

The modulation index is defined as:

\({\mu _a} = \frac{{\max \left| {m\left( t \right)} \right|}}{{{A_c}}}\)

Equation (1) can be written as:

\({P_t} = {P_c}+\frac{P_c\mu ^2}{2}\)

P_t = P_C + P_sb

Calculation:

Given, the amplitude modulated signal as:

x(t) = (20 + 4 sin 500 πt) cos (2π × 10⁵ t)       ---(2)

The general form of an amplitude modulation signal is given by:

x(t) = [A_c + m(t)] cos (ω_ct)     ---(3)

Now, comparing equation (2) and (3), we can write:

A_c = 20

m(t) = 4 sin 500 πt

So, the modulation will be:

\({\mu _a} = \frac{{\max \left| {m\left( t \right)} \right|}}{{{A_c}}}\)

\({\mu _a} = \frac{{\max \left| {4\sin 500\pi t} \right|}}{{{A_c}}}\)

\({\mu _a} = \frac{4}{{20}} = 0.2\)

Now, the total power of the amplitude modulated signal is given by:

\({P_t} = \frac{1}{2}A_c^2\left( {1 + \frac{{\mu _a^2}}{2}} \right)\)

\( = \frac{1}{2} \times {\left( {20} \right)^2}\left[ {1 + \frac{{{{\left( {0.2} \right)}^2}}}{2}} \right]\)

P_t = 204 W

Also, we have the unmodulated carrier power:

\({P_c} = \frac{{A_c^2}}{2} = \frac{{{{\left( {20} \right)}^2}}}{2} = 200\;W\)

The total power of the amplitude modulated signal can be written as:

P_t = P_c + P_sb

Therefore, the power in the sideband signal is obtained as:

P_sb = P_t - P_c

= 204 – 200

P_sb = 4 Watt

Concept:

If a message source produces x_i symbols independently with probability mass function P(X=x_i) then the entropy of the source is defined as:

\(H = \mathop \sum \limits_i P\left( {X = {x_i}} \right){\log _2}\frac{1}{{P\left( {X = {x_i}} \right)}}\) bits

The Nyquist sampling rate is defined as twice the maximum signal frequency, i.e.

N.R. = 2 × f_max

Also, the Information rate is defined as:

\(I = \left( {H \times Symbol\;rate} \right)\) bits/sec

Calculation:

Entropy will be:

\(H = \frac{1}{2}{\log _2}2 + \frac{1}{4}{\log _2}4 + \frac{1}{8}{\log _2}8 + \ldots + \frac{1}{{{2^{N - 1}}}}{\log _2}{2^{N - 1}}\)

\(H = \frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \ldots \frac{{N - 2}}{{{2^{N - 2}}}} + \frac{{N - 1}}{{{2^{N - 1}}}} + \frac{{N - 1}}{{{2^{N - 1}}}}\)   ---(1)

Divide the above equation by 2, we get:

 \(\frac{H}{2} = \frac{1}{4} + \frac{2}{8} + \frac{3}{{16}} + \ldots \frac{{N - 2}}{{{2^{N - 1}}}} + \frac{{N - 1}}{{{2^N}}} + \frac{{N - 1}}{{{2^N}}}\)   ---(2)

Subtract equation (2) from (1), we get:

\(\frac{H}{2} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{{16}} + \ldots + \frac{1}{{{2^{N - 1}}}} + \left\{ {\frac{{N - 1}}{{{2^{N - 1}}}} - \frac{{N - 1}}{{{2^N}}} - \frac{{N - 1}}{{{2^N}}}} \right\}\)

\(H = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots + \frac{1}{{{2^{N - 2}}}} + \left\{ {\frac{{N - 1}}{{{2^{N - 2}}}} - \frac{{N - 1}}{{{2^{N - 1}}}} - \frac{{N - 1}}{{{2^{N - 1}}}}} \right\}\)

\( = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots + \frac{1}{{{2^{N - 2}}}}\)

\(H = \mathop \sum \limits_{n = 0}^{N - 2} \frac{1}{{{2^n}}} = \frac{{1 - {{\left( {\frac{1}{2}} \right)}^{N - 1}}}}{{1 - \frac{1}{2}}}\)

\( = 2 \times \left( {1 - {{\left( {\frac{1}{2}} \right)}^{N - 1}}} \right) = \frac{{{2^{N - 1}} - 1}}{{{2^{N - 2}}}}\)

The maximum frequency of Voice signal = 3.5 kHz

The Nyquist sampling rate will be:

NR = 2 × 3.5 kHz

NR = 7 kHz

Sampling frequency will be:

\({f_s} = 2 \times 7 = 14\;Kb/s\)

The information rate will now be:

\(I = \left( {\frac{{{2^{N - 1}} - 1}}{{{2^{N - 2}}}} \times 14} \right)\) Kbps

For N = 8, the information rate will be:

\(I = \left( {\frac{{{2^7} - 1}}{{{2^5}}} \times 14} \right) \approx 27.78\;\)Kbits/sec

Alternate Method:

\(H = \frac{1}{2}{\log _2}2 + \frac{1}{4}{\log _2}4 + \frac{1}{8}{\log _2}8 + \ldots + \frac{1}{{{2^{N - 1}}}}{\log _2}{2^{N - 1}}\)

\(H = \mathop \sum \nolimits_{n = 1}^{N - 1} \frac{n}{{{2^n}}} + \frac{{N - 1}}{{{2^{N - 1}}}}\)     ---(1)

\(\frac{H}{2} = \mathop \sum \nolimits_{n = 1}^{N - 1} \frac{n}{{{2^{n + 1}}}} + \frac{{N - 1}}{{{2^N}}}\)    ---(2)

Subtracting (2) from (1), we get:

\(\frac{H}{2} = \left( {\frac{1}{2} + \mathop \sum \limits_{n = 2}^{N - 3} \frac{1}{{{2^n}}}} \right)\;\; + \left\{ {\frac{{N - 1}}{{{2^{N - 1}}}} - \frac{{N - 1}}{{{2^N}}} - \frac{{N - 1}}{{{2^N}}}} \right\}\)

\( = \mathop \sum \limits_{n = 1}^{N - 3} \frac{1}{{{2^n}}}\; + 0\)

\(H = \mathop \sum \limits_{n = 1}^{N - 3} \frac{1}{{{2^{n - 1}}}}\)

\(H = \mathop \sum \limits_{n = 0}^{N - 2} \frac{1}{{{2^n}}} = \frac{{1 - {{\left( {\frac{1}{2}} \right)}^{N - 1}}}}{{1 - \frac{1}{2}}} = 2 \times \left( {1 - {{\left( {\frac{1}{2}} \right)}^{N - 1}}} \right)\)

\(H = \frac{{{2^{N - 1}} - 1}}{{{2^{N - 2}}}}\)

The generator Matrix is given by

\(G = \left[ {{I_K}{P^T}} \right]\)

\({P^T} = \left[ {\begin{array}{*{20}{c}}1&1&0\\0&1&1\\1&1&1\\1&0&1\end{array}} \right]\)

The parity check matrix is given by:

H = [P I_{kn – K}]

Syndrome

S = eH^T

\({H^T} = \left[ {\begin{array}{*{20}{c}}{{P^T}}\\{{I_{n - k}}}\end{array}} \right]\)

\({H^T} = \left[ {\begin{array}{*{20}{c}}1&1&0\\0&1&1\\1&1&1\\1&0&1\\1&0&0\\0&1&0\\0&0&1\end{array}} \right]\)

S = eH^T

For error in 7^th Bit

E = [000 0001]

\(S = \left[ {000\;000\;1} \right]\left[ {\begin{array}{*{20}{c}}1&1&0\\0&1&1\\1&1&1\\1&0&1\\1&0&0\\0&1&0\\0&0&1\end{array}} \right]\)

S = [ 0 0 1]

Extra information:

Syndrome for all possible errors

Given the maximum and maximum operating frequencies of the superheterodyne receiver.

F_c,min = 550 kHz

F_c,max = 1650 kHz

Also, the intermediate frequency is:

f_IF = 450 kHz

So, we obtain the range of frequency of local oscillator as:

f_LO,min = f_c,min + f_IF

= 550 + 450 = 1000 kHz

And f_{LO, max} = f_{c, max} + f_IF

= 1650 + 450

= 2100 kHz

The frequency of the local oscillator is given by:

\({f_{LO}} = \frac{1}{{2\pi \sqrt {LC} }}\)

Where L is the inductance and C is capacitance.

Now, for a fixed value of L, we can write:

\(\sqrt C = \frac{1}{{2\pi \sqrt L }f_{LO}}\)

\(C = \frac{1}{{2\pi L}}\frac{1}{{f_{LO}^2}}\)

So, the maximum value of capacitance exists for a minimum value of f_LO, i.e.,

\({C_{max}} = \frac{1}{{2\pi L}}\frac{1}{{f_{LO,\;min}^2}}\)

Similarly, we get:

\({C_{min}} = \frac{1}{{2\pi L}}\frac{1}{{f_{LO,\;max}^2}}\)

Therefore, the ratio of C_max to C_min will be:

\(r = \frac{{{C_{max}}}}{{{C_{min}}}} = \frac{{f_{LO,\;max}^2}}{{f_{LO,\;min}^2}}\)

\( = \frac{{{{\left( {2100} \right)}^2}}}{{{{\left( {1000} \right)}^2}}} = 4.41\)

Also, we obtain the image frequency for f_c = 700 kHz, as:

f_c’ = f_c + 2f_IF

= 700 + 2 × 450

= 1600 kHz

The Capacity of channel

\(C = Blo{g_2}\left( {1 + \frac{S}{N}} \right)\)

\(C = \left( {3000} \right){\log _2}\left( {1 + 10} \right)\)

C = 10,378 bits/sec

Average information content

H(x) = -Σp(x) log₂p(x)

\(p\left( x \right) = \frac{1}{{128}}\)

H(x) = log₂(128) = 7 bits/character.

Average information rate of Source R = r_sH

= 7 r_s

For errorless Transmission

7 r_s < 10,378

r_s < 1482

Given the power spectrum of X(t) as:

\({X_X}\left( \omega \right) = \frac{3}{{49 + {\omega ^2}}}\)

So, the average power in X(t) is obtained as:

\({P_X} = \frac{1}{{2\pi }}\mathop \smallint \limits_{ - \infty }^\infty \frac{3}{{49 + {\omega ^3}}}d\omega \)

\(= \frac{3}{{2\pi }}\left[ {\frac{1}{7}{{\tan }^{ - 1}}\frac{\omega }{7}} \right]_{ - \infty }^\infty \)

\(= \frac{3}{{2\pi }} \times \frac{1}{7}[{\tan ^{ - 1}}\infty - {\tan ^{ - 1}}\left( { - \infty } \right)]\)

\(= \frac{3}{{14}}\)

Option (3) is correct.

Given the power spectrum of input:

\({S_X}\left( \omega \right) = \frac{3}{{49 + {\omega ^2}}}\)

As this passes through the differentiator the output PSD becomes:

\( S_{Y1} (\omega)=\frac{3\omega ^2}{{49 + {ω ^2}}}\)

The transfer function for the second system is:

\(h\left( t \right) = {t^2}{e^{ - 7t}}u\left( t \right)\)

So, we obtain the transfer function in the frequency domain as:

\(H\left( ω \right) = {\cal F} \cdot T \cdot \left\{ {h\left( t \right)} \right\}\)

\(= {\cal F} \cdot T \cdot \left\{ {{t^2}{e^{ - 7t}}} \right\}\)

\(= \frac{2}{{{{\left( {7 + jω } \right)}^3}}}\)

Therefore, the final output spectrum is given by:

SY(ω) = SY1(ω) H(ω)2

\(= \frac{3\omega^2}{{49 + {ω ^2}}} \times \frac{{{{\left( 2 \right)}^2}}}{{{{\left( {{7^2} + {ω ^2}} \right)}^3}}}\)

\(= \frac{{12\omega^2}}{{{{\left( {49 + {ω ^2}} \right)}^4}}}\)