Communications Test 2

Concept:

The generalized AM expression is represented as:

s(t) = Ac [1 + μa mn (t)] cos ωc t

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

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

Pc = Carrier Power

μ = Modulation Index

The above expression can be expanded to get:

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

The total power is the sum of the carrier power and the sideband power, i.e.

\({P_s} = P_c\frac{{{μ^2}}}{2}\)

The power in single sideband will be:

\({P_s} = \frac{1}{2}\times P_c\frac{{{μ^2}}}{2}\)

With \(P_c=\frac{A^2}{2}\), the above can be written as:

\({P_s} = \frac{1}{2}\times \frac{A_c^2}{2}\frac{{{μ^2}}}{2}\)

\({P_s} = \frac{{{A_c^2μ^2}}}{8}\)   ---(1)

Analysis:

Given the modulation index for the AM signal:

μ_a = 0.6

Using Equation (1), the single-sideband power will be:

\(P_{SSB}= \frac{{\mu _a^2\;A_c^2}}{8}\)

Thus, the desired ratio is:

\(\frac{{{P_{SSB}}}}{{{P_T}}} = \frac{{\mu _a^2A_c^2}}{8} \times \frac{1}{{\frac{{A_c^2}}{2}\left[ {1 + \frac{{\mu _a^2}}{2}} \right]}}\)

\( = \frac{{{{\left( {0.6} \right)}^2}}}{8} \times \frac{1}{{\frac{1}{2}\left[ {1 + \frac{{{{\left( {0.6} \right)}^2}}}{2}} \right]}}\)

\(= \frac{9}{{118}} = 0.0763\)

Given the angle modulated signal,

s(t) = A_c cos [ω_c t + 2 cos 60πt + 5 cos 40 πt]

So, we have the instantaneous phase deviation,

θ(t) = 2 cos 60πt + 5 cos 40πt

Therefore, the peak phase deviation is given by

Δθ = max |θ(t)|

= max |θ(t)|

= max |2 cos 60πt + 5 cos 40πt|        

= 2 + 5

= 7 radians.

Solution:

x = ±1

E[x] = 0

Var[x] = 1

y = -x

E[y] = 0

Var[y] = (-1)² var[x] = 1

x + y = 0

var[x + y] = 0

Note:

Var[x + y] ≠ var[x] + var[y]

If x and y are correlated.

Concept:

The bandwidth for a raised cosine pulse with a roll-ff factor of α, is given by:

Bandwidth = \(\frac{{{R_b}}}{2}\left( {1\; + \;α } \right)\)

R_b = Bita/Data rate

Calculation:

Given bandwidth = 75 KHz

R_b = 0.1 Mb/S

\(So,\;75k = \frac{{0.1M}}{2}\left( {1 + α } \right)\)

\(1 + α = \frac{{150K}}{{0.1M}}\)

\(1 + α = \frac{{0.15}}{{0.1}} = \frac{{15}}{{10}} = 1.5\)

α = 1.5 – 1 = 0.5

Hamming codes are linear error correcting code that can detect up to two-bit errors or correct one-bit error.

The total block length n = 2^r – 1

Message bits length k = 2^r – r – 1

i.e. ‘r’ number of bits are added to message bits to get total code word.

(7, 5),

r = 2

But n = 2² – 1 = 3 ≠ 7

(15, 11),

r = 4

n = 2⁴ – 1 = 15

Concept:

Information associated with the event is “inversely” proportional to the probability of occurrence.

Mathematically, this is defined as:

\(I = {\log _2}\left( {\frac{1}{P}} \right)\;bits\)

P and I represent the probability and information associated with the event

Calculation:

Given that the five grades A, B, C, D, and E are given in equal numbers, the probability of each grade will be:

P(A) = P(B) = P(C) = P(D) = P(E) = 1/5

Now the instructor tells a student that his grade is not E. ∴ The probability of this event will be:

\(P(\bar E)=1-\frac{1}{5}=\frac{4}{5}\)

Now, the information in bits in the event will be:

\(P(\bar E)=log_2(\frac{1}{P(\bar E)})\)

\(=log_2(\frac{5}{4})\)

= 0.322 bits

Concept:

In Am-modulated signal,

Total power = carrier power + sideband power

\({p_t} = \frac{{A_C^2}}{2} + \frac{{A_C^2{\mu ^2}}}{4}\)

Where A_C → carrier signal amplitude

μ → modulation index.

Calculation:

When μ = 0.8

\({\left( {{p_t}} \right)_{old}} = \frac{{A_C^2}}{2}\left[ {1 + \frac{{{\mu ^2}}}{2}} \right]\)

\({\left( {{p_t}} \right)_{old}} = \frac{{A_C^2}}{2}\;\left[ {1 + 0.32} \right] = 0.66\;A_C^2\)

When μ = 0.5

\({\left( {{p_t}} \right)_{new}} = \frac{{A_C^2}}{2}\;\left[ {1 + \frac{{0.25}}{2}} \right] = \frac{{0.25}}{4}\;A_C^2\)

\(\frac{{{{\left( {{p_t}} \right)}_{new}}}}{{{{\left( {{p_t}} \right)}_{old}}}} = \frac{{2.25}}{{4 \times 0.66}} = 0.8522\)

Given, the random variable X uniformly distributed on the interval (a, b) such that a = -5, b = 15

So, the probability density function of random variable is

\({f_X}\left( x \right) = \frac{1}{{b - a}}\)

For -5 < x < 15, we get:

\(= \frac{1}{{15 - \left( { - 5} \right)}} = \frac{1}{{20}}\)

\(= \left\{ {\begin{array}{*{20}{c}} {\frac{1}{{20}},\;\;\; - 5 < x < 15}\\ {0,\;\;\;\;\;\;otherwise} \end{array}} \right.\)

So, the expected value of \(Y = {e^{ - \frac{X}{5}}}\) is obtained as:

\(E\left[ Y \right] = E\left[ {{e^{ - \frac{X}{5}}}} \right] = \mathop \smallint \limits_{ - \infty }^\infty e\left( {\frac{{ - x}}{5}} \right){f_X}\left( x \right)dx\)

\( = \mathop \smallint \limits_{ - 5}^{15} e\left( {\frac{{ - x}}{5}} \right) \cdot \frac{1}{{20}}dx = \frac{1}{{20}}\left[ {\frac{{{e^{ - \frac{x}{5}}}}}{{ - \frac{1}{5}}}} \right]_{ - 5}^{15}\)

\( = - \frac{1}{4}\left[ {{e^{ - \frac{{15}}{5}}} - {e^{\frac{5}{5}}}} \right]\)

= 0.667

Given, the angle modulated signal as:

x(t) = 100 cos (2πf_ct + 4 sin 2πf_mt)

Carrier frequency

f_c = 1000 MHz = 10⁹ Hz

Frequency of message signal

f_m = 1000 Hz

So, we have the phase as:

θ(t) = 4 sin 2πf_mt

= 4 sin (2π × 1000t)

Therefore, the maximum frequency deviation is given by

\({\rm{\Delta }}f = \max \left[ {\frac{1}{{2\pi }}\frac{{d\theta \left( t \right)}}{{dt}}} \right]\)

\(= \max \left\{ {\frac{1}{{2\pi }}\left[ {4 \times 2\pi \times 1000\cos \left( {2\pi \times 1000t} \right)} \right]} \right\}\)

= max {400 cos 1000t} = 4000

So, the modulation index of the FM signal is obtained as:

\({\beta _f} = \frac{{{\rm{\Delta }}f}}{{{f_m}}} = \frac{{4000}}{{1000}} = 4\)

Also, we get the bandwidth of FM signal as:

B = 2(β_f + 1)f_m

= 2(4 + 1)1000

= 10 kHz

Again, we have the signal as:

x(t) = 100 cos (2πf_ct + 4 sin 2πf_m t)

= 100 cos [2π × 10⁹t + 4 sin (2π × 1000t)]

So, the peak phase deviation is obtained as

Δθ = max [4 sin (2π × 1000t)] = 4

Therefore, the modulation index of PM is:

β_p = Δθ = 4

Also, we obtain the bandwidth of PM signal as:

B_PM = 2(β_P + 1)f­_m

= 2(4 + 1) × 1000

= 10 kHz

When f_s = 45 × 10³ sample/sec

And each sample is encoded in 8 bits

Bit rate = 45 × 8 × 10³

= 36 × 10⁴ bits/sec

When f_s = 35 × 10³

Number of bits that can be used

\(R \Rightarrow \frac{{36 \times {{10}^4}}}{{35 \times {{10}^3}}} = 10.28 \approx 10\) (As R has to be integer)

Bits increased from 8 to 10 i.e by 2.

Given, the random process

X(t) = A or – A for (n – 1) T < t < nT.

Also, the levels A and -A occur with equal probability. So, we have:

\(P\left\{ {X\left( t \right) = A} \right\} = P\left\{ {X\left( t \right) = - A} \right\} = \frac{1}{2}\)

So, we obtain the mean value as:

E[X(t)] = A × P {X(t) = A} – A × P{X(t) = -A}

\(= \frac{A}{2} - \frac{A}{2}\)

= 0

\(P\left\{ {X\left( t \right) = A} \right\} = P\left\{ {X\left( t \right) = - A} \right\} = \frac{1}{2}\)

Now, we determine the autocorrelation as:

\({R_X}\left( {{t_1},\;{t_2}} \right) = E\left[ {X\left( {{t_1}} \right)X\left( {{t_2}} \right)} \right]\)

Here, t₁ = 0.5 T and t₂ = 0.7 T. Both of the instantaneous time lies in the same interval i.e. 0 < t₁, t₂ < T

So, we have

X(t₁) = X(t₂) = A

X(t₁) = X(t₂) = -A

In both the cases, we get:

X(t₁) X(t₂) = A²

So, we obtain

\({R_X}\left( {{t_1},\;{t_2}} \right) = E\left[ {X\left( {{t_1}} \right)X\left( {{t_2}} \right)} \right]\)

\( = {A^2} P\left\{ {X\left( {{t_1}} \right) = X\left( {{t_2}} \right) = A} \right\} + {A^2} P\left\{ {X\left( {{t_1}} \right) = X\left( {{t_2}} \right) = - A} \right\}\)

\(= \frac{{{A^2}}}{2} + \frac{{{A^2}}}{2}\)

= A²

Bit Rate = 10⁵ bits/sec

Bit Interval = 10^-5 bits/sec

P_e in PAM \(= Q\left[ {\sqrt {\frac{{2Eb}}{{{N_o}}}} } \right]\)

E_b = bit Energy = A² Tb

P_e = 10^-6

\({P_e} = Q\left( {\sqrt {2 \times {A^2}{T_b}} } \right) = {10^{ - 6}}\)

\(\sqrt {\frac{{2{E_b}}}{{{N_o}}}} = 4.75\)

\({E_b} = \frac{{{{\left( {4.75} \right)}^2} \times {N_o}}}{2}\)

E_b = 0.1128

A² T_b = 0.1128

\(A = \sqrt {0.1128 \times {{10}^5}}\)

= 106.21

Given, the bandwidth of the Gaussian noise channel:

B_T = 100 kHz

When \(\frac{sinx}{x}\) pulse waveform is used, the minimum transmission bandwidth for 4 phase PSK system is defined as:

\({B_T} = \frac{{{R_b}}}{{{{\log }_2}M}}\)

Since, for the 4-phase PSK, we have M = 4. Therefore, the maximum data rate for the system is given by:

R_b = (log₂ 4)(100 kHz)

R_b = 200 kHz

Again, we have the channel bandwidth (modulation bandwidth) as:

B_T = 100 kHz

For binary FSK, we have:

M = 2

Therefore, the maximum data rate that can be transmitted through the channel is given by:

\({R_b} = \frac{{{B_T}}}{{{{\log }_2}M}} = \frac{{100}}{{{{\log }_2}2}} \)

R_b = 100 kHz

We have the channel bandwidth (modulation bandwidth)

B_T = 100 kHz

For M-frequency orthogonal signals, the relation between modulation bandwidth and data rate is given by:

\({R_b} = \frac{{2{{\log }_2}M}}{M}{B_T}\)

Since, for 4-frequency orthogonal FSK, we have:

M = 4

Thus, the maximum data rate is obtained as:

\({R_b} = \frac{{2{{\log }_2}4}}{2}\left( {100} \right)\)

R_b = 100 kHz

Given for bit ‘1’ → 5 cos (10⁶t)

⇒ Power of transmitted signal \(= \frac{{{5^2}}}{2} = 12.5 W\)

⇒ Signal energy per bit = P⋅T_b

T_b = 0.5 × 10^-7 S for 20 Mbps data rate.

⇒ E_b = 12.5 × 0.5 × 10^-7 W sec

= 6.25 × 10^-7 W sec

N₀/2 = 0.125 x 10^-7

N₀ = 0.25 x 10^-7

For BPSK \({P_e} = \frac{1}{2}{\rm{er}}{{\rm{f}}_c}\left( {\sqrt {\frac{{{E_b}}}{{{N_o}}}} } \right)\)

\(\begin{array}{l} = \frac{1}{2}{\rm{er}}{{\rm{f}}c}\left( {\sqrt {\frac{{6.25 \times {{10}^{ - 7}}}}{{0.25 \times {{10}^{ - 7}}}}} } \right)\\ = \frac{1}{2}{\rm{er}}{{\rm{f}}c}\left( {\sqrt {25} } \right)\\ \end{array}\)

Number of picture elements per frame = 2.25 x 10⁶

Number of brightness levels = 12 = M

All 12 brightness levels are equiprobable

Number of pictures per minute = 1/3

Information rate (R) = No of message /sec x average information per message

R = r x H

r= 2.25 x 10⁶ / 3 mins = 12500 elements/sec

H = Log₂M = Log₂12

R = 12,500 x Log₂12

R = 44.812 k bits/sec

Using Shannon capacity theorem

C = B log₂(1 + S/N)

44812 = B log₂(1 + 1000)

B = 4.49 kHz