Communications Test 5

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Communications Test 5

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Question 1
1 / -0

A message signal of 20 cos (2π × 10⁴t), sampled at the Nyquist rate is given to a 10-bit PCM system. The resulting binary sequence is represented with rectangular pulses and is transmitted through free space by using ASK binary signaling scheme. The transmission bandwidth is:

A
200 kHz

B
400 kHz

C
600 kHz

D
100 kHz

Solution

Concept:
The transmission bandwidth for ASK scheme is given by:
Bandwidth = 2R_b (for Rectangular pulses)
R_b = Bit rate given by:
R_b = nf_s
n = number of bits used in PCM encoding.
f_s = Sampling frequency
Calculation:
Given n = 10
f_m = Message signal frequency \(= \frac{{2\pi \times {{10}^4}}}{{2\pi }} = 10kHz\)
f_s = Nyquist Rate = 2 f_m = 20 kHz.
R_b = nf_s = 10 × 20 kbps
= 200 kbps.
So the Transmission Bandwidth = 200 k × 2
= 400 kHz.
Question 2
1 / -0

Consider the following statements for M-ary PSK modulation.
I Bandwidth efficiency of 16 PSK is two times that of 4-PSK modulation
II Bandwidth efficiency is inversely Related to Probability of error.
Which of the above statement is true.

A
Both

B
None

C
I only

D
II only

Solution

Bandwidth efficiency of M-ary PSK is
\(p = \frac{{{R_b}}}{B}\)
\(B = \frac{{2{R_b}}}{{{{\log }_2}M}}\)
\({\rm{\rho }} = \frac{{{{\log }_2}M}}{4}\)
for M = 4
\(\rho = \frac{{{{\log }_2}4}}{2}\)
ρ = 1
for M = 16
\(\rho = \frac{{{{\log }_2}{2^4}}}{2}\)
= 4/2 = 2
Hence 1 is true
(b) As Bandwidth efficiency improves, more number of bits are transmitted per second and hence probability of error increases.
Hence (II) is false
Question 3
1 / -0

Consider a continuously operating coherent BPSK receiver with the data rate of 5000 bits/s. The input digital waveforms are
s₁(t) = A cos ω₀ t m V
s₂(t) = -A cos ω₀ t m V
Where A = 1 mV and the signal sided noise power spectral density is N₀ = 10^-11 W/Hz.
Assume that the signal power and energy per bit are normalized relative to a 1 Ω resistive load. What is the expected number of bit errors made in one day by the BPSK receiver?
(Assume \(Q\sqrt {20} = 4.05 \times {10^{ - 6}}\))

A
1740-1760

B
17404-17601

C
17407-17606

D
17409-17602

Solution

R_b = 5000 bits/sec
The amplitude of input waveform is A = 1 mV = 10^-3 V.
Single-sided noise power spectral density N₀ = 10^-11 W/Hz
Since the signal power is normalized relative to 1 Ω resistive load, we have the average signal power as:
\(\frac{{{A^2}}}{2} = \frac{{{{\left( {{{10}^{ - 3}}} \right)}^2}}}{2} = \frac{{{{10}^{ - 6}}}}{2}\;W\)
Therefore, the bit energy is given by:
\({E_b} = \frac{{{A^2}}}{2}{T_b} = \frac{{{A^2}}}{{2{R_b}}}\)
\(= \frac{{{{10}^{ - 6}}}}{{2 \times 5000}} = {10^{ - 10}}\;Joule/bit\)
So, the bit error probability for BPSK receiver is obtained as:
\({P_e} = Q\left( {\sqrt {\frac{{2{E_b}}}{{{N_0}}}} } \right) = Q\left( {\sqrt {\frac{{2 \times {{10}^{ - 10}}}}{{{{10}^{ - 11}}}}} } \right)\)
\(= Q\left( {\sqrt {20} } \right) = Q\left( {4.47} \right)\)
= 4.05 × 10^-16
Now, the average number of errors in one day is given by:
Errors/day = P_eR_b × (86400 sec/day)
= (4.05 × 10^-6) × 5000 × (86400)
= 1750 bits in error