Communications Subject Wise

Given X,Y are Random variables
As per The given data
E[2x + y] = 0
E[x + 2y] = 33

⇒ 2E(X) + E(Y) = 0.........(1)
⇒ E(X) + 2 E(Y) = 33......(2)

Solving (1) and (2);
2E(x) + E(y) = 0
2E(x) + 4E(y) = 66 [equ2 mul by 2 and + to -]
------------------------------
-3E(y) = -66
E(y)=22

From (1); 2E(x) + E(y)= 0
2E(x) = -E(y)
2E(x)= -22
E(x)= -11

By solving (1) and (2)
we got E(Y) = 22
E(X) = −11
∴E(X) + E(Y) = 22 − 11 = 11

As it is given that it is linear hamming code addition of two codes will produce another code.

For two independent random variable
I(X;Y) = H(X)=H(X/Y)
H(X/Y)=H(X) for independent X and Y
Therefore I(X;Y) = 0

In DPCM the difference of the message signal sample value and the output of prediction filter block is quantized

For Inter symbol interference(ISI) free pulse, If P(t) is having spectrum P(f)
Then = constant
R_S= 2.4 kHz
This condition is met by pulse given in option B.

Answer must be 0.25
As X1, X2, X3, and X4 are independent normal random variables with zero mean and unit variance. Then P(X1 is the smallest ) = P(X2 is the smallest ) = P(X3 is the smallest) = P(X4 is the smallest) = 1/4 = 0.25

Given that
Fixed Length of code n=5
Minimum Hamming distance d_min
Without any constraint, 2⁵ = 32 codewords can be formed.
Following table contains all the possible code words having minimum Hamming distance d_min = 2
Hence, total 16 such code words are possible by a 5 bit code with d_min = 2

For memory less binary Symmetric channel
Channel capacity
C=1-H(p)
H(p)=
p Cross over probability
C = 1 + plog₂p + (1-p)log₂(1-p)
At p=0; C=1
At p=1; C=1
At p= C=0

Given
->x(t) =U+Vt
->X(2)=U+2V
->E [x (2)] =E[U+2V]
-> E[U]+2E[V]
->0+2×1=2 

Answer Range : 69.9 to 70.1

Given s(t) is an SSB modulated signal.
Alternate Solution:
It is the canonical representation of a bandpass signal.

Probability that '1' was transmitted when received bit is '0'

It is given that GSM requires 200 kHz for 8 users and is using TDMA to transmit them. Thus for the next 4 users(9^th,10^th,11^th,12^th) will require an extra 200 kHz bandwidth. Thus, total 400 kHz bandwidth is to be used for 12 users.

Given the conditional density function of R as

Decision error probability that transmitted bit is 1 but receiver decides 0 is
f_R/1 (r = 0) =
Again, the decision error probability that transmitted bit is 0 but receiver decides 1 is
f_R/0 (r = 1) =
Hence, the minimum decision error probability is f_R/1 (r = 0) =

Data rate = r_b= 64 kbps M = 4
Minimum bandwidth =

Therefore BW_min = 16 KHz

Given Binary communication channel Information content in receiving y it in given that x_o is transmitted is

Gaussian Minimum Shift Keying (GMSK) is used for GSM mobile terminals.

Bandwidth of PM signal is given by

So, it depends upon f_m and

(m_p = A_m = message signal amplitude)

The upper limit of the total number of user supported simultaneously is given by spreading factor. We defined the spreading factor as

Spread factor (or) process gain and determine to a certain extent the upper limit of the total number of uses supported simultaneously by a station.
Therefore the maximum number of uses. who can be assigned mutually orthogonal signature sequence is 8

For Binary
Bit error probability
E Energy per bit [No. of symbols = No. of bits]

Given Phase response is

∴ t_y= α
Thusis actual signal propagation delay from transmitter to receiver

Given Y(t ) = X(2t −1)
Shifting in time domain does not change Power Spectral Density. Since Power Spectral Density is Fourier transform of autocorrelation function of WSS process, autocorrelation function depends on time difference.

[time scaling property of Fourier transform]

If X(t) has Power Spectral Density(PSD) S_X(f) then X (2t-1) has power spectral density of 1/2 S_X(f/2)

After modulation with cos(16000)

This is obtain the power spectral density Random process y(t), we shift the given power spectral density random process x(t) to the right by f_c shift it to be the left by f_c and the two shifted power spectral and divide by 4.

After band pass filter of center frequency 8 KHz and BW of 2 kHz

Total output power is area of shaded region

For the shown received signal, we conclude that if 0 is the transmitted signal then the received signal will be also zero as the threshold is 1 and the pdf of bit 0 is not crossing 1. Again, we can observe that there is an error when bit 1 is received as it crosses the threshold. The probability of error is given by the area enclosed by the 1 bit pdf (shown by shaded region)


P (error when bit 1 received)
Or
Since, the 1 and 0 transmission is equiprobable:
i.e.,
Hence bit error rate (BER) is

Optimum threshold is the threshold value for transmission as obtained at the intersection of two pdf. From the shown pdf. We obtain at the intersection
Transmitted = 4/5
Received = 1/5
By solving the two linear equation We can obtain the intersection
pdf of received bit 0
pdf of received bit 1
For Solving threshold we have

So Optimum Threshold to achieve Minimum bit error rate (BER) is

and
Is the entire rectangle.
The region in which maximum of {x, y} is less than 1/2 is shown below as shaded region inside this rectangle.

For phase modulator

maximum phase deviation is
…(1)
for frequency modulator




…..(2)
Given

It is not possible to have bandwidth of 1 GHz/Km because for a single mode fibre, the bandwidth range is from 50 GHz / Km to 100 GHZ/Km.

In Binary FSK signal
Mark frequency is fL = 49 kHz
Space frequency is fH = 59 kHz
Rb = 2 KbPs
Now peak frequency deviation is ΔF
∴ 2ΔF = FH – FL
∴ 2ΔF = 51 KHz – 49 KHz
∴ ΔF = 2/2 KHz
∴ ΔF = 1 KHz

Random process X(t) = 2 cos(2π t + y)
X(t) = 2[cos 2πt . cos y – sin 2πt . sin y]
∴ E [x(t)] = 2 cos 2πt E[cosy] – 2 sin 2πt E[siny]

E [x(t)] = cos2πt – sin 2πt
∴ μ_x (t) = cos2πt – sin 2πt
∴ μ_x (1) = 1

Carrier power P_c = 360 W
Modulation index for 1st audio signal μ₁ = 0.55
Modulation index for 2nd audio signal μ₂ = 0.65

Pt = 490.5 watt
∴Total modulation index μ_t = 0.8514
∴Total radiated power P_t = 490.5 W

In PCM system non – uniform quantization leads to higher average SNR.

• If uniform quantization is used, low amplitude signals of voice are quantized to the same value. This problem is eliminated using non-uniform quantization. If non-uniform quantization is used the average SNR will increase.
• Higher average signal to quantization noise power ratio than the uniform quantizer when the signal pdf is non uniform which is the case in many practical situation.
• Non - uniform quantization minimizes the distortion with same no of levels. Hence bandwidth requirement is same.
• Non - uniform quantizer has higher average SNR than uniform quantization
• RMS value of the quantizer noise power of a non – uniform quantizer is substantially proportional to the sampled value and hence

As the number of bits required for encoding an error signal is always less than the number of bits needed to encode the actual input signal, the bit rate (signaling rate ) of DPCM is always less than that of a PCM signal

Four types of primitives are:
1. Request
2. Indication
3. Response
4. Confirm
Request primitive induces an indication primitive.
If an indication primitive requires a reply, a response primitive may be issued.
This response primitive will induce a confirmation primitive.

A fully connected mesh topology can be thought of as a complete graph in which every node is connected to every other node.
So, number of cable links with n nodes = n(n-1)/2
For a fully connected mesh set with n nodes, number of links are

Option 1 : is TRUE Greater quick efficiency than circuit switching because of effective resource utilization in packet switching

Option 2 : is TRUE As packet switching is implemented at network layer of model. Congestion can be handled. So, connections are not blocked.

Option 3 and 4 : False As Packet switching may be classified into connectionless packet switching, so we don’t require connection establishment.

What is Packet Switching ?
Packet switching is different from circuit switching because there is no requirement to establish a channel. The channel is available to users throughout the data network. Long messages are broken down into packets and sent individually to the network.

Advantages of Packet Switching
While packet switching may not be as suited to voice calls as circuit switching, it has a number of advantages that are hard to ignore. The main advantage that packet switching has over circuit switching is its efficiency. Packets can find their own data paths to their destination address without the need for a dedicated channel. In contrast, in-circuit switching network devices can’t use the channel until the voice communication has been terminated. Packet switching is also reliable because it helps to eliminate packet loss. With packet switching, data packets can be resent if they don’t reach their destination. This isn’t the case for circuit switching which doesn’t have the means to send lost packets. As a result, packet switching is the more reliable method of the two because it ensures that packets reach their destination. Packet switching also reduces the costs associated with running the network. Packet switching networks can transfer general network traffic and voice traffic across the network without the need for a dedicated channel. This saves you money because you don’t need to pay to have one channel available for voice communications.

Disadvantages of Packet Switching
The biggest limitation of packet switching is that it is unsuitable for applications that require minimal latency. In a network that uses lots of voice calls circuit switching is a necessity because it is the only setup that delivers a high-quality end call. Packet switching can only provide a voice call experience that results in choppy audio that makes it difficult for the users to understand each other. Similarly, though packet switching is able to resend lost data packets, this isn’t the case if the network becomes overwhelmed by traffic. If there is too much traffic then packets will be dropped in transit. The end result is the loss of important data. This risk is further increased by the lack of security protocols used to protect packets during data transmission. There is no IPsec to give packets that extra barrier of security against damage. Though packet switching reduces costs in a number of ways it is significantly expensive to implement. Packet switching relies on a range of complex protocols that must be managed from deployment onward.

Note: There is a printing mistake. 7T₂ streams are multiplexed to form 1T₃ stream, but given 7T₁
4T₁ = 1T₂
7T₂ = 1T₃
6T₃ = 1T₄
T₁ = 1.544 Mbps
∴ T₄ = 4 × 7 × 6 × 1.544
Therefore the T4 system has a bit rate
∴ T4≈ 274.2 mbps

Option 1 is wrong : GSM uses both FDM and TDM technique whereas CDMA uses CDM technique so both are not similar.
Option 2 is correct : CDMA allows each station to transmit over entire frequency spectrum all the time because it doesn’t use FDM technique.
Option 3 is correct : CDMA consists of RAKE receiver due to which CDMA assumes that multiple signals (Multipath fading signals) add linearly.
Option (c) is correct if the 3rd statement is “multipath signals” instead of “multiple signals”.

Small Cell Cellular Network
“A cellular network is a radio network distributed over land through cells where each cell includes a fixed location transceiver known as base station. These cells together provide radio coverage over a larger geographical area. Thus, principal of cellular systems is to divide a large geographic service area into cells with diameters from 2 to 50 Kms, each of which is allocated a number of radio frequency (RF) channels. Today’s cellular networks give subscribers advanced features over alternative solutions, including increased capacity, small battery power usage, a larger geographical coverage area and reduced interference from other signals. Popular cellular technologies include the Global System for Mobile (GSM) communication, General Packet Radio Service (GPRS), 3GSM and Code Division Multiple Access (CDMA).

Some of the advantages of cellular systems with small cells are brief described in following paragraphs.
1. High Capacity
2. Less transmission power
3. Local Interference only
4. Robustness

Disadvantages:
1. Infrastructure needed
2. Handover need
3. Frequency planning: to avoid interference between transmitters using some frequencies, Frequencies distributed carefully

From Kepler’s law

There are 43,200 seconds in 12 hours and therefore
T = 12hr = 12*hr = 12*60*60sec= 43200 sec
A = perigee distance
a = a_P

∴ a_P = 26525 × 10³ m
Now a_p = a (1 – e)
a = semi major axis =
=
26525000/0.998
Therefore a = 26578 km

Optical spectrum analyzer is useful while measuring the optical power as function of wavelength.
The optical spectrum analyzer (OSA) is used to measure the spectral characteristics of the light. The OSA displays the optical power as a function of wavelength.

Following 3 layers of OPM:
1) Transport monitoring,
2) Signal quality monitoring
3) Protocol monitoring.

Source : https://web.ece.ucsb.edu/publications/Blumenthal/copy.pdf

Angle of elevation = 35^o
Stratiform rain height = 3 km
The physical pathlength, L, through the rain
is given by =

∴ Physical path length L = 5.2 km

The Correct Answer Among All the Options is C
free space path loss =
d =distance in meters.
F= frequency in Hz
c= 3 m/s.
free space path loss(in db) = 10
= 200dB
Refer the Topic Wise Question for Satellite and Optical Communication Communications

The Correct Answer Among All the Options is C
overall C/N ratio in satellite communication =
+ ………………………………………………………..(1)
Given,

→
Similarly,
=14dB
→
Put these values in (1)
+
=


=13dB
Refer the Topic Wise Question for Satellite and Optical Communication Communications

The Correct Answer Among All the Options is B

Correction code-rate = , it means 1 data bit is encoded into ‘2’ bits by adding 1 extra parity bit.
Output rate after channel adding error correction = 5 =10Mbps
Because of 16-QAM {16 =
} , symbol rate = 
Refer the Topic Wise Question for Digital Communication Systems Communications

The Correct Answer Among All the Options is D
The matrix H is called parity check matrix of any code word only if C
Refer the Topic Wise Question for Information Theory Communications

The Correct Answer Among All the Options is C
linear block code,(7,4)=(n,k)
It means 4 bit data is encoded in 7 bit codeword.it will correct any single bit error.

t=no of error correction , here t=1.
Therefore,


Refer the Topic Wise Question for Information Theory Communications

The Correct Answer Among All the Options is C



Symbol energy (E) =
Average energy()= =2
average energy()=
= 5
=
=2.5
Refer the Topic Wise Question for PSK, DPSK and ASK, FSK Communications

The Correct Answer Among All the Options is A
Kepler's three laws of planetary motion can be described as follows:
•The path of the planets about the sun is elliptical in shape, with the centre of the sun being located at one focus. (The Law of Ellipses)
•An imaginary line drawn from the centre of the sun to the centre of the planet will sweep out equal areas in equal intervals of time. (The Law of Equal Areas)
•The ratio of the squares of the periods of any two planets is equal to the ratio of the cubes of their average distances from the sun. (The Law of Harmonies)
Refer the Topic Wise Question for Satellite and Optical Communication Communications

The Correct Answer Among All the Options is B
For maximum entropy, all symbols must have equal probability.
, m=no. of symbols.
Refer the Topic Wise Question for Information Theory Communications

The Correct Answer Among All the Options is A
For bit ‘1’ →
For bit ‘0’ →,
Condition for orthogonality of signals,

Therefore,
On further solving and equating the components to zero we get
Sin[2]] = Sin …………………(1)
Sin[2]= Sin……………………..(2)
Sin[2]= Sin , when
2
n =0 is not possible because it is given that ()
therefore, 2

Refer the Topic Wise Question for PSK, DPSK and ASK, FSK Communications

The Correct Answer Among All the Options is A
Initial mass = m₀
Rate of decrease in mass = βt
In rocket, mass decreases exponentially and depends upon initial velocity of rocket too.
βt = m₀ – m₀e^-u/c
e^-u/c =
u = -c ln
Now applying v(t) = u + at
v(t) = -c ln - gt
Refer the Topic Wise Question for Satellite and Optical Communication Communications

The Correct Answer Among All the Options is C
In this question by analysing the free body diagram of m₁ and m₂. We can get the equation of motion.
Refer the Topic Wise Question for Satellite and Optical Communication Communications

The Correct Answer Among All the Options is C
Critical frequency of ionosphere is the limiting frequency at or below which wave component is reflected by and above which wave penetrates through ionospheric layer .it depends upon electron density of ionosphere.
,
For N=, = 2.8MHz
For N= , = 9MHz
Refer the Topic Wise Question for Satellite and Optical Communication Communications

The Correct Answer Among All the Options is A
Here, x(t) and y(t) are given. Now, we have to determine the graph of A and B that will give us the output Y.
Now, the given x(t) can be expressed in frequency domain as:

Next, see the figure. x₁(t) and 'j' are added to give x₂(t). So, basically the given x(t) in question should be modified by moving the negative spectrum of the signal downwards as follows:

Next in the diagram see signal x₂(t) and x(t) is getting added. So, let us do the same.

Next is block B. So, spectrum of B will be exactly same as shown in figure (a). So, answer will be option (a)
Refer the Topic Wise Question for Spectrums and Receivers Communications

The Correct Answer Among All the Options is A
A closed loop system is shown in the figure.
XOR gates are generally used to implement addition of binary numbers. Ex-Or gates are also regarded as addition modulo-2.
See the first block in the diagram. It can be thought of an summer. So, let us replace it by an Ex-OR gate. Next block can simply be regarded as a register.
So, input to the Ex-Or gate=32bit(Binary Number), and the other input is the output feedback, i.e 32 bit. Now, 'F' is present for both the blocks, which means both have the same synchronized frequency.
Initially, flip flop will be reset. When clock pulses begin, then let 'M' has been locked to a certain value. So, now, frequency of the output will be equal to MF/2³²
So, now let us assume that clock pulse has been applied.
Refer the Topic Wise Question for PSK, DPSK and ASK, FSK Communications

The Correct Answer Among All the Options is D
We have to find=> P(0
So, this can be written as:
dz[Since f(z)=e^-z]
So on solving the integral we get: -e^-2+1=>0.8
Refer the Topic Wise Question for Random Processes and Variables Communications