## Communications Subject Wise

Question 1 |

Fill in the Blank Type Question |

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

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**

Question 2 |

_{c}t + k m(t)). The time t on the x-axis in the figure is in milliseconds. If the carrier frequency is f

_{c}= 50 kHz and k = 10π, then the ratio of the minimum instantaneous frequency (in kHz) to the maximum instantaneous frequency (in kHz) is _________ (rounded off to 2 decimal places).

Fill in the Blank Type Question |

Question 3 |

1111000 | |

0010011 | |

1111111 | |

1100001 |

Question 4 |

Where the threshold θ ∈ [-1, 1] is chosen so as to minimize the probability of error The minimum probability of error, rounded off to 1 decimal place, is ______________.

Fill in the Blank Type Question |

Question 5 |

_{X}(f) = 5 W/Hz. The filter h(t) has a magnitude response given by |H(f)| = 0.5 for -5 ≤ f ≤ 5, and zero elsewhere. Z(t) is a stationary random process, uncorrelated with X(t), with power spectral density as shown in the figure. The power in Y(t), in watts, is equal to _________ W (rounded off to two decimal places).

Fill in the Blank Type Question |

Question 6 |

_{c}t, where f

_{c}= 600 kHz. The AM signal f(t) is to be digitized and archived. This is done by first sampling f(t) at 1.2 times the Nyquist frequency, and then quantizing each sample using a 256-level quantizer. Finally, each quantized sample is binary coded using K bits, where K is the minimum number of bits required for the encoding. The rate, in Megabits per second (rounded off to 2 decimal places), of the resulting stream of coded bits is _________ Mbps.

Fill in the Blank Type Question |

Question 7 |

_{1}and V

_{2}equals radians, is ___.

Fill in the Blank Type Question |

Question 8 |

_{1}, X

_{2}) be independent random variables. X

_{1}has mean 0 and variance 1, while X

_{2}has mean 1 and variance 4. The mutual information I (X

_{1}; X

_{2}) between X

_{1}and X

_{2}in bits is

Fill in the Blank Type Question |

I(X;Y) = H(X)=H(X/Y)

H(X/Y)=H(X) for independent X and Y

Therefore I(X;Y) = 0

Question 9 |

The sum of message signal sample with its prediction is quantized | |

The message signal sample is directly quantized, and its prediction is not used | |

The difference of message signal sample and a random signal is quantized | |

The difference of message signal sample with its predictions is quantized |

Question 10 |

Then = constant

R

_{S}= 2.4 kHz

This condition is met by pulse given in option B.

Question 11 |

In binary frequency shift keying (FSK), the given signal waveform is

U_{0}(t)=5cos(20000πt); 0 ≤ t ≤ T, and

U_{1}(t) =5cos(22000πt); 0 ≤ t ≤ T,

Where T is the bit-duration interval and t is in seconds. Both u_{0} (t)and u_{1} (t)are zero outside the interval 0≤t≤T. With a matched filter filter (correlator) based receiver, the smallest positive value of T (in milliseconds) required to have u_{0}(t)and u_{1}(t) uncorrelated is

0.25ms | |

0.5 ms | |

0.75 ms | |

1.0 ms |

Question 12 |

_{X}(f) as shown in Figure (a), where f is in Hertz (Hz). The random process X(t) is input to an ideal low pass filter with frequency response As shown in figure (b). The output of the lowpass filter is Y(t) Let E be the expectation operator and consider the following statements.

I. E(X(t)) = E(Y(t))

II. E((t)) = E((t))

III. E((t)) = 2

Select the correct option:

Only I is true | |

Only II and III are true | |

Only I and II are true | |

Only I and III are true |

Question 13 |

h(t)=

Let y(t) be the output of this filter. The maximum value of |y(t)| is ______.

Fill in the Blank Type Question |

Question 14 |

If the detection threshold is zero, then the probability of error (correct to two decimal places) is ____________.

Fill in the Blank Type Question |

Question 15 |

The value of amplitude sensitivity of modulator is K

_{a}The ratio (accurate to three decimal places) of the power of the message signal to the power of the carrier signal is ______________

Fill in the Blank Type Question |

Question 16 |

Fill in the Blank Type Question |

**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

Question 17 |

Fill in the Blank Type Question |

**Given that**

Fixed Length of code n=5

Minimum Hamming distance d

_{min}

Without any constraint, 2

^{5}= 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

Question 18 |

then the value of (accurate to two decimal places) is __________.

Fill in the Blank Type Question |

Question 19 |

0.11 | |

0.22 | |

0.33 | |

0.44 |

Question 20 |

0.25 | |

0.5 | |

1 | |

2 |

Question 21 |

Channel capacity

C=1-H(p)

H(p)=

p Cross over probability

C = 1 + plog

_{2}p + (1-p)log

_{2}(1-p)

At p=0; C=1

At p=1; C=1

At p= C=0

Question 22 |

Fill in the Blank Type Question |

**Given**

**->**x(t) =U+Vt

**->**X(2)=U+2V

**->**E [x (2)] =E[U+2V]

**->**E[U]+2E[V]

**->**0+2×1=2

Question 23 |

Fill in the Blank Type Question |

Question 24 |

Fill in the Blank Type Question |

Answer Range :

**69.9 to 70.1**

Question 25 |

The channel is

Lossless | |

Noiseless | |

Useless | |

Deterministic |

Question 26 |

Fill in the Blank Type Question |

Question 27 |

Fill in the Blank Type Question |

Question 28 |

Fill in the Blank Type Question |

Question 29 |

Fill in the Blank Type Question |

Question 30 |

Fill in the Blank Type Question |

Question 31 |

π |

Question 32 |

Where denotes the Hilbert transform of m(t) and the bandwidth of m(t) is very small compared to . The signal s(t) is a

high-pass signal | |

low-pass signal | |

band-pass signal | |

double sideband suppressed carrier signal |

**Alternate Solution:**

It is the canonical representation of a bandpass signal.

Question 33 |

Fill in the Blank Type Question |

Question 34 |

Fill in the Blank Type Question |

^{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.

Question 35 |

Fill in the Blank Type Question |

Question 36 |

and

The minimum decision error probability is

0 | |

1/12 | |

1/9 | |

1/6 |

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) =

Question 37 |

Fill in the Blank Type Question |

Question 38 |

Fill in the Blank Type Question |

_{b}= 64 kbps M = 4

Minimum bandwidth =

Therefore

**BW**

_{min}= 16 KHzQuestion 39 |

_{n}}. an can take one of the two possible values —1 and+1 with equal probability and are statistically independent and identically distributed. This sequence is preceded to obtain another sequence , as . The sequence is used to modulate a pulse g(t) to generate the baseband signal

If there is a null at in the power spectral density of X(t), then k is _____

Fill in the Blank Type Question |

Question 40 |

Fill in the Blank Type Question |

Question 41 |

x

_{o}: a "zero" is transmitted

x

_{1}: a "one" is transmitted

y

_{0}: a "zero" is received

y

_{1}: a "one" is received

The following probabilities are given: The information in bits that you obtain when you learn which symbol has been received (while you know that a "zero" has been transmitted) is

Fill in the Blank Type Question |

_{o}is transmitted is

Question 42 |

Fill in the Blank Type Question |

Question 43 |

*p*1 =

*p*4 = 0.125 and

*p*2 =

*p*3. The information rate (bits/sec) of the message source is_________

Fill in the Blank Type Question |

Question 44 |

Question 45 |

*E*(in mJ/bit) necessary is ____

_{b}Fill in the Blank Type Question |

Question 46 |

^{-5}. If 10

^{5}bits are sent over this channel, then the probability that not more than one bit will be in error is _____.

Fill in the Blank Type Question |

Question 47 |

4-QAM | |

16-PSK | |

Walsh-Hadamard orthogonal codes | |

Gaussian Minimum Shift Keying (GMSK) |

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

Question 48 |

_{m}sin (2πf

_{m}t) is used to modulate the phase of a carrier A

_{c}cos(2πf

_{c}t) to get the modulated signal y(t) = A

_{c}cos(2πf

_{c}t + m(t)). The bandwidth of y(t)

depends on A _{m} but no on f_{m} | |

depends on f _{m} but not on A_{m} | |

depends on both A _{m} and f_{m} | |

does not depend on A _{m} or f_{m} |

Bandwidth of

*PM*signal is given by

So, it depends upon

*f*and

_{m}(

*m*=

_{p}*A*= message signal amplitude)

_{m}Question 49 |

Fill in the Blank Type Question |

Question 50 |

A ^{2}N_{0} | |

Question 51 |

Question 52 |

5 | |

6 | |

7 | |

8 |

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**

Question 53 |

1.0 | |

0.5 | |

1.5 | |

2.0 |

Question 54 |

**α = 4mV**Assume an AWGN channel with two-sided noise power spectral density Using an optimal receiver and the relation du the bit error probability for a data rate of 500 kbps is

Bit error probability

E Energy per bit [No. of symbols = No. of bits]

Question 55 |

The value of the expectation

2 | |

4 | |

6 | |

8 |

Question 56 |

Where f

_{c}is the centre frequency, and are positive constants. The actual signal propagation delay from the transmitter to receiver is

∴ t

_{y}= α

Thusis actual signal propagation delay from transmitter to receiver

Question 57 |

_{c}is

Question 58 |

_{X}(f). If Y(t) is the process defined as Y(t ) = X(2t −1) , the power spectral density S

_{Y}(f) is

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)

Question 59 |

Where f is the frequency expressed in Hz. The signal X(t)modulates a carrier cos16000 and the resultant signal is passed through an ideal band-pass filter of unity gain with centre frequency of 8 kHz and band-width of 2 kHz. The output power (in Watts) is _______.

1.5 | |

2.0 | |

2.5 | |

3.0 |

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

Question 60 |

In a PCM system, the signal V is sampled at the Nyquist rate. The samples are processed by a uniform quantizer with step size 0.75 V. The minimum data rate of the PCM system in bits per second is _____.

50 | |

100 | |

150 | |

200 |

Question 61 |

0.8 | |

1.34 | |

5 | |

6 |

Question 62 |

Droop rate decreases and acquisition time decreases | |

Droop rate decreases and acquisition time increases | |

Droop rate increases and acquisition time decreases | |

Droop rate increases and acquisition time increases |

Question 63 |

2 | |

4 | |

6 | |

8 |

Question 64 |

12 | |

14 | |

16 | |

18 |

Question 65 |

R/10Hz | |

R/10 kHz | |

R/5Hz | |

R/5 kHz |

Question 66 |

F(x) – G(x) ≤ 0 | |

F(x) – G(x) ≥ 0 | |

(F(x) – G(x)).x ≤ 0 | |

(F(x) – G(x)).x ≥ 0 |

Question 67 |

. The entropy H(U+V) in bits is

3/4 | |

1 | |

3/2 | |

log _{2}3 |

Question 68 |

respective received signal for both bits are as shown below

If the detection threshold is 1, the BER will be

None |

*P*(error when bit 1 received)

Or

Since, the 1 and 0 transmission is equiprobable:

i.e.,

Hence bit error rate (BER) is

Question 69 |

respective received signal for both bits are as shown below

The optimum threshold to achieve minimum bit error rate (BER) is

1 | |

None |

**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

Question 70 |

increases | |

Remains the same | |

Increases only if N = 2 | |

Decreases |

Question 71 |

3/4 | |

9/16 | |

1/4 | |

2/3 |

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.

Question 72 |

The ratio kp/kf (in rad/Hz) for the same maximum phase deviation is

8π | |

4π | |

2π | |

π |

maximum phase deviation is

…(1)

for frequency modulator

…..(2)

Given

Question 73 |

The bandwidth of 1 GHz/km. | |

The digital communication rate is excess of 2000 Mbytes/s. | |

More than 100000 voice channels are available | |

The mode field diameter (MFD; spot size) is larger than the core diameter. |

Question 74 |

0.5 kHz | |

1.0 kHz | |

2.0 kHz | |

4.0 kHz |

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

Question 75 |

X(t) = 2cos(2πt + Y)

Where Y is a discrete random variable with and The mean μ

_{x}(1) is

1 |

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

Question 76 |

bit/symbol | |

1 bit/symbol | |

bit/symbol | |

bit/symbol |

Question 77 |

12.2 kW and 20 kW | |

15.2 kW and 20 kW | |

12.2 kW and 25 kW | |

15.2 kW and 25 Kw |

Question 78 |

0.85 and 490.5 W | |

0.65 and 490.5 W | |

0.85 and 450.5 W | |

0.65 and 450.5 W |

_{c}= 360 W

Modulation index for 1st audio signal μ

_{1}= 0.55

Modulation index for 2nd audio signal μ

_{2}= 0.65

Pt = 490.5 watt

∴Total modulation index μ

_{t}= 0.8514

∴Total radiated power P

_{t}= 490.5 W

Question 79 |

40 W | |

50 W | |

60 W | |

80 W |

Question 80 |

25 dB and 41 dB | |

30 dB and 41 dB | |

25 dB and 49 dB | |

30 dB and 41 dB |

Question 81 |

15000 bits/s | |

20000 bits/s | |

25000 bits/s | |

30000 bits/s |

Question 82 |

_{m}(t) = 2cos(2π2000t), the peak phase deviation m will be

1.25 rad | |

2.5 rad | |

5.0 rad | |

7.5 rad |

Question 83 |

increased quantizer noise | |

simplification at the quantization process | |

higher average SNR | |

increased bandwidth |

- 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

Question 84 |

the number of bits per code is reduced resulting in a reduced bit rate | |

the difference signal is larger in amplitude than actual signal | |

more quantization levels are needed | |

the successive samples of signal often differ in amplitude |

Question 85 |

Request, Indication, Response and Confirm | |

Request, Inform, Response and Service | |

Request, Command, Response and Action | |

Request, Confirm, Indication and Action |

**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.**

Question 86 |

35 | |

40 | |

45 | |

50 |

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

Question 87 |

1) Greater link efficiency than circuit switching

2) Connections are not blocked when traffic congestion occurs

3) Direct channel established between transmitter and receiver

4) No time is taken to establish connection

Select the correct answer using the code given below.

1 and 3 | |

1 and 2 | |

2 and 3 | |

3 and 4 |

**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.

Question 88 |

2706 bits | |

2634 bits | |

2554 bits | |

2476 bits |

Question 89 |

211.8 Mbps | |

232.6 Mbps | |

243.4 Mbps | |

274.2 Mbps |

**Note:**There is a printing mistake. 7T

_{2}streams are multiplexed to form 1T

_{3}stream, but given 7T

_{1}

4T

_{1}= 1T

_{2}

7T

_{2}= 1T

_{3}

6T

_{3}= 1T

_{4}

T

_{1}= 1.544 Mbps

∴ T

_{4}= 4 × 7 × 6 × 1.544

Therefore the T4 system has a bit rate

∴ T4≈ 274.2 mbps

Question 90 |

1) It is similar to GSM

2) It allows each station to transmit over the entire frequency spectrum all the time.

3) It assumes that multiple signals add linearly.

Select the correct answer using the code given below.

1 and 2 only | |

1 and 3 only | |

2 and 3 only | |

1, 2 and 3 |

**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”.

Question 91 |

1) Higher capacity and robustness

2) Needless transmission power and have to deal with local interference only

3) Frequency planning and infrastructure needed

4) These require both circuit switching and packet switching

Select the correct answer using the code given below

1, 2 and 4 | |

1, 3 and 4 | |

1, 2 and 3 | |

2, 3 and 4 |

**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

Question 92 |

^{o}, K1 = 66063.17 km

^{2}, μ = 3.99 × 10

^{14}m

^{3}/s

^{2}and the earth’s equatorial radius = 6378.14 km, the semi-major axis will be

34232 km | |

30424 km | |

26612 km | |

22804 km |

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 × 10

^{3}m

Now a

_{p}= a (1 – e)

a = semi major axis =

=

26525000/0.998

Therefore a = 26578 km

Question 93 |

1.6 × 10 ^{-5} | |

2.7 × 10 ^{-5} | |

3.2 × 10 ^{-5} | |

4.9 × 10 ^{-5} |

Question 94 |

Optical power attenuator | |

Optical power meter | |

Optical spectrum analyzer | |

Optical return loss tester |

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.

Question 95 |

transport layer monitoring. optical signal monitoring and protocol performance monitoring | |

physical layer, network layer and application layer monitoring | |

data-link layer, presentation layer and session layer monitoring | |

transport layer, session layer and application layer monitoring |

**1)**Transport monitoring,

**2)**Signal quality monitoring

**3)**Protocol monitoring.

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

Question 96 |

6.3 km | |

5.2 km | |

4.1 km | |

3.0 km |

^{o}

Stratiform rain height = 3 km

The physical pathlength, L, through the rain

is given by =

∴ Physical path length L = 5.2 km

Question 97 |

100 dB | |

150 dB | |

200 dB | |

300 dB |

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

Question 98 |

34 dB | |

6 dB | |

13 dB | |

13.5 dB |

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

Question 99 |

1.25 Msps | |

2.5 Msps | |

5 Msps | |

10 Msps |

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

Question 100 |

Which of the following is a valid codeward?

[1 0 0 1 0 1] | |

[1 1 1 1 0 1] | |

[0 1 0 0 1 0] | |

[1 1 0 1 1 0] |

The matrix H is called parity check matrix of any code word only if C

Refer the Topic Wise Question for Information Theory Communications

Question 101 |

1 | |

2 | |

3 | |

4 |

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

Question 102 |

1 | |

1.25 | |

2.5 | |

5 |

Symbol energy (E) =

Average energy()= =2

average energy()=

= 5

=

=2.5

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

Question 103 |

and | |

and | |

and | |

and |

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

Question 104 |

(1/2, 1/4, 1/4) | |

(1/3, 1/3, 1/3) | |

(1/6, 2/3, 1/6) | |

(1/4, 1/6, 7/12) |

For maximum entropy, all symbols must have equal probability.

, m=no. of symbols.

Refer the Topic Wise Question for Information Theory Communications

Question 105 |

_{1}and f

_{2}for non-coherent, orthogonal FSK?

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

Question 106 |

_{0}is launched from ground level. During flight, fuel burns at a constant rate for seconds and exhaust gases are ejecting from the bottom of the rocket at Kg/sec with speed of cm/s relative to the rocket. Ignoring air resistance and assume acceleration due to gravity, g as constant, which of the following expression represents velocity of rocket v(t).

Initial mass = m

_{0}

Rate of decrease in mass = βt

In rocket, mass decreases exponentially and depends upon initial velocity of rocket too.

βt = m

_{0}– m

_{0}e

^{-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

Question 107 |

In this question by analysing the free body diagram of m

_{1}and m

_{2}. We can get the equation of motion.

Refer the Topic Wise Question for Satellite and Optical Communication Communications

Question 108 |

^{11}to 10

^{12}m

^{3}; the critical frequency changes approximately from:

2.2 MHz to 7 MHz | |

2.5 MHz to 8 MHz | |

2.8 MHz to 9 MHz | |

3.2 MHz to 10 MHz |

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

Question 109 |

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. x

_{1}(t) and 'j' are added to give x

_{2}(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 x

_{2}(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

Question 110 |

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/2

^{32}

So, now let us assume that clock pulse has been applied.

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

Question 111 |

0.368 | |

0.135 | |

0.393 | |

0.865 |

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