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PART A
1. The Nyquist theorem is one of the deciding factor in data communication.
The fibre optics as well as the copper wires are communication mediums. Do
you think the theorem is valid for the fiber optics or for the copper wires.
ANS: The Nyquist theorem is not applicable to copper wire. Because it isapplicable to noiseless channel. The bandwidth of a copper wire depends onthe cable quality, length, and signal to noise ratio of data transmitted. As wesee copper bandwidth is directly depends upon the signal to noise rationwhich concept is given by shannonns in case of noisy channels.Also it is not applicable to fibre optic cable: in fiber optic cable we transmitthe pulse of light and the light is produced due to the motion of moleculesthroughout the system that creates much noise. To calculate the noisecreated per signal we uses the concept of signal to noise ratio that is given
by shanon;s not by nyquiest.2. Noise affects all the signals which are there in air. There are some
communicating modulation techniques. Noise affects which of the
modulation technique the most.
ANS: Techniques of modulation:1 . Pulse amplitude modulation2. Pulse code modulation3. Frequency modulation:Noise affects mostly the pulse code modulation technique. As we much ofnoise is created by the molecules travel in air. In pulse code modulationvoice signal are digitized and combine into a single outgoing digital trunk. AS
we know the molecules of voice traveling in air . An analogy is that supposesender says something to receiver and they are standing upto 10 cm awayfrom each other. the voice signal transmitted by sender travels in air andafter that it is received by the receiver.2nd analogy is that PCM produces a quantization values and the quantizationis produced in the concept of light and again the light is generated when noof molecules collide with each other . due to such collision a noise isproduced.
3. An analog signal carries 4 bits in each signal element. If 10,000 signal
elements are sent per sec, find the Baud Rate and Bit Rate?
ANS:BAUD Rate: it is signal rate or pulse rate: it is denoted by r.it is no of bits a signal carry.Hence baud rate=10000:
No of signal sent / sec= 10,000
Bit rate: s=N*1/r=>N=S*r=10000*4=4000bps
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into a TDM frame which runs at a higher bandwidth. So if the TDM frame consists ofn voice
times, the bandwidth will be n*64 kbit/s
6. While transferring the data from thetransmission medium there are various aspectsof your data getting tempered by other users?What in your opinion is the most secure andinsecure transmission medium. Justify youranswer with an example.
In my opinion fiber optic transmission media is more secure because Due to the degree of
difficulty in tapping fiber optic transmission lines without being detected, fiber optic transmission
media offers a far more secure medium than copper-based or wireless technologies. The result is
that fiber optic transmission media are the media of choice when it comes to long haul applications
such as intercontinental, cross-continental and oceanic (marine) backbone links. It is also thepreferred medium for tier one ISP backbone links..
In other side wireless media is less secure because in this transmission and reception is received
by antennas or satellite . for example when sender sends data to receiver the message is broken
into packets And acc to internet layer of tcp/ip packets are travelling in air.. from these packet some
of packet get lost and receiver cannot find the accurate message.
Directional Point-to-point focused beams employing high frequencies.
Omnidirectional Waves propagating in all directions using signals of lower frequencies.
PART B1.Assume a stream is made of ten 0s .Encode this stream , usingfollowing encoding schemes .How many can you find for each scheme ?
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Unipolar: in unipolor scheme , all the signal levels are on the oneside of the time axis either above or below.
NRZ-L
NRZ-I
RZ
Manchester Differential Manchester
2. Two channels ,one with bit rate of 150kbps and another with abit rate of 140kbps,are to be multiplexed using pulse stuffing TDM withno synchronization bits.Answere the following:
What is the size of frame in bits
What is the frame rate ?Frame rate for ist channel=150,000 bpsFrame rate for 2nd channel=140,000 bps
What is the duration of a frame?Duration of frame for ist channel=1/150,000Duration of frame for 2nd channel=1/140,000
What is the data rate?
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Bit rate=pulse rate X log2L
3. Contrast & compare sampling rate & received signal?Sampling rate: acc to the nyquiest theorm, the sampling rate must be at
least two times the highest frequency.
4. Synchronization is the problem in datacommunication. Explain?
Ans:i. There is no built in mechanism to help the receiving
device adjust bit synchronization midstreamii. The accuracy of received information depends upon the
ability of receiving device.iii. There is no gap between bits in the frames.iv. Decode the separate frame is the headache of receiver
not of senderv. In this we send the bits one after the another without any
start and stop bits.vi. Checksum mechanism has not been used which tells the
begning and ending of bits.5. Can bit rate be less than the pulse rate? Why orwhy not?
ANS:Bit rate is the no of bits to be transmitted in each sample. It depends uponthe level of precession depends. The no of bits chooses such that theoriginal signal can be reproduced with the desired precession in theamplitude..Bit rate= sampling rate X no of bits per secondPulse rate: defines no of pulses per seconds.
If pulse rate =1 then bit rate is also one but if pulse rate>=1 the bit rate isgreater than 1 because bit rate is multiple of no of data levels used in thesignalsExamples: A signal has 4 data levels with duration of 1 ms.Calculate pulse rate and bit rate.
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Pulse rate=1/1* 10 -3=1000 pulses/sec.Bit rate=1000* log24==2000bps.Thus we can say that bit rate> pulse rate
6.A signal is sampled. Each sample represents oneof four levels. How many bits are needed torepresent each sample ?If sampling rate is 8000samples per second, what is the bit rate
Bit per second=4 that is 4 bits are needed for each sampleBit rate= sampling rate X no of bits per second
=8000 X12=96,000=96kbps