Q1. 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. Nyquist theorem is valid for both fibre optics and copper wires this is because the Niquist theorem is purely depend upon the mathematic function and has not any concern with the transmission media for communication.
According to Henry Nyquist, a perfect channel has a finite transmission capacity. He derived an equation expressing the maximum data rate for a finite bandwidth noiseless channel. Nyquist said that if an arbitrary signal has been run through a low-pass filter of bandwidth B, the filtered signal can be completely reconstructed by making only 2Bsamples per second. Sampling the line faster than 2B times per second is pointless because the higher frequency components that such sampling could recover have already been filtered out. If the signal consists of L discrete levels, Nyquist’s theorem states:
Maximum bit rate =2* Bandwidth log2 L.
For noisy channel:
Capacity = bandwidth * log2 (1+SNR), where SNR is signal to noise ratio.
Q2. 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. Noise is defined as the disturbance in the signal. When data is send over the network in the form of signal, due the various natural or unnatural circumstances produce disturbance in the signals which is called noise. The noise is various kinds such as thermal noise, crosstalk, and impulse noise. Noise affects all the signals but there are some modulation techniques which are used to reduce the noise. These techniques are:
The amplitude modulation is affected most by the noise this is because there is the change in only amplitude and the amplitude is affected easily by the noise.
Q3. 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?
Given signal carries r =4
Signal element s=10000
Let N be the bit rate
We know that
S = N*(1/r)
? N= S*r
? N= 10000* 4
? N=40000 bps
Bit rate is equals to 40000bps
Q4. What are the reasons for the imperfection caused in transmission media? How the perfection can be measured?
Ans. There are certain reasons for the imperfection caused by the transmission media. These reasons are describe below:
Attenuation: All the transmission media has some sort of loss of signal i.e. called attenuation which make then imperfect. During the transmission of the signal from one place to another the loss or energy is occurred which reduce the overall strength of the signals which is called attenuation. For this is reason amplifier is used in various place in the transmission.
Noise: Noise is the basically the disturbance which is produced in the signal when the signal send over the transmission media then the natural and un natural climate affect the signal such as pollution etc. So that some kind of distortion is occur in the signal. Every transmission media is affected with the noise which makes the signal imperfect.
Distortion: Distortion means that the change in the signal. Due to some circumstance there must be change in the signal. The signal may deviate from actual one this is called distortion.
The perfection can be measure in the form the above three. A transmission media which has less attenuation, less noise and less distortion that transmission media is called perfect.
Q5. There are numerous multiplexing techniques available. What in your opinion is the most appropriate multiplexing technique for the fiber optics as well as copper wires?
Ans. Multiplexing is the process of converting n signal to the one. There are various technique but two basic technique are
FDM(Frequency Division multiplexing)
TDM(Time Division multiplexing)
wavelength Division multiplexing
In fiber optics multiplexing the best technique used is called WDM Wavelength division multiplexing. The basic principle of WDM on fibers is that n fibers come together at an optical chamber, each with its energy present at a different wavelength. The four beams are combined onto a single shared fiber for transmission to a distant destination. At the far end, the beam is spilt up over as may fibers as there were on the input side. Each output fiber contains a short special- constructed core that filters out all but one wavelength. The resulting signals can be routed to their destination or recombined in different ways for additional multiplexed transport.
In the case of the copper wire Frequency division multiplexing technique is used. First the voice channels are raised in frequency, each by a different amount. Then they can be combined because no two channels now occupy the same portion of the spectrum. Notice that even though there are gaps (guard bands) between the channels, there is some overlap between adjacent channels because the filters do not have sharp edges. But nowadays there is use of the Time division multiplexing is used which is particularly deal with the digital signal. In this technique the conversion must be take place from analog to digital signals.
Q6. While transferring the data from the transmission medium there are various aspects of your data getting tempered by other users? What’s your opinion is the most secure and insecure transmission medium. Justify your answer with an example.
Ans. Transmission media is the aspect which is caused for tempering of data by the other. All transmission media has this problem. When we use the copper wire it can be taped by another one. One of the improve version is the optical fibre which An improved media is fiber optic cabling, this media does not emanate any signals since it uses light sources to transmit the signals from node to node. It has proved to be the most secure media available for use on LANs today and will continue to be the most secure media until the black hats discover a way to tap it undetected. The last medium reviewed is the wireless media which uses the airways as their path from node to node. The only way to guarantee secure transmissions is to use a layered approach, or combination of techniques to try to encrypt the data. The best way currently to encrypt data over LANs is to use the IPSec protocol with any of the discussed media. IPSec security is compatible with all types of media so it is the one thing in common with allmedia that will almost guarantee the security of the pathways between nodes.
Q1.Assume a stream is made of ten 0s .Encode this stream , using following encoding schemes .How many can you find for each scheme ?
Unipolar: it can be drawn as:
NRZ-L: It can be drawn as:
NRZ-I: it can be drawn as:
RZ: it can be drawn as:
Manchester: it can be drawn as:
Differential Manchester: it can be drawn as:
Q2. Two channels ,one with bit rate of 150kbps and another with a bit rate of 140kbps,are to be multiplexed using pulse stuffing TDM with no synchronization bits.Answere the following:
What is the size of frame in bits
What is the frame rate ?
What is the duration of a frame?
What is the data rate?
Given bit rate of 1st channel=150 kbps
Given bit rate of 2nd channel=140kbps
Ans1. We allocate 3 slot for first and 3 slot for 2nd and we each size of frame is six bit.
Q3. Contrast & compare sampling rate & received signal?
Ans. In the conversion of the analog to digital, pulse code modulation is used. Sampling is the first stage of the PCM. The analog signal is sampled every Ts second. Where Ts is the sample interval or period. The inverse of sampling interval is called the sample rate or sampling frequency and denote fs where fs = 1/Ts. To produce the anolog signal one necessary condition is that the sampling rate be at least twice the highest frequency in the original signal. This is according to the Nyquist theorem.
e.g. Wo well-known examples where sampled sound is used are the telephone and audio compact discs. Pulse code modulation, as used within the telephone system, uses 8-bit samples made 8000 times per second. In North America and Japan, 7 bits are for data and 1 is for control; in Europe all 8 bits are for data. This system gives a data rate of 56,000 bps or 64,000 bps. With only 8000 samples/sec, frequencies above 4 kHz are lost.
Received signal strength is a measure of the power present in a received radio signal. RSSI is generic radio receiver technology metric which is usually invisible to the user of device containing the receiver but is directly known to users of wireless networking
Q4. Synchronization is the problem in data communication. Explain?
Ans. Synchronization technologies are designed to synchronize a single set of data between two or more devices, automatically copying changes back and forth. For example, a user’s contact list on one mobile device can be synchronized with other mobile devices or computers. Data synchronization can be local synchronization where the device and computer are side-by-side and data is transferred or remote synchronization when a user is mobile and the data is synchronized over a mobile network. In synchronous communications receiver, this specification discloses a decoder for generating a clock signal to synchronize the receiver with the information data rate of the received signal. The implementation of a decoder apparatus as a periodic finite state machine allows a clock signal to be extracted from the energy or transitions of the encoded signal. Such a decoder can define clock information over a large range of data rates. A band pass filter tuned to the characteristic frequency being received can be coupled to the input of the decoder to limit the synchronization range to that desired by the receiver.
Q5.Can bit rate be less than the pulse rate? Why or why not?
Ans. Yes it is possible. Pulse rate is defined as the number of signals element send per second and the bit rate is defined as the number of data element which may be called bits in one second. Data communication needs to increase the data rate and decrease the pulse rate. So that the speed of transmission may be increased and decrease of bandwidth. But in some of the case it may be possible to the bit rate is less than pulse rate. This is done when the single pulse can carry more bits. When this happen the congestion
Q6. A signal is sampled. Each sample represents one of four levels. How many bits are needed to represent each sample? If sampling rate is 8000 samples per second, what is the bit rate
Given Sampling rate = 8000
We know that bit rate =1/Sample rate
So bit rate =1/8000 =0.000125 bps