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1. The Nyquist theorem is one of the deciding factor in data communication. The fiber 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 related with data rate of a communication channal. The Nyquist formula has been designed for a noiseless channel. It defines the maximum data rate of channel in theory. As per the formula BR=2*bandwidth*logL
Where BR is the Bit Rate, Bandwidth represents the bandwidth of the communication channel and L represents the no of signals levels used to represent data. According to my opinion this theorem is valid if the communication mediums are fibre optic as well the copper wires in both case will be certain bit rate bandwidth and the signal levels which are the required deciding factor for data communication by this theorem
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: - Basically there are three modulation techniques:-
- AM (Amplitude Modulation)
- FM (Frequency Modulation)
- PM (Phase Modulation)
As in amplitude modulation technique, it carries the signal that has the higher frequency than the information signal. In this by imposing the lower frequency information signal on the carrier, the amplitude of the resulting compound signal is made to vary in the form of information signal.
So basically these signals are affected by the noise signals that may add with the information signal and lead to disturbance in the signal.
As the strength of the signal decreases with the distance traveled. It reaches to some minimum level which is unacceptable for adequate information. These signals need to be amplified but the amplifier adds the noise and affects the quality of the information.
But in case of the frequency modulated waves that are only affected by the electrical disturbance.
In other words, AM is more noise susceptible than FM because you hear noise from lightning on AM radio but not on FM radio.
In reality it is the high bandwidth of FM that allows the effects of noise to be minimized.
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: - In This case, r=4, s=10,000 we can find the bit rate i.e. N using the formula
S=N*1/r or N=S*r =10,000*4
(Bit rate) N=40,000 bps
By using formula (S=N*1/r baud) we can find
Baud rate=40,000*1/4=10,000 baud
4. What are the reasons for the imperfection caused in transmission Medias? How the perfection can be measured?
ANS: - The imperfection in transmission media is that the data is not same at destination point as it was at source. There is mismatching in the both signals.
There are three types of impairment usually occur. These are attenuation, distortion and noise.
Attenuation- It is the loss of energy as the signal propagates outward on wires and fibers, the signal falls off as distance is increased. This is due to the resistance of the metal. That is the reason why wires get warm when current is passed from them. The amount of energy lost depends upon the frequency. To solve this problem, devices called amplifiers are used.
Delay Distortion- It is caused by the difference in the operating speed of different carrying components. For digital data, fast components from one bit may overtake show components of bit ahead. This problem results in out of order delivery of data which leads to the errors.
Noise-It is the unwanted energy from the soures other than transmitter. Noises are of three types main types : Thermal noise generated by electron Conductor, crosstalk caused by closed inductive copling and noice caused by spikes on the power line. The following line shows the effect of noise on the signal
5. There are numerous multiplexing techniques available. What in your opinion is the most appropriate multiplexing technique for the fibre optics as well as copper wires?
ANS: - There are three main techniques to implement multiplexing.
- Frequencies division multiplexing.
- Time division multiplexing.
- Wave division multiplexing.
Wavelength Division Multiplexing is done entirely in the optical domain. Input electrical signals are each assigned a wavelength, which are combined on one fiber for transmission and separated before being received.
Wave division frequency multiplexing is according to opinion wave division multiplexing is appropriate for fiber optics as well as copper wires. In this form of multiplexing, various light waves from different sources are combined to form a composite light signal which is transmitted across the channel to the receiver. At the receiver end, the composite signal broken down to different light waves and these are delivered to their corresponding receiver.
One may wonder about the mechanism of a WDM. Although the technology s very complex, the idea is very simple. We want to combine multiple light sources into one single light at the multiplexer and do the reverse as the demultiplexer. Combining and splitting of light sources are easily handled by prism.
6. While transferring the data from the transmission medium there are various aspects of your data getting tempered by other users? What in your opinion is the most secure and insecure transmission medium? Justify your answer with an example.
ANS: - Wireless transmission media is more insecure than wired transmission media because in wired, there is percentage of reaching the data in receiving end is more as compared to wireless, wireless medium is more prone to imperfections, because these mediums uses the waves or analog signal to send the data and these waves are more prone to external noise and tendency of hacking these waves are more as compared to wired because these waves works on some frequency.
Transmission through Ethernet cable would bemore secure as it would be difficult for intruders to access the user resources.
While transferring the data from the transmission medium there is possibility of data getting tempered by others users because these time technology is going to be improve, similarly the hacking activity is also improved, so may be your data is retrieve by another users and sometimes it happens due to the distortion in the transmission medium whether it is wireless or wired transmission medium.
1. Assume a stream is made of ten 0s. Encode this stream, using following encoding schemes. How many can you find for each scheme ?
- Differential Manchester
ANS: - Unipolar line coding-
this encoding uses tow voltages stages to represent data. one of these voltage stages is always zero. in this line coding scheme, 1's are encoded as positive values and 0's are encoded as o's value. The unipolar line coding scheme is simplest and very easy to implement. the unipolar line coding scheme for the stream of ten zero is given as-
2. 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. Answerer 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?
Â· What is the size of frame in bits
Size of frame for 1st channel = 150,000 bits
Size of frame for 2nd channel =140,000 bits
Â· What is the frame rate?
Frame rate for 1st channel= 150kbps
Frame rate for 2nd channel= 140kbps
Â· What is the duration of a frame?
Duration of frame for 1st channel=1/150,000 s
Duration of frame for 2nd channel=1/140,000 s
Â· What is the data rate?
Data rate for 1st channel = 150,000 bps.
Data rate for 2nd channel =140,000 bps.
3. Contrast & compare sampling rate & received signal?
ANS: - comparison between the sampling rate & received signal
Sampling rate is the the number of samples of the audio signal taken per second, and determines the high frequency cutoff the replayed audio of in which received signal is in the distorted form and it is reconstructed again. Mathematically it is formulated as:
Where fs stand for sampling frequency and fm stands modulated frequency
in the sampling rate and simple received signal is the that was received at destination after modulation or demodulation to occur main difference b/t the sampling rate and the received signal is that in sapling encoding to occur because signal is reconstructed and in received signal only decoding to occur
4. Synchronization is the problem in data communication. Explain?
ANS: - Synchronization is a problem since in a communication network there are various devices and terminals connected and they all function or rather work at their own speeds and have different bandwidths. So basically the aim is to bridge the gap between these devices. For example, if a transmitter has high speed and is sending data at its high speed where as the receiver unit is a slow-speed device, it won't be able to accept the data packet/frames at the high speed at which it is being sent. So synchronization basically deals with handling the data packets and seeing through that all that is sent is received correctly at the receiver and there is no data loss. The occurrence of the latter will require re-transmission and the prevention of the same is by using buffers that hold that high speed data packet temporarily and then passing it onto the receiver subsequently. So we can say that synchronization is a problem in data communication
5. Can bit rate be less than the pulse rate? Why or why not?
ANS: - yes, bit rate can be less than the pulse rate it is because of the reason that bit pulse represents the time taken to send the one signal of data. The “bit rate” is used to describe the number of bits sent over a communication line in one second. Signal must be a group of data bits so it is obvious that pulse rate is always greater then the bit rate.
6. 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?
ANS: - According to the formula:-
Bit Rate=sampling rate*no of bits per sample
1 level=2 bit;
In the given signal there are 4 levels
So bit rate is=8000*8=64000 bps