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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: The nyquist theorem gives the different bit rate formulas defines the theoretical maximum bit rate. It says that if you have a function whose Fourier spectrum does not contain any sines or cosines above f, then by sampling the function at a frequency of 2f you capture all the information there is. Thus, the Nyquist theorem is true for all media.
the Nyquist theorem is true for copper wires and fiber optics.
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 affects mostly the pulse code modulation technique.
as we know the molecules of voice traveling in air. An analogy is that suppose sender says something to receiver and they are standing upto 10 cm away from each other. the voice signal transmitted by sender travels in air and after that it is received by the receiver. In general, interferences with continuous spectra are more detrimental than those with line spectra, and non-impulsive types more detrimental than impulsive types. Noise-reducing circuits were ineffective against random fluctuation noise, but in the presence of certain types of impulse interference, limiters and canceling circuits provided such great improvement in performance that it was possible to maintain satisfactory communication despite a 35-decibel reduction in carrier intensity. As a general principle, it appears that, whenever there is a characteristic difference between the wave forms or the spectra of the signal and the interference, the impairment of intelligibility by electrical interference may be reduced by employing amplitude-selective or frequency-selective circuits in the radio receiver.Hence we can say that noise technique is mostly pulse code modulation technique.
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?
Ans : The calculation of bond rate is very siple that are given below.
- Baud rate = number of signal elements=10000bauds per sec.
- Bit rate=baud rate*number of bits per signal elements=1000*4=4000bps.
Q4-What are the reasons for the imperfection caused in tramsmission medias ? how the perfection can be measured?
ANS-Imperfection means the main limitation comes in the transmission media. Imperfection in transmission is because of the Receiving unwanted signal at receiver side. Interference or noise are produce that by signal and Due to higher data rate errors are more generated.
The perfection can be measure to seeing the reliability of a particular transmission media and reliability is measured for checking performance of each transmission media.yhe perfection can be measure by see the tramsmission media, type,imperpection,and after that measure.if we have a imperfection madia is guided and type is coaxial,it cover large distance at higher rate and produce more error.
(measure) It has higher bandwidth, the signals weeks rapidly and repeaters are required for measuring performance.
Q5-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- Multiplexing Techniques
Multiplexing is the process where multiple channels are combined for transmission over a common transmission path.
The techneques are given below:
- 1) FDM 2) TDM
The explanation are given below:
FDM was the first multiplexing scheme to enjoy wide scale network deployment, and such systems are still in use today. However, Time Division Multiplexing is the preferred approach today, due to its ability to support native data I/O channels.
In FDM, multiple channels are combined onto a single aggregate signal for transmission. The channels are separated in the aggregate by their frequency.There are always some unused frequency spaces between channels, known as “guard bands”. These guard bands reduce the effects of “bleedover” between adjacent channels, a condition more commonly referred to as “crosstalk”.In Time Division Multiplexing, channels “share” the common aggregate based upon time timeplex is probably the best in the business at Time Division Multiplexing,. When Timeplex was started by a couple of ex-Western Union guys in 1969 it was among the first commercial TDM companies in the United States. In fact, “Timeplex” was derived from TIME division multiplexing.
Q6-While transferring the data from the transmission medium there are various aspects of your data getting tempered by other users. What in your opinin is the most secure and insecure transmission medium. Justify your answer with an example.
ANS- There are three broad categories of media Wire and fiber. Fiber optic transmission media is more secure because, 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.
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.
Q1-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-Manchester- The idea of rz and nrz-l are combined into the menchester. In Manchester encoding the duration of the bit is divided into the two halves.the voltage remains at the one level during of the bit is divided into two halves.the voltage remains at one level during the first half and moves to the other level in the second half.the transition at the middle of the bit provide synchronization.
Differential Manchester-- Differential Manchester on the other hand combines ideas of RZand NRZ-Lt There is always a transition at the middle of the bit,but the bit values are determine at the begning of the bit.if the next bit is zero,there is transition if he next bit is one there is none.
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?
- acc to me size of bit is 15
- 2) Frame rate for ist channel=150,000 bps Frame rate for 2nd channel=140,000 bps
- 3)Duration of frame for ist channel=1/150,000 Duration of frame for 2nd channel=1/140,000
- 4) Q3-Contrast & compare sampling rate & received signal?
According to the nyquist theorem the sampling rate must be at least 2 times
The highest frequency contained in the signal.the sampling rate is least 2 time the highest frequency not the band width.if the analog signal is low pass the band width is the highest frequency at the same value.if the analog signal is bandpass the bandwidth value is lower than the value of the maximum frequency.
Receiving signal-Receive singnal are the signals having the bitrate which is send by the sender when it sent from one end. What the received bitrate of received data.
Q4-Synchronization is the problem in data communication. Explain?n
Ans:Before we discuss that Synchronization is the problem in data communication firstly we discuss what is data communication.
data communication are the exchance of data between two devices via some form of the tramsmission medium such as a wire cable.for data communication to occur the communication device must be the part of the communication system made up of the combination of the hardware and software.
The data communication is depend on Deliver,accuracy,timeliness,jitter.
The synchronization is faster so it is more use full in the high speed application such as the transmission of data from one computer to the another computer. The Checksum mechanism has not been used which tells the begning and ending of the bit.we can send the the bits one after the another without the star and the end.one thing more is that the accuracy is depend upon the reciving device.
Q5-Can bit rate be less than the pulse rate? Why or why not?
Bit rate: the bit rate is the no of bit transmitted per second.sometime data also called the bit rate.The signal rate also called the pulse rate. the pulse rate
Define the no of pulses per second. If pulse rate =1 then bit rate is also one but if pulse rate>=1 the bit rate is greater than 1 because bit rate is multiple of no of data levels used in the signals.
Exm: A signal has 6 data levels with duration of 1 ms. Calculate pulse rate and bit rate.
Bit rate=1000* log26== 1000* log24=1000*4*log2 =4000bps.
Pulse rate=bir rate/x log2
Pulse rate=1/1* 10 -3=1000 pulses/sec.
From thie example we see that the bit rate is greater then the pulse rate.
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
Ans:Bit rate = 4
The sampling rate is 8000
4 bit are needed the bit rate is given below: