CSE3020 Network Technology Exam Answers

What is modulation 1

Modulation is the process of encoring source data onto a carrier signal with frequency. fc
In another words modulation is the conversion of a binary signal into a form suitable over a PSRN network.
All modulation techniques involve operation on one or more of the three fundamental frequency domain parameters; amplitude, frequency, and phase.
ASK
>>>values represent by different amplitudes by the carrier.
>>>One amplitude is zero absence of signal
>>>suspect able to sudden gain changes
>>>Inefficient
>>>up to 1200bps on voice grade lines.
>>>Used over optical fibre.
FSK
>>>values represent by different frequencies
>>>less suspect able to error than ASK
>>> up to 1200bps on voice grade lines.
>>>High frequency radio
>>>Even higher frequency on LANs using coax cable.
PSK
>>>phase of carrier signal is shifted to represent data.
DPSK
>>> phase shifted relatively to previous transmission rather than some reference signal.
Comparing & evaluating Pulse encoding techniques 2

Signal Spectrum lack of high frequencies reduce required bandwidth
>>>lack of de component allows ac coupling via transformer, providing isolation >>>Concentrate power in the middle of the bandwidth. .
Error detection responsibility of the data link control level of logic above the signalling level. however useful to have some error detection capability built into the physical signalling encoding scheme.
Cost & complexity higher signal rate leads to higher costs >>> Some codes require signal rate generator than data rate.

Main advantage of biphase encoding techniques 3

1.Synchronisation because there is a predictable transition during each bit time, the receiver can synchronize on that transition. For this reason, the biphase codes are known as self clocking codes.
2.No de Component
3.Error detection the absence of an expected transition can be use to detect errors. Noice of the line would have to invert both the signal before and after the expected transition to course an undetected error.
Q2. The bit rate and baud rate 4
Baud rate is known as signalling or modulation rate
Baud is number of signal elements per second. The rate at which signal elements are transmitted.
Bit rate = bid x M M is the number of bits per signal element.
For two level signalling (binary) bit rate is equal to the baud rate.

Describe any two pulse encoding formats, giving an advantage and disadvantage. Your answer should clearly show the encoding rule 5

NRZ L
>>>2 different voltage for 1 and 0 bit
>>> Voltage constant during bit interval
>>> Absence of voltage for zero, constant positive voltage for 1.
ADVANTAGE
>>>makes a good use of bandwidth, easy to engineer
>>used for magnetic recording due to simplicity & low frequency DISADD
>>> DC component
>>> Lack of synchronization capability.
BIPHASE MANCHASTER
>>>Transition in the middle of each bit period
>>>Transition serves as clock of data
>>>Low to high represents one
>>>High to low represents zero
>>>Used by IEEE 802.3 base band coax and twisted pair CSMA/CD bus LANs. Diff Manchester
>>> Mid bit transition is clocking only
>>> Transaction at start of a bit period represents one.
>>>Used by IEEE 802.5 STP token ring LAN.
ADVAN
>>>Self clocking
>>>No D/C component
DISADD
>>>requires more bandwidth.

Pulse coded modulation. 6

If a signal is sampled at regular intervals at a state higher than twice the highest signal frequency, the samples contain all the information of the original signal.
>>>voice data limited to below 4000Hz require 8000 samples sec
>>>Analog samples
>>> Each sample assigned a digital value.
>>> using 8 bit samples give 256 levels –quality comparable with analog transition

Shannon capacity theorem gives a limit for the data transmission rate 7

C is the capacity of the channel in bits per second
B is the bandwidth of the channel in hertz.
SNR signal to noise ratio

What is crosstalk and what is needed to reduce it 8

Is an unwanted coupling between signal paths. It can occur by electrical coupling between nearby twisted pairs, or rarely, coax cable lines carrying multiple signals.
Can reduce by increasing the twist length of the twisted pair cables.

T1. Explain the terms even parity and odd parity. Is one method is better than the other? 9

An odd parity means that the total number of bits that have a value of 1 in a signal, including the parity bit, must be an odd number. An even parity is just the opposite: the total number of 1 bits in a signal must be an even number. Eg. One signal is 1100 and we are using even parity. Since there are two 1 bits, the parity bit will be 0. but if the signal was 1101, there are now 3 1 bits, and the parity bit will be 1 to make the total number of 1 s 4.
>>>Both ways are equally effective.

Consider the transmission of a character of eight bits. If an error occurs transmitting bits 3 and 5, will even parity checking detect the error? If not, give two error detection techniques that can detect such errors. 10

No.
>>>Stop & Wait ARQ.
>>> Go Back N ARQ.

How cyclic redundancy check (CRC) works. What is the main reason for the wide use of CRC? 11

>>for a block of K bits, the transmitter generates a n bit sequence.
>>> Transmit K + n bit which is exactly double by some number.
>>> Receiver divides frame by that number.
>>>If no remainder assume no error. THE MAIN REASON
>>>Extreme error detection capabilities.
>>>little overhead.
>>>Ease of implementation

Compare and contrast Shielded Twisted Pair (STP) and Unshielded Twisted Pair (UTP).. 12

UTP
>>>Ordinary telephone wire.
>>> Cheapest office buildings pre wired with excess
>>>Easiest to install.
>>> Suffers from external EM interference.
STP
>>>Insulated twisted pairs encased in a metal braid or sheathing that reduces interference.
>>> More expensive.
>>> Harder to handle and work with.

Explain the term physical topology in the context of LANs. State three common physical topologies for LANs. 13

PHYSICAL is the way that the work stations are connected to the network through the actual cables that transmit data. Eg: bus, star. Star Ring. LOGICAL is the way that the data passes through the network from and divide to another.

Q3. Describe the three HDLC station types. 14

Primary Station
>>controls the operation of the link.
>> frames issued are called commands
>> maintains a separate logical link to each secondary station.
Secondary Station
>> Under the control of the primary station
>> Frames issued are called responses.
Combined Station
>> May issue commands and responses.

Data transfer mode for HDLC 15

Normal response mode (NRM)
>> Unbalance configuration
>> Primary initiates transfer to secondary
>> Secondary may only transmit data in response to a command from the primary
>> Used on multi drop lined
>> Host computer as primary, terminals as secondary (host polls each terminal for input)
Asynchronous balance mode (ABM)
>> Balance configuration
>> Either station may initiate transmission without receiving permeation.
>> Most widely used (Point to point, full duplex links)
Asynchronous response mode (ARM)
>> unbalance configuration
>> secondary may initiate transmission without permission from primary.
>> Primary is still responsible for the line. (Initialisation, error recovery, disconnection) >> Rarely used.

Three HDLC frame types differ from one another? 16

HDLC frame types differ by the presentation of the control field.
1.Information data bit transmit to user (next layer up) flow and error control piggyback on the information frames.
2.Supervisory ARQ when piggyback is not used.
3.Unnumbered supplementary link control.

Q4 Terrestrial microwave? 17

Is line of sight. The transmit station must be invisible contact with the receiver station. Setting a distance between stations depending of local geograpgy

Advantages of optical fibre over twisted pair and over coaxial cable. 18

There are few advantages in optical fibre when comparing with other two they are:
Greater capacity: Data rate of hundreds of GBs
Greater repeater spacing: Tens of KM at least.
Smaller size and weight. Lower attenuation, Electromagnetic Isolation: No interference, high security.
Twisted Pair For Analog Data: Amplifies every 5 to 6 KMs. For Digital data, repeaters every 2 to 3 KMs. >> Limited Distance >> Limited Bandwidth (1MHz) >> Limited Data rate (100MHz) >> Susceptible to noise and interference.
Coax cable: higher frequency characteristics. Analog data upto 500MHz amplifies every few Kms. Digital Every 1 Km.

Two types of light sources used in optical fiber? 19

Light emitting diode (LED) Advantages cheaper –wide operating temp range. last longer. Injection laser diode (ILD) Advn More efficient greater data rate.

Q5. Name two applications for ADSL (asynchronous digital subscriber line). 20

Video on demand and selected services. >>>Internet (high speed access)

Give two factors that limit the maximum speed of an ADSL connection 21

1. Line distance 2. Quality

Q6. Describe the terms physical topology and logical topology in relation to a LAN. 22

PHYSICAL topology is the way that the workstations are connected to the network through the actual cables that transmit data. Eg: bus/star/starring.
LOGICAL topology is the way that the data passes through the network from one device to another.
Logical Data link control 1.tranmission of link level PDUs between two stations.
2.must support multi access, shared medium.
3.Relived of some link access details by the mac layer.
4.addressing involves specifying source and destination LLC users.

In relation to Ethernet, explain the difference between a cut through switch and a store and forward switch. 23

Store and forward 1.Accept input, buffer it and then output.
Cut through switch 1.take advantage of the destination address being at the start of the frame. 2.Begin repeating incoming frame into the output line as soon as address recognized. 3.Highest possible throughput. 4.may propagate some bad frames(no CRC check performed)

State two differences between the basic FDDI MAC protocol and IEEE 802.5. 24

1.in FDDI station waiting for a token ring seizes the token by aborting. The token transmitted as soon as the token frame is recognised. After the captured token is completely received, the station begin transmitting one or more data frames. The 802.5 technique of flipping a bit to convert a token to the of the data frame was considered impractical because of the high data rate by FDDI.
2. in FDDI, a station that has been transmitting data frames releases a new token as soon as it complete data frame transmission, ever if it has not begun receive ok now. Transaction. This is the same technique as the early token release options of 802.5. again, because of the high data rate, it would be too mesfsdfsd to require the sdfsis wait for its frame t return, as in normal 802.5 operation.

7.What is the difference between baseband and broadband Ethernet? 25

Baseband cabling allowed only one signal carry information at a time. Broadband communication system cabling that allows multiple signals carrying different types of messages to be transfers alone a path simultaneously.
1.carrier extension: append special extensions to the end of short MAC frames.
2.frame must be at least 4096 bit times long (rather thatn512 for 10/100)
3.ensures frame length transmission longer than propagation time at 1Gbps.
4.frame bursting: send multiple short frames consecutively without relinquishing control (avoid carrier extension overhead)

Give an application for Gigabit Ethernet.
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Describe the main functions of the MAC and LLC layers in the IEEE 802 Project.
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Q8. An Ethernet frame should have a minimum data size. What is the reason for this requirement? 28

To ensure frame length of transmission is longer than propagation time.

Suppose there is heavy traffic on both a CSMA/CD LAN and a Token Ring LAN. On which system is a station more likely to wait? Give reasons for your answer. 29

Token Ring LAN, Because under heavy situations, CSMA/CD LAN listen for the idle medium, then transmit immediately. But on token ring LAN round robin is done
under heavy loads.

Give two advantages of a switching hub LAN over a traditional Ethernet LAN 30

1.Each device has a dedicated capacity 2.no changes to software or hardware. 3. unused lines can also be used to switch other traffic.

Q9. The IEEE standard restricts the maximum length of coaxial cable to 500m for 10BASE5. Give a reason for this restriction. 31

The length of the network can be extended by the use of repeaters. A repeater is transparent to the MAC level; as it does no buffering, it does not isolate one segment from another so for example if two stations on deferent segments attempt to transmit at the same time, their transmission will collide. To avoid looping, only one path of segments and repeaters are allowed between two stations. the standard allows maximum of four repeaters in the path between any 2 stations extending the effective length of the medium 2.5Kms.

Briefly explain the CSMA/CD protocol. 32

Commonly used access control for Bus and star topologies. Ineffective with long frames compared to propagation time.
Stations continue to listen whilst transmitting 1. if the medium is idle transmit 2.is busy, listen for idle medium, then transmit immediately. 3.if a collision is detected jamming signal is transmitted and cease transmission 4.after jamming, with a random amount of timing start again 5. after16 unsuccessful attempts the station gives up.

Gigabit Ethernet implements two enhancements to the basic CSMA/CD. Briefly describe these two enhancements. 33

Carrier extensions carrier extension appends a set of physical symbols to the end of short MAC frames so that the resulting block is at least 4096 bit times in duration, up from the minimum 512 bit times imposed at 10 and 100 Mbps. This is so that the frame length of a transmission is longer than the propagation time at 1 Gbps.
Frame Busting this feature allows for multiple short frames to be transmitted consecutively, up to a limit, without relinquishing control for CSMA/CD between frames. Frame bursting avoids the overhead of carrier extension when a single station has a number of small frames ready to send.

Q10. Repeaters and bridges can be used to extend LAN segments. Give two disadvantages of a repeater as a LAN interconnecting device. 34

1.If two stations on different segments sending frames at the same time will collide.
2.only one path of segments and repeaters between only two stations.

Users of an Ethernet LAN of 100 stations were experiencing unsatisfactory levels of performance. So the network manager decided to segment the LAN into two 35

In general performance of a LAN declines with an increase in the number of devices or the length of the wire. A number of smaller LANs will often give improved performance if devices can be clustered; so that intranetwork traffic significantly exceeds internetwork traffic.

Suppose you need to link a 1BASE5 and 10BASE2 network using a bridge or a repeater. Which device is more suitable for this purpose? Give reasons for your answer. 36

Because they have the same data rate, it is possible to combine 10BASE5 and 10BASE2 segments is the same network by using a repeater that conform to 10BASE5 on one side and 10BASE2 on the other side. The only restriction is that a 10BASE2 element should be used to bridge two 10BASE5 segments because a Backbone segment should be as hesitant to noise as the segment connects.

Difference between a passive repeater hub and an active switching hub 37

PAS REP HUB Central Hub
>>>Hub transmits in coming signal to all outgoing lines
>> only one station can transmit at a time.
>>>With 10 Mbps LAN, total capacity is 10Mbps. ACT SWITH HUB Hub acts as a switch.
>>> Incoming frame switches to appropriate outgoing line.
>>> Unused lines can also be used to switch other traffic. with 2 pairs of lines in use, overall LAN capacity is 20Mbps.

State two differences between Token Ring and Ethernet 38

EATHERNET CSMA/CD
>>>random access select one access mediam randomly.
>>>are twisted pair, coaxizl cable, fiber
>>> 10Mbps
>>>bus star tree topology.
Token RING computers connected in a way that the signal can travel around the network in a logical ring.
>>twisted pair
>>> 4 16Mbps.
>>>Star wired ring topology.

How token passing can control access to a Token Ring. What happens if the free token is lost during the transmission? 39

A single electronic token moves around the ring from one computer to the next. If a computer doesn t have information to transmit, it simply passes the token into the next workstation.

Bust Error 40

An error begins and ends with an enoneous bit, although the bit in between may or may not be computed thus an error burst is defined as the number of bits between two successive enoneous bits.

Error detection 41

Error detection is usually done in the data link layer. Provide a check against the
transmitted frame to ensure its integrity.

Name two common methods of error detection 42

PARITY CHECK each character
>>>character code + parity bit
>>>the value of the parity bits such that the character has an even parity or an odd Parity. Number of ones.
>>>even number of bit enter to undetected.
Cyclic REDANDANCY CHECK
>>>for a block of K bits, the transmitter generates a N bit sequence.
>> transmit K + N bit which is exactly double by some number
>>>Receiver divides frame by that number.

guided media and unguided media 43

uses a cabling system that guides the data signals along a specific path. The data signals are bound by the cabling system. 4 basic types twisted pair/ Coaxial cable / optical fibre.
UNGUIDED for the data signals travel but nothing to guide them along a specific path. Quata signals not bound to a cabling media

Data rate & Distance 44

Bandwidth à higher bandwidth gives higher data rate.
>>>transmission impairments
>>>attenuation limits distance
>>>Interference
>>> Number of receivers.

a) Unshielded twisted pair 45

>>>Ordinary telephone wires
>>>Cheapest
>>>Easy Install
>>>Suffers from Em interference

(b) Coaxial cable 46

Most versatile transmission medium
>>>Television distribution
>>> Long distance telephone transmission
>>> can carry 10,000 voice calls simultaneously. (FDM)

(c) Optical fibre. 47

Act as wave guide
>>>Light emition diode cheaper last long
>>>Injection Laser diode more efficient greater data rate.
Differences between (TDM) and (FDM).
1. In TDM time is used to differenceate the datastreams while in FDM you use freaquency. Eg TV radio.

Most widely used LANS. 50

Personal computer LANs
>>>around offices (Printers disks, internet) low cost & limited data rates.
Back end networks and storage aria networks.
>>>Inter connecting large systems.

Describe the data link layer in IEEE Project 802. Your description should include the sublayers and their functions 51

Pysical Encoding/decoding of signals
>>>Preamble generation/removal
>>>Bit transmission/reception
>>>Transmission medium and topology.

Media access control
>>> assembly of data into a frame with addresses & error detection fields >>>Disassembly of frames + perform address recognition + error detection >>>govern access to the transition medium
>>>for the same LLC, several MAC options may be available

Logical Link Control
>>>Provide an interface to the higher levels and perform flow and error control.

ATM protocol reference model 52

User Plane
>>>provides for user information transfer Control Plane
>>>Call and connection control Management Plane whole system functions >>>resources and parameters in protocol entitles.

Assignment 1 Answers

Question 1

a) The ordering and delivery of a pizza with indications of the interaction at each
level.



The guests, hosts, pizza cook and the order clerk are defined as the Source. They are the ones who generate the data to be transmitted. The guests and hosts, particularly are the people who with the enquiry relating to the ordering of the pizza.
The telephone and the delivery van are not transmitted directly, rather the telephone is needed to transform the voice signal to electromagnetic signals. The delivery van takes the pizza to its required destination.
The telephone line and road could be seen as the Transmission System. The line and the road connect two areas, the departure area and the destination.
The Order Clerk, Pizza Cook, Guest and Host can be defined as the Destination System. The data communication models state that a destination system takes incoming data from the receiver. The order clerk deals with the telephone call and gives directions to the Pizza cook. The host accepts the pizza from the pizza deliverer and serves it to the guests.

b) If the delivery van were to have an accident on the road and not reach it’s
destination. The guests might still get their pizza in the following way.
Quite obviously the pizza in the delivery van would not be suitable to be given to the the host. In this case, the order would have to be placed again by the order clerk to the Pizza Cook upon receiving information regarding the accident. In data communication mechanisms usually, a recovery procedure would allow a file transfer to pick up from prior to the point of interruption, or continue and the information will eventually follow.

Question 2

a) If the solid curve in the following figure represents sin (2pt), what does the dotted curve represent? That is, the dotted curve can be written in the form: A sin (2pft + j); what are A, f and j?



The dotted curve represents 2sin(2p(2f)t).
The peak amplitude (A) is the maximum value or strength of the signal over time; usually this value is represented in volts. The amplitude in this case is 2.
The frequency (f) is the rate (in cycles per second, or Hertz (Hz)) at which the signal repeats. An equivalent parameter is the period (T) of a signal, which is the amount of time it takes for one repetition, therefore T = 1/f. The frequency in this case is 2, displaying 1 cycle.
Phase (j) is a measure of the relative position in time within a single period of a signal. For a periodic signal f(t), phase is the fractional part t/P of the period P through which t has advanced relative to an arbitrary origin. In this case is 180

b) If given the following signal: s(t) = 2+4sin(2pft) + 2 sin (2p(3f)t) + 9 sin (2p(2f)t), plot its amplitude frequency-domain function, S(f).


Question 3
a) Study the works of Shannon and Nyquist on channel capacity. Each places an upper limit on the bit rate of a channel based on two different approaches. How are the two related?
With Nyquist, assuming a noise-free channel, the limitation on the data rate is the bandwidth of the signal. If the rate of the signal transmission is 2B, the highest signal rate that can be carried is B and vice versa.
The limitation is due to the effect of the intersymbol interference, such as that produced by delay distortion. For a given bandwidth, the data rate can be increased by increasing the number of different signal elements placing an increased burden on the receiver; instead of distinguishing one of 2 possible signal elements, during each signal time, it must distinguish on of M (the number of discrete signal or voltage levels) possible signals. Noise and other impairments on the transmission line will limit the practical value of M.
In the case of Shannon, it states that an increasing data rate means bits become shorter so that more bits are affected by a given pattern of noise. At a given noise level, the higher the data rate, the higher the error rate. The presence of noise can corrupt one or more bits.
A greater signal strength should improve the ability to receive data correctly in the presence of noise. A key parameter is the SNR (signal-to-noise ratio), which is the ratio of the power in a signal to the power contained in the noise that is present at a particular point in the transmission.
As signal strength increases, so do the effects of non-linearties in the system, leading to an increase in intermodulation noise. As B increases, SNR decreases as the wider the bandwidth the more noise is admitted to the system.
Both rely on increasing the data rate, by increasing the signal strength or bandwidth.

b) Given a channel with an indented capacity of 45 Mbps, and the bandwidth is 4 MHz. According to Shannon, what signal-to-noise ration is required to achieve this capacity?









Question 4
With the aid of sketches, explain the operation of and the differences between the following transmissions modes used with optical fiber:
i) Step-index multimode

Step-index multimode is where light from a source enters the cylindrical glass or plastic core. Rays at shallow angles are reflected and propagated along the fiber; the surrounding material absorbs other rays. Step-index multimode is a form of propagation, which refers to the variety of angles that will reflect.
In a multimode transmission, multiple propagation paths exist. Each of these paths have different path lengths and so have time to traverse the fiber.
This causes signal elements (light pulses) to spread out in time, limiting the rate at which data can be accurately received. In other words, the need to leave spacing between the pulses limits data rate. This type of fiber is best suited for transmission over very short distances.

ii) Graded-index multimode

Graded-index multimode is possible by varying the index of refraction of the core. The higher refractive index at the center makes the light rays moving down the axis advance more slowly than those near the cladding. Rather than zigzagging off the cladding as in a Step-index multimode, light in the core curves helically because of the graded index, reducing its travel distance. The shortened path and higher speed allows light at the periphery to arrive at a receiver at about the same time as the straight rays in the core axis. Graded index fibers are often used in local area networks.

iii) Single mode

When the fiber core radius is reduced, fewer angles will reflect. By reducing the radius of the core to the order of a wavelength, only a single angle or mode can pass: the axial ray. This single mode propagation provides superior performance for the following reason: as there is a single transmission path with single mode transmission, the distortion found in multimode cannot occur. Single mode is typically used for long-distance application including telephone and cable television.


Question 5
For the bit stream 101110010, sketch the waveforms for each of the eight codes of Table 5.2 in Stallings (pg 133). Assume that the signal level for the preceding bit for NRZI was low; the most recent preceding 1 bit (AMI) was a positive voltage; and the most recent preceding 0 bit (pseudoternary) was a positive voltage.

NRZ -L 1 0 1 1 1 0 0 1 0




NRZI 1 0 1 1 1 0 0 1 0




Bipolar -AMI 1 0 1 1 1 0 0 1 0




Pseudoternary 1 0 1 1 1 0 0 1 0





Manchester
1 0
1 1 1
0 0 1 0




Differential Manchester 1 0 1 1 1 0 0 1 0





B8Zs 1 0 1 1 1 0 0 1 0




HDB3 1 0 1 1 1 0 0 1 0




Question 6
a) Why must a modem be used to transmit binary data through a PSTN?
As PSTN is an analog network, a modem is needed to transform from binary data to analog data.

b) Use sketches and additional text to describe the following modulation methods:
i) Amplitude shift keying (ASK)


There are two different t amplitudes of the carrier frequency when a digital signal is converted to an analog signal in order to represent the analog signal in ASK. One binary digit is represented by a 0 and the other by a 1. ASK is known as an inefficient modulation technique as it is prone to sudden gain changes.
The signal as a result is:

ii) Frequency shift keying (FSK)


In FSK two binary values are represented by two different frequencies near the carrier frequency. The signal is:



iii) Phase shift keying (PSK)

A signal burst of the same phase as the previous signal burst sent is represented by a binary 0. Sending a signal burst of the opposite phase to the preceding one represents a binary 1. The signal resulting from this is:



iv) Quadrature Amplitude Modulation (QAM)

Each signal element represents 2 bits. It can have more than one amplitude and can use 8 phase angles. The diagram above indicates that a 9600 bps modem uses 12 angles, and four of them have 2 amplitudes (QAM)
The example shows 4 bits for the signal element: modulation rate is
(bit rate)/ 4: Therefore its 9600bps at 2400baud where D = modulation rate, R = bit rate, L = signal element and b = bit element.


Question 7
a) The analog waveform shown below is to be delta modulated. The grid on the figure indicates the sampling period and the step size. The first DM output and the staircase function for this period are also shown. Show the rest of the stair case function and give the DM output. Indicate regions where slope overload distortion exists.


b) The signal is quantised using 12-bit PCM. Find the signal-to-quantisation noise ratio.


Question 8
Suppose a file of 20,000 bytes is to be sent over a line at 2400 bps.
a) Calculate the overhead in bits and time using asynchronous communication. Assume one start bit and a stop element of length one bit, and 8 bits to send the byte itself for each character. The 8-bit character consists of all data bits, with no parity bit.
File Size = 20,000 bytes
File size in bits = 20,000 x 8
= 160,000 bits
Bits for one frame = 8
Frames needed = 160,000/8
Total frames = 20,000
Start bit size = 1 bit
Stop bit Size = 1 bit
Control bits per frame = 1 + 1= 2 bits
Total control bits for file = 20,000 x 2
= 40,000 bits
Overhead in bits = 40,000

Line speed = 2400 bps.
Overhead in time = 40,000 / 2400
= 16.67 seconds

b) Calculate the overhead in bits and time using synchronous communication. Assume that the data are sent in frames. Each frame consists of 1000 characters (8000 bits), and an overhead of 48 control bits per frame.
File Size = 20,000 bytes
File size in bits = 20,000 x 8
= 160,000 bits
Bits for one frame = 8000 bits
Frames needed = 160,000 / 8,000
Total frames = 20
Control bits per frame = 48 bits
Total control bits for file = 20 x 48
= 960 bits
Overhead in bits = 960

Line speed = 2400 bps.
Overhead in time = 960 / 2400
= 0.4 seconds

c) What would the answers to parts (a) and (b) be for the same file of 20,000 bytes except at a data rate of 9600 bps?
Overhead in bits = 40,000

Line speed = 9600 bps.
Overhead in = 40,000
Overhead in time = 40,000 / 9,600
= 4.17 seconds



Overhead in bits = 960
Line speed = 9600 bps.
Overhead in bytes = 960
Overhead in time = 960 / 9600
= 0.1 seconds





d) Explain the operation of each null modem connection in the below figure.





The purpose of a null-modem cable is to permit two EIA-232-F "DTE" devices tocommunicate with each other without modems or other communication devices (i.e., "DCE"s) between them.
In some situations it id possible to communicate without DCE such as the distance between devices are so close as to allow two DTEs to signal each other directly. A null modem is needed for this scheme to work.
In this null modem situations it interconnect leads in such a way to confuse both DTEs into thinking that they are connected to modems such, the most obvious connection is that the Transmitted Data signal of one device must be connected to the Received Data input of the other device (and vice versa).

According to the picture we can examine to a null modem to work, we have to connct leads like this.
Signal ground 102:Straight connection 102 to 102 of both DTEs
Transmitted data 103 and received data 104:e have to cross connect those two leads, in a way that DTE1’s 103 goes into DTE2’s 104 and DTE2’s 103 goes into DTE1’s 104.
Request to send 105 and clear to send 106:Both DTE’s 105 and 106 leads has to interconnect as showen in the picture.
Rcvd line sig 109:DTE1’s 105 connects with DTE2’s 109 and DTE2’s 105 connects with DTE1’s 109 and
DCE ready 107 and DTE ready 108.2 and ring indicator 125.:DTE1’s 107 connects to DTE2’s 108.2 and DTE2’s 107 connects to DTE1’s 108.2 also DTE1’s 108.2 connects to DTE2’s 107 & 125 and DTE2’s 108.2 connects to DTE1’s 107 & 125
Transmitter timeing 113 and receiver timing 115: DTE1’s 113 connects to DTE2’s 115 and DTE2’s 113 connects to DTE1’s 115

Question 9
With the aid of frame sequence diagrams and assuming a go-back-N error control scheme, describe how the following are overcome:
a) A corrupted I-frame
b) A corrupted ACK-frame
c) A corrupted NAK-frame

A corrupted I-frame

When a corrupted I-frame is received, the loss is detected. All subsequent frames are then rejected until the corrupted frame is sent successfully with no further corruptions the second time. Then all frames following that frame are re-sent.


A corrupted ACK-frame
In the case of a corrupted ACK-frame as shown below, frames 3 and 4 are discarded as it will timeout, and frames 0 – 4 will be retransmitted.




A corrupted NAK-frame
If I-frame is in error, Receiver sends a NAK i. Until the frame in error is correctly transmitted, the receiver discards all incoming frames. When the sender receives a NAK I, it will retransmit the I-frame and all the packets I+1, I+2, which have been sent, but not acknowledged.
.

Question 10
The question relates to the transmission of data using HDLC protocol.
a) It is clear that bit stuffing is needed for the address, data, and FCS fields of an HDLC frame. Is it needed for the control field?

No Because this field is in fixed length


b) Suggest improvements to the bit-stuffing algorithm to overcome the problems of single-bit errors.

Bit – staffing algorithms can help to fix a fixed frames size to match exactly the size of the frame by increasing additional ‘0’ bits. It can help the receiver to recognize any of the stuffed bits and to discard it if necessary.

Bit stuffing can also help in improving the error – checking on the characters count capabilities.

Bit stuffing also helps the receiver to detect errors by checking on the total frame sizes of HDLC protocol. Sizes that they not used will be discarded and ask for retransmission.


c) Using the example bit string below, show the signal pattern on the line using NRZ-L coding. Does this suggest a side benefit of bit stuffing?








Yes it does, a long sequence of 1's or 0's yields a constant output voltage with no transitions. Bit-stuffing at least eliminates the possibility of a long string of 1's.

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