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Question 1

a) For the bit stream 00110011, sketch the waveforms for differential Manchester and MLT-3encodings

Differential Manchester

0 0 1 1 0 0 1 1

MLT-3

0 0 1 1 0 0 1 1

b) Consider the following continuous signal at the senderS(t) = 5 Sin (20t) + Sin(40t) + 20 Sin(80t)

If the medium introduces the following impairments: an attenuation factor of 0.5, a phase

delay of/4 to all components, and a bandwidth limitation of 25Hz, what will be the final

signal at receiver? Draw also the frequency spectrum.

c) Given the following stream 1100 0011 1010, and assume that during the transmission, thefirst and the last bit are corrupted. Show how these errors can be detected by

1. The 1st complements check summing detection with 4 bits per block2. The polynomial generator of value 1010.

Question 2

a) Explain with a diagram how the message digest, with public key algorithm, can preventan intruder from amending the content of a message

b) Consider the network shown in figure 2, and assume that each node initially knows thecosts to each of its neighbours, after pinging. Work out, in an appropriate routing table,

the shortest paths from the router C to all other nodes in the network, using

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1. The link state algorithm2. The distance vector algorithms for 2 iterations, only.

Question 3

Two stations A and B are attached to a hub with 100Base-TX cable. The propagation delay measured

in bits, between stations A and B is 220 bits. The hub also introduces and extra delay of 10-bit.

Stations exchange, on average, frames of 10000bits long (including all headers and preambles).

Assume that both nodes A and B attempt to transmit at time t-0. After the first collision, the nodes

A and B draw K=0 and K=0 respectively from the back off protocol whose interval is multiple of 512

bits. After the second collision the nodes A and B draw K=0 and K=1, respectively. In this scenario

ignore the jamming signal and any other overhead.

a) At what time (in seconds) is As packet completely delivered to B, from t=0? Use adiagram to support your answer, and show your calculations

b) Suppose now that a switch replaces the hub. The switch has zero processing time. Atwhat time, in seconds, is As packet delivered to B, from t=0, if the switch uses Store-

and-Forward scheme? Use a diagram to support your calculations.

c) Discuss why the delay in (a) and (b) are different. Elaborate on your answerQuestion 4

a) Two distant buildings in an airport centre are connected by a 10km silica fiber. The source oflight is a laser with spectral width of 3nm that operates in the spectrum range of 1000nm.

Evaluate the following

I. The carrier transmitting frequencies, and the bandwidth of the transmission.II. The total dispersion on the fiber, if the modal dispersion rate is 5ps/km, the rate of

the material dispersion is 10ps/nm/km, and the rate of polarization-mode dispersion

is 0.4ps/Km

A

B

C

D

I

G

H

3

82

2

6

5

17

6

4

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In order to obtain the total dispersion on the fiber, the modal dispersion rate, material dispersion

and polarization-mode dispersion needs to be converted to nanoseconds

Modal dispersion is 5ps/km

Tm = 5ps/km * 10km = 50ps convert to nanoseconds 50/1000 = 0.05 ns

The material dispersion is 10ps/nm/km

Tc = 10ps*10km*3nm = 300ps convert to nanoseconds 300/1000 = 0.3 ns

Nm * km

Polarization-mode dispersion is 0.4ps/Km

Tp = 0.4ps* 10km = 0.4ps convert to nanoseconds 0.4/1000 = 0.0004ns

10km

Total dispersion

T = 0.05+ 0.3+ 0.0004 = 0.3041 ns

I. The maximum transmission rate that can be achieved with this fiberF = 1

2*0.3041 * 10-9

b) OC-18c carriers data from the same source by concatenation. The concatenation processinterleaves rows from each OC-1 at the time. Draw a diagram for the OC-18 concatenated

frame, and deduced its transmission rate. What is the rate of the SPE?

Oc-1 Oc-1 Oc-1 Oc-1 Oc-1 Oc-1 Oc-1 Oc-1 Oc-1 Oc-1 Oc-1 Oc-1 Oc-1 Oc-1

Oct-18

Sonnet frame has 90 columns and 9 rows

Total sonet frame = 810 byte

Sonet samples/sec = 8000psec

Oct-18 has 90 * 18 = 1620columns

Size of the frame = 1620 *9 = 14580*8 = 1166640 bits

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Data rate = 1166640 bits = 9333.120mbps

125psec

Oc-1 spe = 87 columns

For OC-18 this is 87 *18 = 1566 columns

SPE = 1566 columns * 9 rows * 8 bit = 112752 bits

Data rate = 112752 = 902.016 bits

125

c) Show with a diagram how VT1.5 can be overlaid into a SONET frame and deduce the rate ofthe group, ignoring any overhead

VT1.5 = 8000 frames/s 3 columns 9 rows 8 bits = 1.728 Mbps

VT1.5 = 3 columns and 24 useable cells

T1 = 24/8 = 3

1 2 3

3 columns

3` 3 3

T1 T1 T1