1. Two computers are joined by 300 metres of cable. If a packet travels along the cable at a speed of 1.77×10 8 m/s, how long will it take for a packet to travel between the two computers? t = d / v t = 300 / (1.77 ×10 8 )s = 0.00000169 s = 1.69 µs

2. If the minimum frame length is kept at 512 bits, what is the slot time for 10, 100 and 1000 Mbit/s? The slot time is calculated by working out how long it takes to send 512 bits. At 10 Mbit/s: slot time = 512 / 10 × 10 6 s = 51.2 µs

3. Figure 4 shows a bridge connecting three collision domains, using ports A, B and C. Construct a forwarding table, after the bridge has received the following frames: [to 02 : from 01], [to 07 : from 02], [to 05 : from 08], [to 03 : from 02] The bridge constructs the following forwarding table. It does this by recording the source address against the port number for any packet it receives Destination, port 01, A 02, A 08, B

4. How many unique addresses can be represented by the 32 bit IP address? As each IP address is 32 bits long, there is a theoretical maximum of 2 32 (about 4 billion)

5. Suppose FDMA system with 900 KHz bandwidth, each user use 40 kHz, and there is 5 kHz is allocated as a guard band so the maximum number of user in the system can be. Number of users = 900 / (40+5) = 20 users

6. A repeater introduces a delay of 3.5 µs. What length of cable is this equivalent to, if the propagation speed is 1.77×10 8 m/s? distance = speed × time or d = v × t Before I can do the calculation I have to convert 3.5 µs into seconds (because the speed is in meters per second): 3.5 × 10): 3.5 × 10 -6 seconds. Then d = 1.77 × 10 8 × 3.5 × 10 -6 m = 619.5 m

7. How long does it take an Ethernet network operating at 10 Mbit/s to send 64 bytes? 64 bytes = 64 × 8 bits = 512 bits time taken = number of bits to be transferred / bit rate = 512 / 10 × 10 6 seconds = 0.0000512 s = 51.2 µs

8. Given that radio waves propagation time is 6 μs for a path length of 2 KM, what will be the velocity of the radio waves? Velocity=Distance/Time = 2*1000m/6.10 -6 s = 3.33*10 8 ms -1

9. Suppose that the sampling frequency is 24 KHz, and it is used along with 16 bits representation for each sample. Then what data rate will be generated after digitization. Data rate = 16 (bit/sample) * 24 (Ksamples/sec) = 384 Kbps

10. Suppose that we have an analogue signal covers the frequency range from 2.5 kHz to 14.5 kHz What is the bandwidth of the signal? What is the minimum sampling rate required for a signal? Bandwidth = 14.5 – 2.5 = 12 KHz = 12000 Hz Sampling rate = 2 x Bandwidth = 2 x 12000 = 24000 Hz = 24 KHz

11. Given that radio waves propagate at 3 × 10 8 m/sec, estimate the propagation time for a path length of 1 km Velocity (speed of light) = distance (traveled by the signal) ÷ time (propagation time) for 1 km (that is, 1000 m or 10 3 m) the propagation time is 10 3 ÷ (3 × 10 8 ) s which is approximately 3 µs

12. Convert (D3) 16 hexadecimal system to the equivalent number in Binary system 11010011 2

13. Convert (100011011) 2 binary system to the equivalent number in Denary system 283

14. Assume that we used Run-length coding as a compression algorithms for a stream of 50 bits, the first 15 are 1s and the 20 that follows are zeros and the last 15 are again 1s What is the compressed version of the message? What is the compression ratio? In binary (00001111)1(00010100)0(00001111)1, In decimal 15(1) 20(0) 15(1) The compression ratio is the number of bits in the original message divided by the number in the compressed message which is 50/27=1.85

15. The receiver receive 1111101 binary number then by using hamming code (odd – parity): a- Show the correct binary number that transmitted by the sender b- Show the correct 4-bits number in denary a) BCDX= 1111 (1) ACDY= 1110 (0) ABDZ= 1111 (1) So the correct binary number transmitted by the sender is 1011101 b) The binary number 1011 is equivalent to 11 denary number

16. Given that radio waves propagation time is 6 μs for a path length of 2 KM, what will be the velocity of the radio waves? Velocity=Distance/Time = 2*1000m/6.10 -6 s = 3.33*10 8 ms -1

17. An analogue-to-digital converter has an input voltage range of  1.2 V. If the resolution of the converter is 4 bits, what is the quantization interval? 16 code word for 4 bits. The quantization interval is 2.4/16=0.15 V

18. It is lossless because after decompression of the received message we will recover the original message without any change to the content. a Run-length coding is used in many compression algorithms. This involves detecting sequences of 0s or 1s in a stream of bits and replacing each sequence with a number that specifies the number of zeros or ones that are in sequence. Assume that we have a stream of 200 bits, the first 10 are 1s and the 180 that follows are zeros and the last 10 are again 1s. What is the compressed version of the message The compression ratio is the number of bits in the original message divided by the number in the compressed message, which is 200/27=7.4 In binary (00001010)1(10110100)0((00001010)1 In decimal 10(1) 180(0)10(1) What is the compression ratio? Is it a lossless compression or a lossy compression and if so, why?

19. How long does it take to transfer 20 MBytes of a compressed file over a link with a bandwidth of 800 Kbit/s t=20*10 6 *8bits/800*1024bit/s=195s

20. In an analogue system for mobile communication, each channel is carried by a separate carrier. The carriers are separated by 20 KHz. The total bandwidth allocated to the communication system is 22.825 MHz. How many channels are there? The channels are separated by 20 kHz, so the bandwidth per channel is a notional 20 kHz. (It is ‘notional’ because a gap is usually allowed between the edge of one channel and the edge of the adjacent channels. However, this does not affect the calculation.) Number of channels = (total bandwidth) ÷ (bandwidth per channel) =22.825 MHz ÷ 20 kHz =1142 channels

21. An odd-parity code is to performed by adding a parity check bit at the end of the message code word. What will be the resultant code word for each of the following: 0011, 1011, 1001, 0000? 00111 10110 10011 00001

22. How many links are required in a fully mesh topology in case of 8 computers: Number of links = n(n-1)/2 =8*(8-1)/2 =28

23. What is the maximum round trip delay for an Ethernet frame (512 bits) operating in a 1Gbit/s LAN? Round trip delay=512/10x10 9 =51.2 ns

24. A repeater introduces delay of 3.2 μs. What length of cable is this equivalent to, if the propagation speed is 1.55x10 8 m/s? Distance= speed x time D= 3.2 x 10 -6 x 1.55 x10 8 m= 496m.

25. Given a propagation velocity of 1.66x10 8 m/s and a delay of 3.2 μs introduced by each repeater, what is the round trip delay for two computers at either end of a 1500 m link that includes two repeaters? The signal travels 1500m. Using the formula: t=d/v t=1500/1.66x10 8 = 9 μs The two repeaters add a further 2X3.2 μs delay, which is 6.4 μs. So that the total one-way times is 9+6.4=15.4 μs The round tripe delay is twice this vale: 30.8 μs

27. What does the graph show (explain briefly)? The graph above shows that the samples are converted in 8 bit code words.

28. What is the quantization interval assuming that the signal amplitude varies between +15 and -5 volts? The range is 20 volts, and the number of levels is 256, so the quantization interval is 20/256 volts, or 0.078 volts.

29. If the sample taken is +8.025 volts (see the graph), what will be the corresponding code word? It will be 0000 1000 the same as for a +8 volt sample.

30. How many different samples (see the graph) can be represented by code words? The use of eight bits in this case determines the number of different sample sizes that can be represented in binary form. 256,

31. Part 3: Answer 2 out of 3 of the following questions in the space given below each question. Consider the following diagram: Idealized frequency spectrum of an analogue signal with cut-off frequencies f1 and f2 Let f1=10KHz and f2=5KHz. What is the bandwidth of the above signal? Bandwidth= f1 – f2 =10KHz-5KHz=5KHz What is the minimum sampling frequency in order to convert the above signal into digital form? minimum sampling frequency=2*Bandwidth=2*5KHz=10KHz;

32. Given a propagation velocity of 1.77x10 8 m/s and a delay of 3.5 µS introduced by each repeater, what is the round trip delay for two computers at either end of a 2500 m link that includes three repeaters? The time travel between two computers is equal to the sum of the time required to travel along the cable and the delay introduced by the three repeaters. This is given by: One-way time (t)= (d/v) + (number of repeaters x delay) = (2500/1.77x10 8 ) + (3x3.5x10 -6 )=24.6 µS The round trip delay is twice this value: 49.2 µ S.

33. Consider a sinusoidal signal. Answer the following questions in this regard: What is the general equation for a sinusoidal and a co sinusoidal signal? X(t)=A sin(wt) is a sinusoidal signal Z(t)=B cos(wt) is a co sinusoidal signal Write the general equation for a sinusoidal signal with 0.75 the amplitude of the signal in part a). Y(t)=0.75A sin(wt). Write the general equation for a sinusoidal signal with half the angular frequency of the signal in part a). Y(t)=A sin(ω/2t). Write the general equation for a sinusoidal signal with an advance of π/4 radians. Y(t)=A sin(wt + π/4)

34. –How many links are required to fully mesh: –(a) 4 computers, 5 computers, 6 computers –(b) 100 computers? (See if you can deduce a formula) Suppose that (n) is the number of nodes (computers) number of links = n(n-1)/2 (a) if n = 4 then the number of links = 4(4-1)/2 = 12/2 = 6 (b) For 100 computers (n = 100), the number of connections is: 100 x (100 – 1)/ 2 = 100 x 99 / 2 = 4950

35. Part 4: Answer the following problems: (you should show all the steps required to solve the questions) suppose that we have an analogue signal covers the frequency range from 3 kHz to 11.5 kHz what is the bandwidth of the signal ? bandwidth = 11.5 – 3 = 8.5 kHz = 8500 Hz what is the minimum sampling rate required for a signal ? sampling rate = 2 x bandwidth = 2 x 8500 = 17000 Hz = 17 kHz

36. What is the maximum round trip delay for an Ethernet frame operating at 10 Mbit/s? Delay = number of bits sent / bit rate = 512 / 10 × 10 6 seconds = 1.2 µs

37. an analogue-to-digital converter has an input voltage range of + 5.5 V. if the resolution of the converter is 5 bits. What is the quantization interval? quantization interval = 11 / 2 5 = 11 / 32 = 0.34 volt What is the peak level of quantization noise produced by the converter defined above? peak quantization noise = 0.34 / 2 = 0.17

38. suppose that we have hotels reservation system. this system deals with four famous hotels. the different types of hotels the client wishes to be available (with the price of single room per night ) are: Meridian 55 KD Sheraton 50 KD Safir International 45 KD Crown Plaza 40 KD Whereas the Safir International and Crown Plaza hotels have only a single room, Meridian and Sheraton hotels can have both single room or suite. The suites prices are as following: Meridian 70 KD Sheraton 65 KD The voice recognition interface of this hotels reservation system will provide the following functions: Be friendly and welcoming Ask the user to enter the hotel type Check if they want a single room or suite in case if they choose Meridian or Sheraton hotels Confirm that the hotel reservation is correct Tell the user the price of this hotel If the user agree, then the system print the reservation voucher of this hotel and finish this transaction in a friendly manner Now you should do the following : Clearly draw the flowchart of this hotels reservation system ? Clearly label the symbols of this flowchart ? Shows how errors are handled

39. 39 What is the frequency if period of the waveform (periodic time (T)) = 0.01 seconds So frequency (f) = 1 / 0.01 = 100 Hz Continue