1) By using hamming code (even – parity), Show the correct binary number that transmitted by the sender if the receiver received 1111101 binary number. 1) Answer:   BCDX= 1111 (0) No error ACDY= 1110 (1) Error ABDZ= 1111 (0) No error   So the correct binary number transmitted by the sender is 1111111 The binary number 1111

2) By using hamming code (even – parity), Show the correct 4-bits number in denary number that transmitted by the sender if the receiver received 1111101binary number. Answer:   BCDX= 1111 (0) No error ACDY= 1110 (1) Error ABDZ= 1111 (0) No error   So the correct binary number transmitted by the sender is 1111111 The binary number 1111 is equivalent to 15 denary number

3) By using Run-length coding as a compression algorithms for a stream of 300 bits, the first 120 are 0s and the 100 that follows are 1s and the third 60 are again 0s, and the remaining are 1s. What is the compressed version of the message? 3) Answer:   In binary (01111000)0(01100100)1(00111100)0(00010100)1 In decimal 120(0) 100(1) 60(0)20(1)

4) By using Run-length coding as a compression algorithms for a stream of 300 bits, the first 120 are 0s and the 100 that follows are 1s and the third 60 are again 0s, and the remaining are 1s What is the compression ratio in this transmission? 4) Answer:   In binary (01111000)0(01100100)1(00111100)0(00010100)1, In decimal 120(0) 100(1) 60(0)20(1)   The compression ratio is the number of bits in the original message divided by the number in the compressed message which is 300/36= 8.33

5) Answer:   Velocity (speed of light) = distance (traveled by the signal) ÷ time (propagation time)   for 60 km (that is, 60000 m or 6 * 103 m) the propagation time is 103 ÷ (3 × 108 ) s which is approximately 0.0006 5) Given that radio waves propagate at 3 × 108 m/sec, estimate the propagation time for a path length of 60 km

6) Suppose that we have an analogue signal covers the frequency range from 5.5 kHz to 16.5 kHz What is the bandwidth of the signal? 6) Answer:   (2B)Bandwidth = 16.5 – 5.5 = 11 KHz = 11000 Hz

7) Suppose that we have an analogue signal covers the frequency range from 5.5 kHz to 16.5 kHz What is the minimum sampling rate required for a signal? Answer:   Bandwidth = 16.5 – 5.5 = 11 KHz = 11000 Hz   Sampling rate = 2 x Bandwidth = 2 x 11000 = 22000 Hz = 22 KHz

8) Suppose that the sampling frequency is 10 KHz, and it is used along with 10 bits representation for each sample. Then what data rate will be generated after digitization. 8) Answer:   Data rate = 10 (bit/sample) * 10 (Ksamples/sec) = 100 Kbps

9) Suppose FDMA system with 800 KHz bandwidth, each user use 45 kHz, and there is 5 kHz is allocated as a guard band so the maximum number of user in the system can be Answer:   Number of users = 800 / (45+5) = 16 users

10) Calculate the frequency if the period of the waveform (periodic time (T)) = 0.01 seconds 10) Answer frequency (f) = 1 / 0.01 = 100 Hz

11) Calculate the Period of the waveform if the frequency (f) = 200 Hz 11) Answer Period of the waveform(T) = 1 / 200 = 0.005

12) The following are all multiples of the standard unit hertz (abbreviated Hz), put them into decreasing order. 1)MHz 2)GHz 3)KHz 12) Answer:   1) GHz 2)MHz 3)KHz

13) The following are all fractions of the standard unit the meter (abbreviated m). Put them into increasing order. 1) mm 2) nm 3) µm 13) Answer:   1) nm 2) µm 3) mm

14) Convert (D3)16 hexadecimal system to the equivalent number in Binary system Answer   D316 = 110100112

15) Convert (100011011)2 binary system to the equivalent number in Denary system. 15) Answer   1000110112=28310

MODULE 2

Q: The general equation for a sine wave can be written as: The Answer: y = A × sin(ωt)   Q: Write down the expression for a sine wave of amplitude 4 and frequency 200 Hz. x = A sin(ωt) ω = 2πf, f is the frequency in hertz (Hz) ω = 2 × π × 200 = 400 × π x = 4 × sin(400πt)

Q: What is the minimum Sampling Rate required for a signal with a bandwidth covering frequencies up to 8 kHz? The Answer:   Sampling Rate = Frequency Bandwidth * 2 Sampling Rate = 8 * 2 Sampling Rate = 16 KHz

Q: An analogue-to-digital converter has an input voltage range of ±2 Q: An analogue-to-digital converter has an input voltage range of ±2.5 V. If the resolution of the converter is 12 bits, what is the quantization interval & Peak Level of quantization noise.   Quantization Interval = Voltage Range / Bit QI = 2.5 + 2.5 / Bit QI = 5 / 212 QI = 5 / 4096 QI = 1 millivolt Peak Level = QI / 2 Peak Level = 1 millivolt / 2 Peak Level = 0.5 millivolts

Q32: A converter with 4-bit resolution is used to cover an input range from +2.5 volts to –2.5 volts. What is the quantization interval? Hence find the peak quantization noise. The Answer:   Quantization Interval = Voltage Range / Bit QI = 2.5 + 2.5 / Bit QI = 5 / 24 QI = 5 / 16 QI = 0.3125 millivolt Peak Level = QI / 2 Peak Level = 0.3125 millivolt / 2 Peak Level = 0.16 millivolts

Q26: A signal covers the frequency range from 200 Hz to 3. 7 kHz Q26: A signal covers the frequency range from 200 Hz to 3.7 kHz. What is the bandwidth of the signal? The Answer:   Bandwidth = Highest Frequency – Lower Frequency B= 3700 Hz – 200 Hz B= 3500 Hz B = 3.5 KHz Q27: A sine wave has a period of 50 ms. What is its frequency? F = 1/t F = 1/50 F = 20 Hz

MODUL 3

  Q : Write down the equation of the Data Transfer Rate. The answer: Data rate = quantity of data to be transferred time taken for the transfer Q: Write down the equation of the time taken for the transfer. time taken for the transfer = quantity of data to be transferred Data rate

Q99: Write down the equation of the time taken for the transfer. The answer:   time taken for the transfer = quantity of data to be transferred Data rate Q100: A user of a stand-alone personal computer wishes to transfer a 150 Kbyte file onto another stand-alone personal computer in a nearby building. What is the effective data rate in bit/s of the ‘communication’ created by her journey? Assume that 1 byte = 8 bits The user transfers 150 × 1024 × 8 bits in 5 minutes The data transfer rate = 150 × 1024 × 8 bits / 5 × 60 seconds = Result bit/s Q101: How long would it take to transfer the same 100 Kbyte file over a link with a bandwidth of 10 Mbit/s? The Answer: time taken for a transfer = quantity of data to be transferred data rate = 100 × 1024 × 8 bits = 0.08192 s 10 × 106 bit/s = 81.92 ms

Q: A user of a stand-alone personal computer wishes to transfer a 150 Kbyte file onto another stand-alone personal computer in a nearby building.   What is the effective data rate in bit/s of the ‘communication’ created by her journey? Assume that 1 byte = 8 bits The user transfers 150 × 1024 × 8 bits in 5 minutes The data transfer rate = 150 × 1024 × 8 bits / 5 × 60 seconds = Result bit/s Q: How long would it take to transfer the same 100 Kbyte file over a link with a bandwidth of 10 Mbit/s? time taken for a transfer = quantity of data to be transferred data rate = 100 × 1024 × 8 bits = 0.08192 s 10 × 106 bit/s = 81.92 ms

Q: How many links are required to fully mesh, with four computers. The Answer:   Links = N (N-1)/2 Links = 4 (4-1)/2 Links = 4 (3)/2 Links = 12/2 Links = 6

Q: The round trip delay can be calculated from the minimum frame length (512bits) and the bit rate, what is the propagation delay formula? Delay = number of bits sent / bit rate   Q: What is the maximum round trip delay for an Ethernet frame operating at 10 Mbit/s? The number of bet sent = 512 Ethernet Frame= 10x106 Delay = number of bits sent / bit rate = 512 / 10 × 106 seconds = 51.2 µs So you should be able to see that the round trip delay for an Ethernet system operating at 100 Mbit/s is 5.12 µs Q: 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 × 106 seconds = 0.0000512 s = 51.2 µs

Q : Two computers are joined by 300 metres of cable   Q : Two computers are joined by 300 metres of cable. If a packet travels along the cable at a speed of 1.77×108 m/s, how long will it take for a packet to travel between the two computers? The Answer: T = Distance / Velocity T = 300 / (1.77 ×108 )s = 0.00000169 s = 1.69 µs Q: A repeater introduces a delay of 3.5 µs. What length of cable is this equivalent to, if the propagation speed is 1.77×108 m/s? The Answer: Distance = Speed × Time or D = V × T D = 1.77 × 108 × 3.5 × 10-6 m = 619.5 m

Given a propagation velocity of 1. 77×108 m/s and a delay of 3 Given a propagation velocity of 1.77×108 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?   We will do the calculation by finding the one-way time and then doubling it for the round trip. The signal travels 2500 m. Using the formula T = d / v (as in Activity 21) gives T = 2500 / (1.77×108 ) s = 0.0000141 s = 14.1 µs The three repeaters add a further 3 × 3.5 µs delay, which is 10.5 µs. So the total one-way time is 14.1 + 10.5 µs = 24.6 µs The round-trip delay is twice this value: 49.2 µs, or just under 50 µs

other computer. If it takes her five minutes to walk from one computer to the other, what is the effective data rate in bit/s of the ‘communication’ created by her journey? Assume that 1 byte = 8 bits  The user transfers 100 × 1024 × 8 bits in 5 minutes  So the data transfer rate = 100 × 1024 × 8 bits / 5 × 60 seconds = 2730 bit/s = 2.73 kbits/s

How long would it take to transfer the same 100 Kbyte file over a link with a bandwidth of 10 Mbit/s?  time taken for a transfer = quantity of data to be transferred data rate = 100 × 1024 × 8 bits = 0.08192 s 10 × 106 bit/s = 81.92 ms

 This answer makes no allowance for the time taken to send the header portion of each frame, so in practice the time taken might be longer, perhaps 90 ms  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

What is the maximum round trip delay for an Ethernet frame operating at 10 Mbit/s?  The round trip delay can be calculated from the minimum frame length and the bit rate  Delay = number of bits sent / bit rate = 512 / 10 × 106 seconds = 51.2 μs  You should be able to see that the round trip delay for an Ethernet system operating at 100 Mbit/s is 5.12 μs

 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 × 106 seconds = 0.0000512 s = 51.2 μs

Two computers are joined by 300 metres of cable Two computers are joined by 300 metres of cable. If a packet travels along the cable at a speed of 1.77×108 m/s, how long will it take for a packet to travel between the two computers?  The time taken is the distance traveled divided by the speed. (If you are not sure about this, think of a journey of 120 km at 60 km per hour: it would take 2 hours, which is calculated by dividing 120 by 60.) So  time = distance / speed  which can be written very briefly as  t = d / v  where I have written v for speed because mathematicians often use the term ‘velocity’ and hence use v.  So here  t = 300 / (1.77 ×108 )s = 0.00000169 s = 1.69 μs

A repeater introduces a delay of 3. 5 μs A repeater introduces a delay of 3.5 μs. What length of cable is this equivalent to, if the propagation speed is 1.77×108 m/s?  Here we need to treat the repeater’s delay as a time when otherwise the signal would have been traveling at 1.77 × 108 meters per second, and find out how far it would have traveled. This distance can be found from  distance = speed × time or d = v × t  (Think of a journey of 2 hours at 40 km per hour: you would travel 80 km, which is calculated by multiplying 40 by 2.)  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 × 10-6seconds . Then d = 1.77 × 108 × 3.5 × 10-6 m = 619.5 m

Given a propagation velocity of 1. 77×108 m/s and a delay of 3 Given a propagation velocity of 1.77×108 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?  We will do the calculation by finding the one-way time and then doubling it for the round trip. The signal travels 2500 m. Using the formula  t = d / v  (as in Activity 21) gives  t = 2500 / (1.77×108 ) s = 0.0000141 s = 14.1 μs  The three repeaters add a further 3 × 3.5 μs delay, which is 10.5 μs. So the total one-way time is  14.1 + 10.5 μs = 24.6 μs  The round-trip delay is twice this value: 49.2 μs, or just under 50 μs

An Ethernet frame has a maximum frame size of 1518 bytes (excluding preamble). How many bits is this?  There are 8 bits in a byte, so the maximum sized frame is:  1518 × 8 bits = 12 144 bits  How many different addresses are available to each manufacturer?  Each manufacturer is allocated 24 bits. This provides for a theoretical maximum of 224 = 16 777 216 unique addresses

If the minimum frame length is 512 bits (64 bytes), why is the minimum data field size only 46 bytes?  Looking back at Figure 1 we can see that the data field is the only one with a variable length. We'll add together the fixed parts starting at the left hand end.  We do not include the 64 bit preamble as this is used by the receiver to detect a frame. This leaves a 48 bit destination address, a 48 bit source address, a 16 bit length and a 32 bit frame check sequence. This gives a total of:  48 + 48 + 16 + 32 = 144 bits  With a slot time of 512 bits, this means the data field must have a minimum length of 512 –144 = 368 bits = 46 bytes

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 × 106 s = 51.2 μs  At 100 Mbit/s:  slot time = 512 / 100 × 106 s = 5.12 μs  At 1000 Mbit/s:  slot time = 512 / 1000 × 106 s = 0.512 μs

One way of expressing efficiency is to calculate the proportion of data carried by a packet as a percentage of the overall packet length. Calculate the percentage efficiency of the smallest and largest Ethernet frames.  Each Ethernet frame has 6 bytes for source address, 6 for destination address, 2 for type and 4 bytes for frame checking, giving a total of 18 bytes. Knowing that the data field ranges from 46 to 1500 bytes, we can calculate the minimum and maximum frame sizes to be 64 and 1518 bytes respectively. This gives the following efficiencies:  Minimum efficiency = 46/64 = 72%  Maximum efficiency = 1500/1518 = 98.8%