calculation activites of block 1.pdf

T324 - Block 1- Wireless technologies

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4

Activities Block 1 part[1,2,3,4]Part1

Activity 2 (self-assessment/revision). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Calculate the period of a radio wave whose frequency is 2 GHz.

f= 2 GHz

TT= 1/ (2 x 109 Hz)

= 5 x 10-10 s = 0.5 ns = 500 ps.

Calculate the frequency of a radio wave whose period is 4 ms.T= 4ms

ff= is 1 / (4 x 10-3 s)

= 2.5 x 102 Hz = 250 Hz


Calculate the wavelength of a radio wave whose frequency is 3 GHz.

c= 3 x 108 m/s

= (3 x 108 m/s)/ (3 x 109 Hz) = 0.1 m = 100 mm.

Period T!

T=1/f

Frequency f!

f=1/T

Wavelength ! c = 3 x 108 m/s

2GH = 2x109 (1000 000 000) Hz = > 2000 000 000 Hz5x10-10 s = 5x10-10 * 109 (1000 000 000) ns => 0.5 ns0.5 ns = 0.5x103 (1000) ps => 500ps

4ms = 4x10-3 or (4 1000) s => 4x10-3s or 1/2501/ (1/250) = 250Hz

3GH = 3x109 (1000 000 000) Hz => 3000 000 000 Hz

0.1m = 0.1x103 (1000) mm => 100mm

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4Calculate the frequency of a radio wave whose wavelength is 100 m.

Assume that c= 3 x 108 m/s.

c= 3 x 108 m/s

f 3 x 108 m/s)/ (100 m) = 3 x 106 Hz = 3 MHz.


If each UHF TV channel occupies the same bandwidth in the 470--854MHz frequency range, what is the bandwidth available to each channel? Which frequencies are occupied by

channel 21? Which frequencies are occupied by channel 68?

Frequency range 854 MHz 470 MHz

Channel = 48Lowest frequency at channel 21Highest frequency at channel 68

Bandwidth available to each channel

854 470 = 384

384/48 = 8 MHz

frequencies are occupied by channel 21 470 8

470+8 = 478 MHzfrequencies 470478 MHz

frequencies are occupied by channel 68 854 8

854-8 = 846 MHzfrequencies MHz846--854

Frequency f!

f=c/

Bandwidth BW!

3x106 Hz = 3x106 1000 000 MHz => 3MHz


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4Activity 6 (self-assessment)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .If an antenna in free space receives 16 W of power at a distance of 2 km from an isotropic transmitter, how

much will it receive at 4 km?How much at 8 km?

n

Activity 7 (self-assessment). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

A 1.8 GHz radio wave propagates 16 km through the atmosphere, and then through two brick walls, each of100 mm thickness. Rain leads to an atmospheric loss of 1.5 dB/ km, and brick attenuates at 40 dB/ m at this

frequency .Calculate the total power loss in dB over this path that is due to attenuation by the atmosphereand by the wall. (Ignore the inverse square law.)

Atmospheric loss of 1.5 dB/ km

Path length in (km) at Atmosphere = 16km

Two brick walls attenuates at 40 dB/ m2 brick walls, each of 100 mm

Inverse Square (Free space)!1/d2

: !

)(free space

16 W 2km isotropic

4km8 km

1/d2 ok

4 2+2 2==

2== 0.1 m


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4

The total power loss in dB by the atmosphere + wall atmospheric

1.5 dB/ km x 16 km = 24 dB

Two brick walls

40 dB/ m x 0.1 m = 4 dB 40 dB/ m x 0.1 m = 4 dB

+The total loss is 24 dB + 4 dB + 4 dB = 32 dB


Assuming a 1/d4 relationship, calculate the loss of received power if the reception distance is (a) doubled and(b) tripled. Compare these figures to the corresponding results for the inverse square law.

1/d4 relationship

Distance = doubled => 2Tripled => 3

The loss of received power at distance

Doubled

)a(1/24= 1/16.

Tripled)b(1/34= 1/81.

Compare the results for the inverse square law inverse square law( 1/d2)

1/22

1/4 =1/32

1/9 =

The total power loss in dB!Attenuation in dB/ (km) path length in (km)

The overall loss of signalpower with distance d!


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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Estimate the physical length of a /4 rod to be used for a 100 MHz FM radio station. How long would it be for

a 2.4 GHz Wi-Fi link? (Hint: in each case, assume a free space wavelength.)

/4 4100 MHzf =

f= 2.4 GHz

wavelength100

(3 x 108 m/s)/ (100x106 Hz) = 3m. 4

2.4

(3 x 108 m/s)/ (2.4x109 Hz) = 0.125m = 125 mm.

4

Wavelength

100 MHz = 100 x 106 Hz => 100 000 000 Hz

2.4 GHz = 2.4 x109 Hz => 2.4 000 000 000 Hz

0.125m = 0.125 x 103 mm => 125 mm


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4Part2


What is the rate of data transmission if the signalling rate is 4500 baud, and each symbol represents 4 bits?

signalling rate = 4500 baud

4 bits

Data rate (bit rate)

4500 x 4 = 18 000 bit/s

The data rate is 18 kbit /s


A carrier modulated at a signalling rate 6000 baud. Its used with a modulation system known as 16-QAM, inwhich there are 16 symbols.

signalling rate =6000 baud

16 symbols

)a(How many bits does each symbol represent?

24 = 16 log 16 /log2

Data rate (bit rate) b!Symbol per second bits per second

Signaling rate = number of symbols per secondAlso 1 baud is 1 symbol per second

Number of bits! Log2 (number of symbol)

= log (number of symbol) / log 2 ()Log16/log2

[log10 (number of symbol)] / 0.301

18 000 bit/s = 18 000 x10-3 or (18 000 1000) kbit/s => 18kbit/s


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4(b) What is the data rate of this system in bits per second (or kilobits per second)?

6000 symbols per second

4 bits per second

6000 x 4 = 24 000 bit/s => 24 kbit/s


By what fraction of a cycle do the two waves in figure 26 differ in phase?

The phase difference is one quarter of a cycle


Figure 27 shows sections of sinusoids with differing phase strung together to convey data. The data isindicated by 1s and 0s.

(a) How many complete symbols are transmitted by the wave in figure 27?Three complete symbols are transmitted, shown as representing 1, 0 and 1.

3symbols 101 (b) What item of data is represented by the symbol preceding the first complete symbol in figure27?

Little is shown of the symbol preceding the first complete symbol, but it is clear that there is no changeof phase, so it must be 1.



24 000 bit/s = 24 000 x10-3 or (24 000 1000) kbit/s => 24kbit/s


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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .If the data in figure 30 were DPSK, what would it be decoded to, using the rule given in the T305 extract?

Each time the phase changes the output is a 1. Otherwise it is a 0, giving 101110101.


In what ways are the comments on BPSK and QPSK in the last sentence consistent with the data in table 3?

In the table 3, the modes that use QPSK give twice the data rates of those that use BPSK. That is, BPSK is usedat data rates of 6 Mbit/s and 9 Mbit/s, whereas QPSK is used at the data rates of 12 Mbit/s and 18 Mbit/s.

phase 1 0


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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Suppose that a version of MSK operates at 10 kbaud and one symbol uses a segment of a 10 MHz sinusoid. Ifthe other symbol uses a segment of a higher frequency, what frequency should that higher frequency be?

signalling rate (S) =10 kbaud 10 MHz

What frequency should that higher frequency be The frequency spacing is half the signalling rate

f2 5 kHz (half of 10 kbaud)

= 0.005MHz

0.005MHz + 10.005 MHz


A transmission at 2 GHz has a bandwidth of 4 MHz which of the receiver responses n figure 39 is best suitedfor a receiver of this transmission? (In figure 39, the frequencies on either side of the central frequency arecutoff frequencies.)

10 kbaud( kBd) = 10 kHz

5 kHz = 5 x 10-6 or (5 1000 000) MHz => 0.005MHz

S = frequency spacing


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4

Transmission at 2 GHz has a bandwidth of 4 MHzResponses (a - b - c)

)bandwidth(4

bandwidth

The bandwidth of figure 39(a) is 2.001 GHz 1.999 GHz = 0.002 GHz = 2 MHz

The bandwidth of figure 39(b) is 2.002 GHz 1.998 GHz = 0.004 GHz = 4 MHz

The bandwidth of figure 39(c) is 2.4 GHz 1.6 GHz = 0.8 GHz = 800 MHz

b=4

The response in figure 39 (b) is best suited to a transmission with a bandwidth of 4 MHz


Describe in words the waveform represented by the point -1 on the Q axis.The -1 means it is the inversion of the waveform corresponding to the point +1.so it is an inversion of figure

42(b)


(a) If a QAM system uses the amplitudes 1, -1, 3 and -3 on each of I and Q waves, how many symbols arethere altogether?

There are four states associated with the I wave, and four with the Q wave, giving a total of eight possiblesymbols or states.

(b) How many bits can each symbol represent?bit

8symbols4I4Q

log8/log2 =3

000 001 010 011100 101 110 111

Bandwidth BW!


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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .QPSK could be viewed as a type of QAM. If QPSK were to be described as x-QAM, what would x be?

In figure 45 is the constellation diagram for QPSK. There are four symbols, so if this were referred to as a formof QAM it would be 4-QAM.


(a) In a 64-QAM system, how many bits are there per symbol?

64 = 2x2x2x2x2x2 = 26. So there are 6 bits per symbol.

Alternatively, using

Number of bits = [log10 (number of symbols)] 0.301

Number of bits = [log10 64] 0.301

= 1.806 0.301

= 6

(b) If the symbol rate is 10 000 baud, what is the data rate?

There are 10 000 symbols per second, so there are 60 000 bit/s

10 000 baud (Bd) = 10 000 bit per second (bps)

Data rate = 10 000 x 6 = 60 000 bps

Log64 / log2 = 6

Number of bits! Log2 (number of symbol)

= log (number of symbol) / log 2 ()Log16/log2

[log10 (number of symbol)] / 0.301




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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

What is the spectral efficiency in figure 50 (c)? Hence which of the three transmissions is most spectrallyefficient?

Data rate = 800 kbit/s

Frequency range for (c) = 701- 701.02

Spectral efficiency for (c)

data rate For (c) the bandwidth is (701.02 701) MHz = 0.02 MHz = 20 kHz

=

= 40 kbit/s/kHz = 40 bit/s/Hz

Which of the three transmissions is most spectrally efficient?(c) Is the most spectrally efficient.

a-b 98 a-b c

Spectral efficiency!

Data rate / bandwidth


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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .If the noise power density of a white noise source is as in figure 53, what noise power falls within the channel

shown shaded in figure 53

Noise = 10pW/Hz

Frequency range = 2402 - 2422 MHz

Noise power falls within the channel

20 MHz=2422-2402

10pW/Hz x 20 MHz = 200 W

Noise power!

Noise bandwidth

20 MHz = 20 x106 Hz so 10pW/Hz x 20 x106 Hz = > 200x106 pW

200 x106 pW = 200 x106 x 10-6 or ( 1000 000) W = > 200 W


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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Calculate the signal to noise ratios for devices B and C, and hence state which of A, B and C has the best

signal to noise ratio.

Signal for B, C = 150 pW/Hz

Noise for B, C = 50 pW/Hz

Bandwidth for B = 20

Bandwidth for C = 30

Signal to noise ratios for devices B and C

BDevice Signal noise

The signal power is150 pW/Hz x 20 MHz = 3000 x 10-6 W = 3 mW

Noise power

The noise power is50pW/Hz x 20 MHz = 1000 x 10-6 W = 1 mW

Signal power!

Signal bandwidthNoise power!

Noise bandwidth20 MHz = 20 x106 Hz

So 150pW/Hz x 20 x106 Hz = > 3000 x106 pW3000 x106 pW = 3000 x106 x 10 -12 or ( 1000 000 000 ) W = > 3000 x 10-6 W3000 x 10-6 W = 3000 x 10-6 x 103 mW => 3mW


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4

the signal to noise ratio3 mW 1 mW = 3 (or 3:1). This is the same figure as for device A

Device C

For device C, the passband is 30 MHz wide, but the signal spectrum occupies only 20 MHz ofthis. Hence the signal power is exactly as for devise B, 3 mW

signal powerB

The signal power is150 pW/Hz x 20 MHz = 3000 x 10-6 W = 3 mW

Noise power

The noise power is50pW/Hz x 30 MHz = 1500 x 10-6 W = 1.5 mW

the signal to noise ratio3 mW 1.5 mW = 2 (or 2:1)

Which of A, B and C has the best signal to noise ratio.

Devices A and B equally have the best signal to noise ratio of 3:1

Device A 105


In Figure 56, what power density would be allowed outside the 44 MHz-wide zone (i.e. in the part marked C)for a device with a maximum spectral density of 5 mW/MHz?

Part marked C

Frequency = 44MHz wide

Maximum power density = 5mW/MHz

Power density would be allowed outside the 44 MHz-wide zone

Signal to noise ratio!

Signal power / noise power

Noise power!

Noise bandwidth

Signal power!

Signal bandwidth

Signal to noise ratio!

Signal power / noise power


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4The power level in the C zone must be -50 dB or less relative to the maximum power density.

-50 power ratio

So the power density here cannot be more than:

5 mW/MHz 100 000 = 0.000 05 mW/MHz, or 50 nW/MHz

Activity 23 (exploratory). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

It is not strictly true that channels 1, 6 and 11 do not overlap. In what way do these channels actually overlap?

Only the 22 MHz wide central zones of the spectral masks of these channels do not overlap. Figure 56 showsthat there is 44 MHz-wide zone at -30 dB. The 44 MHz wide zone overlaps in channels 1, 6 and 11.

-50 dB = 10 (-50/10) power ratio => 0.00005 or 1/100 000Power ratio!

P2/P1 = 10 (dB/10)

0.000 05 mW/MHz = 0.000 05 x 1000 000 nW/MHz => 50 nW/MHz


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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .If S/N has a value of 10, and W is 5 MHz, what is the theoretical maximum data rate? Use the graph in Figure

58.

S/N = 10

W = 5MHz

= ? Theoretical maximum data rateC

W log2 (1+ s/n) bit/s

log2 log (number) /log2

5 x log2 (1+10) Mbit/s

5x log11/log2

= 5 x 3.5

=17.5 Mbit/s.

C = maximum data rateW = bandwidth of the dataS/N = signal-to-noise ratio


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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Figure 59 shows the S/N ratio varying from about 106:1 to 10:1 over reasonable distances from the access

point (i.e. ignoring the top point in Figure 59). By what factor can the maximum data rate be expected to vary?

S/N = 106:1 to 10:1 data rate

The maximum data rate variation? C

data rate

When the S/N ratio is 106:1log2 (1+ S/N)log2 (1+ 10

6)Log1000 001/log2

= 20

When the S/N ratio is 10:1

log2 (1+ S/N)log2 (1+ 10)

log11 /log2=3.5

Thus the maximum data rate varies from about 20 W to 3.5 W, a variation of about 5.7:1.

20 3.5 = 5.7


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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .In figure 66 how many bits are transmitted per symbol period?

Each QPSK modulator contributes 2 bits. There are four modulators; hence there are 8 bits per symbol period.Activity 30 (self-assessment)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Does the way of calculating the number of symbols described above preserve the relationship given earlier

that 2(number of bits) = number of symbols? You can use the system in figure 66 to check.

Activity 29 showed that there were 8 bits per symbol period 28 256.This confirms the calculation of the number of symbols (44=256)


By approximately the number of carriers to 180, estimate the frequency spacing of the subcarrier in figure 67,assuming they are evenly spaced along the frequency axis.

For a bandwidth of 9 kHz occupied by approximately 180 carriers, the spacing is approximately9000 Hz 180 = 50 Hz.

9 kHz = 9 x 1000 Hz => 9000Hz


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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Assuming the data in figure 67 is evenly distributed across all 182 subcarriers, what is the approximate data

rate per subcarrier given that the overall data rate of the transmission is approximately 20 kbit/ s?

20 kbit/s distributed across 182 subcarriers gives a data rate per carrier of:20 000 bit/s 182 = 109.89 bit/s, or approximately 110 bit/s


If 16 QAM is used, what is the signaling rate for each subcarrier? Use the approximate data rate persubcarrier of 110 bit/s.

16-QAMData rate = 110bit/s

What is the signaling rate (Symbol?)

bit16 QAM

Log16/log2= 4

110 4

=27.5 baud

20 kbit/s = 20 x 1000 bit/s => 20 000 bit/s

Signaling rate (Symbol)!Data rate bits per second



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4Part3


A signal has a flat power spectrum extending from 3 GHz to 8 GHz. The power density is -20 dBm/MHz. Whatis the total power in the signal?

Frequency range 3 GHz to 8 GHzPower density is -20 dBm/ MHz

Total power

The bandwidth is 8-3 = 5 GHz or 5000 MHz

Power density power ratio-20 dBm => 10 (-20/10)

0.01 mW=

The total power is 0.01 mW x 5000 = 50 mW.

Part4


Based on figure 84 and the last sentence in the extract that precedes it, what is the initial range of the CWmultiplier if aCW min = 63 and aCWmax = 511? What is the range if the first two attempts are unsuccessful?

The initial range is 0 to 63. For the first retransmission the range is 0 to 127 (0 to 27 -1), and for the secondretransmission attempt the range is 0 to 255 (0 to 28 -1).

Total power!

Power density x bandwidth

Power ratio!

P2/P1 = 10 (dB/10)


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4Activity 12 (exploratory)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Given that the slot time is 20 s and the range of multipliers is 0 to 31, calculate the range of values for the

back off time.

The minimum time that can be randomly selected for the back off time is 0 s, and the maximum time is givenby the product of the contention window and the slot time:

31 x 20 x 10-6 = 620 x 10-6 = 0.62ms


Calculate the transmission time for a frame consisting of 2364 bytes over a wireless link operating at a datarate of 2 Mbit/s. Repeat the calculation for a frame consisting of 14 bytes.

Frame size = 2364 bytes

Data rate = 2 Mbit/sAnother frame size = 14 bytes

Transmission time

Time = frame size / data rate

Its important to have compatible units.

The data rate= bit/s,Then frame size should be expressed in bits = 2364 x 8

Time = 2364 x8 / 2 x 106 = 9.456 x 10-3s = 9.456 ms

timeframe size 14 byte

Time = 14 x 8 / 2 x 106 = 5.6 x 10-5 s = 56 s

20 s (microsecond) = 20 x 10 -6 or (20 1000 000) s

620 x 10-6 s = (620 x 10-6) x 103 ms => 0.62msData rate

Frame size / time

Time!

Frame size / data rate

1 byte = 8 bit


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4


Calculate the propagation time for signals over a wireless link of 20 m. you should assume that the speed ofpropagation is 3 x 108 m/s.

Distance = 20 mPropagation speed = 3 x 108 m/s

Time

Time = 20 / 3 x 108 = 6.667 x 10-8 s = 66.67 ns


Given a data rate of 2 Mbit/s, calculate how many bits are transmitted within 66.67 ns. Repeat the calculationfor a data rate of 34 Mbit/s.

Data rate = 2 Mbit/s

Time = 66.67 nsAnother data rate = 34 Mbit/s

Number of bits = 2 x 106 x 66.67 x 10-9 = 0.1333So a station 20 m ) ( away from a station that starts to transmit will receive the signal

before 1 bit has been transmitted.

number of bit34 Mbit/s Number of bits = 34 x 106 x 66.67 x 10-9 = 2.267

JAs expected, the higher the data rate, the greater the proportion of a frame that would be transmitted in the

same time.

Propagation speed

Distance / time

!Time

Distance / Propagation speed

Number of bits!

Data rate x time

2Mbit/s = 2x106 bit/s66.67 ns = 66.67 x10-9 s


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4Activity 16 (exploratory)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Calculate the minimum amount of time that is wasted if a 2364-byte frame is involved in a collision. You

should assume that the data rate is 2 Mbit/s and the period of time a station waits for an acknowledgementis 100s. Repeat the calculation for a 14-byte frame.

Frame size = 2364-byte

Data rate is 2 Mbit/sPeriod of time waits for an acknowledgement= 100s

Another frame size = 14 byte

Minimum amount of timetransmission time 2364-byte

8Time = 2364 x8 / 2 x 106 = 9.456 x 10-3s = 9.456 ms

So the minimum wasted time is 9456+100 = 9556 s 14-byte

transmission timeTime = 14 x 8 / 2 x 106 = 5.6 x 10-5 s = 56 s

So the minimum wasted time is 56 + 100 = 156 s


Let P = 0111010001101000 and K = 0011000100110100 calculate C = P + K and confirm P = C + K.

C = P + KP 0 1 1 1 0 1 0 0 0 1 1 0 1 0 0 0

K 0 0 1 1 0 0 0 1 0 0 1 1 0 1 0 0

C 0 1 0 0 0 1 0 1 0 1 0 1 1 1 0 0P = C + K

C 0 1 0 0 0 1 0 1 0 1 0 1 1 1 0 0

K 0 0 1 1 0 0 0 1 0 0 1 1 0 1 0 0

P 0 1 1 1 0 1 0 0 0 1 1 0 1 0 0 0

0

1

Time!

Frame size / data rate


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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .The initialisation vector is 24 bits long. Assuming that the initialisation vector is incremented by 1 for each

frame transmitted by an access point operating at 11 Mbit /s and frames are, on average, 2000 bytes long,estimate how long it will take for an initialisation vector value to be repeated.

24 bits

Data rate = 11Mbit/s

Frame size = 2000 bytes

With 24 bits, the number of different initialisation vector value is 224 = 16 777 216, which is the number offrames that are transmitted before a value is repeated.

The maximum rate of transmitting frames is given by:Data rate / frame size = 11 x 106 / 2000 x 8 = 687.5 frames/s

The time taken to repeat a value is given:16 777 216 /687.5 = 24 400 s = 6.78 hours

The maximum rate of transmitting frame!

Data rate / frame size

24 400 s = 24 400 3600 hr => 6.78hrs

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calculation activites of block 1.pdf

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