# Solutions to exercise 1 in ETS052 Computer Communication

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1 Solutions to exercise in TS52 Computer Communiction 23 Septemer, 23 If it occupies millisecond = 3 seconds, then second is occupied y 3 = 3 its = kps. kps If it occupies 2 microseconds = 2 6 seconds, then second is occupied y 2 6 =.5 6 its = 5 kps. 2 5 kps If its re produced/received t rte of 3 = 5 per second, then one it occupies 5 = 5 seconds = microseconds. microseconds If its re produced/received t rte of 2 6 =.5 6 seconds =.5 microseconds. 2 6 per second, then one it occupies

2 .5 microseconds 3 In NRZ (NonReturntoZero) modulted trnsmission ones nd zeros re represented y specific output level, constrined to specified durtion, see figure. Figure : NRZ it vs. voltge level ssignment As consequence, unless you know when the trnsmission strted nd the durtion it occupies, you will e unle to distinguish etween multiple consecutive its of the sme sign. In communiction system, oth receiver nd trnsmitter thus need to e synchronized. In Mnchester modulted signl, ech it i demrked y n output trnsition during the llocted it durtion (symol) s opposed to constnt level seen in NRZ modultion. In Figure 2 zero is represented y flling edge, while one is represented y rising edge. 2

3 Figure 2: Mchester it representtion s ccording to I 82.3 As ech it infers output level trnsition, with Mnchester coding, ech it cn e identified without prior clock synchroniztion. Nevertheless, oth receiver nd trnsmitter needs to gree on which trnsition to represent which it. Moreover, In this cse, we re deling with sequence of ll zeros, the signl is thus modulted with ll flling edges for ech it. Mnchester coding hs reltive low throughput compred to more complex modultion schemes nd is tody minly used in systems such s BAST ethernet (I 82.3) nd NFC. c As you will see in prolem 5, Differentil Mnchester coding dels with trnsitions etween symols rther thn output levels within symol. A sequence with ll zeros is represented y the sence of symol trnsitions, nd the wveform is therefore identicl to the Mchester coded one in prolem 3. 4 When pplying the Mnchester modultion scheme presented in Figure 2 to the oserved sequence, we rrive t the it sequence presented in Figures 3, 4, nd 5. 3

4 Following the ove convention yields. Figure 3: Decoded 4 Mnchester wveform Following the ove convention yields. Figure 4: Decoded 4 Mnchester wveform c Following the ove convention yields. Figure 5: Decoded 4c Mnchester wveform 4

5 5 As opposed to Mnchester coding, Differentil Mnchester coding only represents trnsitions etween. See Figure 6. () Without gurd intervl () With gurd intervl, drk Figure 6: Trnsitions As such, severl trnsitions represent just s mny ones. A single trnsition followed y consecutively repeted symol represents one followed y zeros, until the next trnsition. As such, the sence of trnsition signifies zero. See Figure 7 Trnsition Trnsition Figure 7: Sequence with trnsitions When oserving n ritrry wveform, s we do not know whether it is continution of the previous symol or trnsition, the first it is inherently unknown. See Figure 8. 5

6 Oserved wveform? Figure 8: Decoding the first it As you cn see in Figure 8, we do not know if first symol is zero tht trnsitions into one, or the til of sequence of trnsitions representing severl ones. Differentil Mnchester is used ecuse trnsition is less likely to e misinterpreted thn individul symols y the receiver, given noisy chnnel. Differentil Mnchester encoding is predominntly used in mgnetic nd opticl storge. Following the ove convention yields.? Figure 9: Decoded 5 Differentil Mnchester wveform. Following the ove convention yields. 6

7 ? Figure : Decoded 5 Differentil Mnchester wveform. c Following the ove convention yields.? Figure : Decoded 5c Differentil Mnchester wveform. 6 Following the convention estlished in prolem 4. 7

8 Following the convention estlished in prolem 5. The first symol is to illustrte the stte of the signl. 7 Being it wireless, n opticl, or n electricl link, in FDM (Frequency Division Multiple ccess) the entire or portion of tht mediums frequency spectrum ndwidth is divided into chnnels, see Figure 2. Moreover, this is one pproch for ccommodting multiple connections in common link, where the trnsmitter (Tx) nd receiver (Rx) re trnsmitting nd receiving in the sme llocted frequency rnge, chnnel. In most FDM sed communiction systems, frequency gurd intervl is introduced to seprte the chnnels, see Figure 2. This is to ensure tht ny lekge from one chnnel is not interfering with its djcent chnnels. The gurd intervl does not crry ny intentionl informtion. Note tht gurd intervls re only needed to seprte the chnnels in the medium, nd not the chnnels from eyond the frequency rnge of the medium. Unless there is nother communiction link t frequency rnge ove or elow, in which cse, gurd intervl is introduced etween the links. However, this is not tken into ccount when referring the ndwidth of the oserved link. Tx Rx 2 Rx 3 Tx 4 Tx Rx 2 Rx 3 Tx 4 Link ndwidth (Hz) Link ndwidth (Hz) Ch Ch 2 Ch 3 Ch 4 Ch Ch 2 Ch 3 Ch 4 Rx 2 Tx 2 Tx 3 Rx 4 () Without gurd intervl Rx 2 Tx 2 Tx 3 Rx 4 () With gurd intervl, drk Figure 2: Link multiplexed cross 4 frequency chnnels 8

9 In this instnce, it is specified tht ech chnnel needs ndwidth of 4 Hz = 4 KHz with gurd intervl of 2 Hz =.2 khz. SInce there re five chnnels, we need four gurd intervls to seprte them. Moreover, this leves us with totl link ndwidth of 5 4 khz 4.2 khz = 2.8 khz. See Figure khz Link ndwidth (Hz) Ch 5... Ch 4 khz.2 khz Figure 3: Link ndwidth rekdown, gurd intervls re shded (Not to proportion) khz In this exmple we re llocted totl link ndwidth of 79 Hz = 7.9 khz. It is specified tht the link will crry 3 chnnels nd tht the chnnels shll e seprted y unused frequency spce, or gurd intervls of 2 Hz =.2 khz ech. Since there re three chnnels, they will e seprted y totl of 2 gurd intervls. See Figure 4. Link ndwidth (Hz) Ch khz Ch 2... Ch 2.5 khz.2 khz Figure 4: Link ndwidth rekdown, gurd intervls re shded (Not to proportion) Sutrcting the sum of the two gurd intervls of.2 khz ech from the totl llocted link ndwidth leves us with the remining ndwidth for the sum of the three chnnels, 7.9 khz 2.2 khz = 7.5 khz. Furthermore, s there re three chnnels, ech chnnel cn e llowed ndwidth of khz = 2.5 khz. 2.5 khz 9

10 9 In Synchronous TDM (Time Division Multiplexing) system, ech sulink is ssigned reoccurring slot t constnt frequency on the multiplexed link. The cpcity of the multiplexed link is shred proportionlly mongst sulinks. For exmple, s illustrted in Figure 5, if the min link hs cpcity of 3 Mit nd is shred indiscrimintely mongst the three sulinks, they will ech ccess to throughput of Mps. Mit Link 3 Mit Link 2 SDTM Link 3 Figure 5: STDM itrte Furthermore, it is specified tht ech ech of the sulinks hve cpcity of 4,4 kps. In order to ensure tht the ech sulinks cpcity is tlest mintined, the multiplexed link needs to hve minimum cpcity of 4.4 kps =.44 Mps..44 Mps In Synchronous TDM system, unused slots re left unused if sulink hs nothing to trnsmit. See Figure 6. Link Link 2 SDTM Link 3 Figure 6: STDM utiliztion As such, if only 7 out of sulinks hve informtion to trnsmit, then 3 % of the congruent slots re left unused. 3%

11 It is given tht ech chrcter occupies frme. Since the links crry different quntities of dt, there will e vcnt slots. See Figure 7. Link Link 2 Link 3 G T A D C SDTM L Frme 4 Frme 3 Frme 2 Frme I G F D C A T Link 4 L I F Figure 7: STDM exmple Assuming tht the sender on ech link strts to send their letters simultneously, the first frme will contin the first letter trnsmitted on ech link. The susequent frme will contin the second letter, if ny. Moreover, in this instnce 6 out of 6 frmes will go unused, resulting in link utiliztion of 6 = 62.5%. L I G F D C A T In Sttisticl TDM system, the multiplex fills ech susequent nd squll sized slot with the next ville frme from the uffer. In other words, if sulink hs nothing to send in this frme the multiplexer will try to fill tht slot with dt from the next sulink tht hs dt to send in the multiplexers uffer. Consequently, this ensures tht ll slots re utilized, given tht there is dt to send. See Figure 8. Link G T Link 2 Link 3 D A C SDTM L I G F D C A T Link 4 L I F Figure 8: Sttisticl TDM. As such, compred to Synchronous TDM, Sttisticl TDM will not trnsmit the inter

12 medite vcnt frmes. L I G F D C A T 2 The descried Synchronous TDM system comines three sulinks of 3 kps ech. A frme strts with synchroniztion it, followed y three slots of 3 its ech, resulting in frme size of its. See Figure 9. Link Link 2 Link 3 Sync Figure 9: Frme structure The three strems re congruent, s result there re no vcnt slots in the resulting Synchronous TDM trnsmission. When pplying the frme structure in Figure 9 to the specified it strems, ssuming tht the first sync it is, the resulting first frme will hve the following content: Link Link 2 Link 3 Time Figure 2: First frme, with sync it = As you cn see the slot is occupied y the first 3 its from Link, followed y three consecutive its ech from Link 2 nd Link 3 respectively. The following frme will strt with the sync it set to, prefixed to the following three consecutive its from ech Link. It is specified tht ech sulink hs cpcity of 3 kps, s such the multiplexed link should hve cpcity of t lest 3 3 kps = 9 kps. However, the sync it in ech 2

13 multiplexed frme of 9 dt its dd one it of overhed. As result insted of sending 9 its we re now sending its per frme. The link thus needs 9 % extr cpcity in ddition to wht is expect of the sulinks to minting the oncoming trffic from the sulinks nd to ccommodte the sync it. Moreover, pplying the dditionl cpcity needed to the minimum multiplex link cpcity yields, 9 9 kps = kps = Mps. c Mps In 2 we concluded tht the multiplexed link needs to hve minimum ndwidth of Mps. Furthermore, ech one of the its trnsmitted on the multiplex link would hve to occupy t lest 6 = 6 seconds = microsecond. d microsecond Using wht we deducted in 2c, if one it occupies microsecond, then frme of its will occupy microseconds. Consequently, one second is occupied y = 5 = 6 frmes. e As deducted in 2d, if one it occupies microsecond, then frme of its occupy microseconds. 3 microseconds The oserved it strem consists of 24 its nd cn thus e roken down into 2 frmes, ech consisting of 3 slots crrying 4 its ech. See Figure 2. 3

14 Frme 2 Frme Frme 2 Frme Link SDTM Link 2 Time Link 3 Figure 2: Strem demultiplexed into 3 sustrems As there re no synchroniztions its in the strem, ech frme cn trivilly e roken down nd seprted into ech sulinks slots nd then ressemled, s illustrted in Figure 2. Link Link 2 Link 3 The min line is demultiplexed into three congruent su strems. The specified frme only crries dt, we cn thus conclude tht ech resulting sulink is is feed dt t rte of 9 3 = 3 Mps. 3 Mps 4

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