EECS 455 Solutions to Problem Set 8

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Paulina Norman
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1 . Problem 7.55 o text EECS 455 Solutios to Problem Set 8 a) For repeaters i asade, the probability o i out o repeaters to produe a error is give by the biomial distributio P i p i p i However, there is a bit error at the output o the termial reeiver oly whe a odd umber o repeaters produes a error. Hee, the overall probability o error is P P odd i odd p i p i Let P eve be the probability that a eve umber o repeaters produes a error. The ad thereore, P eve P eve P odd i 0 i eve p i p i p i p i p p Oe more relatio betwee P eve ad P odd a be provided i we osider the dieree P eve P odd. Clearly, P eve P odd p i p i i p i p i eve i odd a i eve p p p i p p i i odd p i p i where the equality (a) ollows rom the at that i is or i eve ad whe i is odd. Solvig the system we obtai P eve P odd P eve P odd p P P odd b) Expadig the quatity p, we obtai Sie, p p p p p we a igore all the powers o p whih are greater tha oe. Hee, P p p

2 . Problem 7.56 o text The overall probability o error is approximated by P e KQ Q E 0 b N Thus, with P e 0 6 ad K 00, we obtai the probability o eah repeater P r rom tables to be 0 8. The argumet o the utio Q that provides a value o 0 8 is oud Hee, the required is Problem 7.57 o text 5 6 a) The atea gai or a paraboli atea o diameter D is I we assume that the eiiey ator is 5, the with we obtai m D m G T db b) The eetive radiated power is EIRP P T G T G T 6 6 db ) The reeived power is Note that dbm 0log 0 4πd db 55 3 dbm P T G T atual power i Watts log 0 power i Watts

3 4. Problem 7.58 o text a) The atea gai or a paraboli atea o diameter D is I we assume that the eiiey ator is 5, the with m ad D m we obtai G T db b) The eetive radiated power is EIRP P T G T db ) The reeived power is 4πd db 67 0 dbm P T G T 5. Problem 7.59 o text The wavelegth o the trasmitted sigal is The gai o the paraboli atea is m 6 π0 03 The reeived power at the output o the reeiver atea is db P T G T 4π d db 6. Problem 7.60 o text a) Sie T K, it ollows that kt W/Hz 3

4 I we assume that the reeivig atea has a eiiey 5, the its gai is give by db Hee, the reeived power level is P T G T 4π d b) I 0 db 0, the R 7. Problem 7.6 o text The wavelegth o the trasmissio is db Mbits/se m I MHz is the passbad badwidth, the the rate o biary trasmissio is R b W 0 6 bps. Hee, with 4 0 W/Hz we obtai The trasmitted power is related to the reeived power through the relatio R b P T G T 4π d P T G T Substitutig i this expressio the values G T 0 0 6, 0 5, d ad 75 we obtai P T dbw Alterate Solutio. Sie it metios reliable ommuiatio we assume that the data rate is the apaity ahievig data rate. So (see leture otes ) R W log R W Sie 5 db or 3 6 This orrespodes to R W 8. So R=8Mbps. Now usig R 8 Mbps ad 4 0 we get P r R 3 6 P r 3 6 R P r P r 0 0 4π d 4

5 P T G T 0 3 7dB 4π d π GWatts This is obviously ot a possible solutio. However, rom the problem statemet this is a reasoable approah to take to solve the problem. However modulatio tehiques trasmittig 8Mbps i a MHz badwidth would require a very liear (ad thus very ieiiet) ampliier. So the restritio that the ommuiatio use BPSK with data rate =badwidth is resoable. 8. Problem 7.63 o text Sie T K, it ollows that The trasmittig wavelegth is kt W/Hz Hee, the gai o the reeivig atea is 55 ad thereore, the reeived power level is m db P T G T 4π d db I 6 db 0 0 6, the R Gbits/se 5

De Moivre s Theorem - ALL

De Moivre s Theorem - ALL. Let x ad y be real umbers, ad be oe of the complex solutios of the equatio =. Evaluate: (a) + + ; (b) ( x + y)( x + y). [6]. (a) Sice is a complex umber which satisfies = 0,.