Chapter 2: Sample Questions, Problems and Solutions Bölüm 2: Örnek Sorular, Problemler ve Çözümleri

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1 Chapter : Sample Questios, Problems ad Solutios Bölüm : Örek Sorular, Problemler ve Çözümleri Örek Sorular (Sample Questios): Fourier series What is a badwidth? What is a voice-grade? Nyquist theorem Shao theorem What is a sigal-to-oise ratio? What is a magetic media? What is a twisted pair? What is a coaxial cable? What is a fiber cable? Draw the electromagetic spectrum ad its uses for commuicatio. What is a radio trasmissio? What is a microwave trasmissio? What is a ifrared trasmissio? What is a lightwave trasmissio? What are commuicatio satellites? What is a traspoder? What is PSDN? What is a ed office? What is a local loop? What is a truk? What is a switchig office? Draw a typical circuit route for a medium-distace call. What is a modem? What is atteuatio? What is a distortio? What is a oise? What is a amplitude modulatio? What is a frequecy modulatio? What is a phase modulatio? What is a QPSK? What is a QAM-6? What is a full duplex chael? What is a half duplex chael? What is a simplex? What is a DSL? What is a wireless loop? What is a FDM? What is a DM? What is a Wavelegth Divisio Multiplexer? What is a Delta Modulatio?,, 3, 4 DM formats. What is a SONE/SDN? What is a circuit switchig? What is a message switchig? What is a packet switchig? What is AMPS? What is D-AMPS? Draw a GSM framig structure What is a CDMA?

2 Örek Problemler ve Çözümleri (Sample Problems ad Solutios): (Chapter, Problem 3-) elevisio chaels are 6 MHz wide. How may bits/sec ca be set if four level digital sigals are used? Assume a oiseless chael. Usig the Nyquist theorem, we ca sample millio times/sec. Four level sigals provide bits per sample, for a total data rate of 4 Mbps. Maximum data rate = = *6MHz *log 4 bits/sec =4 Mbps * * log V bits/sec. H = *6MHz * log bits/sec * * 6 = MHz*bits/sec (Chapter, Problem 3-) elevisio chaels are MHz wide. How may bits/sec ca be set if 8 level digital sigals are used? Assume a oiseless chael. Usig the Nyquist theorem, we ca sample 4 millio times/sec. 8 level sigals provide 3 bits per sample. otal data rate is 7 Mbps. (Chapter, Problem 5) What is sigal-to-oise ratio i order to put a carrier o a 5-KHz lie? he data rate of is.544 Mbps. Usig Shao theorem: H*log (+S/N)=.544* 6 5. log (+S/N)=.544* 6 +S/N= S/N= - *log S/N=*log ( -)=*log 3 *log 3 *3 3 db (Chapter, Problem ) A modem costellatio diagram has data poits at the followig coordiates: (,), (,- ), (-,), ad (-,-). How may bps ca a modem with these parameters achieve at baud?

3 here are four legal values per baud, so the bit rate is twice the baud rate. At baud, the data rate is 4 bps (Chapter, Problem 8) Five sigals, each requirig KHz, are multiplexed o to a sigle chael usig FDM. How much miimum badwidth is required for the multiplexed chael? Assume that the guard is 5Hz wide. here are five KHz sigals. We eed four guard bads to avoid ay iterferece. he miimum badwidth required is x5+5x4=5+=53hz=53khz. (Chapter, Problem 3) What is the percet overhead o a carrier; that is, what percet of the.544 Mbps are ot delivered to the ed user? he ed users get 7x4=68 of the 93 bits i a frame. he overhead is therefore 5/93=3%. (Chapter, Problem 34) A sigal is trasmitted digitally over a 4-kHz oiseless chael with oe sample every 5 µ sec. How may bits per secod are actually set for each of these ecodig methods? a) CCI (Comite Cosultatif Iteratioal elegraphique et elegraphique).48 Mbps stadard. b) DPCM (Differetial Pulse Code Modulatio) with a 4-bit relative sigal value. c) Delta modulatio. a) 64 kbps b) 3 kbps c) 8 kbps (Chapter, Problem 53) A CDMA (Code Divisio Multiple Access) receiver gets the followig chips: ( ). Assumig the followig chip sequeces: A: ( ) B: ( ) C: ( ) D: ( )

4 Which statios trasmitted, ad which bits did each oe sed? A: ( ) * ( )/8 =( )/8= B: ( ) * ( )/8=( )/8=- C: ( ) * ( )/8=( )/8= D: ( ) * ( )/8=(++-++-)/8= A set, B set, C ad D did ot sed (Chapter, Page 9) Ca a chael of 3-Hz badwidth with a sigal to thermal oise ratio of 3 dd trasmit much more tha 3 bps? Prove your aswer. No, due to Shao s formula maximum umber of bits/sec=h*log (+S/N), where *log S/N=3 db ad S/N= So, maximum umber of bits/sec=h*log (+S/N)= 3 log (+)=3 log 3bps (Chapter, Problem) A tape ca hold Gigabytes. A box 6x6x6 cm ca hold about of these tapes. What are effective badwidth ad cost if the destiatio is (a) a hour ad (b) 4 hours away by road? otal capacity is Gigabytes * = erabytes = 6 erabits. (a) For destiatio of a hour: 6 erabits / 36 sec = 4 Gbps So, the effective badwidth is over 4 Gbps. No computer etwork ca ever approach this. (b) For destiatio of 4 hours: 6 erabits / 86 4 sec = 9 Gbps So, the effective badwidth is over 9 Gbps. Assume that the cost of tape is $4. Assume that a tape ca be reused at least te times. So the tape cost is $4 per box per usage. Assume that shippig is about $ So we have a cost of roughly $5 to ship B. Or the cost of a Gigabyte is uder 3 cets. (Chapter, Problem) Compute the Fourier coefficiets for the fuctio f(t) (<=t<=; =; f=

5 a = g( t)si( ft) dt = t si(πft) dt π Assume that πt =a a da si( a) = asiada= xsi π π π π xdx x = u dx= du si xdx = dv cosx= v π ( x( cosx) cosxdx) dx= ( cosx* x+ six) π = π b c = π ( cos(πt) * πt+ si(πt)) t π g( t)cos(π ft) t cos(πft) t g( t) tdt = = = = π dx (Chapter, Problem) A oiseless 4-KHz chael is sampled every msec. What is the maximum data rate? A oiseless chael ca carry a arbitrary large amout of iformatio, o matter how ofte it is sampled. Just sed a lot of data per sample. For 4-KHz chael, make samples/sec. If each sample is 6 bits, the chael ca sed 6 Kbps. If each sample is 4 bits, the chael ca sed samples/sec * 4 bits = 4 Mbps he key word here is oiseless. With a ormal 4 KHz chael, Shao limit would ot allow this. For the 4 KHz chael we ca make 8 samples/sec. I this case if each sample is 4 bits this chael ca sed 8. Mbps. (Chapter, Problem) We have a 3 KHz chael whose sigal-to-oise ratio is db. A biary sigal is set by this chael. What is the maximum achievable data rate?

6 logs / N = db S / N = Usig Shao theorem, Maximum umber of bits/sec = H * log (+ S / N) = 3KHz *log(+ ) log () 6, 658 = 3 *6,658= 9,975Kbps Usig Nyquist theorem, Maximum data rate i bits/sec= * H * log V = *3*log = 6Kbp he bottleeck is therefore the Nyquist limit, givig a maximum chael capacity of 6 Kbps. (Chapter, Problem) How much badwidth is there i, micro of spectrum at a wavelegth of micro? c λ We use f =, where λ = 7 meters is the width of a wavelegth ad λ = 6 λ meters is a wavelegth. his gives a badwidth f = 3, GHz. (Chapter, Problem) It is desired to sed a sequece of computer scree images over a optical fiber. he scree is 48x64 pixels, each pixel beig 4 bits. here are 6 scree images per secod. How much badwidth is eeded, ad how may micros of wavelegth are eeded for this bad at,3 micros? he data rate is 48x64x4x6 bps=44 Mbps. For simplicity, let us assume bps per c λ fλ Hz. From f = we fid λ = λ c 8 6 We have f = 4,4x, so λ =,5x micros. As we see the rage of wavelegths used is very short. (Chapter, Problem) Compare the maximum data rate of a oiseless 4 KHz chael usig a) Aalog ecodig (e.g. QPSK) with bits per sample. b) he PCM system. a) I both cases 8 samples/sec are possible. With QPSK ecodig, two bits are set per sample. So the maximum data rate for QPSK is 6 Kbps. b) With, 7 bits are set per period. So the maximum data rate for is 56 Kbps.

ECE 333: Introduction to Communication Networks Fall Lecture 4: Physical layer II

ECE 333: Introduction to Communication Networks Fall Lecture 4: Physical layer II ECE 333: Itroductio to Commuicatio Networks Fall 22 Lecture : Physical layer II Impairmets - distortio, oise Fudametal limits Examples Notes: his lecture cotiues the discussio of the physical layer. Recall,