PSU EE 360 - ee360_hw9_sol (1) (4 pages)

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ee360_hw9_sol (1)

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4
School:
Penn State University
Course:
Ee 360 - Communications Systems I
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1 sin A cos A sin 2 A sin x 2 sin c x x of the sinc function and the definition we may go onsinto x write sin c x sin 2 B 0 t cos 2 B 0 t x sin c 2B 0 t cos 2 B 0 t we may go on to write 2 B 0 t EE 360 Communication Systems I Fall 2017 sin 2 B t cos 2 B t 0 0 sin 4 B 0 t 9 Due 12 05PM Dec 7 2017 sin c 2B 0 t cos 2 B 0 t Homework t 2 B 0 4 B 0 t Collaboration policy You are allowed Howsin to discuss homework problems with your classmates c 4 B 4B 0 t0 tyour sin ever you must write up and submit own solutions representing your own work 2 4 B t Accordingly Eq collaboration 2 in 1 we0policy get Any violationusing of the will be handled seriously 2 t sin c 4B sin c 4B 0 t 0 t 1 p Reading assignment 6 1 6 2 6 3 6 4 6 5 2 2 Eq 2 in 1 we get Accordingly using 1 16B 0t sin c 4B 0 t p t is the which desired 2 2result except for the scaling factor E 2 3 3 6 1 points 16B t Problem 6 9 a d on page 257 in the textbook 0 Solution Problem which is 6 9 the desired result except for the scaling factor E The bandwidth Problem 6 9 B of a raised cosine pulse spectrum is 2B0 f1 where B0 1 2Tb and f1 B0 1 a Thus B B0 1 For a data rate of 56 kilobits per second B0 28 kHz The bandwidth a 0 25 B of a raised cosine pulse spectrum is 2B0 f1 where B0 1 2Tb and f1 B0 1 a B B 28B0kHz 1 25 kHzrate of 56 kilobits per second B0 28 kHz Thus 1 x For a35data b 0 5 a 0 25 BB 28 42 kHz 28kHz kHzxx1 5 1 25 35 kHz c 0 75 b 0 5 BB 28 1 75x 1 5 49 kHz 28xkHz 42 kHz d 1 0 c 0 75 BB 28 56 kHz 28xx21 75 49 kHz d 1 0 Problem B 6 10 28points x 2 56Problem kHz 3 3 2 5 6 10 on page 257 in the textbook The raised 6 10 cosine pulse bandwidth BT 2B0 f1 where B0 1 2Tb For this channel BT 75 kHz Solution Problem For the given bit duration B0 50 kHz Then f1The raised 2B0 cosine BT pulse bandwidth BT 2B0 f1 where B0 1 2Tb For this channel BT 75 kHz For the25given kHz bit duration B0 50 kHz Then f1 12B f01 B BTT 0 5 25 kHz 1 f1 BT 0 5 6 6 1 Bit rate 1 Tb kilobits s 6 4 2 0 1 2 3 Figure 1 Excess bandwidth f kHz 4 5 3 8 points Problem 6 12 on page 257 in the vtextbook Solution Problem 6 12 We are given the following specifications B T 3 kHz 1 4 5 kilobits s Tb a The transmission bandwidth is related to the roll off factor by the formula see Eq 6 21 1 BT B0 1 where B 0 1 2T b Therefore with 1 Tb 4 5 kilobits s we have B 0 2 25 kHz Hence solving Eq 1 for the roll off factor we get B T 1 B0 3 1 2 25 1 3 7 b The excess bandwidth is defined see Eq 6 22 f v B 0 1 2 25 3 0 75 kHz Problem 6 13 According to Eq 6 30 the pulse shaping criterion for zero intersymbol interference is embodied in the relation m m P f constant T 1 where P f is pulse shaping spectrum and 1 T is the signaling rate a The pulse shaping spectrum of Fig 6 13 a is defined by f 0 E 2B 0 for E f P f 1 for 0 f B 0 B 0 2B 0 for f B 0 0 Substituting Eq 2 into 1 leads to the following condition on the signaling rate B 1 Page 2 of 4 0T 2 or equivalently 2 b The excess bandwidth is defined see Eq 6 22 f v B 0 1 2 25 3 0 75 kHzProblem 6 13 on page 257 in the textbook 5 5 5 10 points Problem 6 13 Solution According to Eq 6 30 the pulse shaping criterion for zero intersymbol interference is embodied in the relation m m P f constant T 1 where P f is pulse shaping spectrum and 1 T is the signaling rate a The pulse shaping spectrum of Fig 6 13 a is defined by f 0 E 2B 0 for E f P f 1 for 0 f B 0 B 0 2B 0 for f B 0 0 Substituting Eq 2 into 1 leads to the following condition on the signaling rate B 1 0T 2 or equivalently B0 2 T b The pulse shaping spectrum of Fig 6 12 b is defined by for 0 f f 1 E 2B 0 E f f P f 1 1 for f 1 f B 0 B 0 f 1 2B 0 0 for f B 0 Substituting Eq 3 into 1 leads to the following condition on the signaling rate 1 1 f 1 B 0 2 T Note The solution is not unique For a let T m B0 3 4 m 1 2 or equivalently 1 B0 m 1 2 T m we can achieve zero intersymbol interference Similarly for b let 1 f1 B0 m 1 2 T m we can achieve zero intersymbol interference Page 3 of 4 2 8 6 3 3 6 points Problem 6 14 on page 258 in the textbook Solution a With given bit rate 4 5 Kbps if 8PAM is used the corresponding symbol rate 4 5 1 5 Kbps log2 8 1 T would be 1 0 75 Kbps Thus the minimum bandwidth is B0 2T b For raised cosine pulse spectrum the roll off factor 0 1 When 1 BT 1 B0 1 5KHz 3KHz Thus the maximum roll off factor is 1 and the excess bandwidth is B0 0 75KHz 7 5 5 10 points Problem 6 15 on page 258 in the textbook Solution a The method of data transmission is quaternary i e 4 level PAM and the roll off factor 1 Since BT 1 B0 and 1 BT 13 KHz we have BT B0 6 5KHz 1 1 Since B0 2T where T1 is the symbol rate for 4PAM we have 1 2B0 13 kilo symbols per second T For 4PAM bit rate symbol rate log2 4 26 Kbps b The codeword consists of log2 128 7 bits With an additional bit added for synchronization the overall codeword consists of 8 bits Thus the sampling rate bit rate 8 26 8 3 25 KHz …


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