132A_1_2002finalsol

132A_1_2002finalsol - Final Exam 2002 1(10 points High...

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Final Exam 2002
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1) (10 points) High level communications system In this class we have learned something about every part of a communication system. To demonstrate your understanding of the entire communications system, put the following blocks in order from information source to information destination: A Modulator B Demodulator C Source coding encoder (like Huffman codes) D Source coding decoder E Channel coding encoder (like Hamming codes) F Channel coding decoder Place the appropriate letters in the boxes below. Channel Transmission Information source Information sink (destination) A B C D E F
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2) (20 points total) Huffman code Given the following source distribution: 0.12 E 0.21 D 0.35 C 0.20 B 0.12 A P (X) X a) (5 points) Calculate the entropy of this source. ( 29 [ ] l bits/symbo 2014 . 2 2 log 6627 . 0 12 . 0 log 12 . 0 21 . 0 log 21 . 0 35 . 0 log 35 . 0 20 . 0 log 20 . 0 12 . 0 log 12 . 0 ) ( log ) ( 10 2 2 2 2 2 2 = = × + × + × + × + × - = - = X P X P X H
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b) (5 points) Find a Huffman code for this source. 0 1 1 0 0 1 0 1 0.24 0.12 3 111 E 0.21 2 00 D 0.35 2 10 C 0.20 2 01 B 0.12 3 110 A P (X) Length (bits) Huffman Code X 0.41 C 0.35 D 0.21 B 0.20 A 0.12 E 0.12 0.59
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c) (5 points) What is the average length of the Huffman code for this source? d) (5 points) What is the efficiency of the Huffman code? bits L av 24 . 2 3 12 . 0 2 21 . 0 2 35 . 0 2 20 . 0 3 12 . 0 = × + × + × + × + × = ( 29 % 3 . 98 9828 . 0 24 . 2 2014 . 2 length average entropy efficiency = = = = = av L X H
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( 29 ( 29 ( 29 ( 29 2 1 2 p D P B P p C P A P - = = = = D B A C -2 -1 0 1 2 10 01 00 11 3. Consider the following constellation containing four symbols as shown below. Assume that the noise is additive white Gaussian noise. There are many. One is the following: Assume that the decision region boundaries are exactly midway between symbols. a) (5 points) Provide a gray code labeling for this constellation.
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b) (5 points) Find the probability of symbol error when symbol A is sent. For this calculation, consider only the nearest neighbors and use a union bound approach for those nearest neighbors. ( 29 = 0 2 erfc 2 1 error N d |A P AB 1 = AB d B A -2 -1 AB d ( 29 = 0 2 1 erfc 2 1 error N |A P where d AB is the distance between A and B. So we have Another way of interpreting the union bound is as: where M = 4 and using only nearest neighbors ( 29 = = M 1 0 i 2 erfc 2 1 P i k k ik N d M ( 29 =
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132A_1_2002finalsol - Final Exam 2002 1(10 points High...

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