ch08 - Chapter 8 Advanced Data Communications Topics 8.1...

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Chapter 8 Advanced Data Communications Topics 8.1 Problem Solutions Problem 8.1 Use the relationship R b =(log 2 M ) R s bps where R b is the data rate, R s is the symbol rate, and M is the number of possible signals per signaling interval. In this case, R s = 2000 symbols per second. For M =4 , R b = 2 × 2 , 000 = 4 , 000 bps, for M =8 , R b =6 , 000 bps, and for M =64 , R b =12 , 000 bps. Problem 8.2 (a) 5,000 symbols per second. (b) Recall that y ( t )= A [ d 1 ( t )cos( ω c t )+ d 2 ( t )sin( ω c t )] = 2 A cos [ ω c t θ i ( t )] , θ i ( t )=tan 1 · d 2 ( t ) d 1 ( t ) ¸ Alternating every other bit given in the problem statement between d 1 ( t ) and d 2 ( t ) gives d 1 ( t )=1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 and d 2 ( t , 1 , 1 , 1 , 1 , 1 , 1 which results in θ i ( t π / 4 , 7 π / 4 , 5 π / 4 , 7 π / 4 , 7 π / 4 , 3 π / 4 , 7 π / 4 , ··· . For QPSK the symbol switching points occur each T s seconds. (c) Now the switching instants are each T b seconds. Start with d 1 ( t .T h e n d 2 ( t ) is staggered, or o f set ,by1b itt ime . Sothe f rst phase shift is for d 1 ( t and d 2 ( t or θ 2 ( t π / 4 . After T s =2 T b seconds d 1 ( t ) takes on the second 1-value, but d 2 ( t ) is still 1, so θ 1 ( t π / 4 . At 3 T b seconds, d 2 ( t ) changes to 1 , so θ 2 ( t )=7 π / 4 . 4 T b seconds, d 1 ( t ) changes to 1 ,so θ 2 ( t π / 4 , etc. 1
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2 CHAPTER 8. ADVANCED DATA COMMUNICATIONS TOPICS Problem 8.3 For QPSK, P E, symbol =2 Q ³ p E s /N 0 ´ =10 5 Trial and error using the asymptotic approximation for the Q -function gives [ E s 0 ] req’d 12 . 9 dB =19 . 5 . If the quadrature-carrier amplitudes are A , then the amplitude of the envelope-phase-modulated form of the carrier is 2 A ,and E s 0 = ¡ 2 A ¢ 2 T/ (2 N 0 )= A 2 0 R. Hence, A req’d = q N 0 R [ E s 0 ] req’d = p (10 11 )(19 . 5) R =1 . 4 × 10 5 R. The answers are as follows: (a) 0.0014 V; (b) 0.0031 V; (c) 0.0044 V; (d) 0.0099 V; (e) 0.014 V; (f) 0.0442 V. Problem 8.4 Take the expectation of the product after noting that the average values are 0 because E [ n ( t )] = 0 : E [ N 1 N 2 ]= E ·Z T s 0 n ( t )cos( ω c t ) dt Z T s 0 n ( t )sin( ω c t ) dt ¸ = Z T s 0 Z T s 0 E [ n ( t ) n ( λ )] cos ( ω c t ω c λ ) dtd λ = Z T s 0 Z T s 0 N 0 2 δ ( t λ ω c t ω c λ ) dtd λ = N 0 2 Z T s 0 cos ( ω c t ω c t ) dt = N 0 4 Z T s 0 sin (2 ω c t ) dt =0 Problem 8.5 (a) Use θ i ( t )=tan 1 · d 2 ( t ) d 1 ( t ) ¸ (i) If θ i ( t )=45 o , d 1 ( t )=1 and d 2 ( t ; (ii) If θ i ( t ) = 135 o , d 1 ( t 1 and d 2 ( t ; (iii) If θ i ( t 45 o , d 1 ( t and d 2 ( t 1; (iv) If θ i ( t 135 o ; d 1 ( t 1 and d 2 ( t 1 . (b) Error in detecting d 1 ( t ) :( i ) θ i ( t ) = 135 o ;( i i ) θ i ( t o ; (iii) θ i ( t 135 o iv ) θ i ( t 45 o .
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8.1. PROBLEM SOLUTIONS 3 (c) Error in detecting d 2 ( t ) ;( i ) θ i ( t )= 45 o i i ) θ i ( t 135 o ; (iii) θ i ( t )=45 o iv ) θ i ( t )=135 o . Problem 8.6 (a) Both are equivalent in terms of symbol error probabilities. (b) QPSK is 3 dB worse in terms of symbol error probability for equal transmission bandwidths, but it handles twice as many bits per second. (c) Choose QPSK over BPSK in terms of performance; however, other factors might favor BPSK, such as simpler implementation.
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This note was uploaded on 04/28/2008 for the course EE ELCT 332 taught by Professor Cooper during the Spring '08 term at University of South Carolina Beaufort.

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ch08 - Chapter 8 Advanced Data Communications Topics 8.1...

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