finals08soln - EE 230B Final Examination, Spring 2008 UCLA...

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EE 230B Final Examination, Spring 2008 NAME:_______________________ UCLA Electrical Engineering Department Prof. G. Pottie Instructions: Attempt all five questions; they are of equal weight but not necessarily of equal difficulty. A sheet of useful formulae and other facts is attached. 1. Phase Error in QAM Suppose a persistent phase error of θ occurs. a) By what factor is the effective minimum distance reduced for 4-QAM compared to perfect synchronism? Write an expression for P ( e | θ ). Solution: d 1 d r θ d 2 + d 2 = r 2 ; d = r /2 = r sin( π /4) sin( /4 ) = d 1 / r ; d 1 = r sin( ) P ( e | ) = 2 Q 2 r sin( ) 2 N 0 b) Repeat for the worst-case signal point in 16-QAM. Soln: Here, things are much the same except now there is a new boundary located at 2 r /3.
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d r/3 The coordinates of the desired point are r /2 , r () ; The coordinates of the inner point are r /3 2 , r ; The boundary is thus at 2 r /3; thus d = r sin( π /4) 2 r /3 d 1 = r sin( /4 θ ) 2 r P ( e | ) = 2 Q 2( r sin( ) 2 r /3) 2 N 0 for this point c) From (a) and (b), what do you conclude regarding the relative levels of acceptable phase noise for 4-QAM and 16-QAM? Soln: Obviously 16-QAM is much more sensitive, and so would need tighter synchronism. d) Write, but do not evaluate, an expression for P ( e ) for 4-QAM given Gaussian noise and phase error with distribution p ( ), assuming the phase error is always small enough so that it does not on its own cause errors. Why is the situation more complicated for 16- QAM? Soln: P ( e ) = P ( e | ) p ( ) d 0 2 Things are more complicated for 16-QAM since we would need expressions for three different classes of points which have their own dependence on the phase error. 2. The Gilbert Channel In the Gilbert channel model, a radio channel has two states with transition probabilities among them described by a Markov chain:
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1 2 0.9 0.1 Thus, P (next state=1|current state=1)= P (1|1)=0.9; P (2|1)=0.1; P (1|2)=0.9; P (2|2)=0.1 a) If P ( i ) denotes the prior probability of being in state i , why must it be true in the steady state that P (1|2) P (2)= P (2|1) P (1)? Determine P (2) and P (1). Soln: 0.9P(2)=0.1P(1); P(2)=P(1)/9. (1+1/9)P(1)=1; P(1)=9/10 P(2)=1/10. b) State 1 is the “good” state with P ( e )=10 -6 , and state 2 is the “bad” or “jammed” state with P ( e )=10 -1 . What is the average error probability?
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finals08soln - EE 230B Final Examination, Spring 2008 UCLA...

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