pcshw11_soln

pcshw11_soln - THE STATE UNIVERSITY OF NEW JERSEY RUTGERS...

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Unformatted text preview: THE STATE UNIVERSITY OF NEW JERSEY RUTGERS College of Engineering Department of Electrical and Computer Engineering 332:322 Principles of Communications Systems Problem Set 11 Haykin: 5.1-5.7 Spring 2004 1. Cora and Squirrely Signal Design: Cora the communications engineer has been given a set of signals: s1 t cos c t cos 2c t s2 t cos ct s3 t cos c t cos 2c t s4 t cos 2ct for use on an interval 0 Tb where Tb 2 c . She needs to build a communication system around these signals to transmit her four (4) equally likely messages. However, her boss, Marty the Squirrel (how things change!) want's to know how well they'll work. Specifically, he wants to know the maximum probability of error when using a correlator receiver. (a) Find a set of orthonormal functions to represent this set. SOLUTION: s1 t and s3 t are superpositions of s2 t and s4 t and these latter two are actually orthogonal on 0 Tb . So we just normalize those two to get (b) Rewrite the si t in terms of your orthormal set. SOLUTION: We have: Tb 2 1 Tb 2 2 s2 t s4 t t t s3 t Tb 2 1 t 2 t 1 s1 s2 s3 s4 or 1 1 1 0 1 0 1 1 2 Tb 2 Tb 2 Tb 2 Tb s1 t Tb 2 1 t 2 t 1 t 2 t 2 Tb cos c t 2 Tb cos 2ct (c) Assume zero mean white Gaussian noise, w t , is added to any s i t sent over the channel so that x t si t w t is received. The noise has spectral height N0 2. What is the highest probability of error over all the signals? Find an expression for this error. SOLUTION: The maximum probability of error depends only the minimium distance between signals. The distance squared between s1 and s2 is Tb 2 as is the distance squared between s2 and s3 and s1 and s4 . The noise variance is N0 2 as usual so the maximum probability of error is Pe (d) Suppose Fly-By-Night Industries offers Cora a new and improved set of signals: where g t and h t are proprietary orthogonal waveforms specially designed by FBNI (and they won't actually tell her what they are). The energies in g t and h t over a bit interval are the same and equal to Tb 2. Cora asks whether the probabilities of error are better than he signal set and the company rep smiles as if to say "what do you think?" Assuming equally likely messages i 1 2 3 4 as before, will this set have better worse or the same probabilities of error compared to Cora's original set? SOLUTION: The FBNI signal set is exactly the same constellation as Cora's set when you look in signal space. Therefore their performance, from a Pe point of view must be identical. Of course, they might come in prettier colors.... :) 2. Optimal Detection with a Twist: Here we consider what happens to the decision regions when the noise process does not have the same variance in all orthogonal components. 2 T cos c t and s2 t 2 T sin c t on the interval 0 T Consider two signals s1 t where T 2 c . These signals are rendered in a signal space with orthonormal components 1 t 2 T cos ct and 2 t 2 T sin c t. When these signals are sent over the air, they are corrupted by noise. In signal space we then have x si w where si is the usual signal space representation of the signal and w is the noise vector w1 w2 . Both w1 and w2 are zero mean and Gauusian, BUT have given variances 2 and 2 respectively. That is, THE NOISE VARIANCES FOR EACH 1 2 COMPONENT ARE NOT EQUAL! (a) Plot s1 t and s2 t in signal space and provide expressions for the correponding vectors s1 and s2 . SOLUTION: The signal points are exactly the orthonormal waveforms so s 1 10 and s2 01. 2 t r1 t g r2 t gt r3 t gt r4 t h t ht ht 1 erfc 2 d2 4N0 1 erfc 2 Tb 8N0 (b) Write down the proper expressions for the probability distributions f x s 1 and f x s2 , x x1 x2 , the outputs of the correlators in 1 t and 2 t respectively. HINT: Remember we showed that the outputs of the correlators were independent random variables given si . SOLUTION: Noise is different in different rails of the correlator so 22 1 22 2 f x s2 (c) If s1 is sent with probability p1 and s2 with probability 1 p1 , write down the likelihood ratio test which describes the minimum probability of error receiver. SOLUTION: say 1 1 p1 f x s1 f x s2 p1 say 2 (d) Find an expression which relates the xi to describe the decision region boundary relative the signal points s1 and s2 . Simplify as much as possible and sketch your decision e. boundary for 1 1, 2 2 and p1 1 p1 SOLUTION: Substitute the appropriate distributions and we have Take log of both sides say 1 say 2 2 Expanding and cancelling yields, say 1 say 2 22 1 22 2 So, the decision region boundary is given by 3 x1 1 2 x2 1 2 1 2 2 2 2 1 2x1 1 2x2 1 1 x1 1 22 1 x2 2 22 2 x2 1 2 22 2 x2 1 22 1 e x2 1 2 22 e x2 22 2 1 1 say 2 e x1 1 2 22 e x2 22 1 2 2 say 1 1 e x2 1 2 22 x2 22 2 1 1 1 e 21 2 1 f x s1 e x1 1 2 22 1 e x2 22 1 2 x1 1 2 22 x2 22 1 2 2 1 1 1 e 21 2 which is an offset line. Taking the limit as 2 we see that the decision line becomes x1 1 . This result makes sense if 2 is very large, then you shouldn't pay much 2 attention to the x2 measurement. 4 x1 and when 2 21 2 we have 1 2 x2 1 1 2 4 1 ...
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