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Unformatted text preview: Chapter 12 Spectrum Estimation 12.1 Ergodicity It is needed to estimate the statistical parameters of a random process from real data. Although most parameters to be estimated can be formulated as the expectation of a function of the random process ( ) t X , the problem is that we need to implement the expectation via taking average (ensemble average) over the whole sample space. An example is given in the following. For a specific t , ( ) t X is a random variable. Let { } ( ) ( ) t E t = X . We can estimate the mean of ( ) t X using n observed samples ( , ) i t X of ( ) t X by 1 ( ) ( , ) i i t t n = X . This is a point estimation. It is a consistent estimate of ( ) t if its variance t as n E . In this case, it is a good estimate for large n . But, we usually only have a sample of ( ) t X . Can we use it to estimate { } ( ) E t X ? If { } ( ) E t X depends on t , then this is impossible. If ( ) t X is a regular stationary process, then the time average will approach to { } ( ) E t X as the length t . This is the meaning of ergodicity. Meanergodic process Given a real stationary process ( ) t X , we want to estimate { } ( ) ( ) t E t = X . We form the time average 1 ( ) 2 T T T t dt T = X . Since { } { } 1 ( ) 2 T T T E E t dt T  = = X , T is an unbiased estimator of . 1 If the variance of T , 2 T as T E , then T in the MS sense. We say that ( ) t X is meanergodic, i.e., ( ) t X is meanergodic if the time average T h the ensemble average , as T E . So, meanregodic can be proved via calculating T and showing T as T E . Ex. 12.1 Let c be an RV with mean c and let ( ) t = X c . Then { } { } ( ) c E t E = = = X c . A specific sample ( ) t c = X is a line (constant). In this case, ( ) ( ) T = c (i.e., a specific constant determined by ). Obviously, ( ) T . So ( ) t X is not mean ergodic. Ex. 12.2 Given two meanergodic processes 1 ( ) t X and 2 ( ) t X , their means are 1 and 2 , respectively. We form 1 2 ( ) ( ) ( ) t t t = + X X cX , where c is a RV independent of 2 ( ) t X and = c and 1 with probabilities 0.5, respectively. Then { } { } { } { } { } 1 2 1 2 1 2 ( ) ( ) ( ) ( ) 0.5 E t E t E t E E t = + = + = + X X cX c X But for a specific sample, if ( ) = c , then 1 ( ) ( ) t t = X X and 1 T as T E . If ( ) 1 = c , then 1 2 ( ) ( ) ( ) t t t = + X X X . In this case, 1 2 T = + . Hence, ( ) t X is not meanergodic. To calculate 2 T of T for an RP ( ) t X , we can form 1 ( ) ( ) 2 t T t T t d T + = w X which is a moving average of ( ) t X . Obviously, (0) T = w ....
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 Fall '08
 SinHorngChen

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