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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 lies in the fact 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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