When the source symbol probabilities are equal the ML and MAP generates

# When the source symbol probabilities are equal the ml

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source symbol probabilities or assumes all are equal. When the source symbol probabilities are equal the ML and MAP generates identical decisions. In the figure for an example of ML, the received signal is r , the output alphabet consists of S i ’s and the distances according to a defined metric (usually Euclidean distance) are d i ’s. ML would decide that S 2 is sent because it has the smallest distance from the output signal. On the other hand MAP could have another decision since the symbol probabilities are used in the decision. But it can be said, with a reasonable confidence, that MAP decision would be the same if 2 , ) ( ) ( 2 i S P S P i .
151227621 DIGITAL COMMUNICATIONS 5 6. 25.12.2004 2 nd midterm Describe distance metrics and derive correlation metrics from it. Solution Distance metrics is a measure of the distance/closeness of two points in signal space which is usually represented as Cartesian coordinate system. The square of Euclidean distance between two points, r and s m , in Cartesian coordinate system is given as N k mk k m s r D ! 2 ) ( ) , ( s r where, when comparing, the square and the distance itself do the same jobs. Expanding it, we’d have N n mn N n mn n N n n m s s r r D 1 2 1 1 2 2 ) , ( s r or in norms form 2 2 2 ) , ( m m m D s s r r s r . Since 2 r term is common for all s m , and does not have any effect when comparing distances, we can remove it. The negative of the remaining, 2 2 ) , ( m m m C s s r s r is called the correlation metrics since it involves the correlation (inner product) of r and s m . ) , ( m C s r should be maximized to minimize ) , ( m D s r . 2 m s term represents the signal energy and can also be removed in cases of equal energy signals. 7. 25.12.2004 2 nd midterm A binary antipodal signaling source has P (-1)=0.7 and P (1)=0.3. Find the output ensemble of the BSC through which the output of the source is fed if P e ( s i | s j ; i j )=0.01 by forming the channel transition matrix first. Solution The channel transition matrix for the BSC is formed as 99 . 0 01 . 0 01 . 0 99 . 0 Q . The output probabilities of output symbols are 3 . 0 7 . 0 99 . 0 01 . 0 01 . 0 99 . 0 Qz v and 304 . 0 696 . 0 v . Although the signals would probably be distorted, we can expect that the output symbols are reasonably close to input symbols which are -1 and 1, forming the alphabet as A={-1, 1} and concluding our search for the ensemble (A, v ).
151227621 DIGITAL COMMUNICATIONS 6 8.04.02.2005 Final Exam Show that the following set is orthonormal. Solution For two signals to be orthogonal it is required that ) ( , 0 ) ( ) ( j i dt t t j i 0 ) ( ) ( 2 1 2 1 1 0 2 1 2 1 2 1 2 1 1 0 2 1 2 1 2 1 t t dt dt dt t t (orthogonal) 0 ) ( ) ( 3 1 dt t t (by inspection) (orthogonal) 0 ) ( ) ( 3 2 dt t t (by inspection) (orthogonal) The energies of the signals are 1 ) ( 2 0 2 2 1 2 1 dt dt t 1 ) ( 2 2 dt t (this one is the same as the previous one because of the square) 1 ) ( 3 2 2 3 dt dt t Conclusion : The signals are orthogonal and their individual energies equal to unity. Therefore this set is an orthonormal set. 9. 04.02.2005 Final Exam Determine if the following code is linear. 11011 11 10100 10 01111 01 00000 00 Solution In order for a code to be linear j i c c must also be a symbol in the set.

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