By Edward A. Lee, David G. Messerschmitt
This complement comprises labored out options to the bankruptcy finish challenge units present in Digital verbal exchange, moment Edition, ISBN 0-7923-9391-0.
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Sample text
6-14. (a) From appendix 2-A, the Fourier transform of Re{ z (t) } is Ze UCJ» and the Fourier transform of jIm ( z (I) } is Zo UCJ». Hence the Fourier transform of 1m( z (t) } is - JZo UCJ». So we need to show that H UCJ»ZeUCJ» =-jZoUCJ». 5Z-(-j ro); CJ» CJ) < °°. 324) Hence H UCJ»Ze UCJ» which is the desired result. 326) This follows similarly. 6-15. 327) From the duality property in appendix 2-A Z(jCll) =21th (-(J). 328) Note that Z(j Cll) =0 for Cll < O. so z (t) is analytic. m{HUt») =1.. *Re{HUt»).
If there is one channel error. no detection error is ever made. and this occurs with probability (1 - P )6. If there are two channel errors, if one is in the first three bits and the othec is in the second three bits. no error is made. and this occurs with probability (3p (l - P );2. However. if both errors are in the first three or second three bits. there is an error. If there are three channel errors. thece must always be two errors in either the first three or the second three bits. and there is therefore always a detection error.
And similarly for (111000) or (000111). If(OOOOOO) is ttan~mitted. a detection error occurs whenever two out of the first three bits are in error. or two out of the last three bits. or both. If there is one channel error. no detection error is ever made. and this occurs with probability (1 - P )6. If there are two channel errors, if one is in the first three bits and the othec is in the second three bits. no error is made. and this occurs with probability (3p (l - P );2. However. if both errors are in the first three or second three bits.