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Unformatted text preview: ECE 301 Detailed Solutions - Set 6: Analog Equalizers 1 6-2. A channel transfer function having no interference is given by C s = s s 62 ; 832 A signal x t is sent through this channel and the received signal r t is given by r t = c t ? x t . The received signal r t is then input to an inverse filter equalizer which was designed for the channel C s . The output of this equalizer is the signal y t . Compute the simplest math form for y t as a function of r t . DETAILED SOLUTION: From the Notes we know that y t , the output of the equalizer, is given by y t = h E t ? r t a The s-domain relation corresponding to a is given by Y s = H E s R s b The function H E s is the transfer function of the inverse filter equalizer. Using C s given in the problem statement we have H E s = 1 C s = s 62 ; 832 s = 1 62 ; 832 s c Substitute c into b to see the s-domain expression for Y s : Y s = h 1 62 ; 832 s i R s = R s 62 ; 832 h 1 s R s i d Taking the inverse Laplace transform of both sides of d will give the desired answer. To do this we need the relation between dividing by s in the s-domain and integration in the time-domain. For a general signal g t this relation is as follows: If g t G s e then Z t = g d 1 s G s f Applying f to the specific s-domain function R s in d then gives the desired answer: y t = r t 62 ; 832 Z t = r d g ECE 301 Detailed Solutions - Set 6: Analog Equalizers 2 6-3. A channel transfer function is C s = s 50 . A signal x t is sent through this channel and...
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- Spring '08