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Wequalizers - ECE 301 Detailed Solutions Set 6 Analog...

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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 1 62 832 s R s R s 62 832 1 s R s 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 t τ 0 g τ d τ 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 t τ 0 r τ d τ g
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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 the received signal r t is given by r t c t x t . The received signal r t
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