Solutions Communications technology II WS 2009/2010 - Universität ...

3 WS 2009/2010 “Communications technology II” – SolutionsSolution to exercise 3 (eq09):a) Total pulse response:g(i) = f(i) ∗ e(i) = [1, −α, +α 2 , −α 3 , 0]+[0, α, −α 2 , α 3 , −α 4 ]g(i) = δ(i) − α 4 · δ(i − 4)b)G(z) = 1 − α 4 z −4z 0,ν = |α| · e j· π2 ·ν , ν = 0...3c) ( SN)ISI= 1 α 8Max{∆d(i)} = α 4d) General equalizer of the order n:n∑e(i) = (−1) l α l δ(i − l)l=0f(i) ∗ e(i) = δ(i) − (−1) n · α n · δ(i − n)( ) S 1=NISIα ( 2n)Max{∆d(i)} = α n