1. (4.3) This function has the symmetry property: )()(tyty−=−or sine-like symmetry. Therefore0=nA. ∫∫∫∫+===−2/02/02/2/])sin()()sin()([2)sin()(2)sin()(2TTTTTTndttntydttntyTdttntyTdttntyTBω]|)cos(|)cos()[(2])sin()()sin([22/2/02//TTTTtntnnFTdttnFdttnFT+−=−+=Recall that π2=T. ])1(1[2)]cos(1[2nnnFnnFB−−=−=. e.g. FB41=, 02=B, 343FB=, ... ...)5sin513sin31(sin4)(+++=tttFtywhere lbF1000=, sec/1000002.022radT===2. (5.13) f=50Hz, 95.0)2(112=+τfSolve the above equation to get 00105.0=sec. 3. (5/19) 30.0=ξ, 950=fHz, 1800=nfHz. Error is given by %.2727.01)/2(])/(1[11222==−+−=−nnsdffffPP439.0)/(1/2tan2=−=nnffffφ, o7.23=4. (a) The minimum sampling rate is 4000Hz as dictated by the highest frequency present in the signal.
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This note was uploaded on 01/12/2011 for the course MAE 3800 taught by Professor Feng during the Spring '10 term at Missouri (Mizzou).