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Unformatted text preview: EECS 216 SOLUTIONS TO PROBLEM SET #5 Winter 2008 1.ab #3.21 (a) x ( t ) even → b n = 0. (b) x ( t ) odd → a n = 0 1.cd x ( t ) odd → a n = 0 AND Halfwave odd → b 2 n = 0 1.e x ( t )=odd+ 1 2 → a = 1 2 and a n negationslash =0 = 0. Note you have saved 50% to 75% of your work by noticing these! And see below. 2a. #3.29 H ( jω ) = 1 / ( jωC ) R +1 / ( jωC ) = 1 / [1 + jωRC ]. Note the units here–dimensionless. 2b.  H ( jω )  = 1 / radicalbig 1 + ( ωRC ) 2 and negationslash H ( jω ) = tan − 1 [ ωRC ]. Straight from EECS 215. 2c. 0 . 01 =  y ( t ) − x ( t )   x ( t )  =  H ( jω ) 1  =  jωRC  √ 1+( ωRC ) 2 .Then 0 . 0001[1 + ( ωRC ) 2 ] = ( ωRC ) 2 . → ωRC = radicalBig . 0001 . 9999 → ω = . 01 RC = 10 4 RAD SEC . NOTE: 10 e jωt cancels in  y ( t ) − x ( t )   x ( t )  . 3. #3.9 x ( t ) even & T=2 π → b k =0; a k = 4 T integraltext T/ 2 x ( t ) cos( 2 πkt T ) dt ; a = 2 T integraltext T/ 2 x ( t ) dt ....
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This note was uploaded on 04/03/2008 for the course EECS 216 taught by Professor Yagle during the Fall '08 term at University of Michigan.
 Fall '08
 Yagle

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