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# soln5 - EECS 216 SOLUTIONS TO PROBLEM SET#5 Winter 2008...

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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 Half-wave odd b 2 n = 0 1.e x ( t )=odd+ 1 2 a 0 = 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 ( ) = 1 / ( jωC ) R +1 / ( jωC ) = 1 / [1 + jωRC ]. Note the units here–dimensionless. 2b. | H ( ) | = 1 / radicalbig 1 + ( ωRC ) 2 and negationslash H ( ) = - tan 1 [ ωRC ]. Straight from EECS 215. 2c. 0 . 01 = | y ( t ) x ( t ) | | x ( t ) | = | H ( ) - 1 | = | jωRC | 1+( ωRC ) 2 .Then 0 . 0001[1 + ( ωRC ) 2 ] = ( ωRC ) 2 . ωRC = radicalBig 0 . 0001 0 . 9999 ω = 0 . 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 0 x ( t ) cos( 2 πkt T ) dt ; a 0 = 2 T integraltext T/ 2 0 x ( t ) dt . a k = 4 2 π integraltext π/ 2 0 cos( t ) cos( kt ) dt = 1 π integraltext π/ 2 0 [cos( k + 1) t + cos( k - 1) t ] dt = [note where x ( t )=0] 1 π [ sin( k +1) t k +1 + sin( k 1) t k 1 ] | π/ 2 0 = sin( π 2 k + π 2 ) π ( k +1) + sin( π 2 k π 2 ) π ( k 1) = 1 π cos( π 2 k )[ 1 k +1 - 1 k 1 ] = 2cos( π 2 k ) π (1 k 2 ) .
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