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# ch16_ism - Fourier Series 16 Assessment Problems AP 16.1 av...

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16 Fourier Series Assessment Problems AP 16.1 a v = 1 T Z 2 T/ 3 0 V m dt + 1 T Z T 2 3 ± V m 3 ² dt = 7 9 V m =7 π V a k = 2 T " Z 2 3 0 V m cos 0 tdt + Z T 2 3 ± V m 3 ² cos 0 # = ± 4 V m 3 0 T ² sin 4 3 ! = ± 6 k ² sin 4 3 ! b k = 2 T " Z 2 3 0 V m sin 0 + Z T 2 3 ± V m 3 ² sin 0 # = ± 4 V m 3 0 T ² " 1 cos 4 3 !# = ± 6 k ² " 1 cos 4 3 !# AP 16.2 [a] a v π =21 . 99 V [b] a 1 = 5 . 196 a 2 =2 . 598 a 3 =0 a 4 = 1 . 299 a 5 =1 . 039 b 1 =9 b 2 =4 . 5 b 3 b 4 . 25 b 5 . 8 [c] ω 0 = ± 2 π T ² =50 rad/s [d] f 3 =3 f 0 =23 . 87 Hz [e] v ( t )=21 . 99 5 . 2 cos 50 t + 9 sin 50 t +2 . 6 cos 100 t +4 . 5 sin 100 t 1 . 3 cos 200 t . 25 sin 200 t +1 . 04 cos 250 t . 8 sin 250 t + ··· V AP 16.3 Odd function with both half- and quarter-wave symmetry. v g ( t )= ± 6 V m T ² t, 0 t 6; a v ,a k for all k 16–1

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16–2 CHAPTER 16. Fourier Series b k =0 for k even b k = 8 T Z T/ 4 0 f ( t ) sin 0 t dt, k odd = 8 T Z 6 0 ± 6 V m T ² t sin 0 tdt + 8 T Z 4 6 V m sin 0 = ± 12 V m k 2 π 2 ² sin 3 ! v g ( t )= 12 V m π 2 X n =1 , 3 , 5 1 n 2 sin 3 sin 0 t V AP 16.4 [a] Using the results from AP 16.2, and Equation (16.39), A 1 = 5 . 2 j 9=10 . 4/ 120 ; A 2 =2 . 6 j 4 . 5=5 . 2/ 60 A 3 ; A 4 = 1 . 3 j 2 . 25=2 . 6/ 120 A 5 =1 . 04 j 1 . 8=2 . 1/ 60 θ 1 = 120 ; θ 2 = 60 ; θ 3 not deFned; θ 4 = 120 ; θ 5 = 60 [b] v ( t )=21 . 99+10 . 4 cos(50 t 120 )+5 . 2 cos(100 t 60 ) +2 . 6 cos(200 t 120 )+2 . 1 cos(250 t 60 )+ ··· V AP 16.5 The ±ourier series for the input voltage is v i = 8 A π 2 X n =1 , 3 , 5 ± 1 n 2 sin 2 ² sin 0 ( t + 4) = 8 A π 2 X n =1 , 3 , 5 ± 1 n 2 sin 2 2 ² cos 0 t = 8 A π 2 X n =1 , 3 , 5 1 n 2 cos 0 t 8 A π 2 = 8(281 . 25 π 2 ) π 2 = 2250 mV ω 0 = 2 π T = 2 π 200 π × 10 3 =10
Problems 16–3 · .. v i = 2250 X n =1 , 3 , 5 1 n 2 cos 10 nt mV From the circuit we have V o = V i R +(1 /jωC ) · 1 jωC = V i 1+ jωRC V o = 1 /RC 1 /RC + V i = 100 100 + V i V i 1 = 2250/0 mV ; ω 0 =10 rad/s V i 3 = 2250 9 /0 = 250/0 mV ;3 ω 0 =30 rad/s V i 5 = 2250 25 = 90/0 mV ;5 ω 0 =50 rad/s V o 1 = 100 100 + j 10 (2250/0 ) = 2238 . 83/ 5 . 71 mV V o 3 = 100 100 + j 30 (250/0 ) = 239 . 46/ 16 . 70 mV V o 5 = 100 100 + j 50 (90/0 )=80 . 50/ 26 . 57 mV · v o = 2238 . 33 cos(10 t 5 . 71 ) + 239 . 46 cos(30 t 16 . 70 ) +80 . 50 cos(50 t 26 . 57 )+ ... mV AP 16.6 [a] The Fourier series of the input voltage is v g = 4 A π X n =1 , 3 , 5 1 n sin 0 ( t + T/ 4) =42 X n =1 , 3 , 5 ± 1 n sin ² 2 ³´ cos 2000 nt V From the circuit we have V o sC + V o sL + V o V g R =0 · V o V g = H ( s )= s/RC s 2 +( s/RC )+(1 /LC )

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16–4 CHAPTER 16. Fourier Series Substituting in the numerical values yields H ( s )= 500 s s 2 + 500 s +10 8 V g 1 = 42/0 ω 0 = 2000 rad/s V g 3 = 14/180 3 ω 0 = 6000 rad/s V g 5 =8 . 4/0 5 ω 0 =10 , 000 rad/s V g 7 = 6/180 7 ω 0 =14 , 000 rad/s H ( j 2000) = 500( j 2000) 10 8 4 × 10 6 + 500( j 2000) = j 1 96 + j 1 =0 . 01042/89 . 40 H ( j 6000) = 0 . 04682/87 . 32 H ( j 10 , 000) = 1/0 H ( j 14 , 000) = 0 . 07272/ 85 . 83 Thus, V o 1 = (42/0 )(0 . 01042/89 . 40 )=0 . 4375/89 . 40 V V o 3 . 6555/ 92 . 68 V V o 5 . V V o 7 . 4363/94 . 17 V Therefore, v o . 4375 cos(2000 t +89 . 40 )+0 . 6555 cos(6000 t 92 . 68 ) +8 . 4 cos(10 , 000 t . 4363 cos(14 , 000 t +94 . 17 )+ ...
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ch16_ism - Fourier Series 16 Assessment Problems AP 16.1 av...

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