IRWIN 9e 15_23 - Irwin, Basic Engineering Circuit Analysis,...

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Unformatted text preview: Irwin, Basic Engineering Circuit Analysis, 9/E 1 15.23 U50 PSPIC‘E to determine the FuLIi'icr series 01' the Mit'ct'm'm in Fig. FL 5.23 in the form rim = a” + 25mm;an + an} ra-J =IIl==IIl= Illlfllllllfll Illl’llllll’ll .._. : cu L L 3 U DO 10 20 30 40 50 I30 ?0 BE] 90100 Timeimaj Figure P1523 SOLUTION: Final“ firm: 50mg W” calm}: 50M warm 5% «momdo' we IpbuL m Mom 14(1). w (9rd“ Z : 201/12 __L 5DYY‘ 130): 2+ HMO! u‘MwHioWfi + i664 2130 {Quir'aqfijfi +o~8¥l m‘n (Zwi" IL¢8°)+ ,,,+ 14A _____________________________—_———-—— ; Chapter 15: Fourier Analysis Techniques Problem 15.23 2 Irwin, Basic Engineering Circuit Analysis. 9/E FOURIER COMPONENTS OF TRANSIENT RESPONSE I(I_IS) DC COMPONENT = 1.997500E’06 HARMONIC FRE ENCY FOURIER NORMALIZED PHASE NORMALIZED no HZ) COMPONENT COMPONENT (DEG) PHASE (DEG) 1 2 . oooe+o1 4 . 4015-06 1. 0005+00 1. 049902 0 . 0005+00 2 4.000E+01 1.669e-06 3.79ZE—01 —3.97SE+01 ~2.495E+02 3 6.000E+01 8.8105—07 2.0025-01 -1.480E+02 —4.627E+02 4 8.000E+01 4.518E-07 1.027e-01 3.521E+01 -3.844E+02 5 1.000E+02 3.183E-07 7.233E-02 1.791E+02 -3.4S4E+02 6 1.200E+02 3.001E-07 6.8195-02 2.2125+01 -6.073£+02 7 1.4005+02 2.0465-07 4.648E-02 -1.565E+02 -8.9o7a+02 8 1.6005+02 1.829E—07 4.157e-02 —2.297E+01 -8.621E+02 9 1.800E+02 1.9135—07 4.347E—02 1.6255+02 -7.816E+02 10 2.000E+02 1.5925-07 3.6175-02 -1.8OOE+00 -1.0SlE+03 TOTAL HARMONIC DISTORTION = 4.5973255+01 PERCENT JOB CONCLUDED TOTAL JOB TIME .45 ———__—__—_—__————————_—‘—‘—’ ‘. Problem 15.23 Chapter 15: Fourier Analysis Techniques ...
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This note was uploaded on 10/04/2010 for the course ECE ECE212 taught by Professor Micahstickel during the Fall '10 term at University of Toronto- Toronto.

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IRWIN 9e 15_23 - Irwin, Basic Engineering Circuit Analysis,...

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