IRWIN 9e 15_22 - Irwin, Basic Engineering Circuit Analysis....

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Unformatted text preview: Irwin, Basic Engineering Circuit Analysis. 9/E 1 15.22 LIHe PSPICE tn determine the Fuurier series; of the waveform in Fig. Flt—1.22 in the [mm mm = a” + Ebfifiittmm, + an} rt-l "-Im-I _25 Time {ms} Figure F1522 SOLUTION: Cerutcr ’- l—‘moi hm: low Wklng 69‘ ’nvmonjua :10 W2 VPWL *0 {oaoowce VSM). Wm 09 WP Culling : (cu: V500 : 3'757L 395% w [my WW“) 7" 2.0% 82'n(9w£ ~ 87.7%) + sz 23in ( 3m Wye/35°” H I + v Chapter 15: Fourier Analysis Techniques Problem 15.22 2 Inivin, Basic Engineering Circuit Analysis. 9/E FOURIER COMPONENTS OF TRANSIENT RESPONSE V(V_VS) DC COMPONENT = 3.750000E+00 HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED No (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG) 1 1.000E+02 3.954E+00 1.000E+00 -1.424E+02 0.000E+00 2 2.000E+02 2.016E+00 5.099E-01 -8.774E+01 1.971E+02 3 3.000E+02 1.247E+00 3.154e-01 —8.735£+00 4.185E+02 4 4.000E+02 6.417E~01 1.623e—01 6.9915+01 6.3955+02 S 5.000E+02 - 2.027E‘01 5.126E-02 9.000E+01 8.020E+02 6 6.000E+02 2.852E-01 7.213e-02 1.101E+02 9.645E+02 7 7.000E+02 2.2916—01 5.793e-02 —1.713£+02 8.256E+02 8 8.000E+02 1.260E-01 3.188E-02 -9.226£+01 1.047E+o3 9 9.000E+OZ 4.8835-02 1.2355-02 ~3.759£+01 1.244E+03 10 1.000E+03 1.9936—08 5.040E—09 -1.206E+02 1.303E+03 TOTAL HARMONIC DISTORTION = 6.310255E+01 PERCENT JOB CONCLUDED TOTAL 308 TIME .39 ________—7—————-———-—-—-——‘ Problem 15.22 Chapter 15: Fourier Analysis Techniques ...
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