Signal Processing and Linear Systems-B.P.Lathi copy

1 cos wot hence t he o utput is a sinusoid 10 cos wot

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Unformatted text preview: x 100 = 1.08% a nd D3 = ~ x 100 = 0.527% • Numerical Computation of Dn We can c ompute D n numerically using t he D FT ( the discrete Fourier transform discussed in Sec. 5.2), which uses t he samples of a periodic signal f (t) over one period. T he s ampling interval is T seconds. Hence, there are No = T o/T n umber of samples in o ne p eriod To. To find t he relationship between D n a nd t he samples of f (t), consider Eq. (3.71) expression over the quarter cycle only and then quadrupling the resulting values. Thus e n = an = - f (kT)e-jnOok (3.84) 271' n o = woT = No 8 1o.1024To To a T' (3.85) ~ N o-I L f (kT)e-jnflok (3.86a) No k=O Now, from Eq. (3.85), noNo = 271'. Hence, ejnOo(k+No) (3.86a), it follows t hat ejnOok a nd from Eq. (3.86b) T he powers of t he first harmonic a nd t he t hird h armonic components of Yd(t) are ( 1.04)2/2 = 0.5408 a nd (0.726)2/2 = 0.2635, respectively. Hence, t he first harmonic a nd t he t hird h armonic d istortions are 50 L N _ To 0.865 Dl = N o-l In practice, i t is impossible to make T --> 0 in computing t he r ight-hand side of Eq. (3.84). We can make T small, b ut n ot zero, which will cause t he d ata t o increase without limit. Thus, we shall ignore t he limit on T in Eq. (3.84) with t he implicit understanding t hat T is r easonably small. Nonzero T will result in some computational error, which is i nevitable in any numerical evaluation of a n integral. The error resulting from nonzero T is called t he a liasing e rror, which is discussed in m ore details in C hapter 5. T hus, we can express Eq. (3.84) as 2 Yd (t) dt =~ D tot f (kT)e-jnwokT T where f (kT) is t he k th sample of f (t) a nd 0- = To/4 L T~O No k=O Computing Harmonic Distortion 1 =~ N o-I T~O To k=O = 1.04 cos wot + 0.733 cos 3wat + 0.311 cos 5wat + ... P Yd 217 [10 cos wot - 8] cos nwot dt Clearly, t he Fourier spectrum D n r epeats periodically with period No, which causes spectral overlap due t o r epeating cycles. To u nderstand t he n ature of this overlap, see Fig. 5.12f. Generally, D n decays with n a nd t he s pectral overlap caused by t he periodic repetition will have negligible effect if we use large enough No ( the period). The first cycle intersects t he second cycle a t n = N o/2. Hence, overlapping will be negligible if D n is very small for n 2': N o/2. Because o fthe p eriodicity of D n, we need to evaluate only No values of D n over t he r ange n = - No/2 t o N o/2 - 1. However, the D FT (or F FT) computes D n for n = 0 t o No - 1. T he p eriodicity property Dn+No = D n means, beyond n = N o/2, t he coefficients represent t he values for negative n . For instance, when No = 32, D17 = D -15, D I8 = D -14, . .. , D 3I = D -l. T he cycle repeats again from n = 32 on. We can use t he efficient F FT ( the f ast F ourier t ransform discussed in Sec. 5.3) t o c ompute t he r ight-hand side of t he above equation. We shall use MATLAB to implement t he F FT algorithm. For this purpose, we need samples of f (t) over one period s tarting a t t = O. In this algorithm, it is also preferable (altho...
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This note was uploaded on 04/14/2013 for the course ENG 350 taught by Professor Bayliss during the Spring '13 term at Northwestern.

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