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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)ejnOok (3.84) 271'
n o = woT = No 8 1o.1024To To a T' (3.85) ~ N oI L f (kT)ejnflok (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 ol In practice, i t is impossible to make T > 0 in computing t he r ighthand 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)ejnwokT 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 oI 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 ighthand 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.
 Spring '13
 Bayliss
 Signal Processing, The Land

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