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Unformatted text preview: id 10 cos wot a nd t he o utput
signal y(t). T he signal Yd(t), whose period is To [the same as t hat of y(t)] can be described
over t he first cycle as
10 cos wot  8 Yd(t) = { ~O cos wot +8 o f t he p owers of i ts F ourier c omponents.
.
..
W e c an a pply t he s ame a rgument t o t he e xponential F ourier s enes, w hich IS
a lso m ade u p o f t he o rthogonal c omponents. H ence, t he p ower o f a p eriodic s ignal
F t) c an b e e xpressed a s a s um o f t he p owers of i ts e xponential c omponents. I n
jWot
E~. ( 1.5d), we s howed t hat t he p ower of an e xponential
is 1~21. U sing t~is
r esult we c an e xpress t he p ower of a p eriodic s ignal f (t) I II t erms o f ItS e xponential ?e F ourier s eries coefficients as
00 (3.83a)
n =OO 1f  0.1024To :'0 It I :'0 1f + 0.102 4To everywhere else Observe t hat Yd(t) is a n even function of t a nd its mean value is zero. Moreover, bn
and C n = an. Hence, its Fourier series can be expressed as n =l T his r esult, w hich is a n a lternate f orm o f E q. ( 3.42), is t he P arseval's t heorem
(for F ourier s eries). I t s tates t hat t he p ower o f a p eriodic s ignal is e qual t o t he s um It I :'OO.1024To Yd(t) = LC n = 0, cos nwot n::;::l As usual, we can compute t he coefficients C n (which is equal to an) by integrating
Yd(t) cos nwot over one cycle (and t hen dividing by 2/To). Because Yd(t) has even symmetry, we can find a n by i ntegrating t he expression over half cycle only using Eq. (3.66b).
The straightforward evaluation of the appropriate integral yieldst
t In addition, Yd(t) exhibits halfwave symmetry (see Prob. 3.47), where the second half cycle is
the negative of the first half cycle. Because of this property, all the even harmonics vanish, and
the odd harmonics can be computed by integrating the appropriate expressions over the first half
cycle only (from  To/4 to To/4) and doubling the resulting values. Moreover, because of even
symmetry, we can integrate the appropriate expressions over 0 to To/4 (instead of from  To/4 to
To/4) and double the resulting values. In essence, this allows us to compute e n by integrating the 3 Signal Representation by Orthogonal Sets 216 Gn aQ [ Sin IO.6435(n+l)] _
 n +l 7r { + Sin IO.6435(nl)]]_;g
n l [ Sin(O.6435n)] 71' n o 3.6 Numerical C omputation of Dn n o dd Dn = r f(t)ejnwot dt ~ To j To n even = lim ~ = l im C omputing t he coefficients G l , G2, G3, . .. from this expression, we can write Yd( t ) ~ I n t his case, we can compute t he a mount o f distortion in t he o utput signal b y computing t he power o f t he d istortion component Yd(t). Because Yd(t) is a n e ven function of
t a nd b ecause t he p ower in t he first half cycle is i dentical t o t he power in t he second half
cycle, we c an c ompute t he power by averaging t he energy over a q uarter cycle. T hus j T O/2 a  To/2
1 l a To 4 / y i(t) dt 4 1o.1024To aa (10 cos wat  8)2 dt Dn = T he power of t he desired signal 10 cos wat is ( 10? / 2 = 50. Hence, t he t otal h armonic
distortion is = 0.865
w x 100 = 1.73% 3.6 0.5408 0.2635...
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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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