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Unformatted text preview: of O FT
graphically.
T he choice o f B = 4 r esults in t he s ampling interval T = 2"k = ~. Looking again
a t t he s pectrum i n Fig. 5.16b, we see t hat t he choice of t he frequency resolution Fa = ~
Hz is reasonable. Such a choice gives us four samples in each lobe of F{w). I n t his case
To =
= 4 seconds a nd No = = 32. T he d uration of J{t) is only I second. We must
repeat i t every 4 seconds (To = 4), a s d epicted in Fig. 5.16c, a nd t ake samples every
second. T his c hoice yields 32 samples (No = 32). Also, io ¥ ( b) co J"(Hz) k !k = T J(kT) 16 4 2 16 k  0.5 0 0.5 2  t_ (c) 4 = ~J(kT)
Since J{t) = 8 rect ttl, t he values of Jk are 1, 0, o r 0.5 ( at t he p oints of discontinuity), as
i llustrated in F ig. 5.16c. In this figure, Jk is depicted a s a function of t a s well a s k, for
convenience.
I n t he d erivation of t he O FT, we a ssumed t hat J (t) begins a t t = 0 (Fig. 5.14a),
a nd t hen t ook N o s amples over t he interval (0, To). I n t he p resent case, however, J (t)
begins a t  ~. T his difficulty is easily resolved when we realize t hat t he O FT o btained by
t his p rocedure is actually t he O FT o f !k r epeating periodically every To seconds. Figure
5.16c clearly i ndicates t hat r epeating t he s egment of Jk over t he i nterval from  2 t o 2
seconds periodically is identical t o r epeating t he s egment of !k over t he i nterval from 0 t o
4 seconds. Hence, t he O FT of t he s amples t aken from  2 t o 2 seconds is t he s ame as t hat
of t he s amples t aken from 0 t o 4 seconds. Therefore, regardless of where J (t) s tarts, we
c an always take t he samples of J(t) a nd i ts periodic extension over t he i nterval from 0 t o
To. I n t he p resent example, t he 32 sample values are !k = {~ 0.5 o~ k ~ 3
5 ~ k ~ a nd 29 ~ k ~ Exact F Fr values
(d) 31 31 27 o 2 k = 4,28 Observe t hat t he l ast sample is a t t = 3 1/8, n ot a t 4, because t he signal repetition s tarts
a t t = 4, a nd t he s ample a t t = 4 is t he s ame as t he s ample a t t = O. Now, No = 32 a nd
110 = 27r / 32 = 7r / 16. Therefore [see Eq. (5.18a)] F ig. 5 .16 2 D iscrete Fourier T ransform o f a g ate pulse. J"(Hz) _ 348 5 S ampling 5.2 Numerical C omputation o f t he F ourier T ransform: T he O FT 349 1. Linearity
31 I f / k ~ F r a nd g k ~ G r , t hen k
F r  " fke  jrm
~ k=O * Values of F r are computed according to this equation and plotted in Fig. 5.16d.
T he samples F r are separated by Fo =
Hz. I n this case To = 4, so t he frequency
resolution Fo is i Hz, as desired. T he folding frequency ~ = B = 4 Hz corresponds t o
r=
= 16. Because Fr is Noperiodic (No = 32), the values of Fr for r =  16 to
n =  1 are the s ame as those for r = 16 t o n = 31. For instance, F17 = F  15, F 18 = F 14,
a nd so on. The O FT gives us t he samples of t he s pectrum F(w).
For the sake o f comparison, Fig. 5.16d also shows t he shaded curve 8 sinc (~), which
is t he Fourier transform of 8 rect (t). T he values of F r computed from O FT equation show
aliasing error, which is clearly seen by comparing the two superimposed plots. T h...
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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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