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

# f 128 fi28 t hus we need fr only over t he range r

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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 - jr-m -~ 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 No-periodic (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.

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