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

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This note was uploaded on 04/14/2013 for the course ENG 350 taught by Professor Bayliss during the Spring '13 term at Northwestern.

Ask a homework question - tutors are online