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

638 10 fourier analysis of discrete time signals nals

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Unformatted text preview: if the system is asymptotically stable and if the input signal is DTF-transformable. We shall not belabor this method further because it is clumsier and more restricted than the z-transform method discussed in the next chapter. In the z-transform, we generalize the frequency variable j n t o (7 + j n so t hat the resulting exponentials can grow or decay with k. This procedure is tHere Y(fl) is a function of variable ein . Hence, x = ei n for the purpose of comparison with the expression in Sec. B.5-5. 1 0.6-2 Computation o f Direct and Inverse DTFT S pectral analysis o f d igital signals requires d etermination o f D TFT a nd I DTFT [ determining F (n) from J[k) a nd vice versa). T his d etermination could b e accomplished by using t he D TFT e quations [Eqs. (lO.30) a nd (10.31)) directly on a digital computer. However, t here a re two difficulties i n i mplementation o f t hese equations on a digital computer. 1. E quation (lO.31) involves summing a n infinite n umber o f t erms, which is n ot possible because i t requires infinite computer time. 2. E quation (lO.30) requires integration which can only b e p erformed approximately on a c omputer b ecause a computer approximates a n i ntegral by a s um. T he first problem can be s urmounted e ither by restricting t he analysis only t o a finite length I[k) o r b y t runcating I[k) b y a s uitable window. T he e rror because 642 10 Fourier Analysis o f Discrete-Time Signals II .. .. ... InIIJ. .... r f [k) 643 10.6 Signal processing Using D FT a nd F FT Therefore, NoDr a re t he samples o f F(!1) t aken uniformly a t t he frequency intervals o f !10 . B ut b ecause!1 o = 211"/No, t here a re exactly No n umber of these samples of F(!1) over t he f undamental frequency interval o f 211". According t o Eqs. (10.62) and (10.65), i t follows t hat (a) .. k -- (10.66) where [from Eqs. (10.63) a nd (10.65)] N o-l Fr = . .. 11'J1111 .. 1 I 'oJ f[k]e-jrfl.ok (10.67) k=O Because fNo [k] = f[k] for k = 0, 1, 2, . .. , (No - 1), we can express t he Eq. (10.66) as f[k] = ~ No F ig. 10.10 D FT computation of a finite length signal. of windowing may be reduced by using a wider a nd a t apered window. We shall see t~at if f[k] has a finite duration, t he samples o f F(!1) c an be computed using a fillIte sum (rather t han a n integral). This solves t he second problem. Moreover, f[k] is uniquely determined from these samples of F(!1). In order t o derive appropriate relationships, consider t he signal f[k] s tarting a t k = 0, a nd w ith a finite length No, a s shown in Fig. 1O.lOa. Let us construct a periodic signal f No [k] by repeating f [k] periodically a t intervals of No, a s illustrated in F~g. 1O.lOb. We c an represent t he periodic signal fNo [k] by t he discrete-time FourIer series (DTFS) as [see Eqs. (10.8) a nd (10.9)] L N o-l L Frejrfl.ok k = 0, 1, 2, . .. , No - 1 (10.68) r =O Moreover, we need t o d etermine F r , t he samples of t he D TFT, only over t he interval 211". Therefore, Eq. (10.67) c an be expressed as o :&qu...
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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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