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

# 638 10 fourier analysis of discrete time signals nals

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

This is the end of the preview. Sign up to access the rest of the document.

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...
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