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

I t is therefore not necessary t o show discrete time

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Unformatted text preview: e i ts a mplitude spectrum. T he p hase spectrum, however, is changed by - koO. T his added phase is a linear function of 0 w ith slope - ko. Physical Explanation o f t he Linear Phase T ime delay in a signal causes a linear phase shift in its spectrum. T he heuristic explanation of this result is e xactly parallel t o t hat for continuous-time signals given in Sec. 4.3-4 (see Fig. 4.20). Frequency-Shifting Property Parseval's Theorem If f[k] { :=> F(O) t hen E f , t he energy of f [k], is given by If f[k] { :=> (10.51) F(O) t hen (10.49) In order t o prove this property, we have from Eq. (10.31) T his property is t he d ual of t he time-shifting property. To prove this property, we have from Eq. (10.31) 00 F*( - 0) = L ! *[k]e- jOk (1O.52a) k =-oo 00 00 k =-oo k =-oo T his result shows t hat j*[k] { :=> F *(-O) (1O.52b) Now k~00If[kll2 = k~OO!*[k]f[k] = k~OO!*[k] [2~ 1" F(O)ejnkdO] T ime and Frequency Convolution Property If h[k] { :=> F1 (0) a nd h [k] { :=> F2(0) t hen h [k] a nd * h[k] h [k]h[k] where { :=> { :=> F l(0)F2(0) 1 271' Fl(O) * F2(0) 2~ 1" F(O) [k~OO!*[k]ejm] (10.50a) = (1O.50b) =~ 271' r F(O)F*(O) dO = ~ i2" IF(0)1 2 dO r 271' i27r 636 10 Fourier Analysis o f Discrete-Time Signals lOA 637 Connection with t he C ontinuous-Time Fourier Transform ( FT) f [kl (a) F (Q) (b) (a) ~ o (b) 0) _ _ 1-- k-- -211 -11 0 11 211 Q-- ~(t) 4i[kl ( c) ~[kl=f[2kl 1-- (c) F iQ) A Decimation (Downsampling) (d) 2 -,.:- ":.-.:" f lkl j rIIII -2 0 10 k-- -211 (e) · :11 J -11 0 11 211 Q-- (e) f; [kl k-- ( I) Fig. 1 0.8 C onnection b etween t he D TFT a nd t he F ourier t ransform. 16 24 32 -211 1 0.4 DTFT Connection with the Continuous-Time Fourier Transform C onsider a continuous-time signal fe(t) (Fig. 1O.8a) w ith t he Fourier transform Fe(w). T his signal m ayor may not be bandlimited. For convenience, we shall assume t he signal t o b e bandlimited t o B Hz (Fig. 1O.8b). This signal is sampled with a sampling interval T . T he s ampling r ate m ayor m ay n ot b e above t he N yquist rate. Again, for convenience, we shall assume t hat t he s ampling r ate is a t l east equal t o t he Nyquist rate; t hat is, T ::::: l /2B. T he s ampled signal l e(t) (Fig. lO.8c) c an be expressed as 00 L l e(t) = fe(kT)Ii(t - kT) k =-oo T he c ontinuous-time Fourier transform o f t he above equation yields 00 Fe(w) = L fe(kT)e - jkTw (10.53) k =-oo I n Sec. 5.1 (Fig. 5.le), we have shown t hat Fc(w) is F c(w)IT r epeating periodically with a period w . = 27r I T, as illustrated in Fig. 1O.8d. L et us construct a discretetime signal f[k] such t hat i ts k th element value is equal t o t he value o f t he k th s ample of fc(t), as depicted in Fig. 1O.8e; t hat is, f[k] = f c(kT) Now, F (n), t he D TFT of f[k], is given by F ig. 1 0.9 -11 11 2x Q-- S pectra o f t he d ecimated a nd i nterpolated s ignals. 00 00 k =-oo k =-oo (10.54) C omparison o f (10.54) with (10.53) shows t hat F (n) = Fe (~) (10.55) Thus, F (n) c an b e o bt...
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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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