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

# S plane 1m 1m 2 re 1 2 1 re b a f

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Unformatted text preview: t he u nit circle in t he ,,(-plane. N ext consider t he signal e Ak , where A lies in t he left-half plane in Fig. 8.5a. This means A = a + j b, where a is negative (a < 0). In this case, t he signal decays exponentially. This signal can be expressed as " (k, where 545 jrl o o 2 3 3 (a) a nd because l ejbl "( -k (b) -1 =1 Also, a is n egative (a < 0). Hence, hi = ea < 1. T his result means t hat t he corresponding'Y lies inside t he u nit circle. Therefore, a signal " (k decays exponentially if"( lies w ithin t he u nit circle (Fig. 8.5b). If, in t he above case we h ad selected a t o b e p ositive, (A in t he r ight-half plane), t hen hi > 1, a nd "( lies outside t he u nit circle. Therefore, a signal " (k grows exponentially if "( lies outside t he u nit circle ( Fig.8.5b). To summarize, t he i maginary axis in t he A-plane m aps into t he u nit circle in the ,,(-plane. T he left-half plane in t he A-plane m aps into the inside of t he u nit circle and t he r ight-half of t he A-plane m aps into t he outside of the u nit circle in t he ,,(-plane, as depicted i n Fig. 8.5. This fact means t hat t he signal " (k grows exponentially with k if "( is o utside t he u nit circle (h I > 1), a nd decays exponentially if "( is inside the u nit circle (hi < 1). T he signal is constant or oscillates with constant amplitude if "( is o n t he u nit circle (hi = 1). O bserve t hat (8.4) Figures 8.6a a nd 8.6b show plots of (0.8)k, a nd ( -0.8)k, respectively. Figures 8.6c and 8.6d show plots of (0.5)k, a nd (1.1)k, respectively. These plots verify our earlier conclusions a bout.the location of"( a nd t he n ature of signal growth. Observe t hat a signal (_,,()k a lternates sign successively (is positive for even values of k and negative for o dd values of k , as depicted in Fig. 8.6b). Also, t he e xponential (0.5)k decays faster t han (0.8)k. T he e xponential (0.5)k can also be expressed as 2 - k because ( 0.5)-1 = 2 [see Eq. (8.4)J. ( U)t o E xercise E S.l S ketch s ignals (a) ( l)k (b) ( _l)k (c) (0.5)k (d) ( -0.5)k ( e)(0.5)-k (f) 2 - k (g) ( _2)k. E xpress t hese e xponentials as "Ik, a nd p lot "I i n t he c omplex plane for each case . . Verify t hat "Ik d ecays e xponentially w ith k if "I lies inside t he u nit circle, a nd t hat "(k grows w ith k i f "I is outside t he u nit circle. I f "I is o n t he u nit circle, "Ik is c onstant o r oscillates w ith a c onstant a mplitude. Hint: ( l)k = 1 for all k. However, ( _l)k = 1 for even values of k a nd is - 1 for o dd values of k. T herefore, ( _ l)k s witches back a nd f orth from 1 t o - 1 (oscillates w ith a c onstant a mplitude). Note also t hat E q. (8.4) yields ( 0.5)-k = 2k 'V 2 3 4 5 6 k-- 0 1 2 (c) f::,. 3 4 5 6 k-- (d) F ig. S .6 d iscrete-time e xponentials " (k. E xercise E S.2 ( a) Show t hat (i) ( 025)-k - 4k C ) 4 -k k 2t (0.1353)t = ( 7.389)-t ( v) e3k = ( 20086)k 1(1.) - 1 ';k (0.25) (ii~...
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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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