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

# 1 04 1 f ig p ioi 6 1 01 7 a s ignal fkj is a

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Unformatted text preview: ~ derive t he p air o f e quations defining t he z -transform as s hows JlkJ as a s um of everlasting exponential c omponents Like t he Laplace transform, t he z -transform is a linear operator. If F[z]zk-t dz (11.2) t hen 668 (11.3) 672 11 Discrete- Time S ystems Analysis Using t he Z - Transform T he e xistence of t he z -transform is g uaranteed if IF[zJl ~ ~ 'f;~' &lt; 673 T he Z -Transform Therefore F[zJ = ~ [_Z_ + z - z_ ] _ jP e- 2 z - e jp 00 for s ome Izl. A~y s ignal f [k] t hat grows no faster t han a n e xponential signal r k for s ome ro, satIsfies t his condition. Thus, if 0, for some ro (11.8) z(z - cos ,8) z2 - 2z cos ,8 + 1 ( d) Here f[OJ = f[IJ = f[2J = f[3J according to E q. (11.9) = f[4J 1 t hen IF[zJl~~(ro)k t:o Izl 1 = 1- M Izl&gt; = 1 and 1 F[z] = 1 + :; + ~ 1 f[5] 1 = f[6] = . .. = O. Therefore, 1 + ;J + ? z4 + Z 3 + z 2 + z + 1 Izl &gt; ro z4 T herefore, F[z] e xists for Izl &gt; roo All practical signals satisfy (11.8) a nd a re t herefore z-t~ans~ormable. Some signal models (e.g. ' &quot;(k2) which grow faster t han t he e xponentIal sIgnal r ok (for any ro) d o n ot s atisfy (11.8) a nd t herefore are n ot z transformable. Fortunately, such signals a re o f little practical o r t heoretical interest. • 11.1 We can also express this result in a closed form by summing the geometric progression on the right-hand side of the above equation, using the formula in Sec. B.7-4. Here the common ratio r = ~, M = 0, and N = 4, so t hat E xample 1 1.2 F ind the z-transforms of ( a) a[kJ ( b) u[kJ (c) cos ,8ku[kJ ( d) signal shown in Fig. 11.2. Recall that by definition F[zJ = L JlkJz-k k=O = f[OJ + f[IJ + f[2J + f[3J + ... Z ( a) For f[kJ = a[kJ, f[OJ = 1 and f[2J a[kJ = z3 z2 = (11.9) f[3J = f[4J = . .. = O. Therefore for all z 1 (11.10) ( b) For f[kJ = u[kJ, f[OJ = f[IJ = f[3J = . .. = 1. Therefore F[zJ 1 1 1 = 1 + - + - 2 + - + ... Z z z3 1 1- - z z z -1 I~I &lt; 1 Izl &gt; 1 Therefore u [kJ=-z(11.11) Izl &gt; 1 z-1 ( c) Recall t hat cos,8k = (e jPk + e- jPk ) / 2. Moreover, according to Eq. (11.7), 3 o 4 456789 Fig. 11.3 Fig. 11.2 D. E xercise E ll.l ( a) Find the z-transform of a signal shown in Fig. 11.3. ( b) Using Pair 12a (Table 11.1) find the z-transform of f[k] = 20.65( y '2)k cos ( tk - 1.415) u[k]. A From E q. (11.6) it follows t hat F[zJ= - -1 2 ( nswers: ( ) F a z= ( b) z (3.2z + 17.2) z 2-2z+2 11.1-1 z5 + z4 + + z39 + z2 + z + 1 or z z ( -4 - 10) z -l z -z \l Finding the Inverse Transform As i n t he L aplace transform, we shall avoid t he i ntegration in t he c omplex plane required t o find t he inverse z-transform [Eq. (11.2)] b y using t he ( unilateral) t ransform T able. M any o f t he t ransforms F[z] o f practical i nterest a re r ational f unctions (ratiO o f polynomials in z ). Such f unctions c an b e e xpressed as a s um o f simpler functions using p artial f raction expansion. T his m ethod works because for every t ransformable f [k] defined for k 2: 0, t here is a corresponding unique F[z] defined for Izl &gt; ro (where ro is some c onstant), a nd vice versa. 674 11 D iscrete-Time S ystems A nalysis U sing t he Z -Transfo...
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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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