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

# Thus if 0 for some ro 118 zz cos 8 z2 2z cos 8

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Unformatted text preview: re is explained below. r= 100=t~569'92 = 3.2, (3 = cos-'(~) = 0 .927rad., a nd () = tan-'C:_'sO) = - 2.246rad., s o t hat I [k] = [2 + 3 .2(5)k cos ( 0.927k - 2.246)] u[k] • 678 11 Discrete- Time S ystems A nalysis Using t he Z - Transform T he p rocedure for finding p artial fractions using MATLAB was d emonstrated i n c hapter 6. T he s ame p rogram c an b e used in t his case, except t hat we have t~ ~~d t he modified p artial f ractions here. T his goal is readily accomplished by dlVldm.g F[zJ by z a nd t hen t aking t he p artial fractions. We shall d emonstrate t his proced u re w ith a n example. o 11.1 f[3], . .. , a nd so on. A rational F[zJ c an b e e xpanded i nto a power series of z -I b y dividing i ts n umerator b y t he d enominator. Consider, for example, z 2(7z - 2) F[zJ = (z _ 0 .2)(z - 0.5)(z - 1) 7z 3 - 2z2 C omputer E xampleCll.l Solve Example 11.3a using MATLAB. = z3 - 1.7z 2 + 0.8z - 0.1 T o o btain a series expansion i n powers of z -I, we divide t he n umerator b y t he d enominator a s follows: n um=[S -19); d en=[conv([l -2),[1 -3]) 0); [r, p , k )= r esidue(num,den) i. We c ould a lso e xpress d en=[l - 5 6 0] 7 + 9 .9z- I + 1 1.23z- 2+11.87z- 3 + ... 2 + 0.8z - 0.1 )7z 3 - 2z2 z3 - 1.7z 7z 3 - 11.9z 2+ 5 .60z-0.7 r= 1 .6667 1 .5000 - 3.1667 9.9z 2 - 5.60z+0.7 9 .9z 2 -16.83z+ 7.92 - 0.99z- I 1 1.23z-7.22 + 0 .99z- I p= 3 2 o 1 1.23z-19.09 + 8 .98z- I 11.87 - 7.99z- I k= H ence, F[z] = -3.1667 + 1.5z z -2 + 1.6667z z -3 0 6 . E xercise E ll.2 z(2z - 1) ( a) (z _ 1)(z + 0.5) ( d) r + 1 1.23z- 2 + 1 1.87z- 3 + ... 1 (z - 1)(z + 0.5) flOJ = 7, f[lJ = 9.9, fl2J = 11.23, f[3J = 11.87, . .. , a nd so on. 5z(z - 1) z2 - 1.6z + 0.8 We give here a simple MATLAB p rogram t o find t he first N t erms o f t he inverse z-transform. ( b) -26[kJ + [~+ ~(-0.5)k] u[kJ o - U+ ~(-0.5)k] u[kJ ¥ (Js z 2(7z - 2) = 7 + 9 .9z- I (z - 0.2)(z - 0.5)(z - 1) (d) 9 ( c) 188[kJ - [0.72(-2)k F[zJ = ( b) (z + 2)(z - 0.5)2 Answer: Ca) T hus T herefore F ind t he inverse z- transform of the follOwing functions: ( c) 679 T he Z -Transform + 17.28(0.5)k -14.4k(0.5)kJu[kJ cos (O.464k + 0.464)u[kJ. Hint: v'if.8 = Js. \l Inverse Transform by Expansion o f F[zJ in Power Series o f z -I B y definition 00 F[zJ = L f[kJz- k k=O = f[OJ + f [IJ + z f[2J z2 + f[3J + ... z3 = f[OJzo + f [IJz-I + f[2Jz-2 + f [3Jz-3 + ... T his r esult is a power series in z -I. Therefore, if we c an e xpand F[zJ i nto a power series in z -I, t he coefficients of t his power series can be identified as f[OJ, f [IJ, f[2J, C omputer E xample C 11.2 Using MATLAB, find the first 10 values U[O] through f[9]) of the inverse z-transform of F[z] in t he above example. n um=[7 - 200); d en=[l - 1.7 O.S -0.1); f =dimpulse(num, d en, 1 0) i. We c ould a lso w rite d en=conv(conv([l - 0.2],[1 - 0.5]),[1 - 1]) f = 7 .0000 9 .9000 1 1.2300 11.8710 1 2.1867 1 2.3436 1 2.4218 1 2.4609 12.4805 1 2.4902 0 6~O 11 Discrete-Time Systems Analysis Using t he Z - Transform Although this procedure yields f [k] directly, it does n ot provide a closed-form solution. For this reason, i...
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