HW9 solution - It?').i' ( d) For x[nJ = ( 1/2)n+lu[n + 3J,...

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e UnIt CIrcle. This has four zeros at z = 0 and 8 more zeros distributed on a circle of radius 1/2. The ROC is the entire z plane. (Although from an inspection of the expression for X(z) it seems like there is a pole at 1/2, note that there is also a zero at 1/2 which cancels It?'). ..i' (d) For x[nJ = (1/2)n+lu[n + 3J, X(z) = L 00 x[n]z-n n=-oo = L 00 (1/2)n+l z-n n=-3 00 = I)I/2)n-2 z-n+3 n=O 3 = 4z /(1 - (1/2)z-1), Izl > 1/2 The Fourier transform exists because the ROC . cl d h " . _ _ In U es t --.--- 10.22. (a) Using the z-transform analysis equation, X(z) = (1/2)-4 z4 + (1/2)-3z3 + (1/2)-2z2 + (1/2)-l zl + (1/2)ozo + (1/2)1 z-1 + (1/2)2 z-2 + (1/2)3z-3 + (1/2)4 z-4 This may be expressed as X(z) = (1/2)-4 z 4 [1 - (1/2)9 Z -9] . 1 - (1/2)z-1 with this pole.) Since the ROC includes the unit circle, the Fourier transform exists. (d) The sequence may be written as ei[(21rn/6)+(1r/4)] + e-it(21rn/6)+(1r/4)1 } x[nJ = 4 n u[-n - IJ. { 2
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HW9 solution - It?').i' ( d) For x[nJ = ( 1/2)n+lu[n + 3J,...

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