20095ee113_1_HW4solutions

# 20095ee113_1_HW4solutions - P roblems f rom C hapter 9...

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Problems from Chapter #9 9.3. (a) It is useful to note that nu(n - 1) nu(n). The reason is that nu(n) = 0 at n = O. Since we know that ROC: Izi > lal Hence, z nu (n - 1) f-----> -----c;" ROC: Izi > 1 (z J (b) x(n) = u(n - For u{n 1): z 1 u(n1) ....... z 1 z - 1 1): Since nanu(n 1) nanu(n) o:z Izl> lal Izl > lal Izl> lal Therefore, 1 ROC: {izi > lal 10:1 > 1 X (z) = -z ---1 + --c---:'-::: Izl > 1 lal < 1 z , Izi > a z a z 1 (~) Hu(_n _ 1) ~ Izl <- z Q: O! Therefore, z z 1 . X(z) a < I z I < -, for a < 1 z-a z- a O! (d) In order to find the z-transform of the impulse response sequence, we take the z- transform of both sides with x(n) = 5(n) 3 1 H(z) - 4z-1 H{z) + Sz-2H(z) z-l (1 - ~z-l + ~z-2) H(z) z-l

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Therefore, z H(z) :}z+ 1 4 8 Where ROC: Izl > ~. 9.9. In order to evaluate the summations, we need to express it in terms of the standard z-transform sum 00 X(z) = L x(n)z-n n=-(X) (a) 00 L n 2 '/f(n 2)(3)-n n=-oo Now let x(n) n 2 'u(n - 2), then 00 L x(n)(3)-n x (z) IZ=3 n=-oc We then calculate X(z) as follows
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20095ee113_1_HW4solutions - P roblems f rom C hapter 9...

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