25 4 x z 11z 3 3z 2 1 k 3 z 1 z z 1 z

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Unformatted text preview: 3z 2 + 1 d z( z + 1) = dz 2 z =1 = z =1 11z 4 + 22z 3 + 3z 2 − 2z − 1 z 2 ( z + 1) = 2 z =1 33 = 8.25 4 X ( z) 11z 3 + 3z 2 + 1 K 3 = ( z − 1) = z( z + 1) z z =1 2 Lesson 31 = z =1 15 = 7 .5 2 Challenge 30 Obtain (in residue form) X ( z) = 1 + Inverse 1.75 z 8.25 z 7 .5 z + + ( z + 1) ( z − 1) ( z − 1) 2 x[k] = δ[k] + 1.75(-1)k u[k] + 8.75 u[k] + 7.5k u[k] Lesson 31 Mini Exam Given X(z) = z2/(z-1)2, what is x[k]? X(z)=a0 + a11 z/(z-1) + a12 z/(z-1)2 a0 = X(z)|z→0 = 0 a12 = (z-1)2X(z)|z→1 = 1 a11 = d((z-1)2X(z)/z)/dz|z→1 = dz/dz|z→1 = 1 Check: X(z)= 0 + z/(z-1) + z/(z-1)2 = z2/(z-1)2. Solution x[k] = 0 δ[k] + 1 u[k] + 1 ku[k] = (1+k)u[k] Lesson 31 Preview The Laplace transform was shown to be a viable modeling and analysis paradigm for linear continuous time signals and system. Its discrete-time counterpart is called the z-transform. It is used in a manner that is essentially identical to the application of Laplace transforms. Lesson 31 Example Issues ROC Lesson 31 z-Transforms See Table 5.1 Basic Signals (How are z-transforms actually performed?) Time-domain δ[k] δ[k–m] u[k] ku[k] k2u[k] k3u[k] aku[k] kaku[k] k2aku[k] sin[bk]u[k] cos[bk]u[k] z-transform 1 z–m z/(z–1) z/(z–1)2 z(z+1)/(z–1)3 z(z2+4z+1)/(z–1)4 z/(z–a) az/(z–a)2 az(z+a)/(z–a)3 zsin(b)/(z2-2zcos(b)+1) z(z-cos(b))/(z2-2zcos(b)+1) others …………….. Lesson 31 ROC everywhere everywhere |z|>1 |z|>1 |z|>1 |z|>1 |z|>|a| |z|>|a| |z|>|a| |z|>1 |z|>1 z-Transforms Simple and important primitive signals Z (u[k ]) = ∞ ∑ u [ k ]z k =0 k Z (a u[k ]) = −k = ∞ ∑ z −k = 1 + z −1 + z −2 + ... = k =0 ∞ ∑ a k z −k = 1 + az −1 + a 2 z −2 + ... = k =0 Lesson 31 z z−a z z −1 z-Transforms or, using the slackers friend » syms n a f F » f=1^n; % unit step » F= ztrans(f); » 'f= ', pretty(f) f= 'F= ', pretty(F) F= 1 z ----z-1 It’s not always this friendly Lesson 31...
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