Rational Z-Transforms and Its Inverse Reference

Rational z-Transforms and Its Inverse Reference

Rational z-Transforms and Its Inverse Reference:

Professor Deepa Kundur Sections 3.3 and 3.4 of

University of Toronto John G. Proakis and Dimitris G. Manolakis, Digital Signal Processing: Principles, Algorithms, and Applications, 4th edition, 2007.

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Rational z-Transforms and Its Inverse 3.3 Rational z-Transforms Rational z-Transforms and Its Inverse 3.3 Rational z-Transforms Why Rational?

I X (z) is a rational function iﬀ it can be represented as the ratio of two polynomials in z −1 (or z):

Rational z-Transforms −1 −2 −M b0 + b1z + b2z + ··· + bM z X (z) = −1 −2 −N a0 + a1z + a2z + ··· + aN z

I For LTI systems that are represented by LCCDEs, the z-Transform of the unit sample response h(n), denoted H(z) = Z{h(n)}, is rational

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Let a0, b0 6= 0:

−1 −2 −M B(z) b0 + b1z + b2z + ··· + bM z X (z) = = −1 −2 −N A(z) a0 + a1z + a2z + ··· + aN z I zeros of X (z): values of z for which X (z) =0 −M M M−1 b0z z + (b1/b0)z + ··· + bM /b0 = −N N N−1 a0z z + (a1/a0)z + ··· + aN /a0 I poles of X (z): values of z for which X (z) = ∞ b (z − z )(z − z ) ··· (z − z ) = 0 z −M+N 1 2 M a0 (z − p1)(z − p2) ··· (z − pN ) QM (z − z ) = Gz N−M k=1 k QN k=1(z − pk )

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Rational z-Transforms and Its Inverse 3.3 Rational z-Transforms Rational z-Transforms and Its Inverse 3.3 Rational z-Transforms Poles and Zeros of the Rational z-Transform Poles and Zeros of the Rational z-Transform QM (z − zk ) X (z) = Gz N−M k=1 where G ≡ b0 QN a0 Example: k=1(z − pk ) 2z 2 − 2z + 1 X (z) = z I X (z) has M ﬁnite zeros at z = z1, z2,..., zM 16z 3 + 6z + 5 1 1 1 1 I X (z) has N ﬁnite poles at z = p1, p2,..., pN (z − ( 2 + j 2 ))(z − ( 2 − j 2 )) = (z − 0) 1 3 1 3 1 (z − ( 4 + j 4 ))(z − ( 4 − j 4 ))(z − (− 2 )) I For N − M 6= 0

I if N − M > 0, there are |N − M| zeros at origin, z = 0 1 3 1 I if N − M < 0, there are |N − M| poles at origin, z = 0 poles: z = 4 ± j 4 , − 2 zeros: z = 0, 1 ± j 1 Total number of zeros= Total number of poles 2 2

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unit 0.5 circle

0.5 -0.5 0.5 2z2 − 2z +1 X (z) = z =⇒ 16z3 +6 z +5 -0.5 0.5

-0.5 -0.5

POLE POLE ZERO ZERO

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Rational z-Transforms and Its Inverse 3.3 Rational z-Transforms Rational z-Transforms and Its Inverse 3.3 Rational z-Transforms Pole-Zero Plot and the ROC Pole-Zero Plot and the ROC

I Recall, for causal signals, the ROC will be the outer region of a I Therefore, for a causal signal the ROC is the smallest disk (origin-centered) circle encompassing all the poles.

0.5

I ROC cannot necessarily include poles ( X (pk ) = ∞) ∵ -0.5 0.5

-0.5

ROC

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I Recall, I For stable systems, the ROC will include the unit circle.

LTI system is ⇐⇒ P∞ |h(n)| < ∞ stable n=−∞

I Moreover, ∞ ∞ X −n X −n 0.5 |H(z)| = | h(n)z | ≤ |h(n)z | I For stability of a n=−∞ n=−∞ ∞ causal system, the X -0.5 0.5 = |h(n)| for |z| = 1 poles must lie inside n=−∞ the unit circle. -0.5 I It can be shown: ROC LTI system is ⇐⇒ P∞ |h(n)| < ∞ ⇐⇒ ROC of H(z) contains stable n=−∞ unit circle

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Rational z-Transforms and Its Inverse 3.3 Rational z-Transforms Rational z-Transforms and Its Inverse 3.3 Rational z-Transforms The System Function The System Function of LCCDEs

N M Z X X h(n) ←→ H(z) y(n) = − ak y(n − k) + bk x(n − k) time-domain ←→Z z-domain k=1 k=0 N M impulse response ←→Z system function X X Z{y(n)} = Z{− ak y(n − k) + bk x(n − k)} k=1 k=0 Z N M y(n) = x(n) ∗ h(n) ←→ Y (z) = X (z) · H(z) X X Z{y(n)} = − ak Z{y(n − k)} + bk Z{x(n − k)} k=1 k=0 N M Therefore, X −k X −k Y (z) Y (z) = − ak z Y (z)+ bk z X (z) H(z) = X (z) k=1 k=0

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N M X −k X −k Y (z) + ak z Y (z) = bk z X (z) k=1 k=0 " N # M Inversion of the z-Transform X −k X −k Y (z) 1 + ak z = X (z) bk z k=1 k=0 PM −k Y (z) k=0 bk z H(z) = = h i X (z) PN −k 1 + k=1 ak z LCCDE ←→ Rational System Function

Many signals of practical interest have a rational z-Transform.

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Rational z-Transforms and Its Inverse 3.4 Inversion of the z-Transform Rational z-Transforms and Its Inverse 3.4 Inversion of the z-Transform Inversion of the z-Transform Expansion into Power Series

Three popular methods: Example: 1. Contour integration: I 1 n−1 x(n) = X (z)z dz −1 2πj C X (z) = log(1 + az ), |z| > |a| ∞ ∞ X (−1)n+1anz −n X (−1)n+1an = = z −n −1 n n 2. Expansion into a power series in z or z : n=1 n=1 ∞ X −k X (z) = x(k)z By inspection: n+1 n k=−∞ (−1) a n ≥ 1 x(n) = n and obtaining x(k) for all k by inspection. 0 n ≤ 0

3. Partial-fraction expansion and table look-up.

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1. Find the distinct poles of X (z): p1, p2,..., pK and their Example: Find x(n) given poles of X (z) at p1 = −2 and a double corresponding multiplicities m1, m2,..., mK . pole at p2 = p3 = 1; speciﬁcally, 1 2. The partial-fraction expansion is of the form: X (z) = (1 + 2z −1)(1 − z −1)2 K A A A X (z) z 2 −R X 1k 2k mk = z X (z) = + 2 + ··· + m 2 z − pk (z − pk ) (z − pk ) k z (z + 2)(z − 1) k=1 z 2 A A A = 1 + 2 + 3 where pk is an mk th order pole (i.e., has multiplicity mk ) and R (z + 2)(z − 1)2 z + 2 z − 1 (z − 1)2 is selected to make z −R X (z) a strictly proper rational function. Note: we need a strictly proper rational function. DO NOT FORGET TO MULTIPLY BY z IN THE END. 3. Use an appropriate approach to compute {Aik }

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Rational z-Transforms and Its Inverse 3.4 Inversion of the z-Transform Rational z-Transforms and Its Inverse 3.4 Inversion of the z-Transform Partial-Fraction Expansion Partial-Fraction Expansion

z 2(z + 2) A (z + 2) A (z + 2) A (z + 2) z 2(z − 1)2 A (z − 1)2 A (z − 1)2 A (z − 1)2 = 1 + 2 + 3 = 1 + 2 + 3 (z + 2)(z − 1)2 z + 2 z − 1 (z − 1)2 (z + 2)(z − 1)2 z + 2 z − 1 (z − 1)2 2 2 2 z A2(z + 2) A3(z + 2) z A1(z − 1) 2 = A1 + + 2 = + A2(z − 1) + A3 (z − 1) z − 1 (z − 1) z=−2 (z + 2) z + 2 z=1 4 1 A = A3 = 1 9 3

Professor Deepa Kundur (University of Toronto) Rational z-Transforms and Its Inverse 23 / 26 Professor Deepa Kundur (University of Toronto) Rational z-Transforms and Its Inverse 24 / 26 Rational z-Transforms and Its Inverse 3.4 Inversion of the z-Transform Rational z-Transforms and Its Inverse 3.4 Inversion of the z-Transform Partial-Fraction Expansion Partial-Fraction Expansion Therefore, assuming causality, and using the following pairs: 1 z 2(z − 1)2 A (z − 1)2 A (z − 1)2 A (z − 1)2 anu(n) ←→Z 1 2 3 1 − az−1 2 = + + 2 (z + 2)(z − 1) z + 2 z − 1 (z − 1) az−1 nanu(n) ←→Z z 2 A (z − 1)2 (1 − az−1)2 = 1 + A (z − 1) + A (z + 2) z + 2 2 3 2 2 d z d A1(z − 1) −1 = + A2(z − 1) + A3 4 1 5 1 1 z dz (z + 2) dz z + 2 X (z) = + + z=1 9 1 + 2z −1 9 1 − z −1 3 (1 − z −1)2 5 A = 4 5 1 2 9 x(n) = (−2)nu(n)+ u(n)+ nu(n) 9 9 3 (−2)n+2 5 n = + + u(n) 9 9 3

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