ELC 4351: Digital Signal Processing Liang Dong Electrical and Computer Engineering Baylor University liang [email protected] February 23, 2017 Liang Dong (Baylor University) z-Transform Part 2 February 23, 2017 1 / 38 The z-Transform and Its Application to the Analysis of LTI Systems 1 Rational z-Transform 2 Inversion of the z-Transform 3 Analysis of LTI Systems in the z-Domain 4 Causality and Stability Liang Dong (Baylor University) z-Transform Part 2 February 23, 2017 2 / 38 Rational z-Transforms X (z) is a rational function, that is, a ratio of two polynomials in z−1 (or z). B(z) X (z) = A(z) −1 −M b0 + b1z + ··· + bM z = −1 −N a0 + a1z + ··· aN z PM b z−k = k=0 k PN −k k=0 ak z Liang Dong (Baylor University) z-Transform Part 2 February 23, 2017 3 / 38 Rational z-Transforms X (z) is a rational function, that is, a ratio of two polynomials B(z) and A(z). The polynomials can be expressed in factored forms. B(z) X (z) = A(z) b (z − z )(z − z ) ··· (z − z ) = 0 z−M+N 1 2 M a0 (z − p1)(z − p2) ··· (z − pN ) b QM (z − z ) = 0 zN−M k=1 k a QN 0 k=1(z − pk ) Liang Dong (Baylor University) z-Transform Part 2 February 23, 2017 4 / 38 Poles and Zeros The zeros of a z-transform X (z) are the values of z for which X (z) = 0. The poles of a z-transform X (z) are the values of z for which X (z) = 1. b QM (z − z ) X (z) = 0 zN−M k=1 k a QN 0 k=1(z − pk ) X (z) has M finite zeros at z = z1; z2;:::; zM , N finite poles at z = p1; p2;:::; pN , and jN − Mj zeros (if N > M) or poles (if N < M) at the origin. Poles and zeros may also occur at z = 1. X (z) has exactly the same number of poles and zeros. Liang Dong (Baylor University) z-Transform Part 2 February 23, 2017 5 / 38 Poles and Zeros If a polynomial has real coefficients, its roots are either real or occur in complex-conjugate pairs. That is because e.g., (z − p1)(z − p2) Liang Dong (Baylor University) z-Transform Part 2 February 23, 2017 6 / 38 Poles and Zeros For example, z−1 − z−2 X (z) = 1 − 1:2732z−1 + 0:81z−2 jπ=4 which has one zero at z = 1 and two poles at p1 = 0:9e and −jπ=4 p2 = 0:9e . Liang Dong (Baylor University) z-Transform Part 2 February 23, 2017 7 / 38 Some Common z-Transform Pairs Liang Dong (Baylor University) z-Transform Part 2 February 23, 2017 8 / 38 Poles Locations and Time-Domain Behavior for Causal Signals If a real signal has a z-transform with one pole, this pole has to be real. The only such signal is the real exponential 1 x(n) = anu(n) !z X (z) = ; ROC :jzj > jaj 1 − az−1 Liang Dong (Baylor University) z-Transform Part 2 February 23, 2017 9 / 38 Poles Locations and Time-Domain Behavior for Causal Signals A causal real signal with a double real pole has the form az−1 x(n) = nanu(n) !z X (z) = ; ROC :jzj > jaj (1 − az−1)2 Liang Dong (Baylor University) z-Transform Part 2 February 23, 2017 10 / 38 Poles Locations and Time-Domain Behavior for Causal Signals The case of a causal signal with a pair of complex-conjugate poles. Liang Dong (Baylor University) z-Transform Part 2 February 23, 2017 11 / 38 Poles Locations and Time-Domain Behavior for Causal Signals The case of a causal signal with a double pair of poles on the unit circle. Liang Dong (Baylor University) z-Transform Part 2 February 23, 2017 12 / 38 Poles Locations and Time-Domain Behavior for Causal Signals The impulse response h(n) of a causal LTI system is a causal signal. If a pole of a system is outside the unit circle, the impulse response of the system becomes unbounded and, consequently, the system is unstable. Liang Dong (Baylor University) z-Transform Part 2 February 23, 2017 13 / 38 System Function of a LTI System LTI systems: y(n) = h(n) ⊗ x(n) Y (z) = H(z)X (z) If we know the input x(n) and observe the output y(n) of the system, we can determine the unit sample response (impulse response) by first solving for H(z) from Y (z) H(z) = X (z) and then evaluating the inverse z-transform of H(z). H(z) is called the system function. Liang Dong (Baylor University) z-Transform Part 2 February 23, 2017 14 / 38 System Function of a LTI System When the LTI system is described by a linear constant-coefficient difference equation N M X X y(n) = − ak y(n − k) + bk x(n − k) k=1 k=0 The system function can be calculate: N M X −k X −k Y (z) = − ak Y (z)z + bk X (z)z k=1 k=0 N ! M ! X −k X −k Y (z) 1 + ak z = X (z) bk z k=1 k=0 Y (z) PM b z−k H(z) = = k=0 k X (z) PN −k 1 + k=1 ak z Liang Dong (Baylor University) z-Transform Part 2 February 23, 2017 15 / 38 System Function of a LTI System An LTI system described by a constant-coefficient difference equation has a rational system function H(z). PM b z−k H(z) = k=0 k PN −k 1 + k=1 ak z Liang Dong (Baylor University) z-Transform Part 2 February 23, 2017 16 / 38 System Function of a LTI System (1) All-zero system: If ak = 0 for 1 ≤ k ≤ N, M M X 1 X H(z) = b z−k = b zM−k k zM k k=0 k=0 The system has M nontrivial zeros and M trivial poles (at z = 0). An all-zero system is an FIR system and can be called a moving average (MA) system. Liang Dong (Baylor University) z-Transform Part 2 February 23, 2017 17 / 38 System Function of a LTI System (2) All-pole system: If bk = 0 for 1 ≤ k ≤ M, b b zN H(z) = 0 = 0 PN −k PM N−k 1 + k=1 ak z k=0 ak z where a0 = 1. The system has N nontrivial poles and N trivial zeros (at z = 0). An all-pole system is an IIR system and can be called an auto-regressive (AR) system. Liang Dong (Baylor University) z-Transform Part 2 February 23, 2017 18 / 38 System Function of a LTI System (3) Pole-zero system: In general, the system function contains N poles and M zeros. (Poles and zeros at z = 0 and z = 1 are implied but are not counted explicitly.) Due to the presence of poles, a pole-zero system is an IIR system. Liang Dong (Baylor University) z-Transform Part 2 February 23, 2017 19 / 38 Inversion of the z-Transform Y (z) H(z) = ; H(z) !inv z h(n) X (z) Inverse z-Transform: 1 I x(n) = X (z)zn−1dz 2πj C where the integral is a (counter-clockwise) contour integral over a closed path C that encloses the origin and lies within the region of convergence of X (z). Liang Dong (Baylor University) z-Transform Part 2 February 23, 2017 20 / 38 Methods of Inverse z-Transform (1) Contour integration (2) Power series expansion (using long division) (3) Partial-fraction expansion Liang Dong (Baylor University) z-Transform Part 2 February 23, 2017 21 / 38 Inverse z-Transform by Partial-Fraction Expansion X (z) is rational function. −1 −M B(z) b0 + b1z + ··· + bM z X (z) = = −1 −N A(z) 1 + a1z + ··· + aN z A rational function is proper if aN 6= 0 and M < N. Liang Dong (Baylor University) z-Transform Part 2 February 23, 2017 22 / 38 Inverse z-Transform by Partial-Fraction Expansion An improper rational function (M ≥ N) can always be written as the sum of a polynomial and a proper rational function. B(z) B (z) X (z) = = c + c z−1 + ··· + c z−(M−N) + 1 A(z) 0 1 M−N A(z) The inverse z-transform of the polynomial can easily be found by inspection. We focus our attention on the inversion of proper rational function. Liang Dong (Baylor University) z-Transform Part 2 February 23, 2017 23 / 38 Inverse z-Transform by Partial-Fraction Expansion Let X (z) be a proper rational function. −1 −M B(z) b0 + b1z + ··· + bM z X (z) = = −1 −N A(z) 1 + a1z + ··· + aN z N N−1 N−M b0z + b1z + ··· + bM z = N N−1 z + a1z + ··· + aN Since N > M, N−1 N−2 N−M−1 X (z) b0z + b1z + ··· + bM z = N N−1 z z + a1z + ··· + aN is proper. Liang Dong (Baylor University) z-Transform Part 2 February 23, 2017 24 / 38 Inverse z-Transform by Partial-Fraction Expansion (1) Distinct poles. Suppose that the poles p1; p2;:::; pN are all different. X (z) A A A = 1 + 2 + ··· + N z z − p1 z − p2 z − pN We want to determine the coefficients A1; A2;:::; AN . (z − pk )X (z) (z − pk )A1 (z − pk )AN = + ··· + Ak + ··· + z z − p1 z − pN Therefore, (z − pk )X (z) Ak = ; k = 1; 2;:::; N z z=pk ∗ ∗ (In addition, if p2 = p1, A2 = A1.) Liang Dong (Baylor University) z-Transform Part 2 February 23, 2017 25 / 38 Inverse z-Transform by Partial-Fraction Expansion (2) Multiple-order poles.