EE480.3 Digital Control Systems. Part 2. z-transform

Transcription

1 EE480.3 Digital Control Systems Part 2. z-transform Kunio Takaya Electrical and Computer Engineering University of Saskatchewan January 14, 2008 ** Go to full-screen mode now by hitting CTRL-L 1

2 Contents 1 z-transform 3 2 z-transform of simple functions 3 3 Properties of the z-transform 8 4 Continue to discrete time systems 23 2

3 1 z-transform The definition of the z-transform is X(z) + k x(k)z k. where, x(k) is a discrete time sequence (sampled data). When x(k) is defined for k 0, i.e. causal, one sided z-transform is given by X(z) + x(k)z k. The variable z is complex, so is X(z). 3

4 2 z-transform of simple functions δ function δ(k) 1 if k 0 0 otherwise unit step function Z[u(k)] + Z[δ(k)] u(k) + δ(k)z k 1 1 if k 0 0 if k < 0 u(k)z k z 0 + z 1 + z 2 + +z 3 4

5 Since the sum of n terms of a geometric progression is given by the formula, n 1 Z[u(k)] lim n ar k a 1 rn 1 r 1 z n 1 z z 1. This z-trnasform is valid only for jzj < 1, in the region of convergence in the z-plane. Decaying exponential function f(t) e at t 0, sampled with a sampling period T f(k) e at k, k 0 5

6 Z[f(k)] e at k z k 1 1 e at z 1 (e at z 1 ) k The region of convergence is e at < jzj. If a > 0, the system is stable and the pole at z e at < 1 is inside the unit circle. If a < 0, the system is unstable and the pole at z e at > 1 is outside the unit circle. Damped cosine wave f(k) e at k cos ωt k, k 0 Z[f(k)] e at k cos ωt k z k 6

7 1 2 (e at +jωt z 1 ) k (e at jωt z 1 ) k e at +jωt z e at jωt z 1 1 e at cos ωt z 1 1 2e at cos ωt z 1 + e 2aT z 2 z 2 e at cos ωt z z 2 2e at cos ωt z + e 2aT The region of convergence is e at < jzj. Exercise: Find the z-transform of a discrete sequence f(k) 2 k for k 0 7

8 3 Properties of the z-transform Shift Operations - One step delay with u(k) Z[x(k 1)u(k)] x(k 1)z k k 1 x(k)z (k+1) x(k)z k 1 + x( 1)z 0 z 1 x(k)z k + x( 1) z 1 Z[x(k)u(k)] + x( 1) 8

9 Shift Operations - One step delay with u(k 1) Z[x(k 1)u(k 1)] x(k 1)z k k1 x(k)z (k+1) z 1 x(k)z k z 1 Z[x(k)u(k)] Shift Operations - m step delay with u(k) Z[x(k m)u(k)] x(k m)z k 9

10 k m x(k)z (k+m) x(k)z k m + z m x(k)z k + z m Z[x(k)u(k)] + Shift Operations - m step delay with u(k m) 1 k m x(k)z k m m x( k)z k m k1 m x( k)z k m k1 Z[x(k m)u(k m)] x(k m)z k km x(k)z (k+m) 10

11 z m Z[x(k)u(k)] Shift Operations - m step time advance with u(k + m) Z[x(k + m)u(k + m)] Initial Value theorem Final Value theorem k m x(k + m)z k x(k)z (k m) z +m Z[x(k)u(k)] + lim X(z) lim x(k)z k x(0) z z Z[x(k + 1)u(k)] Z[x(k)u(k)] 11

12 zx(z) zx(0) X(z) x(k + 1)z k x(k)z k zx(z) X(z) zx(0) + x(k + 1)z k lim (z 1)X(z) x(0) + z 1 x( ) x(k + 1) x(k)z k x(k) Derivative dx(z) dz d dz x(k)z k x(k) d dz z k 12

13 Example: x(k)kz k 1 z 1 Z[kx(k)] Z[kx(k)] z dx(z) dz Z[ku(k)] z d 1 dz 1 z 1 z 2 z (1 z 1 ) 2 z 1 (1 z 1 ) 2 z (z 1) 2 Convolution 13

14 Consider multiplying G(z) and X(z). G(z)X(z) g(k)z k x(k)z k [g(0) + g(1)z 1 + g(2)z 2 + g(3)z 3 + ] [x(0) + x(1)z 1 + x(2)z 2 + x(3)z 3 + ] g(0)x(0) + [g(0)x(1) + g(1)x(0)]z 1 + [g(0)x(2) + g(1)x(1) + g(2)x(0)]z 2 + [g(0)x(3) + g(1)x(2) + g(2)x(1) + g(32)x(0)]z 3 n + g(k)x(n k) z n + n0 n g(k)x(n k) z n 14

15 Z[ n g(k)x(n k)] Z[ n x(k)g(n k)] Input Sequence X(z) Impulse Response G(z) Output Sequence Y(z)G(z)X(z) Figure: Convolution and Transfer Function When an input sequence x(k) is applied to a system having an impulse response g(k), the response y(k) of the system is given the convolution sum. k y(k) x(n)g(k n) n0 The equivalent expression in the z-transform is Y (z) G(z)X(z). 15

16 Inverse z-transform Partial Fraction Method The z-transform of an exponential sequence x(k) a k is given by X(z) 1 1 az 1 z z a. Any sequence that starts with a non-zero value at k 0 usually has the same order in the numerator and denominator of the z-transform. To find the inverse z-transform, one must take partial fraction expansion on X(z) instead of X(z) itself. z X (z) X(z) z B(z) (z a 1 ) k (z a 2 )(z a 3 ) c 11 (z a 1 ) k + c 12 (z a 1 ) k c 1k (z a 1 ) 16

17 + c 2 (z a 2 ) + c 3 (z a 3 ) + Where, unknown coefficients are given by c 1n c i (z a i ) X (z)j zai 1 d n 1 (n 1)! dz n 1 (z a 1) k X (z) za1 After partial fraction expansion, find the expanded form of X(z) by multiplying X (z) by z, then find the inverse z-transform term-by-term by table look-up. Example E(z) E(z) z (z 1)(z 2) 1 z 1 z z 2 17

18 E(z) 1 2 z z z z 2 e(k) ( 1 2 δ(k) 1k k )u(k) Exercise: Find the inverse z-transform of Y (z) 1 (1 z 1 )(1 2z 1 ) Use of MATLAB for partial fraction expansion >> [r,p,k]residue([1],[1, -3, 2, 0]) r p

19 k 0 [] Difference Equation y(k) + a 1 y(k 1) + a 2 y(k 2) + + a k y(0) b 0 u(k) + b 1 u(k 1) + + b k u(0) Where, y(k) is output sequence, and u(k) is input sequence. In control systems, b 0 is often 0, as input u(k) does not immediately affect output y(k). Take the z-transform of this difference equation considering u(k) 0 and y(k) 0 for k < 0. Z f[y(k) + a 1 y(k 1) + a 2 y(k 2) + + a k y(0)]u(k)g Z f[b 0 u(k) + b 1 u(k 1) + + b k u(0))]u(k)g 19

20 Since Z[y(k i)] z i Y (z) and Z[u(k i)] z i U(z), Y (z) + a 1 z 1 Y (z) + a 2 z 2 Y (z) + + a k z k Y (z) b 0 U(z) + b 1 z 1 U(z) + b 2 z 2 U(z) + + b k z k U(z) This leads to a transfer function of the difference equation. G(z) Y (z) U(z) b 0 + b 1 z 1 + b 2 z b k z k 1 + a 1 z 1 + a 2 z a k z k The output Y (z) for a given input U(z) is given by Example Y (z) G(z)U(z) b 0 + b 1 z 1 + b 2 z b k z k 1 + a 1 z 1 + a 2 z 2 U(z) + + a k z k A difference equation, m(k) e(k) e(k 1) m(k 1) 20

21 is driven by an input sequence e(k) 1, k 0 even 0, k 0 odd Using the z-transform, M(z) z 1 z + 1 E(z) E(z) 1 z2 1 z 2 z 2 1 Thus, the solution in the z-transform is M(z) z 2 z 2 + 2z + 1 z 2 (z + 1) 2 1 2z 1 + 3z 2 4z

22 Assignment No. 2 Solve following problems related to the z-transform fundamentals in the textbook pp (a)(b) (c) (a)(e) Do only the partial fraction expansion method. 22

23 4 Continue to discrete time systems 23