EL 625 Lecture 11. Frequency-domain analysis of discrete-time systems

Transcription

1 EL 625 Leture 11 1 EL 625 Leture 11 Frequeny-domain analysis of disrete-time systems Theorem: If S is a fixed linear disrete-time system, the zero-state response, S{z k } = H(z)z k (1) Proof: S{z k } = S{z k } z k }{{} H(z,k) S{z k k 0 } = H(z, k k 0 )z (k k 0).z k = H(z, k)z k (2) (3) = z k 0 S{z k } = z k 0 H(z, k)z k (4) = H(z, k) = H(z, k k 0 ) for any k 0 (5) Thus, H(z, k) is independent of k = H(z, k) = H(z). Thus, S{z k } = H(z)z k The response of a linear time-invariant disrete-time system to a omplex exponential input, z k is the same omplex exponential with a hange of omplex magnitude (magnitude and phase).

2 EL 625 Leture 11 2 z-transform: F (z) = Z{f(k)} = k=0 f(k)z k Inverse z-transform: f(k) = 1 2π C F (z)zk 1 dz Output, y(k) = S{u(k)} = 1 2π C U(z)S{zk 1 }dz (6) = 1 2π C U(z)H(z)zk 1 dz (7) Y (z) = H(z)U(z) H(z) is the disrete transfer funtion matrix. The element at the (i, j) position is the disrete transfer funtion from the j th input to the i th output. Properties of z-transform:

3 EL 625 Leture Linearity: Z{α 1 f 1 (k) + α 2 f 2 (k)} = α 1 Z{f 1 (k)} + α 2 Z{f 2 (k)} (8) 2. Convolution: f(k) = k j=0 3. Time translation: f 1 (k j)f 2 (j) = Z{f(k)} = Z{f 1 (k)}z{f 2 (k)}(9) Z{f(k + m)} = z m Z{f(k)} m 1 i=0 Z{f(k m)} = z m Z{f(k)} + m i=1 f(i)z i for m 0 f( i)z i for m 0 (10) 4. Initial value: 5. Final value: f(0) = lim z F (z) = lim z (1 z 1 )F (z) (11) lim f(n) = lim (1 N z 1 z 1 )F (z) (12) provided (1 z 1 )F (z) has no poles for z Time multipliation: df (z) Z{kf(k)} = z dz (13)

4 EL 625 Leture 11 4 Solution of state equations: Consider the fixed disrete-time linear system below with onstant A, B, C and D matries. x(k + 1) = Ax(k) + Bu(k) y(k) = Cx(k) + Du(k) (14) Taking the z-transform of both sides, Z{x(k + 1)} = Z{Ax(k) + Bu(k)} (15) z[z{x(k)} x(0)] = AZ{x(k)} + BZ{u(k)} (16) = (zi A)X(z) = zx(0) + BU(z) (17) where X(z) and U(z) are the z-transforms of x(k) and u(k) respetively. Thus, X(z) = z(zi A) 1 x(0) + (zi A) 1 BU(z) (18) = Y(z) = Z{y(k)} = CX(z) + DU(z) = Cz(zI A) 1 x(0) With zero input, + [C(zI A) 1 B + D] U(z) (19) }{{} H(z) X(z) = z(zi A) 1 x(0) (20) = Z{φ(k)} = Z{A k } = z(zi A) 1 (21)

5 EL 625 Leture 11 5 A system of differene equations an be solved similarly through the use of z-transforms. Sampled signals and sampled data systems: Many real physial systems are hybrid ommonly with an analog system ontrolled by a digital ontroller. Hybrid systems have both analog and disrete signals. The onversion between analog and digital signals is performed using A to D (analog to digital) and D to A (digital to analog) onverters. A to D e(t) e(kt ) or e(t) e(kt ) Ideal Sampler Figure 1: A to D onverter: ideal sampler m(kt ) D to A m a (t) m(kt ) m a (t) or f T (t) Hold Unit Figure 2: D to A onverter Zero Order Hold (ZOH): ZOH is a D to A onverter whih holds the output value onstant between sampling instants as shown in the figure below. m a (t) is the analog signal reonstruted by the zero order hold when the disrete signal, m(kt ) is passed through it.

6 EL 625 Leture 11 6 b b. b. m(kt ) b. b b. b. m a (t) b. 0 T 2T 3T 0 T 2T 3T t Figure 3: Zero Order Hold The zero-order hold is haraterized by the extrapolation funtion, f T (t) = 1 for 0 t < T (22) 0 for all other t 1 f T (t) 0 T Figure 4: Extrapolation funtion for a zero order hold. m a (t) = k= m(kt )f T (t kt ) (23) = M a (s) = m(kt )F T (s)e kt s k= = F T (s) m(kt )e kt s (24) Here, F T (s) is the laplae transform of the extrapolation funtion, f T (t).

7 EL 625 Leture 11 7 F T (s) = L{f T (t)} = 1 e st s (25) The fator, m(kt )e kt s is the z-transform of m(kt ) evaluated at e st. m(kt )e kt s = M(z) z=e st M (s) (26) m (t) = L 1 {M (s)} = = = m(t) m(kt )L 1 {e kt s } m(kt )δ(t kt ) (27) = m(t) δ(t kt ) (28) δ T (t) (29) δ T (t) is an impulse train. m (t) whih is a produt of m(t) and an impulse train is alled an impulse-modulated signal. δ T (t) = = 1 T = m (t) = 1 T = M (s) = 1 T = 1 T δ(t kt ) (30) n= n= n= n= e j2πn T t (31) m(t)e j2πn T t (32) L{m(t)e j2πn T } M(s + j2πn T ) (33)

8 EL 625 Leture 11 8 Hene, M (s) is periodi in the frequeny domain. To obtain the Z-transform from the L-transform: It an be proved that M (s) = M(z) z = e st (34) = M(z) = M (s) e st = z (35) M (s) = M(w) residues of 1 e st e wt at the poles of M(w) (36) = M(z) = M(s) residues of 1 z 1 e st at the poles of M(s) (37) provided lim s sf (s) is finite for all s with Re{s} < Important properties of Impulse modulated signals: {G(s)U (s)} = G (s)u (s) (38) where {} is the impulse-modulation operation. Proof: Let Y (s) = G(s)U (s). Y (s) = 1 T = 1 T n= n= Y (s + j2πn T ) (39) G(s + j2πn T )U (s + j2πn T ) (40)

9 EL 625 Leture 11 9 But, sine U (s) is an impulse modulated signal, Hene, U (s) = U (s + j2πn ) for all n. (41) T Y (s) = 1 G(s + j2πn T n= T )U (s) (42) = G (s)u (s) (43) {G(s)U (s)} = G (s)u (s) 2. (GU) (s) = {G(s)U(s)} = G (s)u (s) From the relations between M(z) and M (s) derived previously in this leture, the operation of sampling and holding is equivalent to the operation of impulse modulating and filtering. m(t) m(kt ) m A to D a (t) D to A M(s) M(z) M a (s) Figure 5: Sampling and holding. The operation of sampling, Z-transforming and holding is equiva-

10 EL 625 Leture m(t) M(s) IM m (t) M (s) m F T (s) a (t) M a (s) Figure 6: Impulse modulating and filtering lent to the operation of impulse modulating, transforming and filtering. e(t) e(kt ) m(kt ) m A to D D(z) a (t) D to A E(s) E(z) M(z) M a (s) Figure 7: Sampling, Z transforming and holding. e(t) E(s) IM e (t) D (s) E (s) m (t) 1 e st m a (t) M (s) s M a (s) Figure 8: Impulse modulating, transforming and filtering

11 EL 625 Leture Example: G 2 (s)x(s) IM G 3 (s) R(s) m + G 1 (s) E(s) H(s) + m C(z) C(s) C(s) = G 3 (s)x (s) + G 1 (s)e(s) (44) E(s) = R(s) H(s)C(s) (45) = R(s) H(s){G 3 (s)x (s) + G 1 (s)e(s)} (46) = E(s) = R(s) H(s)G 3(s)X (s) 1 + H(s)G 1 (s) (47) = A(s)R(s) A(s)H(s)G 3 (s)x (s) (48) where A(s) = H(s)G 1 (s) (49) X(s) = G 2 (s)e(s) (50) = G 2 (s){a(s)r(s) A(s)H(s)G 3 (s)x (s)} (51) = X (s) = (G 2 AR) (s) (G 2 AHG 3 ) (s)x (s) (52) = X (G 2 AR) (s) (s) = (53) 1 + (G 2 AHG 3 ) (s) = C(s) = G 3 (s)x (s) + G 1 (s)e(s) (G 2 AR) (s) = G 3 (s) 1 + (G 2 AHG 3 ) (s) +G 1 (s){a(s)r(s) A(s)H(s)G 3 (s)x (s)}(54)

12 EL 625 Leture C (s) = G (G 2 AR) (s) (s) 1 + (G 2 AHG 3 ) (s) +(G 1 AR) (s) (G 1 AHG 3 ) (s)x (s) (55) G 2 AR(z) C(z) = G(z) 1 + G 2 AHG 3 (z) +G 1 AR(z) G 1 AHG 3 (z)x(z) (56)