ECEN 420 LINEAR CONTROL SYSTEMS. Lecture 6 Mathematical Representation of Physical Systems II 1/67


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1 1/67 ECEN 420 LINEAR CONTROL SYSTEMS Lecture 6 Mathematical Representation of Physical Systems II
2 State Variable Models for Dynamic Systems u 1 u 2 u ṙ. Internal Variables x 1, x 2 x n y 1 y 2. y m Figure 1: A conceptual model The general model of Figure 1 is applicable to dynamic systems as well. The only difference is that time derivative of the internal variables x i (t) also enter into the mathematical description of the behavior of the system. For linear time invariant (LTI) systems it 2/67
3 State Variable Models for Dynamic Systems (cont.) will be possible, in many cases, to write the system equations in the standard form ẋ(t) = A x(t) + B u(t) y(t) = C x(t) + D u(t) (1a) (1b) where the matrices A, B, C, D have appropriate dimensions and y, u and x denote the output, input and state vectors respectively. Note that the static model from the previous lecture is a special case of (1), when ẋ(t) = 0. In general the internal or state variables should be chosen to provide a complete description of the system. We illustrate the use of the model (1) with examples in the next section. 3/67
4 Solution of State Space Equations X (s) = (s I A) 1 x(0 ) + (s I A) 1 B U(s). (2) }{{}}{{} X o(s) X u(s) Y (s) = C (s I A) 1 x(0 ) }{{} Y o(s) [ + The transfer function matrix is defined to be ] C (s I A) 1 B + D U(s). (3) } {{ } Y u(s) G(s) := C(sI A) 1 B + D (4) so that With Y u (s) = G(s)U(s). (5) y 0 (t) := L 1 {Y 0 (s)} y u (t) := L 1 {Y u (s)} (6a) (6b) 4/67
5 Solution of State Space Equations (cont.) the total response is y(t) = y 0 (t) + y u (t) (7) that is, it is the sum of the zero input response y 0 (t) and the zero state response y u (t). The transfer function matrix G(s) is m n, with entries g ij (s) = c i (si A) 1 b j + d i j (8) which are proper rational functions with denominator (before any cancellations), si A = s n + a n 1 s n a 1 s + a 0 (9) the characteristic polynomial of A. 5/67
6 Solution of State Space Equations (cont.) Example Consider the state space model ẋ(t) = x(t) + 0 u(t) a 0 a 1 a 2 1 y(t) = [ c 0 c 1 c 2 ] x(t) + du(t). (10) Find the transfer function g(s). 6/67
7 Solution of State Space Equations (cont.) Example (cont.) Solution. 1 s 1 0 (si A) 1 = 0 s 1 Adj[sI A] = si A a 0 a 1 s + a 2 s 2 + a 2 s + a 1 s + a 2 1 a 0 s 2 + a 2 s s sa 0 a 1 s a 0 s 2 = s 3 + a 2 s 2. + a 1 s + a 0 (11) 7/67
8 Solution of State Space Equations (cont.) Example (cont.) Therefore c(si A) 1 b + d = [ ] s 2 + a 2 s + a 1 s + a c 0 c 1 c 2 a 0 s 2 + a 2 s s 0 + d sa 0 a 1 s a 0 s 2 1 = c 0 + c 1 s + c 2 s 2 s 3 + a 2 s 2 + a 1 s + a 0 + d. (12) 8/67
9 Examples of State Space Models R L C circuits Inductors and capacitors have dynamic voltagecurrent relationships as shown below: i L L + v L Figure 2: v L = L d i L d t C + v C Figure 3: i C = C d v C d t It is possible to show, based on the above vi relationships that an arbitrary R L C circuit can be completely described by choosing all inductor currents and capacitor voltages as components of the state vector x. i C 9/67
10 Example (R L C circuits) Consider the series R L C circuit: + u R 1 L x 1 + C x 2 + y Figure 4: An R L C circuit 10/67
11 Example (R L C circuits (cont.)) Choosing x 1 and x 2 as the inductor current and capacitor voltage respectively we see that the vector (x 1 (t), x 2 (t)) provides a complete description of all currents and voltages in the circuit. The system equations are obtained from Kirchhoff s laws: x 1 = C ẋ 2 (current summation) u = R x 1 + L ẋ 1 + x 2 (voltage summation) y = x 2 (output). (13) 11/67
12 Example (R L C circuits (cont.)) These can be placed in the standard form (1): ] [ [ẋ1 R = L 1 ] [ ] [ 1 ] L x1 1 + L u ẋ 2 C 0 x 2 0 }{{}}{{} A B y = [ 0 1 ] [ ] x1 + [ 0 ] u. }{{} x 2 }{{} C D (14) 12/67
13 Example (Two input Two output R L C circuit) L 1 x 1 R 1 L 2 x 2 + x 3 y 2 C + u 1 + R 3 y 1 R 2 + u 2 Figure 5: 2 input 2 output R L C circuit 13/67
14 Example (Two input Two output R L C circuit (cont.)) Choosing inductor currents x 1, x 2, and capacitor voltage x 3 as state variables we can write Kirchhoff s equations: current summation x 1 = x 2 + C x 3 voltage summation u 1 = L 1 ẋ 1 + R 1 x 1 + x 3 + R 3 C ẋ 3 + u 2 L 2 ẋ 2 + R 2 x 2 = x 3 + R 3 C x 3 + u 2 y 1 = x 3 + R 3 C x 3 + u 2 y 2 = x 1 x 2. (15a) (15b) (15c) (15d) (15e) 14/67
15 Example (Two input Two output R L C circuit (cont.)) Equation (15) can be rewritten in the standard state variable form: ẋ 1 (R 1+R 3 R 3 L ẋ 2 1 L L 1 x 1 = R 3 L 2 (R L 2+R 3 ) [ ] L 1 x L 2 L u1 2 L 2 u ẋ 3 1 C 1 C 0 x }{{}}{{} A B [ ] [ ] x y1 R3 R = [ ] [ ] x y u u }{{} x 2 3 }{{} C D (16) 15/67
16 Example Consider the system, + R 1 R 2 u(t) + + x 1 L x 2 C y(t) Let x 1 be the current of the inductor L and let x 2 be the voltage across the capacitor C. 16/67
17 Example (cont.) Then, u = R 1 x 1 + L ẋ 1 (17) u = R 2 C ẋ 2 + x 2 (18) y = x 2. (19) Thus, ] [ẋ1 = ẋ 2 [ R 1 L y = [ 0 1 ] [ x 1 x 2 R 2 C ] [ ] x1 x 2 [ 1 ] + L 1 u (20) R 2 C ] + [ 0 ] u. (21) 17/67
18 Example (cont.) The transfer function is G(s) = [ 0 1 ] [ s + R L 0 0 s + 1 = 1 R 2 C s + 1. R 2 C R 2 C ] 1 [ 1 ] L R 2 C (22) 18/67
19 Example Consider the system, + u(t) R 1 i 1 + C R 2 i 2 x 2 + L x 1 i 3 R 3 y(t) 19/67
20 Example (cont.) From Kirchhoff current law, we have i 1 = i 2 + x 1 (node on the left) (23) i 2 + C ẋ 2 = i 3 (node on the right) (24) x 1 + i 3 = i 1 + C ẋ 2 (current from/to u(t)). (25) Rewriting above with respect to i 1, i 2 and i 3, we get i 1 = u L ẋ 1 R 1 (26) i 2 = u L ẋ 1 R 1 x 1 (27) i 3 = u L ẋ 1 R 1 x 1 + C ẋ 2. (28) 20/67
21 Example (cont.) From Kirchhoff voltage law, we have ( ) u L ẋ1 u L ẋ 1 + R 2 x 1 x 2 = 0 (Top loop) (29) R 1 ( ) ( ) u L ẋ1 u L ẋ1 L ẋ 1 = R 2 x 1 + R 3 x 1 + Cẋ 2 (30) R 1 R 1 (Bottom loop). Thus, ( ) u L ẋ1 u L ẋ 1 + R 2 x 1 R 1 x 2 = 0 (31) u + R 2 u R 2 x 1 x 2 = L ẋ 1 + R 2 L ẋ 1 (32) R 1 R ( ( 1 L 1 + R )) ( 2 ẋ 1 = R 2 x 1 x R ) 2 u (33) R 1 R 1 21/67
22 Example (cont.) ẋ 1 = R 2 L(R 1 +R 2 ) R 1 x 1 = R 1 R 2 L(R 1 + R 2 ) x 1 R 1 L(R 1 + R 2 ) x L u R 1 L(R 1 + R 2 ) x L u (35) R 1 L ẋ 1 = R 2 u R 2 L ẋ 1 R 1 R 2 x 1 + R 3 u R 3 L ẋ 1 R 1 R 3 x 1 + R 1 R 3 C ẋ 2 (36) R 1 (R 2 + R 3 )x 1 (R 2 + R 3 )u + L(R 1 + R 2 + R 3 )ẋ 1 = R 1 R 3 C ẋ 2 (37) 22/67
23 Example (cont.) { R1 R 2 R 1 R 3 C ẋ 2 = (R 1 + R 2 + R 3 ) x 1 R } 1 x 2 + u R 1 + R 2 R 1 + R 2 + R 1 (R 2 + R 3 )x 1 (R 2 + R 3 )u (38) ( ) R2 (R 1 + R 2 + R 3 ) R 3 C ẋ 2 = + (R 2 + R 3 ) x 1 R 1 + R 2 + (R 1 + R 2 + R 3 ) R 1 + R 2 x 2 + u (39) R 3 C ẋ 2 = R 1 R 3 R 1 + R 2 x 1 R 1 + R 2 + R 3 ẋ 2 = R 1 C(R 1 + R 2 ) x 1 x 2 + u (40) R 1 + R ( 2 ) 1 R 3 C + 1 x 2 + u C(R 1 + R 2 ) (41) 23/67
24 Example (cont.) ( ) u L ẋ1 y = R 3 i 3 = R 3 x 1 + C ẋ 2 (42) R 1 = R 3 x 1 + R 3 u R ( 3 R1 R 2 L R 1 R 1 L(R 1 + R 2 ) x R 1 1 L(R 1 + R 2 ) x ) L u { ( ) R R 3 C C(R 1 + R 2 ) x 1 R 3 C + 1 x } C(R 1 + R 2 ) R 3 C u (43) ( = R 3 + R 2 R 3 + R ) ( 1 R 3 R3 x R ) 3 x 2 R 1 + R 2 R 1 + R 2 R 1 + R 2 R 1 + R 2 ( R3 + R ) u (44) R 1 R 1 = x 2 + u. (45) 24/67
25 Example (cont.) (45) can also be obtained from the fact that The state variable equations are ] [ [ẋ1 R 1 R 2 = ẋ 2 u = x 2 + R 3 i 3. (46) L(R 1 +R 2 ) R 1 ( L(R 1 +R 2 ) R 1 C(R 1 +R 2 ) 1 R 3 C + 1 C(R 1 +R 2 ) [x1 ] )] x 2 [ 1 ] + L 1 u R 3 C (47) y = [ 0 1 ] [ ] x 1 + [ 1 ] u. (48) x 2 25/67
26 Example (cont.) The transfer function is G(s) = [ 0 1 ] [ R1 R2 s + L(R 1+R 2) ( R1 C(R 1+R 2) s + R 1 L(R 1+R 2) 1 R 3 C + 1 C(R 1+R 2) ) ] 1 [ 1 L 1 R 3 C ] + 1 (49) 26/67
27 Example Consider the system, + L C u(t) R + y(t) 27/67
28 Example (cont.) By Kirchhoff current law, we have By Kirchhoff voltage law, we have The state variable form is ] [ [ẋ1 R 1 = L ẋ 2 x 1 = C ẋ 2. (50) u = L ẋ 1 + x 2 + R x 1. (51) 1 L 1 L 0 y = [ R 0 ] [ x 1 x 2 ] [ ] x1 x 2 + [ 1 L 0 ] u (52) ] + [ 0 ] u. (53) 28/67
29 Example (cont.) The transfer function is G(s) = [ R = = 0 ] [ s + R L 1 L [ R 0 ] [ s + R L 1 L s 2 + R L s + 1 R L s s 2 + R L s + 1. L C ] 1 1 [ 1 ] L L + 0 s 0 ] 1 1 [ 1 ] L L s 0 L C (54) 29/67
30 Mechanical Systems The common elements of mechanical systems are masses, springs and dampers. If a linear spring is stretched (compressed) from its relaxed position by a certain amount, it produces a restoring force K x in the opposite direction. A linear damper as shown below produces a force B ẋ opposing the motion, ẋ B ẋ Figure 6: A linear damper (dashpot) 30/67
31 Mechanical Systems with ẋ being the relative velocity of the plunger with respect to the housing. The main law of physics governing the behavior of mechanical systems is Newton s equation of motion which can be stated as follows, for linear motion: The algebraic sum of forces acting on a mass m, in the positive x direction = m ẍ, where ẍ is the acceleration of m in the positive x direction. A similar law holds for rotational motion with force, mass and linear acceleration replaced by torque, moment of inertia and angular acceleration, respectively. 31/67
32 Mechanical Systems Example (Two input, two output mechanical system) Consider the mechanical system below: u 1 u 2 K m 1 B m 2 x 1 = y 1 x 2 = y 2 Figure 7: 2 input 2 output mechanical system I 32/67
33 Mechanical Systems Example (Two input, two output mechanical system (cont.)) where x 1, x 2 denote the displacements from the equilibrium positions of the masses m 1 and m 2 respectively. A free body diagram of each mass is shown below: K x 1 u 1 B(ẋ 1 ẋ 2 ) u 1 m 1 m 2 B(ẋ 1 ẋ 2 ) x 1 x 2 Figure 8: Free Body Diagrams 33/67
34 Mechanical Systems Example (Two input, two output mechanical system (cont.)) From Figure 8, Newton s equations for the two masses are: u 1 K x 1 B(ẋ 1 ẋ 2 ) = m 1 ẍ 1 u 2 + B(ẋ 1 ẋ 2 ) = m 2 ẍ 2. (55a) (55b) To rewrite (55) in the standard first order state variable form we introduce two additional variables x 3 := ẋ 1 x 4 := ẋ 2. (56) 34/67
35 Mechanical Systems Example (Two input, two output mechanical system (cont.)) We may now write the system equations in the standard first order state variable form, ẋ x [ ] ẋ 2 ẋ 3 = x 2 K /m 1 0 B /m 1 B/m 1 x u1 1/m 1 0 u 2 ẋ B/m 2 B /m 2 x 4 0 1/m 2 }{{}}{{} A B [ ] [ ] x 1 [ ] [ ] y = x y x 3 + u1 0 0 u 2 }{{} x }{{} C 4 D (57) with matrices {A, B, C, D} as shown. 35/67
36 Mechanical Systems Example Consider the system in Figure 9, u 1 u 2 K 1 K 2 m 1 m 2 x 1 = y 1 x 2 = y 2 Figure 9: 2 input 2 output mechanical system II 36/67
37 Mechanical Systems Example (cont.) The free body diagrams for m 1 and m 2 are: K 1 x 1 u 1 K 2 (x 1 x 2 ) u 1 m 1 m 2 K 2 (x 1 x 2 ) x 1 Figure 10: The free body diagram for m 1 x 2 Figure 11: The free body diagram for m 2 37/67
38 Mechanical Systems Example (cont.) Newton s equation of motion are: u 1 K 1 x 1 K 2 (x 1 x 2 ) = m 1 ẍ 1 (58) u 2 + K 2 (x 1 x 2 ) = m 2 ẍ 2 (59) They can be cast in the standard state variable form: ẋ 1 =: x 3 (60) ẋ 2 =: x 4 (61) ẋ 3 = K 1 + K 2 m 1 x 1 + K 2 m 1 x m 1 u 1 (62) ẋ 4 = K 2 m 2 x 1 K 2 m 2 x m 2 u 2 (63) 38/67
39 Mechanical Systems Example (cont.) ẋ 1 ẋ 2 ẋ 3 = ẋ 4 [ y1 y 2 ] = x K 2 x 2 m 1 m x 3 + m x 4 ] x 1 [ ] [ ] x x 3 + u1 0 0 u 2 x 4 K 1+K 2 K 2 m 2 K 2 [ m m 2 [ u1 u 2 ] (64) 39/67
40 Mechanical Systems Example Consider the system in Figure 12, u 1 u 2 B 1 B 2 m 1 m 2 x 1 = y 1 x 2 = y 2 Figure 12: 2 input 2 output mechanical system III 40/67
41 Mechanical Systems Example (cont.) The free body diagrams and Newton s equations are: B 1 ẋ 1 u 1 m 1 B 2 (ẋ 1 ẋ 2 ) x 1 Figure 13: The free body diagram for m 1 u 1 B 1 ẋ 1 B 2 (ẋ 1 ẋ 2 ) = m 1 ẍ 1 (65) 41/67
42 Mechanical Systems Example (cont.) B 2 (ẋ 1 ẋ 2 ) u 1 m 2 x 2 Figure 14: The free body diagram for m 2 u 2 + B 2 (ẋ 1 ẋ 2 ) = m 2 ẍ 2 (66) 42/67
43 Mechanical Systems Example (cont.) The standard state variable form is: ẋ 1 =: x 3 (67) ẋ 2 =: x 4 (68) ẋ 3 = B 1 + B 2 m 1 x 3 + B 2 m 1 x m 1 u 1 (69) ẋ 4 = B 2 m 2 x 3 B 2 m 2 x m 2 u 2 (70) 43/67
44 Mechanical Systems Example (cont.) ẋ x 1 ẋ 2 ẋ 3 = B 1+B 2 B 2 x 2 m 1 m 1 x 3 + ẋ B m 2 B 2 x m 2 4 [ ] [ ] x 1 [ ] [ ] y = x y x 3 + u1 0 0 u 2 x m m 2 [ u1 u 2 ] (71) 44/67
45 Electromechanical System An armature controlled, separately excited DC motor is shown below driving a rotational load with a resistive load torque T L. + u R i d L d a + Field + e b (t) ω J B T m, θ T L Figure 15: Separately excited DC motor 45/67
46 Electromechanical System The armature resistance, inductance and current are denoted R a, L a and i a, e b denotes the back electromotive force (emf), T m the motor torque, θ and ω the motor shaft angular position and velocity and J and B the moment of inertia of the load (reflected to the motor drive shaft) and the viscous damping constant seen by the load. The motor equations are T m = K t i a e b = K b ω (72a) (72b) where K t, K b are constants provided by the manufacturer. We write the system equations by summing voltages around the loop (electrical balance) and sum of mechanical torques resulting in acceleration (mechanical balance). 46/67
47 Electromechanical System Electrical Balance Mechanical Balance u = R a i a + L a i a + e b. (73) T m = T L + B θ + J θ. (74) Assuming that angular position and velocity are the outputs of interest we have y 1 = θ y 2 = ω := θ. (75a) (75b) 47/67
48 Electromechanical System We may now choose x 1 = θ 1, x 2 = ω and x 3 = i a as the state variables to obtain the DC motor model in the standard form: ẋ x [ ] ẋ 2 = 0 B /J K t /J x /J u T ẋ 3 0 Kb /L a Ra /L a x 3 1/L a 0 L }{{}}{{} A B (76) [ ] [ ] x y [ ] [ ] = x y u T }{{} x L 3 }{{} C D 48/67
49 Example (Inverted Pendulum on a Cart) Consider the inverted pendulum below: y l sin θ x m l cos θ θ mg l 0 u M x Figure 16: Inverted Pendulum 49/67
50 Example (Inverted Pendulum on a Cart (cont.)) There are two balance equations; 1) force balance and 2) torque balance. Firstly we consider the force balance. Since the force in the horizon is the force of the cart and the force of the bob, we have M d 2 d t 2 x + m d 2 d t 2 x m = u (77) where the coordinate of the point mass, bob, is given by (x m, y m ). Using the reference coordinate, length of massless rod and the angle we have x m = x + l sin θ y m = l cos θ. (78) 50/67
51 Example (Inverted Pendulum on a Cart (cont.)) Thus, we can rewrite (77) as Noting that M d 2 d t 2 x + m d 2 (x + l sin θ) = u (79) d t2 d 2 d t 2 sin θ = (sin θ)( θ) 2 + (cos θ) θ we get (M + m) ẍ m l (sin θ)( θ) 2 + m l (cos θ) θ = u. (80) 51/67
52 Example (Inverted Pendulum on a Cart (cont.)) Secondly, for the torque balance, refer to Figure 17 for the notation of directions. y F ym m F xm θ mg Figure 17: Torque directions x 52/67
53 Example (Inverted Pendulum on a Cart (cont.)) Since the torque due to the acceleration force is balanced by the torque due to the gravity force we have F xm cos θ F ym sin θ = m g sin θ (81) where the forces are F xm = m d 2 d t 2 x m = m F ym = m d 2 d t 2 y m = m ( ) ẍ l (sin θ)( θ) 2 + l (cos θ) θ, ( l (cos θ)( θ) ) 2 + l (sin θ) θ. (82) 53/67
54 Example (Inverted Pendulum on a Cart (cont.)) Substituting (81) in (82) gives m g sin θ = F xm cos θ F ym sin θ { } = m ẍ l (sin θ)( θ) 2 + l (cos θ) θ cos θ { m l (cos θ)( θ) } 2 + l (sin θ) θ sin θ = m ẍ cos θ + m l θ. (83) 54/67
55 Example (Inverted Pendulum on a Cart (cont.)) Using small signal approximation about θ = 0, we have cos θ = 1, sin θ = θ, and θ 2 = 0; thus the inverted pendulum equations become (M + m)ẍ + m l θ = u (84) m ẍ + m l θ = m g θ. (85) Now we define the state vector as θ z = θ x. (86) ẋ 55/67
56 Example (Inverted Pendulum on a Cart (cont.)) Since ż has elements ẍ and θ in the vector we derive these expressions using (84) and (85). First, from (85), we have and substituting this into (84), m l θ = m g θ m ẍ (87) (M + m)ẍ + m l θ = u (M + m)ẍ + m g θ m ẍ = u M ẍ + m g θ = u ẍ = m M g θ + 1 u. (88) M 56/67
57 Example (Inverted Pendulum on a Cart (cont.)) Next, substituting (88) back into (87) we get m l θ = m g θ m ẍ { m l θ = g θ + M g θ 1 } M u θ = g θ + 1 { m l l M g θ 1 } M u θ = g ( 1 + m ) θ 1 u. (89) l M l M Now fit these to the state space representation of the form: ż = A z + B u (90) y = C z + D u. (91) 57/67
58 Example (Inverted Pendulum on a Cart (cont.)) Then, the final state space representation is ż z 1 0 ż 2 ż 3 = g l (1 + m M ) z z 3 + ż 4 g θ( m M } ) {{ z 4 } A [ ] [ ] z 1 y = z 2 y z 3 + [ 0 0 ] u }{{}}{{} z C 4 D 1 /l M 0 1/M } {{ } B u (92) where ż 1 ż 2 ż 3 ż 4 θ = ż = θ ẋ ẍ and z 1 z 2 z 3 z 4 θ = z = θ x. (93) ẋ 58/67
59 Exercise Exercise 1 y + 2 R 1 + R 2 u 1 + R 3 y 1 + u 2 59/67
60 Exercise (cont.) a) Write equations for the circuit in the matrix form : Ax + Bu = 0, Cx + Du = y. (94) b) Solve for the gain matrix G, where in terms of R i, i = 1, 2, 3. y = Gu (95) 60/67
61 Exercise (cont.) Exercise 2 A photovoltaic (PV) array is connected to a load R L as shown: I + V R L 61/67
62 Exercise (cont.) The nonlinear V I characteristic of the PV panel is shown I P : Panel current V P : Panel voltage Figure 18: V I characteristic of the PV array a) Find the current and voltage in the circuit as a function of R L. b) Find R L to obtain the maximum power output. 62/67
63 Exercise (cont.) Exercise 3 R 1 + L C + u y R 2 R 3 63/67
64 Exercise (cont.) a) Write state variable equations for the circuit in the standard form: ẋ(t) = Ax(t) + Bu(t), (96) y = Cx(t) + Du(t). b) Find the transfer function using the state space model c) Solve for the state x(t), and output y(t), assuming zero initial conditions on the capacitor voltage and inductor current, u(t) = U(t) and R i = 1, i = 1, 2, 3, L = 1, C = 1. d) Verify that the steady state values for a constant input u satisfy Ax + Bu = 0, (97) Cx + Du = y. 64/67
65 Exercise (cont.) Exercise 4 u 1 K m 1 viscous friction coefficient µ x = y 1 ẋ = y 2 a) Write state variable equations for the massspring system. b) Calculate the transfer function y 1 /u and y 2 /u using the state space model. c) Find the poles of the system (or eigenvalues of A) as a function of m, k, µ. 65/67
66 Exercise (cont.) d) Solve for y 1 (t), y 2 (t), for zero initial condition and u(t) = U(t), as functions of m, k, µ. e) Comment on the solution in d) considering the cases µ 0, and K 0. 66/67
67 Exercise (cont.) Exercise 5 u 1 u 2 K m 1 B m 2 x 1 = y 1 x 2 = y 2 a) Write state variable equations for the system. b) Find the transfer functions Y i (s), i = 1, 2, j = 1, 2. (98) U j (s) c) Find the impulse response matrix G(t) 67/67
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