Solutions: Homework 2 Biomedical Signal, Systems and Control (BME )

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1 Solutions: Homework 2 Biomedical Signal, Systems and Control (BE ) Instructor: René Vidal, rvidal@cis.jhu.edu TA: Donavan Cheng, donavan.cheng@gmail.com TA: Ertan Cetingül, ertan@cis.jhu.edu April, (20 points) One of the simplest and most fundamental of all physiological control systems is the muscle stretch reflex. The most notable example of this kind of reflex is the knee jerk, which is used in routine medical examinations as an assessment of the state of the nervous system. A sharp tap to the patellar tendon in the knee leads to an abrupt stretching of the extensor muscle in the thigh to which the tendon is attached. This activates the muscle spindles, which are stretch receptors. Neural impulses, which encode information about the magnitude of the stretch, are sent along afferent nerve fibers to the spinal cord. Since each afferent nerve is synaptically connected with one motorneuron in the spinal cord, the motorneurons get activated and, in turn, send efferent neural impulses back to the same thigh muscle. These produce a contraction of the muscle, which acts to straighten the lower leg. Figure (a) shows the basic components of this reflex. Construct a block diagram similar to the one shown in Figure (b) to represent the major control mechanisms involved in the muscle stretch reflex. Clearly identify the physiological correlates of the controller, the plant, and the feedback element, as well as the controlling, controlled, and feedback variables. (a) Schematic illustration of the muscle stretch reflex (b) Closed-loop control system Figure : The muscle stretch reflect as a control system Answer: Consider the block diagram representation of this reflex, as shown in Figure 2. Comparing this configuration with the general closed-loop control system of Figure (b), one can see that the thigh muscle now corresponds to the plant or controlled system. The disturbance, x, is the amount of initial stretch produced by the tap to the knee. This produces a proportionate amount of stretch, y, in the muscle spindles, which act as the feedback sensor. The spindles translate this mechanical quantity into an increase in afferent neural traffic (z) sent back to the reflex center in the spinal cord, which corresponds to our controller. In turn, the controller action is an increase in efferent neural traffic (u) directed back to the thigh muscle, which subsequently contracts in order to offset the initial stretch. Although this closed-loop control system differs in some details from the canonical structure shown in Figure (b), it is indeed a negative feedback system, since the initial disturbance (tap-induced stretch) leads to a controller action that is aimed at reducing the effect of the disturbance.

2 Figure 2: Block diagram of the reflex 2. (20 points) A two-input, single-output system is described by the equation ÿ(t) + sin(y(t))ẏ(t) + u 2 (t)y(t) u (t) + u 2 (t). () Compute the linearized state equation that describes this system about the constant operating point corresponding to u 0 0 and u 0 2. Answer: Let x y and ẏ x ẏ and x 2 ÿ. Replacing the state variables into the nonlinear 2 nd order differential equation (), we get x 2 (t) + sin(x (t)) (t) + u 2 (t)x (t) u (t) + u 2 (t). Then the nonlinear state equations are written as: x (t) f (x, u, t) (t), x 2 (t) f 2 (x, u, t) sin(x (t)) (t) u 2 (t)x (t) + u (t) + u 2 (t). At the constant operating point, we are given u 0 0 and u 0 2. Now let s find x 0 and x 0 2 that vanish, together with {u 0, u 0 2}, the nonlinear state equations, i.e., f (x, u) 0 and f 2 (x, u) 0, where we omit t for brevity, and x and u denote the vectors x and u u 2, respectively. f (x, u, t) x 0,x 0 0 2,u0,u0 2 x0 2 0, f 2 (x, u, t) x 0,x 0 0 2,u0,u0 2 x0. The linearized state equation can then be written by computing the Jacobians and evaluating them at the constant operating point, i.e., f f x A 0 0, cos(x ) u 2 sin(x ) sin() B x f f u u 2 u u x x 0,x0 2,u0,u x 0,x0 2,u0,u0 2 By defining the residuals x. x x 0 x2 0. and u u u 0, u 2 u2 0 the linearized state equation has the following form x A x + B u x + u. sin() 0. (20 points) Write the equations of motion for the double-pendulum system shown in Figure (a). Assume the displacement angles of the pendulums are small enough to ensure that the spring is always horizontal. The pendulum rods are taken to be massless, of length l, and the springs are attached / of the way down. 2

3 mgl sin θ k l (sin θ sin θ 2 )cosθ l ml2 θ 206 CHAPTER 2. DYNAIC ODELS 8. InFigure many 2.5: mechanical Doublepositioning pendulumsystems Solution: there is flexibility between one part of the system and another. An example is shown in Figure 2.6 where there is flexibility of the solar panels. Figure 2.6 depicts such a situation, where a force u is applied to the mass and another l mass θ θ m is connected to it. The coupling between the objects 2 is often modeled byaspringconstantk with a damping coefficient b, although the actual situation is usually much more complicated than this. (a) Double pendulum (a) The FBDFigure for the : Analysis system isof the double pendulum 2006 CHAPTER 2. DYNAIC ODELS. Write the equations of motion for the double-pendulum system shown in Fig Assume the displacement angles of the pendulums are small enough to ensure that the spring is always horizontal. The pendulum rods are taken to be massless, of length l, and the springs are attached / of the way down. (a) Write the equations of motion governing this system. (b) Find the transfer function between the control input, u, and the output, y. Solution: m k m l sinθ l sinθ2 If we write the moment equilibrium about the pivot point of the left pendulem from the free body (b) diagram, mgl sin θ k l (sin θ sin θ2)cosθ l ml2 θ Answer: By analyzing Figure (b), the equations of motion can be written as Figure 2.5: Double pendulum ml 2 θ mgl sin(θ ) k l(sin(θ ) sin(θ 2 )) cos(θ ) l, ml 2 θ2 mgl sin(θ 2 ) + k l(sin(θ ) sin(θ 2 )) cos(θ 2 ) l. which results in the equations For small θ, sin(θ) θ and cos(θ). Thus the equations of motion become mẍ k (x ml θ mgθ k 9 y) b (ẋ ẏ) ÿ u + k (x 6 l(θ y)+b θ 2 (ẋ ), ẏ) or ml θ 2 mgθ 2 + k 9 6 l(θ θ 2 ).. (20 points) In many mechanical positioning systems there is flexibility between one part of the system and another. Figure depicts such a situation, ẍ + k where m x + b amẋ force k um is y applied b mẏ to the 0 mass and another mass m is connected to it. The coupling between the objects is often modeled by a spring constant k with a damping coefficient b, although the actual situation k x is b usually ẋ +ÿ much + k more y + b complicated ẏ than u this. a) Write the equations of motion governing this system and put them in state space form. b) Find the transfer function Figurebetween 2.6: Schematic the controlof input a system u and the with output flexibility y. Figure : Schematic of a system with flexibility Answer: a) The equations of motion of the system in Figure are ÿ u k(y x) b(ẏ ẋ), (2) mẍ k(y x) + b(ẏ ẋ). ()

4 Let us define the state vector x as x. x x x x ẋ y. ẏ Then the state space representation of the system is ẋ x 0 ẋ 2 ẋ k b k b m m m m x u, ẋ x k b k y b) Assuming y(0) 0, ẏ(0) 0, x(0) 0, ẋ(0) 0, the Laplace transforms of the equations of motion Eqns. (2),() are From Eqn. (5), we get X(s) x x x b. s 2 Y (s) U(s) k(y (s) X(s)) bs(y (s) X(s)), () s 2 mx(s) k(y (s) X(s)) + bs(y (s) X(s)). (5) (k+bs)y (s) ms 2 +bs+k, and replace it into Eqn. () to obtain the transfer function H(s) Y (s) U(s) ms 2 + bs + k ms + ( + m)bs + ( + m)ks (20 points) Consider the block diagram shown in Figure 5. Note that a i and b i are constants. Compute the transfer function for this system. Put also the equations in state space form. Figure 5: Block diagram for Problem Answer: From the block diagram, we can write the following equations sx (s) U(s) a X (s) a 2 X 2 (s) a X (s), sx 2 (s) X (s), sx (s) X 2 (s), Y (s) b X (s) + b 2 X 2 (s) + b X (s). By putting the first two equations above into the last two and writing in terms of X (s), we get U(s) (s + a )s 2 X (s) + a 2 sx (s) + a X (s), Y (s) b s 2 X (s) + b 2 sx (s) + b X (s).

5 Then the transfer function H(s) can be written as H(s) Y (s) U(s) b s 2 + b 2 s + b s + a s 2 + a 2 s + a. (6) To obtain the state space representation, take the inverse Laplace transform of the equations and put into the following form ẋ ẋ 2 ẋ ẋ u a x a 2 a x, ẋ 2 x, ẋ, y b x + b 2 + b x, a a 2 a 0 0 x + 0 u, y b b 2 b x x. x 5