1 Closed Loop Systems

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1 Harvard University Division of Engineering and Applied Sciences ES 45 - Physiological Systems Analysis Fall 2009 Closed Loop Systems and Stability Closed Loop Systems Many natural and man-made systems exhibit some form of feedback or closed-loop behavior. What is feedback? It implies feeding back information from later to earlier parts of the block diagram. Figure illustrates the basic concept, where the feedback element can be one part of a much larger block diagram. Figure : The Feedback Element To understand the utility of feedback mechanisms in natural systems, we first look at manmade systems in which feedback control strategies are implemented. Why might feedback be implemented? If the open-loop system is stable, and behaves in the way that we want it to, and there are no real disturbances that can affect the performance, then feedback would not be necessary. More realistically, however, this is not the case. Closing the loop, i.e. providing feedback, can be used to ) stabilize an unstable system or 2) achieve some performance criteria that you desire. Examples of Feedback Systems Before we move to the quantitative nature of feedback in system dynamics, first consider the following physiological examples, in which we see naturally occurring feedback mechanisms. Cardiovascular System. Many aspects of the mammalian cardiovascular system exhibit feedback mechanisms that help to maintain oxygen transport in the face of extremely varying conditions and large disturbances. Consider the baroreflex mechanism. This is a mechanism for regulating blood pressure in response to transients. For example, when you move from lying down to standing up, blood rushes to your lower extremities, lowering the blood pressure. One of the primary objectives of the cardiovascular system is to maintain adequate oxygen transport to the cells in the brain, which requires the maintenance of arterial blood pressure. A sudden deficit in oxygen supply to the brain can cause dizziness and/or blackout very quickly (not to mention worse things over the course of just seconds to minutes).

2 ES45 - Closed Loop Systems and Stability 2 To react to transients in blood pressure, pressure sensors (baroreceptors) are located in the walls of large arteries. Changes in pressure induce mechanical deformation in the vessel wall, causing these cells to respond. These neurons then project this information back to the heart through the baroreflex arc, as simplistically illustrated in Figure 2. An increase Figure 2: Baroreflex Mechanism Const disturbance Desired BP HR BP BP Baro recep. in the pressure serves to decrease heart rate, thereby decreasing the pressure. Similarly, a decrease in the pressure serves to increase heart rate, which increases the pressure. The additive constant is to indicate that there is a maintained baseline of heart rate in the absence of perturbation. This mechanism, therefore, is a negative feedback loop, which will be discussed in more detail in later lectures. Temperature Control. Thermoregulatory mechanisms in warm-blooded animals serve to maintain relatively constant body temperatures in the face of internal and external disturbances. As we have mentioned previously, when the body temperature drops two primary responses are initiated. There is a constriction of blood vessels to shunt blood away from the periphery, and muscles are activated to generate metabolic heat, which is exhibited through shivering. When the body temperature increases, blood vessels are dilated to move blood towards the periphery for increased heat exchange across the skin, and sweating is initiated to further facilitate heat transfer. We can summarize this in the simplistic block diagram shown in Figure 3. The additive constant represents the maintenance of tonic activity in Figure 3: Temperature Control Const disturbance Desired Temp Constr/ Dilation Effects on Temp Temp Shiver/ Sweat Temp Recept the absence of perturbation. Block Diagram Manipulation Consider the system shown in Figure 4. We already know how to find the transfer function

3 ES45 - Closed Loop Systems and Stability 3 Figure 4: Block Diagram Manipulation u e y K H between U(s) and Y (s). Recall that: Y (s) = K(s)E(s) E(s) = U(s) H(s)Y (s) So we have: Y (s) = K(s)[U(s) H(s)Y (s)] = K(s)U(s) K(s)H(s)Y (s) which leads us to: Y (s) U(s) = K(s) + K(s)H(s) We can find the transfer function between U(s) and E(s) in the same manner: E(s) = U(s) H(s)Y (s) Y (s) = K(s)E(s) So we have: E(s) = U(s) H(s)K(s)E(s) which leads to: E(s) U(s) = + K(s)H(s) The point of this exercise is to illustrate that the transfer function between any two measures in the block diagram can be determined through these simple manipulations. As you may have guessed, feedback elements have a variety of effects on the dynamics of the system, in terms of both the stability and performance. In the following, we will discuss several properties that are critical in system design, such as ) stability, 2) sensitivity, 3) disturbance rejection, and 4) error.. Stability Suppose that the system under consideration is naturally unstable. Obviously naturally occurring systems don t exhibit unbounded outputs, since they eventually reach physical boundaries as the mechanical system breaks apart or the electrical system saturates and/or

4 ES45 - Closed Loop Systems and Stability 4 a Figure 5: Stability b u y u y burns out circuit components. However, these are still undesirable outcomes, and we will thus consider the effects of the feedback on the stability of the system. Consider the system shown in Figure 5a. Suppose that this transfer function (s) has the form: (s) = s 2 This system is unstable, since it has a pole at +/2, resulting in exponential growth in the output. Suppose we simply close the loop with a negative feedback mechanism, by feeding back the measured output and subtracting it from the input, as shown in Figure 5b. We then have: Y (s) U(s) = (s) + (s) = s 2 + s 2 = s + 2 We have effectively moved the pole from the right half of the complex plane to the left, resulting in a stable system..2 Sensitivity Often times we build models of physical systems where our estimates of the physical constants are inaccurate, or that the properties of the system are changing over time. The fundamental issue here is that we would like our closed loop system to be insensitive to changes in the system properties over time. Consider the system shown in Figure 6. u e Figure 6: Sensitivity c y H Let T represent the closed-loop transfer function (CLTF): T = c + c H = f + f H

5 ES45 - Closed Loop Systems and Stability 5 where f = c represents the forward cascade. We wish to determine the sensitivity of the dynamics to changes in f. We can define sensitivity as: S = T T f f = f T = = f T = [ ] f f = f [ f ( )( + f H) 2 H + ( + f H) ] T f T f + f H T [ ] f H ( + f H) 2 + = [ ] f f H + + f H + f H T ( + f H) 2 = [ ] f T ( + f H) 2 + f H The interpretation is that the sensitivity represents the percent change in the CLTF in response to a percent change in the forward cascade. Suppose that f has the form: f = 2500 s(s + 5)(s + 50) and H =. The magnitude Bode plot of the corresponding sensitivity function is shown in Figure 7. Note that the system is relatively insensitive to changes in the forward cascade Magnitude (db) Figure 7: Bode Plot of Sensitivity Frequency (rad/sec) The sensitivity of the closed-loop transfer function to changes in the forward cascade. at low frequencies, but is maximally sensitive at around 0 rad/sec, and remains fairly sensitive at higher frequencies..3 Disturbance Rejection How do uncontrolled disturbances affect a system s behavior? Feedback is often implemented to provide robust performance in the face of disturbances. Consider the following example of speed control of an engine, as illustrated in Figure 8. The term ω r represents the desired speed. The controller acts on the difference between the actual and desired speed. The actual speed ω is shown with and without feedback in Figure 9. Here we have assumed that there is a step input in disturbance, such as a sudden increase in friction (have you checked your oil lately?). We can see that without the feedback, the response to the disturbance is pretty drastic, and the actual speed is very far from the desired. With feedback, however, the system is able to recover after a very short dip away from the desired speed. This is disturbance rejection.

6 ES45 - Closed Loop Systems and Stability 6 Figure 8: System with Disturbance d ωr Control Engine ω Speed Transducer Figure 9: Disturbance Rejection d d ω r ω r Without Feedback With Feedback Suppose the system has the specific form shown in Figure 0. The goal is to find E(s)/D(s). Figure 0: Disturbance Rejection Example d ωr e c 2 ω H We can think of this system as having multiple inputs, U(s) and D(s). To find the transfer function from one input, D(s), to the error E(s), we simply hold ω r equal to zero. which yields: E(s) = H( 2 (D(s) + c E(s))) = H 2 D H 2 c E(s) If H 2 c, then we have: E(s) D(s) = E(s) D(s) H 2 + H 2 c c So for frequencies at which c, the disturbance has little effect on error.

7 ES45 - Closed Loop Systems and Stability 7.4 Steady State Error For example, suppose that we wish the output of a system to follow a prescribed trajectory. Any corresponding deviations from this desired trajectory would be considered error; the nature of the error is greatly affected by feedback mechanisms and is in fact one of the critical reasons that engineers use feedback in system design. Consider the system shown in Figure. u Figure : Steady State Error e y H We know, from the final value theorem introduced in the Laplace handout, that the steady state error can be written as: We have that: e ss = lim t e(t) = lim s 0 se(s) E(s) = U(s) HE(s) which yields: The steady state error becomes: Suppose that H has the form: We can then write: e ss = lim s 0 E(s) = e ss U(s) + H = lim s 0 su(s) + H H = K(a ms m + a m s m ) s p (b n s n + b n s n ) su(s) + K(amsm +a m s m +...+) s p (b ns n +b n s n +...+) su(s) s p+ U(s) = lim s 0 + K = lim s 0 s p + K s p where p represents the type number of the system. The type number of the system, therefore, represents the number of pure integrators. For example, if p = 0, and u(t) is a unit step (U(s) = s ), then we have: e ss s 0 = lim s 0 s 0 + K = lim s 0 + K

8 ES45 - Closed Loop Systems and Stability 8 which leads to a nonzero steady state error of /( + K). For a type system (p = ) with a unit step input, we have: e ss = lim s 0 resulting in a zero steady state error. s p s p + K = lim s 0 s s + K = 0 For a type system (p = ) with a ramp input (u(t) = t;u(s) = s 2 ), we have: e ss = lim s 0 s + K = K resulting in a steady state error of K. 2 Stability We have previously touched on the topic of stability, but now we will explore it much more in-depth. What do we really mean by stability? Definition: a system is stable if the transient solution decays to zero, and unstable if it grows. This also implies that for a bounded input, the system exhibits a bounded output. This implies that a system is stable if and only if all of the poles lie in the left-half of the s-plane, as shown in Figure 2. Figure 2: Stable region of the S-plane imag s plane stable region real 2. Effects of Closing the Loop on System Poles With the open loop system, as shown in Figure 3a, the stability is a function of the physical dynamics of the system. As with performance issues, like steady state error, disturbance rejection, etc., the stability can be changed using carefully designed controllers. One possibility is that a controller, c, can be placed in cascade with the original system so that we might effectively cancel out any unwanted dynamics, and replace them with more desirable ones, as shown in Figure 3b. This turns out to be a bad idea, because it is often the case that there are discrepancies between our model and the actual system, making it difficult to actually cancel the undesirable dynamics. An alternative is to introduce feedback, so we close the loop, as shown in Figure 4. The CLTF is:

9 ES45 - Closed Loop Systems and Stability 9 Figure 3: Open Loop Dynamics a b u(t) y(t) u(t) c y(t) Figure 4: Closed Loop Dynamics u e K y H Y (s) U(s) = K(s) + KH(s) Note that the poles of the open loop system are the roots of the denominator of (s), and will thus be called open-loop poles, or OLP s. Correspondingly, the poles of the closed loop system are the roots of the denominator of the CLTF, and are called closed-loop poles, or CLP s. CLP s are values of s for which + KH(s) = 0, or equivalently, KH(s) = = 80. The new characteristic equation can be written: + K (s a )(s a 2 )... (s a m ) (s b )(s b 2 )... (s b n ) = 0 What are s a i and s b i? We can write these as: s a i s b i = A i e jα i = B i e jβ i which can be illustrated graphically, as shown in Figure 5. So we have that KH(s) = and KH(s) = 80. The values of s that satisfy this are the CLP s. The magnitude condition becomes: KH = K A A 2... A m B B 2...B n = which implies that: K = B B 2...B n A A 2... A m The angle condition becomes: KH = α + α α m β β 2... β n

10 ES45 - Closed Loop Systems and Stability 0 Figure 5: Poles and Zeros in Complex Plane α i s Bi 0 0 A i ai β i b i and therefore: α α m β... β n = ±(2p + )80 where p is any integer. So for a particular value of K (assume that we have a big knob that we can turn, changing the gain), particular values of s satisfy the magnitude and angle constraints. The root loci are the set of s values that satisfy the constraints for different values of K. Incidentally, this same procedure can be performed for any parameter for which we have access. 2.2 Constructing the Root Locus Consider that: which we can write as: + KH = + K (s a )(s a 2 )... (s a m ) (s b )(s b 2 )...(s b n ) = 0 (s b )(s b 2 )... (s b n ) + K(s a )(s a 2 )... (s a m ) = 0 If K = 0, the roots are b,b 2,...,b n, the OLP s. a,a 2,...,a m, which are the open-loop zeros (OLZ s). As K, the roots approach Here are some general rules that apply to the root locus: ) For K = 0, the CLP s are the OLP s. 2) For K, the CLP s approach the OLZ s. 3) There are as many locus branches as OLP s. 4) Loci are symmetric about the real axis, since complex OLP s and OLZ s occur in complex conjugate pairs.

11 ES45 - Closed Loop Systems and Stability 5) The section on the real axis to the left of an odd # of open loop poles and zeros forms part of the locus, since the angle conditions are inherently satisfied. 6) If there are fewer OLZ s than poles (m < n), the branches for which there are no OLZ s go to infinity along asymptotes. The number of asymptotes, therefore, is n m. 7) The directions of the asymptotes are found from angle conditions. The asymptote angles α must satisfy: where i is any integer. (2i + )80 α = ± n m 8) The asymptotes intersect the real axis at a distance ρ 0 from the origin: ρ 0 = ( OLP) ( OLZ) #OLP #OLZ A Simple Example Consider the following simple example, with the open loop relationship shown in Figure 6a. Let the transfer function (s) have the form: a Figure 6: Root Locus Example b u(t) y(t) u(t) K y(t) (s) = (s )(s + 3) Alone the system is unstable, as one of the roots is in the right-half plane. Suppose we can add a pure gain, and a unity feedback, as shown in Figure 6b. The CLTF is: CLTF = K + K = K (s )(s+3) + K s s + 3 So the characteristic equation becomes: + K (s )(s + 3) = 0 Here is how we can draw the locus, as shown in Figure 7:

12 ES45 - Closed Loop Systems and Stability 2 ) First we draw the complex plane, and indicate the OLP s with x s, and OLZ s with o s. 2) On the real axis, the root locus lies to the left of an odd # of poles, so we fill in the area between -3 and +. 3) When K = 0, the CLP s will be located at the OLP s. As we increase K from 0, the CLP s move in and away from the OLP s. For example, if K =, we get that the roots are ± 3. 4) If we continue to increase K, we can show that the roots eventually become complex, which means that the roots leave the real axis. Recall that complex roots show up in conjugate pairs, so we have one with a positive imaginary part, and one with a negative imaginary part. The locus, therefore, follows a trajectory towards n m = 2 0 = 2 asymptotes. The angles α of the asymptotes can be computed: (2i + )80 α = ± 2 = ±90 The asymptote intersects the real axis at the location: ρ 0 = ( 3) 0 2 = Figure 7: Root Locus Construction k=5 s plane 3 + k= Reading Required reading: Chapters 8 and 9 from Ogata [] References [] K. Ogata. System Dynamics. Prentice-Hall, New Jersey, fourth edition, 2004.

MAK 391 System Dynamics & Control. Presentation Topic. The Root Locus Method. Student Number: Group: I-B. Name & Surname: Göksel CANSEVEN

MAK 391 System Dynamics & Control Presentation Topic The Root Locus Method Student Number: 9901.06047 Group: I-B Name & Surname: Göksel CANSEVEN Date: December 2001 The Root-Locus Method Göksel CANSEVEN