Lecture 2 Spring 2007

PAGE 1

Second order SystemsIn this set of notes we address the properties of the second order, constant coefficient differential equation:

.(1)

This equation describes the relation between an input f(t) and an output y(t). We will restrict our attention to the solution of (1) via Laplace transforms. Furthermore, we will assume here that all initial conditions are zero. The reader will address the case in which they are not in the homework problems. Taking the Laplace transform of (1), we obtain: .

It follows that the system transfer function is

.(2)For any input the solution of (1) under zero initial conditions is, in the s-domain, . The Unit Impulse Response of a Second order SystemForwe have . Notice that the units of (2) is the ratio of the output to the input. Hence, while is mathematically correct, one must note that the units of as the solution for a unit impulse are those of the output. The system (2) is a stable system if t he impulse responseas . This requires that the two roots of the characteristic polynomial both be in the Left Half Plane (LHP). A first order system has only one root, and it must be a real number. In contrast, a second order system can have either (i) two real roots, or (ii) a pair of complex-conjugate roots. We now consider these two cases. Furthermore, we will assume that the system is stable.Case 1 The roots of are real:Here, we will assume that the two real roots are not equal. The reader will address the case where they are equal in the homework problems. The roots of are the values of s that make it zero. We will express this as:

.(3)

The roots are and . Stability requires that both roots be real and negative, and so the parameters and are positive real numbers. Furthermore, they are the inverses of the system time constants, and . QUESTION: What does it mean to say that a system has two time constants?

To answer this question, we will use the table of Laplace transforms in the book to recover . This will necessitate transform pairs #13 and #16, which are:#13: and #16: .

However, before we go any further, lets consider the transform pairs #17 and #18: #17: and #18: .

Because the Laplace transform is a linear operation, from #17 we have

.

Recall that multiplication by s is equivalent to differentiation in time. And so:

.

Hence, we have proven that #18 follows from #17. Now lets look again at #16.

QUESTION: Does #16 follow from #13?

ANSWER: Clearly, there is a typo in this formula. It should be:

The moral of the story is: Dont blindly accept what is in print- even if it is from an expert in the field. Always use a bit of thought related you your own knowledge of the subject, as well. Having made this correction, lets proceed to write .We now have the denominator of in the appropriate form for both #13 and #16, but we have a ways to go. Write

.The far right term in this expression is .

The other term is:

.

Hence,

This can be rearranged into a cleaner form:

.(4)

And so, we see that the system impulse response is the sum of two decaying exponential functions. The relative contribution of each one will be determined both by which one decays slower and by the relative weights. Often, it is presumed that the slower one will dominate the shape of (4). However, if the faster one has a much larger weight, then this will not be the case. The nice thing about having (4) is that it shows you the exact nature of the weights

.(5)Conclusion: A second order system whose characteristic polynomial as negative real roots has an impulse response that is the sum of two decaying exponentials whose relative weights are given by (5).Definition 1. A second order stable system having two real roots is called an overdamped system. If the roots are real and equal, it is called a critically damped system. If the roots are a complex-conjugate pair, then it is called an underdamped systemExample 1. Consider the system described by the following differential equation:

(a) What is the system transfer function ?

(b) Without actually computing the system poles (i.e. the roots of the polynomial ) determine whether the system is overdamped or underdamped.Recall that, from the quadratic formula, the roots will be complex, if . In relation to P(s) we have . Hence, the poles are complex, and hence the system is underdamped

(c) You should have found that the system poles are complex. Compute the system static gain, undamped natural frequency, damping ratio, damped natural frequency, time constant, settling time, percent overshoot, and the rise time.The static gain is simply

The above polynomial, P(s), can be identified with . Hence, we easily obtain, by inspection:

.For the following quantities, see the discussion in the book:

It follows that rad/sec

The system time constant is sec.

The 4- settling time is, therefore sec

The overshoot is or 97%.

The rise time is sec.

(d) Use Matlab to verify the information in part (c).The terms overshoot and rise time are in relation to the system STEP RESPONSE.

To obtain the step response using Matlab, we first define the system transfer function:>> sys=tf(1,[1 .1 25])

Transfer function:

1

----------------

s^2 + 0.1 s + 25To get the unit step response for this system, we type

>> step(sys)

Figure 1. The unit step response of the given system.Unity Feedback ControlFor the block diagram below show that the closed loop transfer function between the input, yR , and the output, y, is given by

.

IS IT POSSIBLE TO USE UNITY FEEDBACK PROPORTIONAL CONTROL TO REDUCE THE AMOUNT OF OVERSHOOT?The closed loop system transfer function is

Since the closed loop system natural frequency is larger than that of the original system, but the middle term, which is is unchanged, it follows that the closed loop system damping ratio is smaller. Hence, the overshoot will actually INCREASE using proportional control.

IS IT POSSIBLE TO USE UNITY FEEDBACK DERIVATIVE CONTROL TO REDUCE THE AMOUNT OF OVERSHOOT?

The controller transfer function is . Hence, the closed loop transfer function corresponding to the unity feedback control system is now

We see that the undamped natural frequency is unchanged (=5 rad/sec). Hence, we have

Hence, it IS possible to increase the damping ratio, and, consequently reduce the overshoot using derivative control.

THE PROBLEM WITH THIS TYPE OF CONTROL IS THAT NOW THE CLOSED LOOP SYSTEM HAS A STATIC GAIN WHICH IS ZERO!

Suppose that we set (so that the closed loop system has optimal damping = 0.707). Then the closed loop system transfer function becomes

.

The Matlab command to obtain the unit step response of this system is

>> sys = tf([0.607 0],[1 0.707 25])

Transfer function:

0.607 s

------------------

s^2 + 0.707 s + 25>> step(sys)

Figure 2. Closed loop system step response for derivative unity feedback control.COMMENT: Here, overshoot is MEANINGLESS, since overshoot is related to the final (steady state) value of the system response to a step. Since this final response is ZERO, ANY overshoot is technically speaking, INFINITE.QUESTION: WHAT WOULD HAPPEN IF WE PUT THE DERIVATIVE CONTROLLER IN THE FEED BACK LOOP?

To answer this question, we need to develop the closed loop system transfer function for the case of NON-UNITY feedback. The student should be able to carry out this development on an exam. It is a small generalization of the development of the system closed loop transfer function for a unity-feedback system.

How about using Proportional-Plus-Devivative (i.e. PD) Control?The controller transfer function is . Hence, the closed loop transfer function corresponding to the unity feedback control system is now

.

We are now in a position to specify TWO dynamic parameters. Lets specify (i) . A reasonable second parameter to specify is (ii) . QUESTION: What is the smallest that we can achieve?

ANSWER: Since , we can make arbitrarily small by making Kp arbitrarily big. Lets try Kp = 75 (Oh, how convenient (!) Then we obtain a closed loop time constant value

To find Kd: . Hence, Kd = 14.044.And so, the closed loop transfer function is: .

The closed loop unit step response is shown below.

QUESTION: Do you see any issues related to this command response?

Answer 1: One issue might be the amount of overshoot. This was a consequence of choosing . We could have chosen . But then, why is it that is so often considered the best? It is because it gives the most linear phase in the system FRF. OK. So whats so important about linear phase? Answer: Suppose that a command input signal is run through the closed loop transfer function, and that the output is . Now, suppose that the phase shift is a linear function of ; that is, . Then, . In words, what this says is that a sinusoid of any frequency is delayed by an amount of time c. Because so many signals can be represented by a sum of sinusoids, the linear phase property ensures that they are all delayed by the same amount of time. In other words, the signal structure is preserved.Answer 2: ???ADDING INTEGRAL CONTROL TO ACHIEVE UNITY STATIC GAIN:

The controller transfer function is now

.

Hence, the open loop transfer function becomes:

Hence, the closed loop transfer function corresponding to the unity feedback control system becomes:

.

Notice that now, if we assume that the closed loop system is stable for an appropriate numerical value of Ki then the system static gain is W(s=0) = 1.0 (QUESTION: What price needed to be paid in order to achieve unity static gain?

ANSWER: For one thing, we needed a more sophisticated controller. But more importantly, the closed loop system went from being a 2nd order system to being a 3rd order system.

So what???

Well, the 2nd order system had only two poles, and was guaranteed to be stable for ANY positive PD controller gains. The 3rd order system has THREE poles, and will, in general, NOT be stable for any and all values of the controller gain parameters.

To finish this example, lets retain the numerical values Kp = 75 and Kd = 14.044. It should be noted that NOW these values will NOT result in the desired damping ratio and time constant that motivated them. With these numerical values, the closed loop transfer function is:

.QUESTION: For what, if any numerical values of Ki will W(s) be stable?

ANSWER: It will be stable as long as all three of the system poles that are the roots of the characteristic polynomial are in the Left Half Plane (LHP).QUESTION: Whats the best way to find the Ki values for which the system will be UNSTABLE?

ANSWER: There are many ways.

Method 1: Brute Force

%PROGRAM NAME: PID.m%Relates to 'Second Order System' lecture notes% METHOD 1: Brute ForceKmax = 1000;K=.1:.1:Kmax;nK = length(K);r = zeros(nK,3);for n = 1:nK p = [1 14.144 100 K(n)]; r(n,:) = roots(p);endrreal = real(r);rimag = imag(r);plot(rreal,rimag,'*')hold onplot(rreal(1,:),rimag(1,:),'xk','LineWidth',3)gridxlabel('real')ylabel('imaginary')title('Plot of Closed Loop Poles for K_i = 0.1 : 1000')

Method 2: Routh Array

Hence, the system will become unstable for

Method 3: Root Locus (Matlab command rlocus.m) To use this command, we need to reconfigure p(s) into the form of an open loop transfer function:

Hence,

where

Hence, in order for a candidate value of s to be a root of p(s) it must satisfy two conditions:

The magnitude condition: .

Well, for pretty much any s-value, one can find a Ki-value to satisfy this. (boring!)

The angle condition: To arrive at this condition in a clear notational form, we need to express p(s) in polar form: To begin, let p0 = 0. Then

Hence: .Hence, the angle condition is: .Aha! Unlike the magnitude condition that depends on Ki , the angle condition does NOT! (To demonstrate this condition, lets prove that the candidate s = -10 +i10 does NOT satisfy this condition (hence, is not a closed loop system pole for any value of Ki):

.

Hence: .The geometry associated with the above algebra is shown along with the plot below.

The following root locus plot was generated by the Matlab commands:

G = tf(1,[1 14.144 100 0]);rlocus(G)gridThe data cursor was then used to identify specific closed loop poles for specific Ki-values.

FINAL QUESTION: So, OK: The rlocus.m does the same thing that Method 1 does. But it seems to be a lot more complicated. Why cant I just use Method 1? Its so simple!!!

ANSWER: The root locus angle criterion not only can be used to determine whether or not a desired s-value is/is not a closed loop pole for some value of K. By computing the angle, we see how far from satisfying the angle criterion that value is. This allows us to incorporate more open loop poles and/or zeroes so that it WILL be a closed loop pole. The above proof was not simply a mathematical exercise. We found that in order to make the value s = 10 + i10 a closed loop pole, we will need to add controller poles/zeroes that contribute a net angle of either +167o or -193o. EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

_1325481513.unknown

_1452497407.unknown

_1452622416.unknown

_1452667615.unknown

_1452668950.unknown

_1452678332.unknown

_1452678413.unknown

_1452669612.unknown

_1452669627.unknown

_1452669590.unknown

_1452669585.unknown

_1452667761.unknown

_1452668473.unknown

_1452667626.unknown

_1452625087.unknown

_1452666615.unknown

_1452666908.unknown

_1452625320.unknown

_1452623308.unknown

_1452624801.unknown

_1452624989.unknown

_1452623542.unknown

_1452622561.unknown

_1452497780.unknown

_1452497847.unknown

_1452622304.unknown

_1452497807.unknown

_1452497708.unknown

_1452497749.unknown

_1452497650.unknown

_1326089908.unknown

_1452496432.unknown

_1452496845.unknown

_1452497049.unknown

_1452496751.unknown

_1452495473.unknown

_1452496273.unknown

_1452496240.unknown

_1452495573.unknown

_1452495379.unknown

_1325579813.unknown

_1325581952.unknown

_1325582261.unknown

_1325582419.unknown

_1325581966.unknown

_1325582231.unknown

_1325580429.unknown

_1325580797.unknown

_1325580221.unknown

_1325481572.unknown

_1325578904.unknown

_1325481526.unknown

_1325444946.unknown

_1325446229.unknown

_1325480226.unknown

_1325480476.unknown

_1325481145.unknown

_1325481220.unknown

_1325480915.unknown

_1325480367.unknown

_1325479589.unknown

_1325480153.unknown

_1325446260.unknown

_1325445989.unknown

_1325446099.unknown

_1325446113.unknown

_1325446015.unknown

_1325445057.unknown

_1325445618.unknown

_1325444978.unknown

_1230914332.unknown

_1325444197.unknown

_1325444473.unknown

_1325444518.unknown

_1325444265.unknown

_1230914508.unknown

_1325443750.unknown

_1325444035.unknown

_1230914998.unknown

_1325443246.unknown

_1230915087.unknown

_1230914692.unknown

_1230914382.unknown

_1230914425.unknown

_1230914354.unknown

_1230912009.unknown

_1230912239.unknown

_1230912416.unknown

_1230914063.unknown

_1230914213.unknown

_1230912711.unknown

_1230912319.unknown

_1230912153.unknown

_1230911370.unknown

_1230911453.unknown

_1230911047.unknown

_1230911354.unknown

_1230489964.unknown