Systems with Real Components and Saturating Signals ◾ 67 For the phase plane, instead of plotting position against time we plot velocity against position. An extra change has been made in www.esscont.com/6/PhaseLim.htm. Whenever the drive reaches its limit, the color of the plot is changed, though this is hard to see in the black-and-white image of Figure 6.3. For an explanation of the new plot, we will explore how to construct it without the aid of a computer. Firstly, let us look at the system without its drive limit. To allow the computer to give us a preview, we can “remark out” the two lines that impose the limit. By putting // in front of a line of code, it becomes a comment and is not executed. You can do this in the “step model” window. Without the limit there is no overshoot and the velocity runs off the screen, as in Figure 6.4. Phase plane—limit at 1.0 Figure 6.3 Phase plane with limit. Phase plane—limit at 1.0 Figure 6.4 Phase-plane response without a limit. 91239.indb 67 10/12/09 1:42:00 PM 68 ◾ Essentials of Control Techniques and Theory To construct the plot by hand, we can rearrange the differential equation to set the acceleration on the left equal to the feedback terms on the right:   xxx=− −56. (6.1) If we are to trace the track of the (position, velocity) coordinates around the phase plane, it would be helpful to know in which direction they might move. Of particular interest is the slope of their curve at any point, dx dx  . We wish to look at the derivative of the velocity with respect to x, not with respect to time, as we usually do. ese derivatives are closely related, however, since for the general function f, df dx dt dx df dt x df dt == 1  . So, we have: Slope == dx dx x x    1 . (6.2) But Equation 6.1 gives the acceleration as   xxx= −−56 so, we have Slope == 1 56 56    x xx x x () .−− −− (6.3) e first thing we notice is that for all points where position and velocity are in the same proportion, the slope is the same. e lines of constant ratio all pass through the origin. On the line x = 0, we have slope –5. On the line x =  x , we have slope –11. On the line x = −  x , we have slope +1, and so on. We can make up a spider’s web of lines, with small dashes showing the direc- tions in which the “trajectories” will cross them. ese lines on which the slope is the same are termed “isoclines.” (Figure 6.5). 91239.indb 68 10/12/09 1:42:03 PM Systems with Real Components and Saturating Signals ◾ 69 An interesting isocline is the line 650xx+=  . On this line the acceleration is zero, and so the isocline represents points of zero slope; and the trajectories cross the line horizontally. In this particular example, there are two isoclines that are worth even closer scrutiny. Consider the line  xx+=20 . Here the slope is –2—exactly the same as the slope of the line itself. Once the trajectory encounters this line, it will lock on and never leave it. e same is true for the line  xx+=30 , where the trajectory slope is found to be –3. Q 6.3.1 Is it a coincidence that the “special” state variables found in Section 5.5 could be expressed as  xx+=20 and  xx+=30 , respectively? Having mapped out the isoclines, we can steer our trajectory around the plane, following the local slope. From a variety of starting points, we can map out sets of trajectories. is is shown in Figure 6.6. e phase plane “portrait” gives a good insight into the system’s behavior, with- out having to make any attempt to solve its equations. We see that for any starting point, the trajectory homes in on one of the special isoclines and settles without any oscillatory behavior. Q 6.3.2 As an exercise, sketch the phase plane for the system   xx x++ =60, x x . Slope = –11 Slope = 1 Slope = –5 Infinite slope Figure 6.5 Isoclines. 91239.indb 69 10/12/09 1:42:06 PM 70 ◾ Essentials of Control Techniques and Theory i.e., for the same position control system, but with much reduced velocity feedback. You will find an equation similar to Equation 6.3 for the slopes on the isoclines, but will not find any “special” isoclines. e trajectories will match the “spider’s web” image better than before, spiraling into the origin to represent system responses which now are lightly damped sine-waves. Scale the plot over a range of –1 to 1 in both position and velocity. Q 6.3.3 is is more an answer than a question. Go to the book’s website. Run the simula- tion www.esscont.com/6/q6-3-2.htm, and compare it with your answer to question Q 6.3.2. 6.4 Phase Plane for Saturating Drive Now remember that our system is the simple one where acceleration is proportional to the input.  xu= . e controlled system relies on feedback alone for damping, with the input determined by Isocline slope = –11 Isocline slope = 1 Zero slope isocline “Special” isoclines Figure 6.6 Example of phase plane with isoclines. 91239.indb 70 10/12/09 1:42:08 PM Systems with Real Components and Saturating Signals ◾ 71 uxxx d = 65()−−  or uxx= −−65  if the demanded position is zero. On the lines −−65 1xx  =± the drive will reach its saturation value. Between these lines the phase plane plot will be exactly as we have already found it, as in Figure 6.7. Outside the lines, the equations are completely different. e trajectories are the solutions of  x = −1 to the right and  x = 1 to the left. To fill in the mystery region in this phase plane, we must find how the system will behave under saturated drive. Figure 6.7 Linear region of the phase plane. 91239.indb 71 10/12/09 1:42:11 PM 72 ◾ Essentials of Control Techniques and Theory Equation 6.2 tells us that the slope of the trajectory is always given by   xx/ , so in this case we are interested in the case where  x has saturated at a value of +1 or –1, giving the slope of the trajectories as 1/x or –1/x. Now we see that the isoclines are no longer lines through the origin as before, but are lines of constant  x , parallel to the horizontal axis. If we want to find out the actual shape of the trajectories, we must solve for the relationship between x and  x . On the left, where u = 1, we can integrate the expression for  x twice to see  x = 1 so  xta=+, and xt at b=++ 2 2/ from which we can deduce that xx ba=+  22 22//,− i.e., the trajectories are parabolae of the form xx c=+  2 2/ with a horizontal axis which is the x-axis. Similarly, the trajectories to the right, where u = –1, are of the form xx c=+−  2 2/. e three regions of the phase plane can now be cut and pasted together to give the full picture of Figure 6.8. We will see that the phase plane can become a powerful tool for the design of high performance position control systems. e motor drive might be pro- portional to the position error for small errors, but to achieve accuracy the drive must approach its limit for a small displacement. e “proportional band” is small, and for any substantial disturbance the drive will spend much of its time saturated. e ability to design feedback on a non-linear basis is then of great importance. 91239.indb 72 10/12/09 1:42:15 PM Systems with Real Components and Saturating Signals ◾ 73 Q 6.4.1 e position control system is described by the same equations as before,   xxxw++=566. is time, however, the damping term 5  x is given not by feedback but by pas- sive damping in the motor, i.e.,   xxu=+−5 where ux= −−6( ).demand Once again the drive u saturates at values +1 or −1. Sketch the new phase plane, noting that the saturated trajectories will no longer be parabolae. e answer can be found on the website at www.esscont.com/6/q6-4-1.htm. Q 6.4.2 Consider the following design problem. A manufacturer of pick-and-place machines requires a robot axis. It must move a one kilogram load a distance of one meter, bringing it to rest within one second. It must hold the load at the target position with sufficient “stiffness” to resist a disturbing force, so that for a deflection of one millimeter the motor will exert its maximum restoring force. u = –1 u = +1 Isoclines Figure 6.8 Phase plane with linear and saturated regions. 91239.indb 73 10/12/09 1:42:18 PM 74 ◾ Essentials of Control Techniques and Theory Steps in the design are as follows: (a) What acceleration is required to achieve the one meter movement in one second? (b) Multiply this by an “engineering margin” of 2.5 when choosing the motor. (c) Determine feedback values for the position and velocity signals. You might first try pole assignment, but you would never guess the values of the poles that are necessary in practice. Consider the “stiffness” criterion. An error of 10 –3 meters must produce an acceleration of 10 m/s 2 . at implies a position gain of 10,000. Use the phase plane to deduce a suitable velocity gain. 6.5 Bang–Bang Control and Sliding Mode In the light of a saturating input, the gain can be made infinite. e proportional regional has now shrunk to a switching line and the drive always takes one limiting value or the other. is is termed bang–bang control. Clearly stability is now impos- sible, since even at the target the drive will “buzz” between the limits, but the quality of control can be excellent. Full correcting drive is applied for the slightest error. Suppose now that we have the system  xu= , where the magnitude of u is constrained to be no greater than 1. Suppose we apply a control law in the form of logic, if ((x+xdot)<0){ u=1; }else{ u=-1; } then Figure 6.3 will be modified to show the same parabolic trajectories as before, with a switching line now dividing the full positive and negative drive regions. is is the line xx+=  0. Now at the point (–0.5, .5), we will have u = 1. e slope of the trajectory will be:   x x 91239.indb 74 10/12/09 1:42:19 PM Systems with Real Components and Saturating Signals ◾ 75 which at this point has value 2. is will take the trajectory across the switching line into the negative drive region, where the slope is now –2. And so we cross the trajectory yet again. e drive buzzes to and fro, holding the state on the switching line. is is termed sliding. But the state is not glued in place. On the switching line, xx+=  0. If we regard this as a differential equation, rather than the equation of a line, we see that x decays as e –t . Once sliding has started, the second order system behaves like a first order system. is principle is at the heart of variable structure control. 91239.indb 75 10/12/09 1:42:20 PM This page intentionally left blank [...]... 10/12/09 1:42:21 PM 78  ◾  Essentials of Control Techniques and Theory 7.2  Sine-Wave Fundamentals Sine-waves and allied signals are the tools of the trade of classical control They can be represented in terms of their amplitude, frequency, and phase If you have ever struggled through a page or two of algebraic trigonometry extracting phase angles from mixtures of sines and cosines, you will realize... on the other hand, we take the real part of (a + jb )e jt we get a cos(t ) − b sin(t ) So we can express a mixture of sines and cosines with a simple complex number This number represents both amplitude and phase On the strict understanding that we take the real part of any expression when describing a function of time, we can now deal in complex amplitudes of e jt Algebra of addition and subtraction... any complications The real part of the sum of (a + jb)e jt and (c + jd) e jt is seen to be the sum of the individual real parts, i.e., we can add the complex numbers (a + jb) and (c + jd) to represent the new mixture of sines and cosines (a + c)cos(t) − (b + d) sin(t) Beware, however, of multiplying the complex numbers to denote the product of two sine-waves For anything of that sort you must go back... solutions with sinusoidal forcing functions In the “knife and fork” approach we would have to assume a result of the form A cos(t) + B sin(t), substitute this into the equations and unscramble the resulting mess of sines and cosines Let us use an example to see the improvement 91239.indb 79 10/12/09 1:42:26 PM 80  ◾  Essentials of Control Techniques and Theory Q  7.3.1 The system described by the second... led to rules of thumb, then they became an art, and now a science We can put the methods onto a firm foundation of mathematics 7.7  Poles and Polynomials Analysis of the servomotor problem from the state-space point of view gave us a list of first order differential equations A “lumped linear system” of this type will have a set of state equations where each has a simple d/dt on the left and a linear... same function of time Since the derivative of e st is the same function, but multiplied by s, all the time derivatives simply turn into powers of s We end up with: 91239.indb 85 ( s n + a1s n−1 + a2 s n−2 +  + an ) y = (b0 s m + b1s m−1 +  + bm )u (7.3) 10/12/09 1:42:38 PM 86  ◾  Essentials of Control Techniques and Theory The gain, the ratio between output and input, is now the ratio of these two... power therefore drops by four, giving a fall of 6 decibels for each octave Keep the slogans “three decibels down” and “six decibels per octave” safe in your memory! 7.10 Frequency Plots and Compensators Let us return to simpler units, and look again at the example of 91239.indb 89 G(s ) = 1 s+2 10/12/09 1:42:44 PM 90  ◾  Essentials of Control Techniques and Theory We have noted that the low frequency... not a simple proportion of the output, 91239.indb 91 10/12/09 1:42:47 PM 92  ◾  Essentials of Control Techniques and Theory but instead contains some “compensator” with gain function F(s), then the stability will be dictated by the product of the two gains, F(s)G(s) The closed loop gain will be G(s ) 1+ F ( s )G ( s ) and so the poles of the closed loop system are the roots of 1 + F ( s )G ( s ) =... logarithm to base e First of all, we are likely to want to work out a frequency response, by substituting the value jω for s, and we are faced with a set of logarithms of complex expressions Now a complex number can be expressed in polar form as re jθ as in Figure 7.2 (Remember that e jθ = cosθ + jsinθ.) Here r is the modulus of the number, the square root of the sum of the squares of real and imaginary parts,... magnitude of the output, while the imaginary part determines the phase In the early days of electronic amplifiers, phase was hard to measure The properties of the system had to be deduced from the amplitude alone 91239.indb 87 10/12/09 1:42:41 PM 88  ◾  Essentials of Control Techniques and Theory Imaginary axis (x, y) represents x + jy y=r sin θ Real axis θ x = r cos θ Figure 7.2  Illustration of polar . between output and input magnitude, and a phase shift. –1 4j –1 2j –1 0j –1 –2j –0.5 4j –0.5 2j –0.5 0j –0.5 –2j 0 –2j 0.5 –2j 1 –2j –1 –4j –0.5 –4j 0 –4j 0.5 –4j 1 –4j 0 4j 0.5 4j 1 4j 1 2j 1 0j 0.5. feedback.) 91239.indb 83 10/12/09 1 :42 : 34 PM 84 ◾ Essentials of Control Techniques and Theory Now we have: v G kv in out =       1 + which can be rearranged to give the gain, the ratio of output to input,. but never the products of variables, whether states or inputs. 91239.indb 81 10/12/09 1 :42 :31 PM 82 ◾ Essentials of Control Techniques and Theory Now the derivative of a sine-wave is another