Particular Solutions Again PDF

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Longman 9/21/05 EEME E3601 Handout Number 6 Finding Particular Solutions to Differential Equations -- Response of a Control System to Commands and Disturbances

(A) A Quick Way to Find Particular Solutions for Special Forcing Functions Given an nth order non-homogeneous linear differential equation with constant coefficients

The general solution is the sum of the general solution to the homogeneous equation, found in the previous handout, plus a particular solution , i.e. any function which when substituted into the nonhomogeneous equation, satisfies it:

For control problems we will be creating a particular solution for the command, , and a particular solution for any disturbance, , and adding them together when both are present in . For certain common forcing functions, there is an easy way to find a particular solution given below. If the (i) (ii) (iii)

in the equation above is on the table below then Substitute the corresponding guess from the right hand column of the table into the non-homogeneous differential equation. Adjust the constants in the guess to make it satisfy the equation. The result is a particular solution. If this fails, i.e. if, when you plug in the guess, the left hand side of the equation comes out zero, then multiply the guess by t and try again. Repeat if necessary. You will always get an answer eventually. Forcing Function Guess for Particular Solution ________________________________________________________

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NOTE: The next to the last entry on the table has inputs that are pure sinusoids. The response to a pure sinusoid, considered for a wide range of frequencies of the input, is called the frequency response of the system. When we cover the topic of frequency response, we will develop a different method that is easier to use than the guess given above. But both work. (B) Example of a Proportional Control System with Constant Command and Constant Disturbance

The proportional controller equation is

, and suppose that the plant

differential equation is

. Using the fact that

, combining the equations, and collecting terms that depend on the unknown on the left side of the equation, and collecting terms that are given functions of time on the right hand side of the equation, produces

(1) Consider for a constant command : To find a particular solution associated with the command, set the disturbance to zero. Then the right hand side of the equation is which is a constant. Hence, the first column of the above table applies, and it says we should substitute a constant as a particular solution. Call this constant . Substituting into the equation gives

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So, we asked for and we got (after the transients have become negligible), and this number is less than what we asked for -- by the factor . If we can make K large, we will get something closer to what we asked for. What is the steady state error in response to the command?: It is what we wanted (the command) minus what we got (steady state usually just means after the transients are gone)

And the error is smaller when K is larger. Note that if we are designing a regulator, and we know it will always be too small by this factor, we can fix the problem by asking for more than we really want, i.e. ask for times what we really want, and we will get what we want – i.e. we calibrate the input to account for the behavior of the control system. (2) Consider for a constant command : To find a particular solution associated with the disturbance, set the command to zero. Suppose the disturbance is a constant . According to the table we look for a constant particular solution . Substituting into the equation gives

What is the steady state error in response to the disturbance?: It is what we wanted the disturbance to do -- which is have zero effect on the output -- minus what the disturbance actually did to the output (after the transients are gone)

And the error is smaller when K is larger. NOTE THE DIFFERENCE: (1) For the error in response to a command we take the command minus the response. (2) For the error in response to a disturbance, we take 0 minus the response. (3) Comments on the solution to the homogeneous equation : Both of the above parts of the solution get better as the gain K gets larger. The characteristic polynomial is . Of course, the roots move around as the value of K is changed. We could plot the root locations as a function of K, and this is called a root locus plot. We will have special methods developed for this purpose. But we can understand some properties of the solution by the following reasoning. Suppose the roots are .

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Then the polynomial could be factored into , which when multiplied produces . Equating coefficients we have that and . Let us interpret this result. As we increase K, the sum of the roots always adds to the same value equal to -3, i.e. the average root location (-3 divided by the number of roots which is 3) is fixed no matter what we do with K. But the product of the roots will go to infinity as K goes to infinity. The implication is that some roots must get large, and if one root gets large going toward negative values, there must be other roots that get large going to positive values to keep the average location fixed at -1. Hence, once K becomes sufficiently large, some root or roots must have a positive real part, and the system necessarily will go unstable. SUMMARY: It is a common phenomenon that the particular solution associated with the command gets better as K gets larger, and the particular solution associated with disturbances gets better as K gets larger, but the transient response will most often go unstable when K gets too large which makes an unusable system. Hence, the control system designer is forced to make a compromise between getting the desired behavior of the particular solutions for command and for disturbances and the desired behavior for the transients. (C) Examples of Using the Table (1) Consider the problem

The table says try

. Plug into the differential equation to get

Collect terms with the same power of t together (i.e. collect the coefficients of the linearly independent functions 1 and t into separate groups):

For this to equal zero for all times, for example for time and then for time clear that the coefficient of and the coefficient of must be zero

Solve these equations for the values of C and D that make the guess satisfy the nonhomogeneous equation

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, it is

Let us interpret this particular solution in a control system context. If the t on the right of the differential equation we are solving were the command term, then the response is t times a slope that is a little too small (but bigger K makes the slope closer to right), and it is a little late, actually seconds late, getting going (again, the larger the K the less behind it is). (2) Consider the problem

The table says, try a solution of the form equation produces

. Substituting into the

Again, collect the like terms together, for the linearly independent sine and cosine functions:

For this to be zero for all time, the terms in square brackets must be zero (of course it must be true when which establishes that the first square bracket must be zero, and it must be true when which establishes that the second square bracket must be zero). Therefore, and and therefore the desired particular solution is . (3) Consider the same problem, with the right hand side changed to . When you try the guess it does not work, because the guess is a solution of the homogeneous equation. The rules above say, when this happens change your guess by multiplying by t, i.e. , and try again. The reader can demonstrate that it works. (D) The Unit Step Response We can now use our knowledge of how to get a particular solution and how to find the solution of the homogeneous equation, to find the unit step response. For any control system it should be reasonable to suddenly change the command from one constant value to another. We don’t expect the system to be able to actually perform

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this sudden change in command, and how the system reacts to the sudden change tells us a lot about the behavior of the system in more generality. We specialize this somewhat to make the unit step response problem. We agree to start from zero, with the system totally at rest. With no input on the right hand side of the equation, is obviously a solution to the differential equation, and to be on that solution you start with all zero initial conditions. Starting from this rest situation, at time zero we suddenly command the system to go to one. SUMMARY: To find the unit step response, set all initial conditions to zero, and then at time zero give the command . Example: Find the unit step response of the following system

Using the table above, a particular solution associated with command is , and adding the general solution of the homogeneous equation produces

Setting the initial conditions to zero gives the following equations

Solving for these two constants, produces the unit step response

The first two terms are the transients, and they disappear with time. The settling time according to our definition is 4 seconds. Once this much time has passed, the response to the step is essentially complete (within 1.6%), and we are left with 0.9. We asked for one, and we eventually got 0.9, which is called the steady state response. And the steady state error is after the transient terms are negligible, which is for a 10% error.

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