Routh Stability Criteria To Determine Kc Range - Example Problem

So this is an example where we use the root stability criteria to determine the range of controller parameters to stabilize a process. Now this is typical for a cascade control system. Here you see that we have the primary process having a second order dynamics with the right half plane. Zero. We have the first order process with the time delay and a first order dynamics. For this case, we have a pi controller for the secondary loop and a proportion only controller for the primary loop. So the.

Objective is to determine the range of the controller gain for which the process will be stable. Now, as you can see from this diagram that the output is affected by both set point and the disturbance, so both regulatory and tracking control is involved for the disturbance dynamics. We see that the pole is at has a value of negative 0.05.

So having a negative pole, the output will be stable with change in the disturbance. If the loop transfer function is stable, okay, meaning that they will have the same. Characteristic equation, meaning the relation between y and r and y and d will have the same characteristic equation. So we can show that the YM r has a transfer function that has a characteristic equation, that's stable that will determine that the process will also be stable with respect to change in the disturbance. So to find the transfer function between the output and the set and to determine the range of kc for which it will stable that's the objective here. And we can use the root stability. Analysis method to determine the range of KSC.

Now to start with we have learned about this four-step procedure to use the root criteria to analyze stability. The first step is to formulate the characteristic equation. So we'll go step by step first to obtain the characteristic equation to find the transfer function between y and r. We see that there are two loops here, the inner loop and the outer loop. So what we'll do first we'll try to simplify this in a loop to represent it by one simple, transfer. Function to do.

So if you look at the relation between WS, and we can simply use the relation that w. S, over p, s, it will be all the transfer function between w. S. And p, s, meaning this. And this transfer function that goes in the numerator, and the denominator will be 1, plus all the transfer function in the loop again, the same to transfer function. And if we simply do some mathematical manipulation, we see that we'll get a simple first order model as this one. Okay. So replace the inner loop by this closed. Loop transfer function to simplify the overall block diagram and doing. So we get this diagram with that involves only one loop.

Now we can find the relation between y and r again. We follow the same procedure. The numerator gets all the transfer function between y and r. So between y and r, we have this transfer function this transfer function and this transfer function. So these three goes in the numerator. And then in the denominator, we have one plus all the transfer function in the loop.

So in the. Loop what additional we have been this transfer function for this measurement device. So you see this 4 is not appearing in the numerator it's appearing in the number simply it does not appear anywhere between y and r. So again, if we simplify this equation, we'll end up getting this term in the numerator and the denominator.

We have just simply multiply. The overall thing by 9 is squared, plus 2.5 is plus 1 times 6 is plus 1. We end up within the denominator to be this.

Okay. So we have the characteristic. Equation we get by setting the denominator to 0. We get the characteristic equation. Okay.

Now to express the characteristic equation to be applicable for the root method. We need to express it in to a polynomial form. We see that we have a time delay form here. So we need to approximate the time delay.

So we can do it using the either using the Taylor series approximation or the party approximation. We use here, the Taylor's approximation up to the first term to get e to the negative 2. S, equals 1 minus.2 s, and if you plug in this value, we end up getting the characteristic equation to be you see that we have just replaced the time delay term by this polynomial term here. And if we further simplify this equation, we will get the characteristic equation to be in the polynomial form of the decreasing order of s. As this we see that the first term here is already positive. So we don't need to multiply by negative one if it was negative, then you would have to multiply by 81 to get it.

The first value to. Be positive okay, now we got the characteristic equation in the polynomial form. The second step is to check whether the necessary conditions are made the necessary condition requires that all the coefficients should be positive. If not the system is not stable. And if yes you need to go to the next step.

So we got the characteristic equation. We see that the first element that's a positive number so that's positive. I don't need to have all the coefficients to be positive.

So the necessary condition. Is that all of these coefficients should be positive to meet that requirement. We need this 24, plus 32 kc to be greater than 0.

This requires 32 kc should be greater than negative 24 or KFC should be greater than negative 0.75. So the second coefficient will need 8.5 minus 24 kc to be greater than 0 or 24 kc to be less than 8.5 or kc should be less than 0.35. Okay. And the last coefficient 1, plus 4 kc should be again greater than 0. This requires that 4k c should be greater than negative 1 or kc. Should be greater than negative 0.25.

So these are the necessary conditions that kc should be greater than 0.25. It should be less than 0.35. And again, it should be greater than negative 0.75 next.

So we see that the necessary conditions can be met next stable to formulate the root array. So for the root array, we'll have n, plus one column n, plus one rows, meaning, we have four rows and n, plus one over two columns, meaning, two columns. So let's formulate the root array. Then so we'll have one two three. Four rows and one two columns. So the first element it directly goes to the first position.

Here second element, directly goes to the second position. Here. The third element again, comes to the first row to the second column. So and this fourth element again, goes to the second row. Second column 1, plus 4 k, c. Now. So the first element in the third row in the denominator, it will be this term 24, plus 32 kc in the numerator.

It will be this time. This minus this time this okay. So 24, kc 24, plus 32 KFC times.8.5 minus 24 kc, minus 54 times 1, plus 4kc. And the second element there is nothing here. So it should be 0. Okay. The fourth column you see, this element is zero.

So in the denominator, it will be this entire term b1. Okay. And then on the numerator, it will be b1 times one, plus four c, minus this element times 0, so it's 0. So you see that b 1, b, 1 gets cancelled out.

So it should be simply 1 plus 4 k, c now to analyze the stability we have to have all the elements in the first column to be positive. The. Recurrent is that all the elements in the first column should be greater than zero. Now, the first element being a positive number it's, always greater than zero. We see the recurrent will be these three elements be greater than zero. And to meet this requirement, we'll have 24, plus 32 kc greater than zero that will be one that we got as the necessary conditions.

So kc should be greater than negative. Zero point, seven. So you see the last element is also the common with that necessary condition.

So. That will give us kc should be greater than negative 0.25. Okay. Now to deal with this element. Here we see that this entire thing should be greater than zero.

So this individually this term should be greater than zero. So if the numerator term is greater than zero, so that satisfies the requirement, so we'll have 24, plus 32 kc times 8.5, minus 24 kc, minus this 54 times 1, plus 4 kc that should be greater than 0 so that's the requirement. So if we simplify this we'll end up getting 384 PCs squared, minus.260 KFC, plus 75 that should be greater than 0. Okay. Now, if we look at this function, this is a quadratic function in k. C. So a quadratic function with a positive coefficient would simply go in the positive direction and intersecting the x-axis at two points. Now you plot this function, it will be. It will look like something like this just got two roots being a quadratic equation.

You see that these two roots will be one is negative. Zero point, eight, nine, six and the positivity zero point, two, one, eight. Okay, what does simply mean that between these two values, the function will be greater than 0. So the limit of kc for which the element will have a value greater than 0 it's bounded by its root roots.

So kc should be greater than negative 0.896 and kc should be less than 0.218. So these two along with these two formulate, the requirement for the sufficient condition. Now to think about a little that the sufficient condition should also include the necessary condition because the sufficient. Condition is by itself, it's mean by sufficient and necessary. So you see that kc being less than 0.218 will fulfill the condition that kc being less than 0.35.

So if the substitution condition is made the necessary condition is also made. Also, you see that these two conditions are same as the necessary conditions. So if all of these force sufficient conditions are met the necessary condition will also be met. So you see that the necessary condition requires that kc should be greater than 0. Negative.0.75, it should be less than 0.35, and it should be greater than negative 0.25 and the sufficient conditions required that it should be greater than negative 0.75. It should be greater than negative 0.25. It should be greater than negative 0.9, and it should be less than positive 0.21.

Now, if you try to see all of these values on an axis, so we have this negative, three negative values and kc should be greater than these three values. And we have these two positive values and then case should be less.Then those two positive values. Now, if you have three numbers and the variable needs to be greater than all three. So if the variable is greater than the largest number, then that meets all the requirement, meaning that if it's greater than the largest number. So it should be also greater than the other numbers in the same way for these two positive numbers. It requires that the variable should be less than these two numbers. So if it's less than that, the smallest number, then it should mean the.

Requirements that it's smaller than both of these numbers. So you see that the range of kc here, if it meets the requirement that it's greater than the largest of these three and is smaller than the smallest of these two, then it will meet all the requirements. So we have the range of kc to be negative 0.25 to positive 0.21.


Dated : 18-Apr-2022

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