Lecture 9 the Root Locus Concept

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Lecture 9 The Root Locus Concept

Consider a feedback control system with open loop. The closed loop poles are the values of s that solve the equation: . This equation is exactly the equation:

.(1)

This equation can be replaced by the following two equations:

Hence, s will be a closed loop pole if and only if it satisfies the left equation (the RL magnitude condition) and the right equation (the RL angle condition).

The magnitude condition allows one to compute the required value of K to solve it for almost any chosen value of s. It is the angle condition that governs the behavior of the root locus as . Consequently, the key to understanding the behavior of the root locus is to understand the angles involved in the right equation. To this end, consider

.(2)

This is a difference vector. Specifically, it is the difference between the vector s and the vector .

We will now sketch out this difference vector IN CLASS.

Now that we have done this, it should be clear that for any ‘candidate’ s to be a closed loop pole, the following general angle criteria must hold:

.

Moreover, remember, the closed loop poles solve the equation: . And so for the closed loop poles are very close to the open loop poles. As the closed loop poles move toward the open loop zeros. Let’s look at this behavior in the root locus plot at right.

There are two loci. On begins at and moves toward . The other begins at and moves toward minus infinity. Hence, we say that there is a second zero .

Now let’s prove that this behavior must be as shown.

We went through the proof, not as an exercise in mathematics, but to demonstrate the procedure for carrying out controller design.

Suppose that we desire to place a closed loop pole at . This is not on our current root locus because it does not satisfy the angle criterion. Let’s find out how far away from satisfying this criterion it is:

We are above . By using a controller that has a pole the will contribute to the angle criterion, we will guarantee that will be a closed loop pole for the appropriate value of K. To find the location of this controller pole we first obtain x=2/tand(180-101.3) = 0.4. So, the controller pole is , and the controller transfer function is then . The open loop becomes .

The corresponding root locus is shown at right. From the data cursor we have . We also used the cursor to determine where the added 3rd closed loop pole is.

Comment This pole-placement method is nothing like the PID controller design method we used. In fact, the controller

has no PID elements in it.