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Lecture 16 - Root Locus Based Controller Design Using MATLAB - Electrical Engineering (EE) PDF Download

Lecture 16 - Root Locus Based Controller Design Using MATLAB, Control Systems

 

1 Root Locus Based Controller Design Using MATLAB 

In this lecture we will show how the MATLAB platform can be utilized to design a controller using root locus technique.
Consider the closed loop discrete control system as shown in Figure 1. Design a digital controller such that the closed loop system has zero steady state error to step input with a reasonable dynamic performance. Velocity error constant of the system should at least be 5.

Lecture 16 - Root Locus Based Controller Design Using MATLAB - Electrical Engineering (EE)

Figure 1: A discrete time control system

Gp(s) = Lecture 16 - Root Locus Based Controller Design Using MATLAB - Electrical Engineering (EE)T = 0.1 sec

Gh0Gp(z) 
Lecture 16 - Root Locus Based Controller Design Using MATLAB - Electrical Engineering (EE)

The MATLAB script to find out Gh0Gp(z) is as follows.

>> s=tf(’s’);

>> Gp=10/((s+1)*(s+2));

>> GhGp=c2d(Gp,0.1,’zoh’);

Lecture 16 - Root Locus Based Controller Design Using MATLAB - Electrical Engineering (EE)

Figure 2: Root locus of the uncompensated system

The root locus of the uncompensated system (without controller) is shown in Figure 2 for which the MATLAB command is

>> rlocus(GhGp)

Lecture 16 - Root Locus Based Controller Design Using MATLAB - Electrical Engineering (EE)

Figure 3: Pole zero map of the uncompensated system

Pole zero map of the uncompensated system is shown in Figure 3 which can be generated using the MATLAB command

>> pzplot(GhGp)

One of the design criteria is that the closed loop system should have a zero steady state error for unit step input. Thus a PI controller is required which has the following transfer function in z-domain when backward rectangular integration is used.

Lecture 16 - Root Locus Based Controller Design Using MATLAB - Electrical Engineering (EE)

The parameter Ki can be designed using the velocity error constant requirement.

Lecture 16 - Root Locus Based Controller Design Using MATLAB - Electrical Engineering (EE)

Above condition will be satisfied if Ki ≥ 1. Let us take Ki = 1. With K= 1, the characteristic equation becomes

(z − 1)(z − 0.9048)(z − 0.8187) + 0.004528(z + 0.9048) + 0.04528Kp(z − 1)(z + 0.9048) = 0

Lecture 16 - Root Locus Based Controller Design Using MATLAB - Electrical Engineering (EE)

Now, we can plot the root locus of the compensated system with Kp as the variable parameter.

The MATLAB script to plot the root locus is as follows.

>> z=tf(’z’,0.1);
>> Gcomp=0.04528*(z-1)*(z+0.9048)/(z^3 - 2.724*z^2 + 2.469*z - 0.7367);
>> zero(Gcomp);
>> pole(Gcomp);
>> rlocus(Gcomp)

The zeros of the system are 1 and −0.9048 and the poles of the system are 1.0114 ± 0.1663i and 0.7013 respectively. The root locus plot is shown in Figure 4.

Lecture 16 - Root Locus Based Controller Design Using MATLAB - Electrical Engineering (EE)

Figure 4: Root locus of the system with PI controller

Lecture 16 - Root Locus Based Controller Design Using MATLAB - Electrical Engineering (EE)

Figure 5: Root locus of the system with PI controller

It is clear from the figure that the system is stable for a very small range of Kp. The stable portion of the root locus is zoomed in Figure 5. The figure shows that the stable range of Kp is 0.239 < Kp < 6.31. The best achievable overshoot is 45.5%, for Kp = 1, which is very high for any practical system. To improve the relative stability, we need to introduce D action. Let us modify the controller to a PID controller for which the transfer function in z-domain is given as below.

Lecture 16 - Root Locus Based Controller Design Using MATLAB - Electrical Engineering (EE)

To satisfy velocity error constant, K≥ 1. If we assume 15% overshoot (corresponding to ξ ≌ 0.5) and 2 sec settling time (corresponding to ωn ≌ 4), the desired dominant poles can be calculated as,

Lecture 16 - Root Locus Based Controller Design Using MATLAB - Electrical Engineering (EE)

Thus the closed loop poles in z-plane

Lecture 16 - Root Locus Based Controller Design Using MATLAB - Electrical Engineering (EE)

The pole zero map including the poles of the PID controller is shown in Figure 6 where the red cross denotes the desired poles.

Lecture 16 - Root Locus Based Controller Design Using MATLAB - Electrical Engineering (EE)

Figure 6: Pole zero map including poles of the PID controller

Let us denote the angle contribution starting from the zero to the right most pole as θ1, θ2, θ3, θ4 and θ5 respectively. The angles can be calculated as θ1 = 9.5o, θ2 = 20o, θ3 = 99.9o, θ= 115.7o and θ5 = 129.4o.

Net angle contribution is A = 9.5o − 20o − 99.9o − 115.7o − 129.4= −355.5o. Angle deficiency is −355.5o + 180o = −175.5o

Thus the two zeros of PID controller must provide an angle of 175.5o. Let us place the two zeros at the same location, zpid.

Since the required angle by individual zero is 87.75o, we can easily say that the zeros must lie on the left of the desired closed loop pole

Lecture 16 - Root Locus Based Controller Design Using MATLAB - Electrical Engineering (EE)

Lecture 16 - Root Locus Based Controller Design Using MATLAB - Electrical Engineering (EE)

The controller is then written as GD (z) =  Lecture 16 - Root Locus Based Controller Design Using MATLAB - Electrical Engineering (EE) The root locus of the compensated system (with PID controller) is shown in Figure 7. This figure shows that the desired closed loop pole corresponds to K = 4.33.

Thus the required controller is GD (z) = Lecture 16 - Root Locus Based Controller Design Using MATLAB - Electrical Engineering (EE) If we compare the above

Lecture 16 - Root Locus Based Controller Design Using MATLAB - Electrical Engineering (EE)

Figure 7: Root locus of compensated system

transfer function with the general PID controller, Kp and Kd can be computed as follows.

Kd/T = 0.5761 ∗ 4.33 ⇒ K= 0.2495
Kp + Kd/T = 4.33 ⇒ Kp = 1.835
KiT − Kp − 2Kd/T = −1.518 ∗ 4.33 ⇒ Ki = 2.521

Note that the above Ki satisfies the constraint Ki ≥ 1. One should keep in mind that the design is based on second order dominant pole pair approximation. But, in practice, there will be other poles and zeros of the closed loop system which might not be insignificant compared to the desired poles. Thus the actual overshoot of the system may differ from the designed one.

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FAQs on Lecture 16 - Root Locus Based Controller Design Using MATLAB - Electrical Engineering (EE)

1. What is root locus in control system?
Ans. Root locus is a graphical representation of the possible locations of the system's poles as a parameter, usually the gain, is varied. It helps in understanding the stability and transient response of a control system.
2. How can MATLAB be used for root locus based controller design?
Ans. MATLAB provides a built-in function called "rlocus" that can be used to plot the root locus of a control system. By analyzing the root locus plot, one can design controllers to achieve desired system performance.
3. What are the advantages of using root locus in controller design?
Ans. Root locus provides a visual representation of how the system's poles move as the gain is varied, allowing engineers to easily analyze and design controllers for stability and performance. It helps in understanding the effects of different controller gains on the system's response.
4. How does root locus help in controller design?
Ans. Root locus helps in determining the range of gain values for which the system remains stable. It also helps in identifying the optimal gain value that provides the desired system response. By analyzing the root locus plot, engineers can design controllers to meet specific requirements.
5. Can root locus be used for systems with multiple inputs and outputs?
Ans. Yes, root locus can be used for systems with multiple inputs and outputs. The root locus plot shows the possible locations of the system's poles for different variations of the system's parameters, such as gains or time constants. It can be used to analyze and design controllers for multivariable systems as well.
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