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Continuous to Discrete Conversion
PID Controller
In this page, we will consider the digital control version of DC motor speed problem. A digital DC motor model can be obtained from conversion of the analog model, as we will describe. The controller for this example will be designed by a PID method.
From the Modeling: a DC Motor, the open-loop transfer function for DC motor's speed was derived as:
The design requirements for 1 rad/sec step input are
From the design requirement, let the sampling time, Ts equal to 0.12 seconds, which is 1/10 the time constant of a system with a settling time of 2 seconds. Let's create a new m-file and enter the following commands:
R=1; L=0.5; Kt=0.01; J=0.01; b=0.1; num = Kt; den = [(J*L) (J*R)+(L*b) (R*b)+(Kt^2)]; Ts = 0.12; [numz,denz] = c2dm(num,den,Ts,'zoh')
Running this m-file should return the following:
numz =
0 0.0092 0.0057
denz =
1.0000 -1.0877 0.2369
From these matrices, the discrete transfer function can be written as:
First, we would like to see what the closed-loop response of the system looks like without any control. If you see the numz matrices shown above, it has one extra zero in the front, we have to get rid of it before closing the loop with the Matlab cloop command. Add the following code into the end of your m-file:
numz = [numz(2) numz(3)]; [numz_cl,denz_cl] = cloop(numz,denz);After you have done this, let's see how the closed-loop step response looks like. The dstep command will generate the vector of discrete output signals and stairs command will connect these signals. Click here for more information. Add the following Matlab code at the end of previous m-file and rerun it.
[x1] = dstep(numz_cl,denz_cl,101);
t=0:0.12:12;
stairs(t,x1)
xlabel('Time (seconds)')
ylabel('Velocity (rad/s)')
title('Stairstep Response:Original')
You should see the following plot:
There are several ways for mapping from the s-plane to z-plane. The most
accurate one is 
. We cannot obtain PID transfer
function in this way because the discrete-time transfer function would have
more zeroes than poles, which is not realizable. Instead we are going to use
the bilinear transformation shown as follows:
Thus we can derive the discrete PID controller with bilinear transformation mapping. For more detail derivation of discrete PID controller, see Discrete PID Controller. Equivalently, the c2dm command in Matlab will help you to convert the continuous-time PID compensator to discrete-time PID compensator by using the "tustin" method in this case. The "tustin" method will use bilinear approximation to convert to discrete time of the derivative. According to the PID Design Method for the DC Motor page, Kp = 100, Ki = 200 and Kd = 10 are satisfied the design requirement. We will use all of these gains in this example. Now add the following Matlab commands to your previous m-file and rerun it in Matlab window.
% Discrete PID controller with bilinear approximation Kp = 100; Ki = 200; Kd = 10; [dencz,numcz]=c2dm([1 0],[Kd Kp Ki],Ts,'tustin');Note that the numerator and denominator in c2dm were reversed above. The reason is that the PID transfer function is not proper. Matlab will not allow this. By switching the numerator and denominator the c2dm command can be fooled into giving the right answer. Let's see if the performance of the closed-loop response with the PID compensator satisfies the design requirements. Now add the following code to the end of your m-file and rerun it. You should get the following close-loop stairstep response.
numaz = conv(numz,numcz);
denaz = conv(denz,dencz);
[numaz_cl,denaz_cl] = cloop(numaz,denaz);
[x2] = dstep(numaz_cl,denaz_cl,101);
t=0:0.12:12;
stairs(t,x2)
xlabel('Time (seconds)')
ylabel('Velocity (rad/s)')
title('Stairstep Response:with PID controller')
As you can see from the above plot, the closed-loop response of the system is unstable. Therefore there must be something wrong with compensated system. So we should take a look at root locus of the compensated system. Let's add the following Matlab command into the end of your m-file and rerun it.
rlocus(numaz,denaz)
title('Root Locus of Compensated System')
From this root-locus plot, we see that the denominator of the PID controller has a pole at -1 in the z-plane. We know that if a pole of a system is outside the unit circle, the system will be unstable. This compensated system will always be unstable for any positive gain because there are an even number of poles and zeroes to the right of the pole at -1. Therefore that pole will always move to the left and outside the unit circle. The pole at -1 comes from the compensator, and we can change its location by changing the compensator design. We choose it to cancel the zero at -0.62. This will make the system stable for at least some gains. Furthermore we can choose an appropriate gain from the root locus plot to satisfy the design requirements using rlocfind.Enter the following Matlab code to your m-file.
dencz = conv([1 -1],[1.6 1])
numaz = conv(numz,numcz);
denaz = conv(denz,dencz);
rlocus(numaz,denaz)
title('Root Locus of Compensated System');
[K,poles] = rlocfind(numaz,denaz)
[numaz_cl,denaz_cl] = cloop(K*numaz,denaz);
[x3] = dstep(numaz_cl,denaz_cl,101);
t=0:0.12:12;
stairs(t,x3)
xlabel('Time (seconds)')
ylabel('Velocity (rad/s)')
title('Stairstep Response:with PID controller')
The new dencz will have a pole at -0.625 instead of -1, which almost cancels
the zero of uncompensated system. In the Matlab window, you should see the command
asking you to select the point on the root-locus plot. You should click on the
plot as the following:
Then Matlab will return the appropriate gain and the corresponding compensated poles, and it will plot the closed-loop compensated response as follows.
The plot shows that the settling time is less than 2 seconds and the percent overshoot is around 3%. In addition, the steady state error is zero. Also, the gain, K, from root locus is 0.2425 which is reasonable. Therefore this response satisfies all of the design requirements.
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