Drawing the original Bode plot
Adding proportional gain
Plotting the closed-loop response
Adding a lag controller
From the main problem, the dynamic equations and the open-loop transfer function of the DC Motor Speed Control example are:
and the system schematic looks like:
For the original problem setup and the derivation of the above equations, please refer to the Modeling a DC Motor page.
With the 1 rad/sec step input, the design criteria are:
Create a new m-file and type in the following commands (refer to the main problem for the details of getting those commands).
J = 0.01; b = 0.1; K = 0.01; R = 1; L = 0.5; num = K; den = [(J*L) ((J*R)+(L*b)) ((b*R)+K^2)]; motor = tf(num,den);
The main idea of frequency-based design is to use the Bode plot of the open-loop transfer function to estimate the closed-loop response. Adding a controller to the system changes the open-loop Bode plot, therefore changing the closed-loop response. Let's first draw the Bode plot for the original open-loop transfer function. Add the following code to the end of your m-file, and then run it in the MATLAB command window.
bode(motor)You should get the following Bode plot:
From the bode plot above, we see that the phase margin can be greater than about 60 degrees if w is less than 10 rad/sec. Let's add gain to the system so the bandwidth frequency is 10 rad/sec, which will give us a phase margin of about 60 degrees. To find the gain at 10 rad/sec, you can try to read it off the Bode plot (it looks to be slightly more than -40 dB, or 0.01 in magnitude). The bode command, invoked with left-hand arguments, can also be used to give you the exact magnitude:
[mag,phase,w] = bode(motor,10)
mag =
0.0139
To have an open-loop gain of 1 at 10 rad/sec, we need to add a
control gain of 1/0.0139 or
approximately 72.
contr=72; bode(contr*motor);and rerun your m-file. You should have the following Bode plot:
From the plot above we see that the phase margin is now quite large. Let's see what the closed-loop response look like. Add a % in front of the bode commands to comment them out, and add the following code to the end of your m-file:
sys_cl=feedback(contr*motor,1); t=0:0.01:10; step(sys_cl,t)You will see the following plot:
The settling time is fast enough, but the overshoot and the steady-state error are too high. The overshoot can be reduced by reducing the gain a bit to get a higher phase margin, but this would cause the steady-state error to increase. A lag controller is probably needed.
We can add a lag controller to reduce the steady-state error. At the same time, we should try to reduce the overshoot by reducing the gain. Let's reduce the gain to 50, and try a lag controller of
which should reduce the steady-state error by a factor of 1/0.01 = 100 (but could increase the settling time). Go back and change your m-file so it looks like the following:
num = K; den = [(J*L) ((J*R)+(L*b)) ((b*R)+K^2)]; motor = tf(num,den); Zo = 1; Po = 0.1; contr = 50*tf([1 Zo],[1 Po]); bode(contr*motor)Rerun the file and you will get this plot:
The phase margin looks good. The steady-state error is predicted to be about 1/40dB or 1%, as desired. Close the loop and look at the step response. Add the following lines of code to the end of you m-file and rerun.
sys_cl=feedback(contr*motor,1); t=0:0.01:10; step(sys_cl,t)
Now you have a step response that meets the design requirements. The steady-state error is less than 1%, the overshoot is about 5%, and the settling time is about 2 seconds.
