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 DC Motor Speed 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:

• Settling time less than 2 seconds
• Overshoot less than 5%
• Steady-state error less than 1%

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)];

Drawing the original Bode plot

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(num,den)

Adding proportional gain

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(num,den,10)

mag =

    0.0139

num = 70*num

Plotting the closed-loop response

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 and add the following code to the end of your m-file:

[numc,denc]=cloop(num, den, -1);
t=0:0.01:10;
step(numc,denc,t)

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.

Adding a lag controller

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)];
num=50*K;	
   
z=1;
p=0.1;
numa=[1 z];
dena=[1 p];
numb=conv(num,numa);
denb=conv(den,dena);

bode(numb,denb)

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.

[numc,denc]=cloop(numb, denb, -1);
t=0:0.01:10;
step(numc,denc,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.

Frequency Response Examples

Cruise Control | Motor Speed | Motor Position | Bus Suspension | Inverted Pendulum | Pitch Controller | Ball and Beam

Motor Speed Examples

Modeling | PID | Root Locus | Frequency Response | State Space | Digital Control: PID

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