1.      Dr. Abner Mallity has a problem.  He has a motor driving a pump and he needs to control the pump speed.  The motor-pump combination has this transfer function.  (The output is pump speed in rpm, and the input is the voltage applied to the pump.)  (Apparently there is a little "springiness" in the shaft that gives the small resonant peak in the Bode' plot.)

Here is the Bode' plot for the motor-pump combination.  Note, on this plot the frequency axis is in Hz.

Mallity needs to control the pump speed and proposes using a standard control loop to do that.  Here is the system he wants to use.

He has a few specifications that the system needs to meet.

• He wants the system to have a good speed of response, and needs to have the zero-db crossing at f = 0.25Hz - or higher.
• He needs the system to have at least 40^o phase margin.  (The lines on the plot are at -140^o and -180^o.)

(5 pts) If he adjusts the gain for a zero db crossing at .25 Hz, what would the phase margin be?  Write your answer to the right and show your work below. f_m = _0_degrees

(5 pts) If he adjusts the gain for a f_m = 40^o  where would the zero db crossing occur?  Write your answer to the right and show your work below.   f_zero = _0.19_ Hz.

And, here is the Bode' plot for just the compensator.  Note that the compensator has been designed so that the compensator has zero db (gain = 1.0) DC gain.

Willie and Milly argue that this compensator can be used to produce the required phase margin.

(10 pts) Determine the phase margin that can by achieved with a zero db crossing of .25 Hz by using this compensator, and determing the gain, K, that is needed to get that phase margin.

• Willie and Milly have constantly reminded Dr. Mallity of the following two points.
• The total open loop gain for the system (in db) - at any frequency - is the sum of the gain of the system (the first Bode' plot) and the gain of the compensator (the second Bode' plot) and the gain that you add (K).
• The total phase for the system - at any frequency - is the sum of the phases of all three components - the system, the compensator and the gain that you add.
• Give your answers here.
• (5pts) f_m = _40_degrees
• (5pts) K = _-15_db, and K = _0.178   .

        Here is the Bode' plot with f = 0.25 marked with a vertical black line.

It is clear that there is virtually no phase margin, i.e., the phase margin is right about zero degrees. Note that the gain at f = 0.25 is about 10 db, so you would need to drop the gain by that amount.  Now, if we adjust things for a 40^o phase margin, we get the plot below.

Here, we need to adjust the gain yet.  At f = 0.18 or thereabouts, we have a phase of -140^o.  At that frequency, the gain is about 16 db, and we would need to drop the gain by that amount to put the zero db crossing at that frequency (although that answer wasn't required).

        Now, if we add the compensator, we might be able to improve things here.  Looking at the compensator at f = 0.25 Hz, we have the plot below.

At 0.25 Hz, we can see that we can get +40^o from the compensator.  We also will have 5 db gain added by the compensator.  Since we had -180^o without the compensator, we can get a 40^o phase margin by adding the compensator and adjusting the gain.  Examining the gain, we see that we have the following:

• 10 db gain from the system
• 5 db gain from the compensator.
• A total of 15 db.