Problem 2 (10 points) - 2nd order system PD controller...

Problem 2 (10 points) - 2nd order system PD controller design We want to design a motor angular velocity controller including the motor inductance. A design of the PD controller is shown in Figure 1. We want make the percent overshoot (%OS) smaller than 10% and the settling time (Ts) as 1 second. Fig. 1 PD controller Block diagram Hint) if %OS < 10%, damping ratio must be bigger than 0.59 a) Find and plot the open-loop poles on the complex plane (label all graphs). What is the current settling time? (2 points) b) Find the closed-loop transfer function including the PD controller. (2 points) c) Choose minimum Kp and Kd to satisfy both design requirements. (3 points) d) With the Kp and Kd values found, if the input is a constant number, R(t) = 5, what is the final steady state value? (3 points)

Transcript

00:01 So here we are considering a unity feedback control system with pd controller with the maximum overshoot which is less than 10 % is shown below.

00:10 Let's say here rs it is positive cross here we are given the value of kp which is s plus 9 which from here is given that is 10 divided by s plus 12 which from here is 1 divided by s raised to the power 2 plus 6s plus 15 from here we get the value of output that is equals to c of s and this from here is approaching to this point.

00:46 So from here we have to find out the value of kp.

00:50 We have the first file so to find out this we have to first find out the transfer function which is cs divided by the rs which can be calculated as cs divided by rs is equals to g of s multiplied by the h of s which is divided by 1 plus g of 1 multiplied by the h of 1 sorry it is s not 1.

01:15 So c of s divided by the r of s from here become equals to g of s which is equals to kp multiplied by the s plus 9 multiplied by the 10 multiplied by the 1 which is divided by s plus 12 multiplied by the s raised to the power 2 plus 6s plus 15 which is further divided by 1 plus 10 of kp multiplied by the s plus 9 which is divided by s plus 12 multiplied by the s raised to the power 2 plus 6s plus 15.

01:49 So solving this term from here this from here is equals to ts which is equals to cs divided by the rs that is equals to 10 of kp which is divided by s plus 12 multiplied by the s raised to the power 2 plus 6s plus 15 plus 10 of kp.

02:10 So the characteristic equation from here is 1 plus gs multiplied by the hs is equals to 0.

02:18 So this equation from here is that is s plus 12 multiplied by the s raised to the power 2 plus 6s plus 15 plus 10 of kp is equals to 0.

02:28 So this term from here become equals to s raised to the power 3 plus 18 of s raised to the power 2 plus 87 of s plus 10 of kp that is equals to 0...

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