Quiz 5 & Quiz 6

There is a minor discrepancy in Quiz 6. Equation 1 should be

$$
H(s) = \frac{1}{s^3 + 4s^2 + ms + 6}
$$

 

This makes it consistent with Equation 2 and Figure 1.

13 thoughts on “Quiz 5 & Quiz 6

  1. For part 4b, is it acceptable to find any kp that is larger than the one found in part 4a, as long as the overshoot is less than 5%, or should we be finding the kp for which the overshoot is exactly 5%?

    • To clarify, I have found the butterworth damped condition which does indeed have an overshoot of less than 5%, would this suffice, or do I need to keep increasing kp?

  2. To be clear, Task 4 does not specify any *exact* overshoot. Increasing $k_p$ over the critically damped case should have a clear effect on the settling time (therefore an increase of, say, 1% would be meaningless). You are right that the overshoot for the ITAE-optimal case (my nickname: “Butterworth-damping”) is less than 5%, so you could go even higher, but not much. All of these cases are valid solutions.

  3. Dr. Haidekker,

    Is it possible to go over problem 4 from quiz 5 at the beginning of class tomorrow? I think it would benefit at least a handful of students in the class. It was slightly confusing.

    • Yes, I see it. The problem is probably with the dynamic response in general (what we tried today with the PCR). I’ll be sure to put some focus on this topic on March 14.

  4. Dr. Haidekker,

    I had a question regarding the solutions to quiz 6. Is there a reason why we couldn’t have chosen an inverting amplifier, making kp negative. There didn’t appear to be a restriction that kp > 0. Also there was no description of the system involved either. If Y(s) was a voltage, an amplifier could easily invert it in the feedback loop.

    • Not a bad question… Actually, there is a restriction, but it is possible that I never really emphasized it enough. Our coefficients (e.g., mass, friction, resistance, pressure, …) are generally positive. Implicitly, we assume the same for the control elements. Specifically, if we chose $k_p < 0$, we would instead use $|k_p|$ and change the sign at the summation point, which keeps us within the convention of $k_{something} > 0$.

      Strictly, if the restriction is not given (and you are right, it is not explicitly given) and you chose a $k_p < 0$ you would get a stable system for a limited range of values for $k_p$.

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