4

The value of the error at a given point of time does not really give you insight into the derivative. If the error is constant with time (and is zero, or something other than zero) then the derivative with respect to time is zero. If it happens to be crossing zero at a given time the derivative will depend on the rate of change of the error, and could be ...


3

Because the system is unstable and if you execute rlocus(sys) you will see this The reason you do not see Inf or NaN in the step data is because it runs over a finite time period (50seconds in this case) while the stepinfo solves the system equation to determine what these quantities are and because such a system potentially keeps rising at an infinite rate ...


3

I found the answer in "Fundamentals of Power Electronics" by Robert W. Erickson and Dragan Maksimovic (Pages 365-367). The book works through an example of testing the stability of a system with three crossover frequencies and states the following. Hence the simple phase margin test is ambiguous, and it is necessary to sketch the Nyquist plot to ...


3

I have explained operation of a PI controller in my answer to Understanding the flow of a PI Controller?. You might find it useful to read that as it gives a simple example of what happens the P and I terms in a cruise control system. *Figure 1. The classic PID control function. Source: Wikipedia - PID controller. I mentioned in my linked answer that it had ...


3

I believe your problem is the 940F capacitor, with a time constant of about 600s in combination with the 0.61Ω resistor. What you are seeing here is a very very slow charging capacitor - check out the x axis units of kiloseconds! Did you mean picofarads, or nanofarads? Since the capacitor really is that large (which is ridiculous), then what you are seeing ...


2

To answer my own question: I think the solution is as follows (it took me incredible long to find it out, although the problem seems quite simple): Assuming to be at the edge of just having no complex solution, there must be one double root and one single root: So the ansatz is: $$Ts(1+s)^2+1 = T(1+s_1)^2(1+s_2)$$ This gives by coefficient comparison $$s_1=1/...


1

If the gain crosses zero db multiple times, it's the last (highest frequency) crossing that counts for stability purposes.


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