I have asked these 2 questions in some other forums before, but it always have been buried deep with my other questions, resulting to it being overlooked.
However, I think that these 2 questions deserve its own title here. I feel that the answer to these questions should be mandatorily written in every control system textbook talking about critically damped PID control.
The situation is we have a 2nd order ode:
$$
\ddot{x}+\left[ 2-p \right]\dot{x}+\frac{\left[ 2-p \right]^{2}}{4}x=u
$$
Clearly, it is critically damped ($a^{2}=4b$). But u is:
$$
u=2\dot{r}+\frac{\left[ 2-p \right]^{2}}{4}r
$$
where r is some constant. Notice that $\dot{r} = 0$, when r is a constant.
Hence we can write the transfer function from r to x is:
$$
\frac{x(s)}{r(s)} = G(s) = \frac{2s+\frac{\left[ 2-p \right]^{2}}{4}}{s^{2}+\left[ 2-p \right]s+\frac{\left[ 2-p \right]^{2}}{4}}
$$
Notice that it has zeros on the numerator of G(s). This zeros will cause an overshoot, even though $\dot{r} = 0$.
The 2 questions:
- How to explain the zeros exist, even when $\dot{r}=0$? If you 'properly' do the Laplace transform with $\dot{r}$, you indeed get G(s). However, if you substitute $\dot{r}=0$ in the beginning (before you do the Laplace transform), you get G(s) without the 2s term in the numerator, resulting to no zeros.**
- How to explain, without doing simulation, that the zeros is the culprit for the overshoot? I thought that zeros have no effect on final steady state value. What I meant by without doing simulation is, I don't want people to just show me using a computer simulation that when system has no zeros, it has no overshoot. I already know that, what I don't know is how to "rigorously" proof that it is the zeros that is causing the overshoot.
Best Answer
The answer to your first question, there is no zero in the transfer function when you set $\dot{r}=0$, and this is mostly certainly due to some confusion.
When you differentiate the reference and consider the term $\dot{r}(t)$ in the control input, the corresponds in the Laplace domain to a term $sR(s)$ where $R(s)$ is the Laplace transform of the reference. As a result, you will introduce a zero in the dynamics. This zero will not be there if you do not consider that derivative term.
However, it will be there if you consider an ODE of the form
$$\ddot{x}+\left[ 2-p \right]\dot{x}+\frac{\left[ 2-p \right]^{2}}{4}x=2\dot{u}+\frac{\left[ 2-p \right]^{2}}{4}u.$$
If you consider a step input, then you may have some overshoot which is due to the presence of the zero. I think that there is a confusion here between the two models.
For the second question, I am assuming that we are working with the transfer function with the zero, which can be rewritten as
$$G(s)=\dfrac{2s+\lambda^2}{(s+\lambda)^2}$$
where $\lambda=(p-2)/2$ and I am assuming that $p$ is such that $\lambda<0$.
It can be proven analytically for such a simple system that the zero may yield an overshoot but we can first get some intuition about that.
The zero is given here by $-\lambda^2/2$ and it is known that zeros introduce some amplification (check the Bode diagram). Therefore, if the zero acts on the lower frequency range before the poles attenuate the response, then we will have an overshoot. That is this will be the case if $\lambda^2/2<<\lambda$ or, in other words, if $\lambda<<2$. In fact, since the system is critically damped, this will happen if and only if $\lambda<2$.
Conversely, if the zero kicks in after the poles have started attenuating the response, then the effect of the zero will be negligible and there will be no overshoot. This will be the case if $\lambda>2$ and, similarly, since the system is critically damped, this will happen if and only if $\lambda\ge2$. Note that in the case where $\lambda=2$, we have a pole-zero cancellation and no overshoot since the resulting system is of first-order.
Another, more analytic, approach would be to notice that due the linearity of the system, the response of the system is the sum of the response to the transfer function
$$G_d(s)=\dfrac{2s}{(s+\lambda)^2},$$
and that of
$$G_p(s)=\dfrac{\lambda^2}{(s+\lambda)^2}.$$
We now that the step response of the second one is critically damped. Therefore, if there is an overshoot, it is necessarily due to the presence of the zero.
We have that
$$y_d(t)=2te^{-\lambda t}$$
and
$$y_p(t)=1 - \lambda te^{-\lambda t} - e^{-\lambda t}.$$
Note that $y_d$ increases first and decreases then to zero. In this regard, the only term that can cause the overshoot is the term that depends on $te^{-\lambda t}$ which is given for $y_d+y_p$ by
$$(2-\lambda)te^{-\lambda t}.$$
Since the system is critically damped, then we need this term to be nonpositive, otherwise it will create an overshoot. So, we need that $2-\lambda\le0$.