Any particular reason you are considering implementing a D term?
The reason I ask is while they are great in theory, providing a higher component for high error steps and smaller component for smaller error steps THEY are very prone to noise.
They are at the end of the day, differentiators. As a result they are usually not considered for practical implementations of control loops. If they are, it is small component or in the form of a lead-lag compensator.
For you specific example with an actuator hitting an end-stop, the issue isn't so much with hitting the end-stop, but with hitting it with "high velocity".
One option to consider is at the output of your PI (maybe D) position loop, the output which is a speed demand into the next loop, is to include a dynamic saturation block.
The saturation limits are then dependent on the absolute position of the actuator. 100% rate for say... 90% of the stroke. then as the actuator comes closer and close to the end-stop, saturation limits are decreased so at... 98% IF there is any high speed demand out of the position loop for whatever reason (instability, noise, BAD command...) the actual speed limit is 10% of maximum
CLTF is: $$\small \frac{Y(s)}{R(s)}=\frac{K(s+a)(s+b)}{s^3+(2+K)s^2+(2+K(a+b))s+abK}$$
To produce a 1st order TF, we would like \$\small (s+a)(s+b)\$ to cancel two of the roots in the denominator. Exact cancellation is not normally attainable, so let's go for approximate.
Thus, write the 'ideal' denominator as \$\small(s+a)(s+b)(s+c)\$, and find the value of \$\small c\$ that approximates this denominator when \$\small K\$ is large.
Expanding \$\small(s+a)(s+b)(s+c)\$ and comparing denominators:
\$\small s^3+(a+b+c)s^2+(ab+bc+ca)s+abc\: \approx\: s^3+(2+K)s^2+(2+K(a+b))s+abK\$
Comparing constant terms, \$\small abc=abK\$, hence \$\small c\rightarrow K\$, and substituting:
\$\small s^3+(a+b+c)s^2+(ab+bc+ca)s+abc\: \approx\: s^3+(a+b+K)s^2+(ab+K(a+b))s+abK\$
Now, let \$\small K\$ become large, so that:
\$\small [(a+b)+K]\approx [2+K]\$ and \$\small [ab+K(a+b)]\approx [2+K(a+b)]\$
Hence we have the denominator:\$\small\ (s^2+(a+b)s+ab)(s+K)\$, and the required 1st order CLTF is:
$$\small \frac{Y(s)}{R(s)}\approx \frac{K}{s+K}$$
To illustrate, consider \$\small a=1\$; \$\small b=1\$; \$\small K=100\$.
The CLTF is:
$$\small \frac{Y(s)}{R(s)}= \frac{100(s^2+2s+1)}{s^3+102s^2+202s+100}$$
It can be seen that \$\small (s^2+2s+1)\$ is an approximate factor of \$\small (s^3+102s^2+202s+100)\$ [it's an exact factor of \$\small (s^3+102s^2+201s+100)\$], giving the 1st order CLTF: $$\small\frac{Y(s)}{R(s)}\approx\frac{100}{s+100}$$
Analysis for small \$\small K\$ can proceed similarly.
Best Answer
Damping means the real part of the complex conjugate pole is larger or smaller, relative to the magnitude of the pole. To be fair, it means much more, but mathematically that's where it gets to. A 1st order system has only real poles, for example:
$$H(s)=\frac{1}{s+1}\quad\Rightarrow\quad s=-1$$
There can be 2nd order systems, and higher, with real poles, and in those cases, damping no longer applies:
$$H(s)=\frac{1}{s^2+3s+2}=\frac{1}{(s+1)(s+2)}\quad\Rightarrow\quad s_1=-1\,,\,s_2=-2$$
But for a 2nd order system, or higher, where there are complex conjugate poles, the realpart is (considering a unity magnitude) \$\Re(s)=-\sqrt{1-\Im(s)^2}\$. With this you get the usual pictures.
In short: you don't see the forest because of the trees, i.e. you got stuck in the namings and can't see the meaning of damping.