Weird condition for a transfer function I am reading a paper that presents the following system, represented by a second-order transfer function:
$G(s) = \frac{K\times(1+0.036s)}{(1+0.0018s)(1+as)}$, 
where the gain $K$ is a known constant. They explicitly declare $a<0.0018$, but I don't understand why. What particular property or behavior needs this condition?
I do not come from a solid background with skills of control or differential equations, so please bear with me if this is too simple. I'm not even sure if math.stackexchange is the correct stackexchange place for this question.
I have tried to analyse this by reading a lot about this second-order systems and the only thing that seems interesting is that the damping factor $\zeta >1$ for all values of $a$ except $a=0.0018$, where the damping factor is 1. 
 A: From doing a little bit of background reading, it seems as they want to create a sort of band-pass filter. I cannot locate the original 1965 paper on which the model is based (reference 35 in your link), but it seems as though they select $a$ in such a way as to define the frequency range that gets passed. In this case, they want $a < c_1$, $a < c_2$, where $c_1 = 0.0018$ and $c_2 = 0.0036$. Hence, the condition. Since $a$ and $c_1$ define the frequency range to be "passed", the restriction on $a$ gives the desired filter behavior.
A: The transfer function
$$G (s) = \displaystyle\frac{K \, (1 + 0.036 s)}{(1 + 0.0018 s) (1 + a s)}$$
can be rewritten in the following form
$$G (s) = \displaystyle\frac{\tilde{K} (a) \, (s + \frac{1}{0.036})}{(s + \frac{1}{0.0018}) (s + \frac{1}{a})} \approx \displaystyle\frac{\tilde{K} (a) \, (s + 27.8)}{(s + 555.6) (s + \frac{1}{a})}$$
where $\tilde{K} (a) = \frac{0.036 K}{0.0018 \, a} = \frac{20 K}{a}$. Note that the zero at approximately $s = -27.8$ creates a rising ramp in the Bode plot starting at frequency $\omega = 27.8$ rads / second. The pole at approximately $s = -555.6$ will flatten the Bode plot after frequency $\omega = 555.6$ rads / second. The condition $a < 0.0018$ (I will assume that $a > 0$ to ensure stability) ensures that the pole at $s = - \frac{1}{a}$ is "faster" than the one at $s = -555.6$, which will produce a falling ramp in the Bode plot starting at frequency $\omega = \frac{1}{a} > 555.6$ rads / second. Here is a possible Bode plot:

where I chose gain $K = 1$ and used $a = 0.0009$. This could be a bandpass filter, but it looks more like some sort of lead-lag compensator. Here is the MATLAB code that generates the Bode plot:
% system parameters
K = 1;
a = 0.5 * 0.0018;
K_tilde = 20 * K / a;

% create transfer function
sys = zpk([-inv(0.036)],[-inv(0.0018),-inv(a)],K_tilde);

% Bode plot
figure; bode(sys);

A: It is hard to say exactly without reading the paper, but the answer may well be related to stability of the system.  Stability in various senses can be inferred by looking at the poles of $G(s)$, which are defined as the zeros of the denominator.
For instance a system is asymptotically stable (if you excite the system at time zero, the output will eventually settle back to zero) if and only if the real parts of all poles are negative.  How close these real parts are to zero determines how slowly the output settles back to zero after being excited.  The condition $a<0.0018$ says that the pole at $s = \frac{-1}{a}$ is loser to the imaginary axis than the one at $s=\frac{-1}{0.0018}$, and so the former is the one relevant to stability.
