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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.

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up vote 2 down vote accepted

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:

Bode plot of $G (s)$

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);
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Sorry to have taken weeks to accept this answer. I have been following some control engineering videos and it seems that you are right on the spot; the system is a lead-lag compensator – YuppieNetworking Aug 30 '12 at 10:28

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.

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Interesting @Ed, thank you for doing that background check. Me neither, I cannot find the 1965 paper. Now, I've seen some bode plots of this function and I wouldn't say its a bandpass filter since the form looks more like a lowpass. I'll recheck that and get back to you on monday, I'm away for the weekend. – YuppieNetworking Aug 10 '12 at 10:17
@YuppieNetworking A low-pass filter is typically realized by a first order ODE, which would have a single-pole in the transfer function, and is equivalent to a RC circuit. Band-pass filters are typically analogous to RLC circuits, which are second order ODEs, and have two poles in the transfer function. That's not to say that you can't construct an RLC that works more like a low-pass filter, though. Have you looked at a Bode plot of the system when $a > 0.018$? How is it different? – Emily Aug 10 '12 at 13:32

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.

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Here is the paper, hope you can access. If you can't and you are interested please send a private message. Back to the question: since both poles are real negative numbers (assuming $a>0$, but not necessarily $a<0.0018$), the system is asymptotically stable. So i still don't see the point of the condition $a<0.0018$ when considering stability – YuppieNetworking Aug 9 '12 at 18:18
Hmm. It is not at all clear to me glancing at the paper what the relevance of $a<0.0018$ is. It could just be a minor piece of side information. – Noah Stein Aug 9 '12 at 18:50

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