Courses on automatics define usually the second order transfer function as

$$ H(s) = \frac{1}{\frac{s^2}{w_n^2} + \frac{2z}{w_n} s + 1} $$ (sse here for instance).

However, it assumes that the coefficents $a,c$ of the corresponding second order differential equation $$ a\ddot x(t) + b \dot x(t) + c x(t)= u(t) $$ have the same sign.

Why is that ? If I consider the inverted pendulum linearized at the unstable equilibrium, then these coefficients won't have the same sign since the dynamics of the angle is: $$ m\ddot \theta(t) - mg\theta(t) = u(t) $$


1 Answer 1


The classical form of a second order system, as I have usually seen it, is:

$$ T(s) =K \frac{ω_n^2}{s^2+2ζω_ns+ω_n^2} $$

  • $K \ \rightarrow $ DC Gain of the system
  • $ζ \ \ \ \rightarrow $ Damping Ratio of the system
  • $ω_n \rightarrow $ Natural Frequency of the system

By definition the damping ratio of a system is $ζ \geq 0$. As its name suggests this term is responsible for the exponential decay of any oscillation the system is subject to. Imagine it being equal to $ζ=0$, this means that there is nothing to "slow down" the system when it oscillates and it will oscillate forever. It is obvious that it can't be negative since this would mean that instead of "slowing down" the system, it would enhance its oscillation. For a more physical approach you can think of the damping as being the friction, the air resistance, water resistance etc. Anything out there in the physical world which adds resistance to the movement of the system.

As far as the natural frequecny goes there is a very good definition at Wikipedia. Imagine that there isn't any damping force acting on a system ($ζ=0$). Then the natural frequency is the frequency at which the system oscillates when no damping forces are applied to it. For physically realizable systems, this frequency can never be less than $0$ (I think this is obvious).

The second order differential $a\ddot{x(t)}+b\dot{x(t)}+cx(t) = u(t)$ is transformed into the $s$-domain as follows (assuming $0$ initial conditions):

$$ s^2aX(s)+sbX(s)+cX(s)=U(s) \Rightarrow \frac{X(s)}{U(s)} = \frac{1}{s^2+\frac{b}{a}s+\frac{c}{a}} $$

The denominator of the transfer function is the characteristic equation of the system. In order for a second order system to be stable, the coefficient of $s$ and the constant coefficient need to be greater than $0$ (derived from Routh-Hurwitz criterion). So, suppose $a>0$ then $b$ and $c$ should also be positive in order for this system to be stable. Same goes for the case of $a$ being negative.

Now for the inverted pendulum, as you wrote, the straightforward linearized system is unstable. This means that the coefficients of the characteristic equation don't have the same sign. Although I can't speak french I managed to understand that the link you provided assumes that the coefficients are positive. This goes for the case of the stable systems and also for the systems represented using the damping ratio and natural frequency terms, which is indeed the truth.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.