5
\$\begingroup\$

What's the definition of the undamped natural frequency? I've looked and I cannot find it.

A second order system has the following transfer function $$ H(s) = \frac{A_o \omega_n^2}{s^2 + 2 \zeta \omega_n s + \omega_n^2} $$ where \$\zeta\$ and \$\omega_n\$ are the damping coefficients and the undamped natural frequency respectively.

Let's suppose I have a transfer function as follows $$ T(s) = \frac{200}{s^2 + 10s - 50} $$ What would the undamped natural frequency be? I don't know if it'd be \$\sqrt{50}\$ or \$j\sqrt{50}\$ or something else... If \$\omega_n\$ were complex, this would also imply that \$\zeta\$ be complex. Is it even possible to have a complex valued damping coefficient? Also, since it's \$\omega_n^2\$, does this imply that there are always two undamped natural frequencies? A positive and a negative one?

\$\endgroup\$
7
  • \$\begingroup\$ You answered your own question. If you have a system with that transfer function, the undamped natural frequency is defined to be \$\omega_n\$. \$\endgroup\$
    – The Photon
    Jun 17, 2013 at 16:26
  • \$\begingroup\$ Yes I know that. What I'm trying to say is, if I have the system, \$T(s)\$ with those numbers, what would the value of \$\omega_n\$ be? Surely there are some restrictions on \$\omega_n\$, for example, I find it hard to believe that every system has two \$\omega_n\$ values, a positive and negative one; I'm also not so sure about \$\omega_n\$ and \$\zeta\$ being complex valued. \$\endgroup\$
    – user968243
    Jun 17, 2013 at 16:29
  • \$\begingroup\$ Your T(s) can be written as \$\dfrac{200}{(s-3.66)(s+13.66)}\$. Both of it's poles are real poles. So it doesn't have a resonant frequency. \$\endgroup\$
    – The Photon
    Jun 17, 2013 at 16:32
  • \$\begingroup\$ What do you mean? I'm pretty sure real-valued poles can have a resonant frequency: \$\omega_n = \sqrt{\omega_{p1}\omega_{p2}}\$. I understand that conceptually, there will be no frequency at which it appears to resonate; however, I still the natural frequency surely still has some value... What are you suggesting \$\omega_n\$ be? Surely it does not vanish when the poles are real. \$\endgroup\$
    – user968243
    Jun 17, 2013 at 16:38
  • 1
    \$\begingroup\$ @user968243, see my answer from the perspective of the time domain. An overdamped system does not have a natural frequency in any meaningful sense since the impulse response is monotonic. \$\endgroup\$ Jun 18, 2013 at 13:13

3 Answers 3

3
\$\begingroup\$

Rather than the frequency domain, let's look at this in the time domain and particularly, the characteristic equation associated with a linear homogeneous 2nd order differential equation for some system:

\$r^2 + 2 \zeta \omega_n r + \omega^2_n = 0\$.

If the roots of the characteristic equation are real (which is the case if \$\zeta \ge 1\$), the general solution is the sum of real exponentials:

\$Ae^{\sigma_1 t} + Be^{\sigma_2t} \$

where

\$\sigma_1 = -\zeta \omega_n + \sqrt{(\zeta ^2 - 1)\omega^2_n} \$

\$\sigma_2 = -\zeta \omega_n - \sqrt{(\zeta ^2 - 1)\omega^2_n} \$

Since these are real exponentials, there is no oscillation in these solutions.


If the roots are complex conjugates (which is the case if \$\zeta < 1\$), the general solution is the sum of complex exponentials:

\$e^{\sigma t}(Ae^{j\omega t} + Be^{-j\omega t})\$

where

\$\sigma = -\zeta \omega_n\$

\$\omega = \sqrt{(1 - \zeta ^2)\omega^2_n}\$

This solution is a sinusoid with angular frequency \$\omega\$ multiplied by a real exponential. We say the system has a "natural frequency" of \$\omega\$ for a reason that I think is obvious.

Finally, setting \$\zeta = 0\$ (an undamped system) , this solution becomes:

\$Ae^{j\omega_n t} + Be^{-j\omega_n t}\$

which is just a sinusoid of angular frequency \$\omega_n\$.

In summary, a system may or may not have an associated natural frequency. Only systems with \$\zeta < 1\$ have a natural frequency \$\omega\$ and only in the case that \$\zeta = 0\$ will the natural frequency \$\omega = \omega_n\$, the undamped natural frequency.

\$\endgroup\$
6
  • \$\begingroup\$ Brilliant! So there actually is an \$\omega_n\$ regardless, it's just that it doesn't really have an oscillatory meaning in the time domain if the system is underdamped (as it fully contributes to the rise). In the time domain, the step response will oscillate at a particular frequency (if the system is underdamped)---we were told it would oscillate at the natural frequency. Looking at the above equations, this doesn't seem as though it would be true! Could you perhaps also shed some light on this? Looks to me like it will oscilate at \$\sqrt{(1 - \zeta^2)\omega_n^2}\$ rad/s. \$\endgroup\$
    – user968243
    Jun 18, 2013 at 13:58
  • 1
    \$\begingroup\$ @user968243, again, the natural frequency \$\omega\$ is not the same thing as the undamped natural frequency \$\omega_n\$. The underdamped system has a natural frequency that is less than the undamped natural frequency. \$\endgroup\$ Jun 18, 2013 at 14:03
  • \$\begingroup\$ Yes, I realise that. I'm saying in the time domain, the step response will oscillate of the system is underdamped. Will the oscillations occur at a rate of \$\sqrt{(1 - \zeta^2)\omega_n^2}\$ rad/s ? \$\endgroup\$
    – user968243
    Jun 18, 2013 at 14:04
  • 1
    \$\begingroup\$ @user968243, yes the oscillations occur at the natural frequency \$\omega = \sqrt{(1 - \zeta ^2)\omega^2_n}\$ \$\endgroup\$ Jun 18, 2013 at 14:06
  • \$\begingroup\$ Thanks. I get it now! Would you be able to perhaps recommend a good book which goes further with this stuff? \$\endgroup\$
    – user968243
    Jun 18, 2013 at 14:09
1
\$\begingroup\$

Undamped natural frequency occurs when zeta is less than 1. This is, as far as I'm aware the only condition that produces a peak in the frequency spectrum, jw. I used jw because the derivation of your formula,

$$ H(s) = \frac{A_o \omega_n^2}{s^2 + 2 \zeta \omega_n s + \omega_n^2} $$

produces a formula

$$ s = - \zeta \omega_n +/- \omega_n \sqrt{\zeta^2-1} $$

And, if zeta is less than 1 then the sqrt part of the formula yields a complex number that produces a pole that has some magnitude on the jw axis of the pole zero diagram: -

enter image description here

For completeness, the red line in the lower part of the picture, in magnitude is the undamped natural frequency and you can simply prove that by pythagorous.

Your last formula, as the previous answer pointed out, does not produce a complex answer to s and does not therefore have a zeta less than 1.

\$\endgroup\$
1
\$\begingroup\$

Your first equation for H(s) is meant to give the simplest formula for a 2-pole system with complex-conjugate poles, without using complex numbers. \$\omega_n\$ is defined by its use in this formula. If your system doesn't have complex-conjugate poles, then this formula doesn't apply and there is no natural frequency.(edit: see Alfred's answer)

Your transfer function T(s) can't be written in the form of H(s) in your first equation.

The reason is that it is not a resonant system. You can see this by factoring the denominator:

\$T(s) = \dfrac{200}{(s-3.66)(s+13.66)}\$

This system has two real poles, rather than a pair of conjugate complex poles.

Also, notice that the first pole is in the right half-plane, indicating this is also not a stable system.

\$\endgroup\$
2
  • \$\begingroup\$ Doesn't this imply that the equation of \$H(s)\$ I put is incorrect? Or rather, it is only correct if the poles are complex? I don't think this is correct. There must be a frequency despite the poles being purely real. I'm pretty sure if both poles are real, \$\zeta = 1\$ and \$\omega_n = \sqrt{\omega_{p1}\omega_{p2}} = \frac{\omega_{p1} + \omega_{p2}}{2}\$ \$\endgroup\$
    – user968243
    Jun 17, 2013 at 16:41
  • \$\begingroup\$ Yes, the H(s) you gave is the H(s) for a system with a pair of complex poles. \$\endgroup\$
    – The Photon
    Jun 17, 2013 at 16:49

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