What is the bandwidth for a 2nd order system?

Question

I know that for a first-order system, the bandwidth can be computed known the time constant, tau, where the bandwidth is equal to 1/τ.

Is there an equivalent formulation for a second-order system, particularly an overdamped system? I have done a search but nothing obvious popped up. I need a formula and derivation if such a thing exists rather than a call to a Matlab function for example. I did come across this on page 12: http://engineering.nyu.edu/mechatronics/Control_Lab/Criag/Craig_RPI/2002/Week2/Second-Order_System_2002.pdf

but it doesn't say where the formulation comes from but suggests a good approximation is the natural frequency $\omega_n$.

Answer 1

This question is related to:

Contradiction of bandwidth and damping

We know that the transfer function for a second-order system is:

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

Switching s to $j\omega$ and dividing top and bottom by $\omega_n^2$ and setting $u = \omega/\omega_n$ we obtain:

$$ H(j\omega) = \frac{1}{1 + j 2 \zeta u - u^2} $$

From this, we can derive the amplitude A:

$$ A = \frac{1}{\sqrt{(1-u^2)^2 + (2\zeta u)^2}} $$

Since the bandwidth is $1/\sqrt{2}$ of the value of $A$ at DC, we can find $u$ at this point which we call $u_c$:

$$ \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{(1-u_c^2)^2 + (2\zeta u_c)^2}} $$

Solving for $u_c$ gives us:

$$ u_c = \sqrt{1 - 2 \zeta^2 + \sqrt{(2 \zeta^2 - 1)^2 + 1}} $$

Since $u_c$ was defined as $\omega/\omega_n$ where $\omega$ is now equal to the bandwidth, $\omega_c$, we can rewrite the above as:

$$ \omega_c = \omega_n \sqrt{1 - 2 \zeta^2 + \sqrt{(2 \zeta^2 - 1)^2 + 1}} $$

This is the bandidth for as second-order system.