# 2nd order transfer function

I know that the standard form of a second-order transfer function is as follows, $$T(S) = \frac{\omega_n^2}{S^2+2\zeta\omega_nS+\omega_n^2}$$

Now I have two transfer functions $$F(S) = \frac{25}{S^2+2S+25}$$ $$G(S) = \frac{25+3S}{S^2+5S+25}$$

F(S) is clearly 2nd order and I can calculate natural frequency and damping ration by comparing it with standard form. But is G(S) second order? If yes, How can I calculate the damping ratio and natural frequency of it?

• The coefficient of s in the TF numerator can have a large influence on overshoot, so the damping ratio calculated from the denominator polynomial might be misleading.
– Chu
Aug 16, 2020 at 6:27

$$\G(s)\$$ is second order. The denominator polynomial has degree 2 (and is not cancelled by a factor of the numerator polynomial).
The natural frequency and damping ratio can be defined in terms of the roots of the denominator polynomial. The presence of the numerator doesn't affect its calculation. So $$\\omega_n^2 = 25\$$ and $$\2\cdot \zeta \cdot \omega_n = 5\$$.
One thing to remember is that the numerator in a transfer function $$\\frac{N(s)}{D(s)} = \frac{Y(s)}{X(s)}\$$ multiplies to the input $$\X(s)\$$. The transfer function is defined for a system with zero initial conditions. But, for the same system starting from non-zero initial conditions, the dynamics of the system is still governed by the denominator polynomial of $$\G(s)\$$. So the natural frequency and damping ratio calculated from the denominator polynomial still hold good (again, assuming that the numerator and denominator do not contain common factors which can be cancelled out).
You have to rewrite the denominator to adopt the following normalized form: $$\D(s)=1+\frac{s}{\omega_0Q}+(\frac{s}{\omega_0})^2\$$. You can then easily swap $$\Q\$$ the quality factor and $$\\zeta\$$ the damping ratio with $$\\zeta=\frac{1}{2Q}\$$.
In your 1st case, you can factor 25 in both the numerator and the denominator to end-up with $$\F(s)=\frac{1}{1+\frac{2}{25}s+(\frac{s}{5})^2}\$$. It is now easy to identify the resonant frequency and the damping ratio or quality factor with the upper normalized expression.
For the second case, you have a zero in the numerator that needs factorization: $$\G(s)=\frac{1+\frac{3}{25}s}{1+\frac{1}{5}s+(\frac{s}{5})^2}\$$. You can now identify a zero and other components such as $$\\omega_0\$$ and $$\Q\$$ or $$\\zeta\$$. In your notations, make sure you use $$\s\$$ and not $$\S\$$.