# Does DC gain affect the unit response characteristics of a 2nd order system?

Given a system with transfer function: $$G(s)=\frac{Κω_n^2}{s^2+2ζω_ns+ω_n^2}$$ Where ω is the natural frequency, K is the DC gain (I'm pretty sure that's what it's called, please correct me if I am mistaken), and ζ is the damping ratio.

Does K not matter when it comes to calculation the system characteristics like the settling time or the overshoot? All the formulae I can find, including those in my professor's notes, make no mention of K.

For example, my professor's notes have this one: $$T_{settling}\approx\frac{4}{ζω_n}$$

Also, I can't find information on such systems with real poles. Pretty much everyone assumes there are going to be two complex poles. If the poles are real, do they have a different frequency each or something? And how do I estimate these metrics in this case?

Specifically, I'm trying to solve an old exam for practice and came upon a transfer function with $K=1/2$, $ω_n=\sqrt2$ and $ζ=\frac{3\sqrt{2}}{4}$, which has two real poles (-1, -2) and the students were asked to find the $\lim_{t\to \infty}y(t)$ (I assume using the final value theorem), the maximum of the output y(t) and the percent overshoot (I assume both imply a unit step input). However, there are no formulae involving K and all the analysis is about complex poles. Am I supposed to approximate to a first order system? The poles aren't that far from the origin.

Settling time is usually defined as when when the step response is within 2% of the steady state response. Even though the difference between the step response and the steady state response gets bigger for all times when you use a bigger gain $K$, the percentage difference remains the same. That is why $K$ does not affect settling time.

The poles of your system are,

$$p = \omega_n \left(-\zeta \pm \sqrt{\zeta^2 - 1}\right).$$

The transient response of such a system will look like,

$$y_t(t) = C_1\, e^{\omega_n \left(-\zeta + \sqrt{\zeta^2 - 1}\right) t} + C_2\, e^{\omega_n \left(-\zeta - \sqrt{\zeta^2 - 1}\right) t},$$

where $C_1$ and $C_2$ depend on the initial conditions.

For $\zeta<1$ the poles will be each others complex conjugate and the transient can also be written as,

$$y_t(t) = e^{-\omega_n\, \zeta\, t} \left(C_3 \cos\left(\omega_n\,\sqrt{1 - \zeta^2}\, t\right) + C_4 \sin\left(\omega_n\,\sqrt{1 - \zeta^2}\, t\right)\right).$$

For a step response the total response can be written as the sum of the steady state and the transient response ($y(t)=y_{ss}(t)+y_t(t)$), with the steady state equal to $K$ (which can be found by setting $s$ to zero). For a step response it is assumed that the system is initial at rest, so position and velocity are zero for all $t\leq0$. Applying these constraints yields $C_3=-K$ and $C_4=0$,

$$y(t) = K \left(1 - e^{-\omega_n\, \zeta\, t} \cos\left(\omega_n\,\sqrt{1 - \zeta^2}\, t\right)\right).$$

Using an upper bound of one for the cosine, then the settling time can be approximated by only considering the exponential,

$$\left|\frac{y(t)-y_{ss}(t)}{y_{ss}(t)}\right| \approx e^{-\omega_n\, \zeta\, t}.$$

It can be noted that if you work out the left hand side of this equation yourself you will see that all $K$'s cancel.

The settling time for this approximation can be found by solving when $e^{-\omega_n\, \zeta\, t}=0.02$. This has the solution,

$$T_{settling} = \frac{-\ln(0.02)}{\zeta\,\omega_n},$$

$-\ln(0.02)$ is indeed close to $4$.

Solving for the constants for two real poles (when $\zeta\geq 1$) you get,

$$C_1 = K \frac{\sqrt{\zeta^2 - 1} + \zeta}{2 \sqrt{\zeta^2 - 1}},$$

$$C_2 = K \frac{\sqrt{\zeta^2 - 1} - \zeta}{2 \sqrt{\zeta^2 - 1}}.$$

For $\zeta$ equal or slightly bigger than 1 you will have to solve the following equation for $t$ in order to get the settling time,

$$\left|e^{-\omega_n\, \zeta\, t} \left(\cosh\left(\omega_n \sqrt{\zeta^2 - 1}\, t\right) + \frac{\zeta}{\sqrt{\zeta^2 - 1}} \sinh\left(\omega_n \sqrt{\zeta^2 - 1}\, t\right)\right)\right| = 0.02.$$

When $\zeta\gg1$ then $\zeta\, (\zeta^2 - 1)^{-1/2} \approx 1$ and $\cosh(t)+\sinh(t)=e^t$ can be used. Solving the above equation then has the solution,

$$T_{settling} = \frac{-\ln(0.02)}{\left(\zeta - \sqrt{\zeta^2 - 1}\right) \omega_n}.$$

This can also be seen as taking the biggest (least negative) pole, instead of the real part of the complex conjugate poles. Because when $\zeta$ is large the difference between the two real poles is large and only the "slowest" pole will dominate the transient response for the settling time.

• @Grigorios How did you end up solving for the settling time, because in your example case $\zeta$ is only slightly bigger than one, so neither of the approximations can be used? – Kwin van der Veen Jan 19 '17 at 16:02
• I didn't... I was studying for an upcoming exam that was cancelled about an hour after I posted the question so I lost my concentration and gave up, until yesterday. I'd probably use the 4/ζω approximation in an exam setting since this is the only formula in the notes, or I'd go the long way and get that formula you gave if I could. – Grigorios Jan 20 '17 at 10:15