How pole positions set the time response

You will be able to read the damping ratio and natural frequency of a second-order system from its poles, and predict its peak time and overshoot. You will know what the $4/(\zeta {\omega }_n)$ settling rule really approximates. You will be able to tell when a far pole can be ignored. You will find the steady-state error of a unity-feedback loop from its type.

The damping ratio $\zeta$ and natural frequency ${\omega }_n$ fix the standard second-order step response, and both can be read off its poles. Feedback is negative, initial conditions are zero, inputs start at $t=0$, $j$ is the imaginary unit and $s=\sigma +j\omega$; transforms are unilateral from $0^{-}$ (the bilateral transform gives the same result for signals that are zero before $t=0$). Take $T(s)=C(s)/R(s)={\omega }_n^2/D(s)$, $D(s)=s^2+2\zeta {\omega }_{n}s+{\omega }_n^2$, with $0<\zeta <1$ (dimensionless) and ${\omega }_n$ in rad/s. Its DC gain is $T(0)=1$. Completing the square, $D(s)=(s+\alpha {)}^2+{\omega }_d^2$: the poles are $-\alpha \pm j{\omega }_d$, with $\alpha =\zeta {\omega }_n$ (s⁻¹) and damped frequency ${\omega }_d={\omega }_n\sqrt{1-{\zeta }^2}$ (rad/s).

The sine in $c(t)$ never exceeds 1 in size, so the response stays between the decaying curves $1\pm e^{-\alpha t}/\sqrt{1-{\zeta }^2}$, the envelope. Their time constant is $\tau =1/\alpha =1/(\zeta {\omega }_n)$, in s. The settling time $t_s$ is the time after which the response stays inside a band around its final value (2% here). Once the envelope is inside the band, so is the response. That bounds $t_s$ from above. The usual $4\tau$ rule rounds that bound, so it is an estimate, not a bound.

Now add a third pole. Cascade a real pole at $s=-p$ ($p>0$) with the standard form and keep the DC gain at 1: $T(s)=\frac{{\omega }_n^2}{D(s)}\cdot \frac{p}{s+p}$, so a unit step gives $C(s)=\frac{{\omega }_n^{2}p}{sD(s)(s+p)}$. The pole adds a term $r_p{e}^{-pt}$, $r_p$ being the residue of $C$ at $s=-p$. A pair pole $s_k=-\alpha \pm j{\omega }_d$ has residue ${\lim }_{s\to s_k}(s-s_k)C(s)$; the factor $p/(s+p)$ is finite there, so it multiplies the old residue by its value at $s_k$. Both effects are small when $p\gg {\omega }_n$. The pair then decays slowest, as $e^{-\alpha t}$, and carries the large residues, so it is the dominant pair.

A change of about 11 percent is not negligible: $p=10{\omega }_n$ is the edge of the dominant-pole approximation, not deep inside it. The change falls roughly as ${\omega }_n/p$, so set $p$ by the accuracy you need.

The error is $e(t)=r(t)-c(t)$, and the steady-state error $e_{ss}$ is its limit as $t\to \infty$, when that limit exists. Take unity feedback: the feedback block is $H=1$. The summing junction gives $E=R-C$ and the forward path $C=GE$, so $E=R/(1+G)$. With $H=1$, $G$ is also the open-loop transfer function. The loop is stable when every closed-loop pole lies in the LHP (the open left half-plane; the $j\omega$ axis is excluded). The type $N$ of the loop is the number of factors $1/s$ in $G$, that is, the number of integrators in the loop.

Why does an integrator remove step error? Write $G=G_1/s$ with $G_1(0)$ finite and not zero, and let the closed loop be stable and the input a step. Every loop signal then settles to a constant. An integrator fed a nonzero constant grows without limit, so the integrator's input must settle to zero. That input is $G_1$ acting on the error, which settles to $G_1(0)e_{ss}$, so $e_{ss}=0$.

Each infinite entry is a case where $sE$ keeps a pole at $s=0$: it comes from that growing term in $e(t)$, not from the theorem.