The step response of a system is how a system $H(s)$ responds to a step input.

$y(t) = \mathcal{L}^{-1}\left\{ \frac{H(s)}{s} \right\}$

$H(s) = \frac{s+\alpha}{s+\beta}.$

$Au(t) + Be^{-\beta t}u(t).$

Thus, the larger $\beta$ is (i.e the deeper in the left half plane it is), the faster the system will “settle”.

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

$\omega_n$ is known as the natural frequency, and $\zeta$is known as the damping factor.

$s = \frac{-2\zeta\omega_n \pm \sqrt{4\zeta^2\omega^2_n-4\omega^2_n}}{2} = -\zeta\omega_n \pm \omega_n\sqrt{\zeta^2 - 1}.$

There are four cases of interest based on $\zeta$.

When $\zeta=0$, the poles are $s = \pm \omega_n j$. Because they are purely imaginary, the step response will be purely oscillatory.

$Y(s) = \frac{1}{s}\frac{\omega_n^2}{s^2+\omega_n^2} \leftrightarrow y(t) = u(t) - \cos(\omega_n t)u(t)$

When $\zeta\in(0, 1)$, the poles are $s = -\zeta\omega_n\pm j\omega_n\sqrt{1-\zeta^2}$. They are complex and in the left-half plane, so the step response will be a exponentially decaying sinusoid. We define the damped frequency $\omega_d = \omega_n\sqrt{1-\zeta^2}$ so that the poles become $s=-\zeta\omega_n \pm \omega_dj$. Notice that $\omega_d < \omega_n$. If we compute the time-response of the system,

$y(t) = \left[ 1 - \frac{e^{-\zeta\omega_nt}}{\sqrt{1-\zeta^2}}\cos\left(\omega_d t - \arctan\left( \frac{\zeta}{\sqrt{1-\zeta^2}} \right)\right)\right]u(t)$

When $\zeta=1$, both poles are at $s=-\omega_n$. The poles are both real, so the time-response will respond without any overshoot.

When $\zeta>1$, the poles are $-\zeta\omega_n\pm \omega_n\sqrt{\zeta^2-1}$. Both of these will be real, so the time-response will look similar to a first-order system where it is slow and primarily governed by the slowest pole.

$\begin{aligned} sY(s) &= \frac{\omega_n^2}{s^2+2\zeta\omega_ns+\omega_n^2} = \frac{\omega_n^2}{\omega_d} \frac{\omega_d}{(s+\zeta\omega_n)^2+\omega_d^2}\\ \therefore \frac{d^{}y}{dt^{}} &= \frac{\omega_n^2}{\omega_d}e^{-\zeta\omega_nt}\sin(\omega_d t)u(t) \qquad (8)\end{aligned}$

The Time to Peak ($T_p$) of a system is how long it takes to reach is largest value in the step response.

Using Equation 8, we see that the derivative is first equal to 0 when $t = \frac{\pi}{\omega_d}$.

$\therefore T_p = \frac{\pi}{\omega_d}$

The Percent Overshoot ($\% O.S$) of a system is by how much it will overshoot the step response.

The percent overshoot occurs at $t = \frac{\pi}{\omega_d}$, so

$\% O.S = e^{-\zeta\omega_n \frac{\pi}{\omega_d}} = e^{\frac{-\zeta\pi}{\sqrt{1-\zeta^2}}}.$

The Settling Time ($T_s$) of a system is how long it takes for the system to start oscillating within 2\% of its final value.

$\begin{aligned} |y(T_s) - 1| < 0.02 \implies \frac{e^{-\zeta\omega_nT_s}}{\sqrt{1-\zeta^2}} = 0.02\\ \therefore T_s = -\frac{1}{\zeta\omega_n} \ln(0.02 \sqrt{1-\zeta^2})\end{aligned}$

$\begin{aligned} r = \omega_d^2 + \zeta^2 + \omega_n^2 = \omega_n^2(1-\zeta^2)+\zeta^2\omega_n^2 = \omega_n^2\\ \cos(\pi-\theta) = \frac{-\zeta\omega_n}{\omega_n} = -\zeta\\\end{aligned}$

What this tells us is that if we search along the vector at angle $\pi-\theta$, we get a constant $\zeta$.

$H(s) = \frac{bc}{(s+c)(s^2+2as+b)}.$

$\begin{aligned} Y(s) &= \frac{1}{s}+\frac{D}{s+c}+\frac{Bs+C}{s^2+as+b}\\ B &= \frac{c(a-c)}{c^2+b-ca}\quad C = \frac{c(a^2-ac-b)}{c^2+b-ca} \quad D = \frac{-b}{c^2-ac+b}.\end{aligned}$

$\lim_{c\to\infty} D = 0 \quad \lim_{c\to\infty} B = -1 \lim_{c\to\infty} C = -a.$

In other words, as the additional pole moves to infinity, the system acts more and more like a second-order. As a rule of thumb, if $Re\{c\}\geq5Re\{a\}$, then the system will approximate a second order system. Because of this property, we can often decompose complex systems into a series of first and second order systems.

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

Thus if $a$ is small, then the effect of the zero is similar to introducing a derivative into the system, whereas if $a$ is large, then the impact of the zero is primarily to scale the step response. One useful property about zeros is that if a zero occurs close enough to a pole, then they will “cancel” each other out and that pole will have a much smaller effect on the step response.

$\mathbf{x}(t) = e^{At}\mathbf{x}(0) + \int_{0}^{t}e^{A(t-\tau)}B\mathbf{u}(\tau)d\tau.$

A system is bounded-input, bounded output(BIBO) stable if $\exists K_u, K_x < \infty$ such that $|\mathbf{u}(t)| < K_{u} \implies |\mathbf{x}(t)| < K_x$.

$\lim_{t\to\infty}\mathbf{x}(t) = \boldsymbol{0}.$

If instead $\lim_{t\to\infty}\mathbf{x}(t) = \infty$, then the system is unstable.

A system is called marginally stable if the zero-input response does not converge to $\boldsymbol{0}$.

A system is marginally stable if there is exactly one pole at $s=0$ or a pair of poles at $s=\pm j\omega_0$.

Consider the unity feedback loop depicted in Figure 4 where we put a system $G(s)$ in unity feedback to control it.

$\lim_{t\to\infty}x(t) = \lim_{s\to0} sX(s).$

$E(s) = \frac{R(s)}{1+G(s)}.$

$K_p = \lim_{s\to0}G(s) \qquad (9)$

$\lim_{t\to\infty} e(t) = \lim_{s\to0} s \frac{1}{s} \frac{1}{1+G(s)} = \frac{1}{1+K_p}$

$K_v = \lim_{s\to0}sG(s) \qquad (10)$

$\lim_{t\to\infty} e(t) = \lim_{s\to0} s \frac{1}{s^2} \frac{1}{1+G(s)} = \frac{1}{K_v}$

$K_a = \lim_{s\to0}s^2G(s) \qquad (11)$

$\lim_{t\to\infty} e(t) = \lim_{s\to0} s \frac{1}{s^3} \frac{1}{1+G(s)} = \frac{1}{K_a}$

Notice that large static error constants mean a smaller error. Another observation we can make is that if a system has $n$ poles at $s=0$, it can perfectly track an input whose laplace transform is $\frac{1}{s^{n-k}}$ for $k\in[0, n-1]$. We give $n$ a formal name.

$\lim_{t\to\infty}\frac{d^{}\mathbf{x}}{dt^{}} = \boldsymbol{0} \implies \lim_{t\to\infty}\mathbf{x} = \mathbf{x}_{ss}.$

$\frac{d^{}\mathbf{x}}{dt^{}} = \boldsymbol{0} = A\mathbf{x}_{ss} + B\cdot I \implies \mathbf{x}_{ss} = -A^{-1}B \qquad (12)$

$\mathbf{e}(t) = \mathbf{r}(t) - \mathbf{y}(t) = 1 - C\mathbf{x}_{ss} = 1 + CA^{-1}B.$

If we take a complex exponential and pass it into a causal LTI system with impulse response $g(t)$, then

$y(t) = e^{j\omega t} * g(t) = \int_{-\infty}^{\infty}g(\tau)e^{j\omega(t-\tau)}d\tau = e^{j\omega t} \int_{0}^{\infty}g(\tau)e^{-j\omega \tau}d\tau.$

This shows us that $e^{j\omega t}$ is an eigenfunction of the system.

$G(j\omega) = \int_0^{\infty}g(\tau)e^{-j\omega\tau}d\tau \qquad (13)$

Suppose we put a linear system $G(s)$ in negative feedback. We know that if $\angle G(j\omega) = (2k+1)\pi$ for some $k\in\mathbb{Z}$, then the output of the plant will be $-|G(j\omega)|e^{j\omega t}$. If $|G(j\omega)| \geq 1$, then this will feed back into the error term where it will be multiplied by $|G(j\omega)|$ repeatedly, and this will cause the system to be unstable because $|G(j\omega)|\geq1$ and thus will not decay.

The gain margin $G_m$is the change in the open loop gain required to make the closed loop system unstable.

The phase margin $\phi_m$is the change in the open loop phase required to make the closed loop system unstable.

$1 + G_me^{-j\phi_m}G(s) = 0.$

At the gain margin frequency $\omega_{gm}$,

$|G_m||G(j\omega_{gm})| = 1 \implies |G_m| = \frac{1}{|G(j\omega_{gm})|}.$

where the gain margin frequency is $\angle G(j\omega_m) = (2k+1)\pi$ for $k\in\mathbb{Z}$. Likewise, at the phase margin frequency $\omega_{pm}$,

$1 + G_me^{-j\omega_m}G(j\omega_{pm}) = 0 \implies -\phi_m + \angle G(j\omega_{pm}) = (2k+1)\pi.$

where the phase margin frequency is $|G(j\omega_{pm})| = 1$.

Notice that if there is a time delay of $T$ in the system, the phase margin will remain unchanged since the magnitude response will be the same, but the gain margin will change because the new phase will be

$\angle G(j\omega) - \omega T.$