System Performance

Definition 14

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

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

First Order Systems

Definition 15

A first order system is one with the transfer function of the form

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

After applying partial fraction decomposition to them, their step response is of the form

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

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

Second Order Systems

Definition 16

Second order systems are those with the transfer function in the form

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

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

Notice that the poles of the second order system are

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

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

1. Undamped

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

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

   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,

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

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

   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.

The Underdamped Case

If we analyze the underdamped case further, we can first look at its derivative.

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

Definition 17

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 $t = \frac{\pi}{\omega_d}$.

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

Definition 18

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

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

Definition 19

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.

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

Since our poles are complex, we can represent them in their polar form.

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

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

Additional Poles and Zeros of a Second Order System

Suppose we added an additional pole to the second order system so its transfer function was instead

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

Then its step response will be

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

Notice that

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

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.

If we instead add an additional zero to the second order system so its transfer function looks like

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

and its step response will look like

$sY(s) + aY(s).$

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.

Stability

Recall Equation 7 which told us the time-domain solution to state-space equations was

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

Definition 20

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

Following from Definition 20, Equation 7, this means that

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

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

Theorem 3

If all poles are in the left half plane and the number of zeros is less than or equal to the number of poles, then the system is BIBO stable.

Definition 21

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

Theorem 4

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

In all other cases, the system will be unstable.

Steady State Error

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

Figure 4: Unity Feedback Loop

We want to understand what its steady state error will be in response to different inputs.

Theorem 5

The final value theorem says that for a function whose unilateral laplace transform has all poles in the left half plane,

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

Using this fact, we see that for the unity feedback system,

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

Using these, we can define the static error constants.

Definition 22

The position constant determines how well a system can track a unit step.

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

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

Definition 23

The velocity constant determines how well a system can track a ramp.

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

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

Definition 24

The acceleration constant determines how well a system can track a parabola.

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

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

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.

Definition 25

The system type is the number of poles at 0.

This also brings another observation.

Definition 26

The internal model principle is that if the system in the feedback loop has a model of the input we want to track, then it can track it exactly.

If instead we have a state-space system, then assuming the system is stable,

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

Applying this to the state space equations for a step input,

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

Looking at the error between the reference and the output in the 1D input case,

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

Margins

Figure 5: Frequency Response

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.

Definition 27

The frequency response of the system determines how it scales pure frequencies. It is equivalent to the Laplace transform evaluated on the imaginary axis.

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

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.

Definition 28

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

Definition 29

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

We can imagine the gain and phase margin like placing a “virtual box” before the plant as shown in Figure 6.

Figure 6: The Gain and Phase Margin virtual system

The characteristic polynomial of the closed loop transfer function is

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

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

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