# System Performance

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”.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.

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

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

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 occurs at

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

, so

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

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

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 $\zeta$

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

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

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.In all other cases, the system will be unstable.

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.

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.

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.This also brings another observation.

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,

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

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.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.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

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

Last modified 11mo ago