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

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