System Performance

Definition 14
The step response of a system is how a system H(s)H(s)H(s)
 responds to a step input.
​y(t)=L−1{H(s)s}y(t) = \mathcal{L}^{-1}\left\{ \frac{H(s)}{s} \right\}y(t)=L−1{sH(s)​}
​
First Order Systems
Definition 15
A first order system is one with the transfer function of the form
​H(s)=s+αs+β.H(s) = \frac{s+\alpha}{s+\beta}.H(s)=s+βs+α​.
​
After applying partial fraction decomposition to them, their step response is of the form
​Au(t)+Be−βtu(t).Au(t) + Be^{-\beta t}u(t).Au(t)+Be−βtu(t).
​
Thus, the larger β\betaβ
 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)=ωn2s2+2ζωns+ωn2.H(s) = \frac{\omega_n^2}{s^2+2\zeta\omega_ns+\omega_n^2}.H(s)=s2+2ζωn​s+ωn2​ωn2​​.
​
​ωn\omega_nωn​
 is known as the natural frequency, and ζ\zetaζ
is known as the damping factor.
Notice that the poles of the second order system are
​s=−2ζωn±4ζ2ωn2−4ωn22=−ζωn±ωnζ2−1.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}.s=2−2ζωn​±4ζ2ωn2​−4ωn2​​​=−ζωn​±ωn​ζ2−1​.
​
There are four cases of interest based on ζ\zetaζ
.
1.Undamped
When ζ=0\zeta=0ζ=0
, the poles are s=±ωnjs = \pm \omega_n js=±ωn​j
. Because they are purely imaginary, the step response will be purely oscillatory.
​Y(s)=1sωn2s2+ωn2↔y(t)=u(t)−cos⁡(ωnt)u(t)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)Y(s)=s1​s2+ωn2​ωn2​​↔y(t)=u(t)−cos(ωn​t)u(t)
​
2.Underdamped
When ζ∈(0,1)\zeta\in(0, 1)ζ∈(0,1)
, the poles are s=−ζωn±jωn1−ζ2s = -\zeta\omega_n\pm j\omega_n\sqrt{1-\zeta^2}s=−ζωn​±jωn​1−ζ2​
. They are complex and in the left-half plane, so the step response will be a exponentially decaying sinusoid. We define the damped frequency ωd=ωn1−ζ2\omega_d = \omega_n\sqrt{1-\zeta^2}ωd​=ωn​1−ζ2​
 so that the poles become s=−ζωn±ωdjs=-\zeta\omega_n \pm \omega_djs=−ζωn​±ωd​j
. Notice that ωd<ωn\omega_d < \omega_nωd​<ωn​
. If we compute the time-response of the system,
​y(t)=[1−e−ζωnt1−ζ2cos⁡(ωdt−arctan⁡(ζ1−ζ2))]u(t)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)y(t)=[1−1−ζ2​e−ζωn​t​cos(ωd​t−arctan(1−ζ2​ζ​))]u(t)
​
3.Critically Damped
When ζ=1\zeta=1ζ=1
, both poles are at s=−ωns=-\omega_ns=−ωn​
. The poles are both real, so the time-response will respond without any overshoot.
4.Overdamped
When ζ>1\zeta>1ζ>1
, the poles are −ζωn±ωnζ2−1-\zeta\omega_n\pm \omega_n\sqrt{\zeta^2-1}−ζωn​±ωn​ζ2−1​
. 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.
​sY(s)=ωn2s2+2ζωns+ωn2=ωn2ωdωd(s+ζωn)2+ωd2∴dydt=ωn2ωde−ζωntsin⁡(ωdt)u(t)(8)\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}sY(s)∴dtdy​​=s2+2ζωn​s+ωn2​ωn2​​=ωd​ωn2​​(s+ζωn​)2+ωd2​ωd​​=ωd​ωn2​​e−ζωn​tsin(ωd​t)u(t)(8)​
​
Definition 17
The Time to Peak (TpT_pTp​
) 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=πωdt = \frac{\pi}{\omega_d}t=ωd​π​
.
​∴Tp=πωd\therefore T_p = \frac{\pi}{\omega_d}∴Tp​=ωd​π​
​
Definition 18
The Percent Overshoot (%O.S\% O.S%O.S
) of a system is by how much it will overshoot the step response.
The percent overshoot occurs at t=πωdt = \frac{\pi}{\omega_d}t=ωd​π​
, so
​%O.S=e−ζωnπωd=e−ζπ1−ζ2.\% O.S = e^{-\zeta\omega_n \frac{\pi}{\omega_d}} = e^{\frac{-\zeta\pi}{\sqrt{1-\zeta^2}}}.%O.S=e−ζωn​ωd​π​=e1−ζ2​−ζπ​.
​
Definition 19
The Settling Time (TsT_sTs​
) of a system is how long it takes for the system to start oscillating within 2\% of its final value.
​∣y(Ts)−1∣<0.02  ⟹  e−ζωnTs1−ζ2=0.02∴Ts=−1ζωnln⁡(0.021−ζ2)\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}∣y(Ts​)−1∣<0.02⟹1−ζ2​e−ζωn​Ts​​=0.02∴Ts​=−ζωn​1​ln(0.021−ζ2​)​
​
Since our poles are complex, we can represent them in their polar form.
​r=ωd2+ζ2+ωn2=ωn2(1−ζ2)+ζ2ωn2=ωn2cos⁡(π−θ)=−ζωnωn=−ζ\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}r=ωd2​+ζ2+ωn2​=ωn2​(1−ζ2)+ζ2ωn2​=ωn2​cos(π−θ)=ωn​−ζωn​​=−ζ​
​
What this tells us is that if we search along the vector at angle π−θ\pi-\thetaπ−θ
, we get a constant ζ\zetaζ
.
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)=bc(s+c)(s2+2as+b).H(s) = \frac{bc}{(s+c)(s^2+2as+b)}.H(s)=(s+c)(s2+2as+b)bc​.
​
Then its step response will be
​Y(s)=1s+Ds+c+Bs+Cs2+as+bB=c(a−c)c2+b−caC=c(a2−ac−b)c2+b−caD=−bc2−ac+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}Y(s)B​=s1​+s+cD​+s2+as+bBs+C​=c2+b−cac(a−c)​C=c2+b−cac(a2−ac−b)​D=c2−ac+b−b​.​
​
Notice that
​lim⁡c→∞D=0lim⁡c→∞B=−1lim⁡c→∞C=−a.\lim_{c\to\infty} D = 0 \quad \lim_{c\to\infty} B = -1 \lim_{c\to\infty} C = -a.limc→∞​D=0limc→∞​B=−1limc→∞​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}≥5Re{a}Re\{c\}\geq5Re\{a\}Re{c}≥5Re{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)=s+as2+2ζωn+ωn2H(s) = \frac{s+a}{s^2+2\zeta\omega_n+\omega_n^2}H(s)=s2+2ζωn​+ωn2​s+a​
​
and its step response will look like
​sY(s)+aY(s).sY(s) + aY(s).sY(s)+aY(s).
​
Thus if aaa
 is small, then the effect of the zero is similar to introducing a derivative into the system, whereas if aaa
 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
​x(t)=eAtx(0)+∫0teA(t−τ)Bu(τ)dτ.\mathbf{x}(t) = e^{At}\mathbf{x}(0) + \int_{0}^{t}e^{A(t-\tau)}B\mathbf{u}(\tau)d\tau.x(t)=eAtx(0)+∫0t​eA(t−τ)Bu(τ)dτ.
​
Definition 20
A system is bounded-input, bounded output(BIBO) stable if ∃Ku,Kx<∞\exists K_u, K_x < \infty∃Ku​,Kx​<∞
 such that ∣u(t)∣<Ku  ⟹  ∣x(t)∣<Kx|\mathbf{u}(t)| < K_{u} \implies |\mathbf{x}(t)| < K_x∣u(t)∣<Ku​⟹∣x(t)∣<Kx​
.
Following from Definition 20, Equation 7, this means that
​lim⁡t→∞x(t)=0.\lim_{t\to\infty}\mathbf{x}(t) = \boldsymbol{0}.limt→∞​x(t)=0.
​
If instead lim⁡t→∞x(t)=∞\lim_{t\to\infty}\mathbf{x}(t) = \inftylimt→∞​x(t)=∞
, 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 0\boldsymbol{0}0
.
Theorem 4
A system is marginally stable if there is exactly one pole at s=0s=0s=0
 or a pair of poles at s=±jω0s=\pm j\omega_0s=±jω0​
.
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)G(s)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→∞x(t)=lim⁡s→0sX(s).\lim_{t\to\infty}x(t) = \lim_{s\to0} sX(s).limt→∞​x(t)=lims→0​sX(s).
​
Using this fact, we see that for the unity feedback system,
​E(s)=R(s)1+G(s).E(s) = \frac{R(s)}{1+G(s)}.E(s)=1+G(s)R(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.
​Kp=lim⁡s→0G(s)(9)K_p = \lim_{s\to0}G(s) \qquad (9)Kp​=lims→0​G(s)(9)
​
​lim⁡t→∞e(t)=lim⁡s→0s1s11+G(s)=11+Kp\lim_{t\to\infty} e(t) = \lim_{s\to0} s \frac{1}{s} \frac{1}{1+G(s)} = \frac{1}{1+K_p}limt→∞​e(t)=lims→0​ss1​1+G(s)1​=1+Kp​1​
​
Definition 23
The velocity constant determines how well a system can track a ramp.
​Kv=lim⁡s→0sG(s)(10)K_v = \lim_{s\to0}sG(s) \qquad (10)Kv​=lims→0​sG(s)(10)
​
​lim⁡t→∞e(t)=lim⁡s→0s1s211+G(s)=1Kv\lim_{t\to\infty} e(t) = \lim_{s\to0} s \frac{1}{s^2} \frac{1}{1+G(s)} = \frac{1}{K_v}limt→∞​e(t)=lims→0​ss21​1+G(s)1​=Kv​1​
​
Definition 24
The acceleration constant determines how well a system can track a parabola.
​Ka=lim⁡s→0s2G(s)(11)K_a = \lim_{s\to0}s^2G(s) \qquad (11)Ka​=lims→0​s2G(s)(11)
​
​lim⁡t→∞e(t)=lim⁡s→0s1s311+G(s)=1Ka\lim_{t\to\infty} e(t) = \lim_{s\to0} s \frac{1}{s^3} \frac{1}{1+G(s)} = \frac{1}{K_a}limt→∞​e(t)=lims→0​ss31​1+G(s)1​=Ka​1​
​
Notice that large static error constants mean a smaller error. Another observation we can make is that if a system has nnn
 poles at s=0s=0s=0
, it can perfectly track an input whose laplace transform is 1sn−k\frac{1}{s^{n-k}}sn−k1​
 for k∈[0,n−1]k\in[0, n-1]k∈[0,n−1]
. We give nnn
 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→∞dxdt=0  ⟹  lim⁡t→∞x=xss.\lim_{t\to\infty}\frac{d^{}\mathbf{x}}{dt^{}} = \boldsymbol{0} \implies \lim_{t\to\infty}\mathbf{x} = \mathbf{x}_{ss}.limt→∞​dtdx​=0⟹limt→∞​x=xss​.
​
Applying this to the state space equations for a step input,
​dxdt=0=Axss+B⋅I  ⟹  xss=−A−1B(12)\frac{d^{}\mathbf{x}}{dt^{}} = \boldsymbol{0} = A\mathbf{x}_{ss} + B\cdot I \implies \mathbf{x}_{ss} = -A^{-1}B \qquad (12)dtdx​=0=Axss​+B⋅I⟹xss​=−A−1B(12)
​
Looking at the error between the reference and the output in the 1D input case,
​e(t)=r(t)−y(t)=1−Cxss=1+CA−1B.\mathbf{e}(t) = \mathbf{r}(t) - \mathbf{y}(t) = 1 - C\mathbf{x}_{ss} = 1 + CA^{-1}B.e(t)=r(t)−y(t)=1−Cxss​=1+CA−1B.
​
Margins
Figure 5: Frequency Response
If we take a complex exponential and pass it into a causal LTI system with impulse response g(t)g(t)g(t)
, then
​y(t)=ejωt∗g(t)=∫−∞∞g(τ)ejω(t−τ)dτ=ejωt∫0∞g(τ)e−jωτdτ.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.y(t)=ejωt∗g(t)=∫−∞∞​g(τ)ejω(t−τ)dτ=ejωt∫0∞​g(τ)e−jωτdτ.
​
This shows us that ejωte^{j\omega t}ejωt
 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(jω)=∫0∞g(τ)e−jωτdτ(13)G(j\omega) = \int_0^{\infty}g(\tau)e^{-j\omega\tau}d\tau \qquad (13)G(jω)=∫0∞​g(τ)e−jωτdτ(13)
​
Suppose we put a linear system G(s)G(s)G(s)
 in negative feedback. We know that if ∠G(jω)=(2k+1)π\angle G(j\omega) = (2k+1)\pi∠G(jω)=(2k+1)π
 for some k∈Zk\in\mathbb{Z}k∈Z
, then the output of the plant will be −∣G(jω)∣ejωt-|G(j\omega)|e^{j\omega t}−∣G(jω)∣ejωt
. If ∣G(jω)∣≥1|G(j\omega)| \geq 1∣G(jω)∣≥1
, then this will feed back into the error term where it will be multiplied by ∣G(jω)∣|G(j\omega)|∣G(jω)∣
 repeatedly, and this will cause the system to be unstable because ∣G(jω)∣≥1|G(j\omega)|\geq1∣G(jω)∣≥1
 and thus will not decay.
Definition 28
The gain margin GmG_mGm​
is the change in the open loop gain required to make the closed loop system unstable.
Definition 29
The phase margin ϕm\phi_mϕm​
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+Gme−jϕmG(s)=0.1 + G_me^{-j\phi_m}G(s) = 0.1+Gm​e−jϕm​G(s)=0.
​
At the gain margin frequency ωgm\omega_{gm}ωgm​
,
​∣Gm∣∣G(jωgm)∣=1  ⟹  ∣Gm∣=1∣G(jωgm)∣.|G_m||G(j\omega_{gm})| = 1 \implies |G_m| = \frac{1}{|G(j\omega_{gm})|}.∣Gm​∣∣G(jωgm​)∣=1⟹∣Gm​∣=∣G(jωgm​)∣1​.
​
where the gain margin frequency is ∠G(jωm)=(2k+1)π\angle G(j\omega_m) = (2k+1)\pi∠G(jωm​)=(2k+1)π
 for k∈Zk\in\mathbb{Z}k∈Z
. Likewise, at the phase margin frequency ωpm\omega_{pm}ωpm​
,
​1+Gme−jωmG(jωpm)=0  ⟹  −ϕm+∠G(jωpm)=(2k+1)π.1 + G_me^{-j\omega_m}G(j\omega_{pm}) = 0 \implies -\phi_m + \angle G(j\omega_{pm}) = (2k+1)\pi.1+Gm​e−jωm​G(jωpm​)=0⟹−ϕm​+∠G(jωpm​)=(2k+1)π.
​
where the phase margin frequency is ∣G(jωpm)∣=1|G(j\omega_{pm})| = 1∣G(jωpm​)∣=1
.
Notice that if there is a time delay of TTT
 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
​∠G(jω)−ωT.\angle G(j\omega) - \omega T.∠G(jω)−ωT.
​
Modeling Systems
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Contents
First Order Systems
Second Order Systems
The Underdamped Case
Additional Poles and Zeros of a Second Order System
Stability
Steady State Error
Margins