ECE 486 Control Systems
Lecture 6
Second Order System (Continued)
Last time we learned how to visualize system dynamics with block diagrams, especially all-integrator diagrams and basic system interconnections. We also began our discussion of prototype second order transfer function. We shall continue our investigation.
Recall from last time, the unit step response of a prototype underdamped second order transfer function H(s)=ω2ns2+2ζωns+ω2n=ω2n(s+σ)2+ω2d can be obtained as y(t)=1−e−σt(cos(ωdt)+σωdsin(ωdt)), where σ=ζωn and ωd=ωn√1−ζ2 is the damped natural frequency.
We will see that the damping ratio ζ and natural frequency ωn determine certain important features of the transient part of the above step response. We will also learn how to pick ζ and ωn in order to shape these features according to given specifications.
Transient Response Specifications
Rise Time
We will start by taking a look at first order step response though.
H(s)=as+a, with a>0.It has a stable pole s=−a since a is positive. By FVT, its DC gain is 1. (How?)
We can write down its unit step response as
y(t)=L−1{Y(s)}=L−1{H(s)1s}=L−1{as(s+a)}=L−1{1s−1s+a}=1(t)−e−at.We define rise time as the time it takes to get from 10% to 90% of steady-state value (of a step response). Rise time is denoted tr. Figure 1 shows the rise time of step response of a first order transfer function.
Figure 1: Rise time of a first order system
To compute tr analytically in this example for step response y(t)=1(t)−e−at, we follow the above definition: denote t0.1 and t0.9 as the time instances when it reaches 10% and 90% of its steady-state value respectively (for the first time), then
1−e−at0.1=0.1,e−at0.1=0.9,t0.1=−ln0.9a;1−e−at0.9=0.9,e−at0.9=0.1,t0.9=−ln0.1a.tr=t0.9−t0.1=ln0.9−ln0.1a=ln9a≈2.2a.Overshoot and Settling Time
Now let’s consider the more interesting case of a second order step response. When underdamped,
H(s)=ω2ns2+2ζωns+ω2n=ω2n(s+σ)2+ω2d,where σ=ζωn, ωd=ωn√1−ζ2 with ζ<1.
We can graph the step response y(t)=1−e−σt(cos(ωdt)+σωdsin(ωdt)) as shown in Figure 2.
Figure 2: Rise time of a second order system (underdamped)
In addition to rise time, we also introduce two more specs: overshoot and settling time,
- Rise time tr: time to get from 0.1y(∞) to 0.9y(∞)
- Overshoot Mp and peak time tp (note Mp could a percentage overshoot)
- Settling time ts: the first time for transients to decay to within a
specified small percentage of y(∞) and stay in that range. We will
usually worry about 5% settling time; the default threshold for
stepinfo()
inMatlab
is 2%.
In general, the desired situation is to have fast rising, quickly settled step responses with low overshoot,
- tr, small
- Mp, small
- tp, small
- ts, small
but it is impossible because of the trade-off that decreasing tr leads to increased Mp.
Formulas for Time Domain Specifications
Rise Time
Rise time tr is hard to calculate analytically in general, but empirically, on the normalized time scale t→ωnt, rise times are approximately the same as
wntr≈1.8.(exact when ζ=0.5)
So, we will work with tr≈1.8ωn. It is a good approximation when ζ≈0.5.
Overshoot and Peak Time
In Figure 2, peak time tp is the first time t>0 when y′(t)=0. By
y(t)=1−e−σt(cos(ωdt)+σωdsin(ωdt))y′(t)=(σ2ωd+ωd)e−σtsin(ωdt)=0 when ωdt=0,π,2π,…,we get tp=πωd.
To find overshoot Mp, plug this value into y(t),
Mp=y(tp)−1=−e−σπωd(cos(ωdπωd)+σωdsin(ωdπωd))=−exp(−σπωd)(−1+0)=exp(−πζ√1−ζ2).The formula for Mp is exact.
Settling Time
The settling time is the last moment the step response enters the (5%) error strip and never leaves from that point on, ts=min
In Figure 2, y(\infty) = 1. Therefore, the error strip can be bounded according to
\begin{align*} |y(t) - 1| &= e^{-\sigma t} \left| \cos(\omega_d t) + \frac{\sigma}{\omega_d} \sin(\omega_d t)\right|, \end{align*}where the decaying exponential e^{-\sigma t} is what matters since \sin and \cos functions are bounded by 1. Hence e^{-\sigma t_s} \le 0.05 gives t_s = - \dfrac{\ln 0.05}{\sigma} \approx \dfrac{3}{\sigma}.
In conclusion, for an underdamped second order transfer function
\begin{align*} H(s) &= \frac{\omega^2_n}{s^2 + 2\zeta\omega_n s + \omega^2_n} \\ &= \frac{\sigma^2 + \omega_d^2}{(s+\sigma)^2 + \omega^2_d}, \end{align*}we can apply the following formulas to calculate its time domain specifications,
\begin{align*} t_r & \approx \frac{1.8}{\omega_n}, \\ t_p &= \frac{\pi}{\omega_d}, \\ M_p &= \exp\left( - \frac{\pi\zeta}{\sqrt{1-\zeta^2}}\right), \\ t_s &\approx \frac{3}{\sigma}. \end{align*}Time Domain Specifications Related to Poles on Complex Plane
By the formulas at the end of the previous section, we may visualize time-domain specs in terms of admissible pole locations for an underdamped second order system.
Constraints on Rise Time
Suppose we want t_r \le c, where c is some desired upper bound on rise time.
\begin{align*} t_r & \approx \dfrac{1.8}{\omega_n} \le c \\ \implies \omega_n &\ge \dfrac{1.8}{c}. \end{align*}Geometrically, we want poles to lie in the shaded region as in Figure 3.
Figure 3: Possible pole locations in shaded region when t_r \le c
Constraints on Overshoot
Suppose we want M_p \le c, by M_p = \underbrace{\exp\left( - \dfrac{\pi\zeta}{\sqrt{1-\zeta^2}}\right)}_{\text{decreasing function}} \le c,
we need a large damping ratio. Furthermore, recall from last time,
\begin{align*} \frac{\zeta}{\sqrt{1-\zeta^2}} &= \frac{\omega_n\zeta}{\omega_n\sqrt{1-\zeta^2}} \\ &= \frac{\sigma}{\omega_d} \\ &= \cot \varphi, \end{align*}therefore \varphi shall be small. Geometrically, we want poles to lie in the shaded region as in Figure 4.
Figure 4: Possible pole locations in shaded region when M_p \le c
Intuition: A good damping ratio implies a good decay in roughly half a period.
Constraints on Settling Time
Suppose we want t_s \le c, then
\begin{align*} t_s &\approx \dfrac{3}{\sigma} \le c \\ \implies &\sigma \ge \dfrac{3}{c}. \end{align*}Therefore the real part of poles {\rm Re}(s) = - \sigma shall be large in terms of magnitude, i.e., far away from origin. Geometrically, we want poles to lie in the shaded region as in Figure 5.
Figure 5: Possible pole locations in shaded region when t_s \le c
Intuition: Poles that are deep into the Left Half Plane (LHP) implies transients decay fast, i.e., overall t_s is smaller.
Combination of Constraints on Rise Time, Overshoot, Settling Time
If we have specifications for combinations of t_r, M_p, t_s, we can easily relate them to allowed pole locations as we did for each of them, only overlaying them on a single sheet.
Figure 6: Possible pole locations in overlaid shaded region when t_r, M_p, t_s are constrained
The shape and size of the region for admissible pole locations will change depending on which specs are more severely constrained.
Note: Drawing sketches is very appealing to engineers since it is easy to visualize things, even though no such crisp visualization is easily available in time domain. But sketches are not very rigorous, let alone proofs. Also since we used approximations for prototype second order system, which has only two poles and no zeros, these geometric drawings are only valid in a very narrow range of scenarios.