A first order system is one with the transfer function of the form
After applying partial fraction decomposition to them, their step response is of the form
Second Order Systems
Definition 16
Second order systems are those with the transfer function in the form
Notice that the poles of the second order system are
Undamped
Underdamped
Critically Damped
Overdamped
The Underdamped Case
If we analyze the underdamped case further, we can first look at its derivative.
Definition 17
Definition 18
Definition 19
Since our poles are complex, we can represent them in their polar form.
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
Then its step response will be
Notice that
If we instead add an additional zero to the second order system so its transfer function looks like
and its step response will look like
Stability
Recall Equation 7 which told us the time-domain solution to state-space equations was
Definition 20
Following from Definition 20, Equation 7, this means that
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
Theorem 4
In all other cases, the system will be unstable.
Steady State Error
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,
Using this fact, we see that for the unity feedback system,
Using these, we can define the static error constants.
Definition 22
The position constant determines how well a system can track a unit step.
Definition 23
The velocity constant determines how well a system can track a ramp.
Definition 24
The acceleration constant determines how well a system can track a parabola.
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,
Applying this to the state space equations for a step input,
Looking at the error between the reference and the output in the 1D input case,
Margins
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.
Definition 28
Definition 29
We can imagine the gain and phase margin like placing a “virtual box” before the plant as shown in Figure 6.
The characteristic polynomial of the closed loop transfer function is
The step response of a system is how a system H(s) responds to a step input.
y(t)=L−1{sH(s)}
H(s)=s+βs+α.
Au(t)+Be−βtu(t).
Thus, the larger β is (i.e the deeper in the left half plane it is), the faster the system will “settle”.
H(s)=s2+2ζωns+ωn2ωn2.
ωn is known as the natural frequency, and ζis known as the damping factor.
s=2−2ζωn±4ζ2ωn2−4ωn2=−ζωn±ωnζ2−1.
There are four cases of interest based on ζ.
When ζ=0, the poles are s=±ωnj. Because they are purely imaginary, the step response will be purely oscillatory.
Y(s)=s1s2+ωn2ωn2↔y(t)=u(t)−cos(ωnt)u(t)
When ζ∈(0,1), the poles are s=−ζωn±jωn1−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 so that the poles become s=−ζωn±ωdj. Notice that ωd<ωn. If we compute the time-response of the system,
When ζ=1, both poles are at s=−ωn. The poles are both real, so the time-response will respond without any overshoot.
When ζ>1, the poles are −ζωn±ωnζ2−1. 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.
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}, 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.
H(s)=s2+2ζωn+ωn2s+a
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.
x(t)=eAtx(0)+∫0teA(t−τ)Bu(τ)dτ.
A system is bounded-input, bounded output(BIBO) stable if ∃Ku,Kx<∞ such that ∣u(t)∣<Ku⟹∣x(t)∣<Kx.
limt→∞x(t)=0.
If instead limt→∞x(t)=∞, then the system is unstable.
A system is called marginally stable if the zero-input response does not converge to 0.
A system is marginally stable if there is exactly one pole at s=0 or a pair of poles at s=±jω0.
Consider the unity feedback loop depicted in Figure 4 where we put a system G(s) in unity feedback to control it.
limt→∞x(t)=lims→0sX(s).
E(s)=1+G(s)R(s).
Kp=lims→0G(s)(9)
limt→∞e(t)=lims→0ss11+G(s)1=1+Kp1
Kv=lims→0sG(s)(10)
limt→∞e(t)=lims→0ss211+G(s)1=Kv1
Ka=lims→0s2G(s)(11)
limt→∞e(t)=lims→0ss311+G(s)1=Ka1
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 sn−k1 for k∈[0,n−1]. We give n a formal name.
limt→∞dtdx=0⟹limt→∞x=xss.
dtdx=0=Axss+B⋅I⟹xss=−A−1B(12)
e(t)=r(t)−y(t)=1−Cxss=1+CA−1B.
If we take a complex exponential and pass it into a causal LTI system with impulse response g(t), then
This shows us that ejωt is an eigenfunction of the system.
G(jω)=∫0∞g(τ)e−jωτdτ(13)
Suppose we put a linear system G(s) in negative feedback. We know that if ∠G(jω)=(2k+1)π for some k∈Z, then the output of the plant will be −∣G(jω)∣ejωt. If ∣G(jω)∣≥1, then this will feed back into the error term where it will be multiplied by ∣G(jω)∣ repeatedly, and this will cause the system to be unstable because ∣G(jω)∣≥1 and thus will not decay.
The gain margin Gmis the change in the open loop gain required to make the closed loop system unstable.
The phase margin ϕmis the change in the open loop phase required to make the closed loop system unstable.
1+Gme−jϕmG(s)=0.
At the gain margin frequency ωgm,
∣Gm∣∣G(jωgm)∣=1⟹∣Gm∣=∣G(jωgm)∣1.
where the gain margin frequency is ∠G(jωm)=(2k+1)π for k∈Z. Likewise, at the phase margin frequency ωpm,
1+Gme−jωmG(jωpm)=0⟹−ϕm+∠G(jωpm)=(2k+1)π.
where the phase margin frequency is ∣G(jωpm)∣=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