. The poles are both real, so the time-response will respond without any overshoot.
4.
Overdamped
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.
The Underdamped Case
If we analyze the underdamped case further, we can first look at its derivative.
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.
If we instead add an additional zero to the second order system so its transfer function looks like
H(s)=s2+2ζωn+ωn2s+a
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
x(t)=eAtx(0)+∫0teA(t−τ)Bu(τ)dτ.
Definition 20
A system is bounded-input, bounded output(BIBO) stable if
∃Ku,Kx<∞
such that
∣u(t)∣<Ku⟹∣x(t)∣<Kx
.
Following from Definition 20, Equation 7, this means that
limt→∞x(t)=0.
If instead
limt→∞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
.
Theorem 4
A system is marginally stable if there is exactly one pole at
s=0
or a pair of poles at
s=±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)
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,
limt→∞x(t)=lims→0sX(s).
Using this fact, we see that for the unity feedback system,
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=lims→0G(s)(9)
limt→∞e(t)=lims→0ss11+G(s)1=1+Kp1
Definition 23
The velocity constant determines how well a system can track a ramp.
Kv=lims→0sG(s)(10)
limt→∞e(t)=lims→0ss211+G(s)1=Kv1
Definition 24
The acceleration constant determines how well a system can track a parabola.
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.
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,
limt→∞dtdx=0⟹limt→∞x=xss.
Applying this to the state space equations for a step input,
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.
Margins
Figure 5: Frequency Response
If we take a complex exponential and pass it into a causal LTI system with impulse response
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)
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.
Definition 28
The gain margin
Gm
is the change in the open loop gain required to make the closed loop system unstable.
Definition 29
The phase margin
ϕm
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.
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