After applying partial fraction decomposition to them, their step response is of the form
Thus, the larger
is (i.e the deeper in the left half plane it is), the faster the system will “settle”.
Notice that the poles of the second order system are
There are four cases of interest based on
- 1.UndampedWhen, the poles are. Because they are purely imaginary, the step response will be purely oscillatory.
- 2.UnderdampedWhen, the poles are. They are complex and in the left-half plane, so the step response will be a exponentially decaying sinusoid. We define the damped frequencyso that the poles become. Notice that. If we compute the time-response of the system,
- 3.Critically DampedWhen, both poles are at. The poles are both real, so the time-response will respond without any overshoot.
- 4.OverdampedWhen, the poles are. 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.
If we analyze the underdamped case further, we can first look at its derivative.
Using Equation 8, we see that the derivative is first equal to 0 when
The percent overshoot occurs at
Since our poles are complex, we can represent them in their polar form.
What this tells us is that if we search along the vector at angle
, we get a constant
Suppose we added an additional pole to the second order system so its transfer function was instead
Then its step response will be
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
, 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
and its step response will look like
is small, then the effect of the zero is similar to introducing a derivative into the system, whereas if
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.
Recall Equation 7 which told us the time-domain solution to state-space equations was
Following from Definition 20, Equation 7, this means that
, then the system is unstable.
In all other cases, the system will be unstable.
Consider the unity feedback loop depicted in Figure 4 where we put a system
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.
Using this fact, we see that for the unity feedback system,
Using these, we can define the static error constants.
Notice that large static error constants mean a smaller error. Another observation we can make is that if a system has
, it can perfectly track an input whose laplace transform is
. We give
a formal name.
This also brings another observation.
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,
Figure 5: Frequency Response
If we take a complex exponential and pass it into a causal LTI system with impulse response
This shows us that
is an eigenfunction of the system.
Suppose we put a linear system
in negative feedback. We know that if
, then the output of the plant will be
, then this will feed back into the error term where it will be multiplied by
repeatedly, and this will cause the system to be unstable because
and thus will not decay.
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
At the gain margin frequency
where the gain margin frequency is
. Likewise, at the phase margin frequency
where the phase margin frequency is
Notice that if there is a time delay of
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