First Order Transient and Steady State Analysis

When analysing a control system we want to know how it responds to the inputs we're likely to give it when it's running. Unfortunately we often don't know ahead of time exactly what the inputs are going to be. For instance a thermostat system that uses a thermometer and a set ideal temperature to control an air conditioner is going to have all kinds of combinations of ambient temperature and ideal temperature to deal with. We can't possibly model every scenario ahead of time. So what do we do? When designing and analysing a control system we use "Test Signals" which represent different kinds of signals we're likely to encounter. Your typical test signals are an impulse pulse (like a delta function), a step function, a ramp and maybe a parabola. By observing how your system responds to these test inputs we can get a pretty good idea of how it'll work when given a complex signal out in the real world.
To analyse how your system responds to these test signals we look at two parts of its response, the transient response and the steady state response. The steady state response is what value your system eventually settles at, its final value. The transient response is everything your system does until it settles down.
If a system has transient response \(c_{tr}(t)\) and steady state response \(c_{ss}(t)\) then its total response is given by: $$ c(t) = c_{tr}(t) + c_{ss}(t) $$
The most important part of a control system is whether or not it is stable. Stability means that your system will either settle on a value when not given any inputs, or will stay between two values when not given any inputs.
A system can be unstable, stable or critically stable. Unstable: The system's output goes to \(\infty\) or \(-\infty\) Stable: The system's output settles on a single value when not given a changing input Critically Stable: The system's output oscillates between two non-infinite bounds forever
If a system is shown to be stable (we'll get to how in a little bit), the next thing we care about is the relative stability and the steady-state error of the system.
Relative stability is a measure of how "close" the system is to being unstable. Since real parts used to build a system have tolerances we want a large relative stability to know that even if a resistor is off by 10% our system will still be stable.
The steady-state-error of a system is the difference between the input and the output as time goes to infinity.

Find the steady state error of the system shown below: The red line is the input and the blue line is the system output

We can see that the output settles at 0.8 while the input holds steady at 1 so our steady state error is: \(e_{ss} = 1 - 0.8 = 0.2\)
It's possible to have a finite steady state error even when neither the input nor the output settle to any value, so long as their different converges to some number as time goes to infinity

Find the steady state error of the system shown below: The purple line is the input and the yellow line is the system output

We can see that the input and output are parallel lines so their difference is going to remain the same out to infinity. \(e_{ss} = \text{input} - \text{output} = 0.1\)
Now we're going to see how to actually analyse first order systems by finding their ouputs when given some of these typical test signals. In all the discussion that follows remember that the output of a system is given by \(C(s)\) and the input is \(R(s)\) making the transfer function \(\frac{C(s)}{R(s)}\)
A first order system is one whose input-output equation is a first order differential equation
First order systems are the simplest of the system's we'll be looking at and serve a great point of introduction into some computational techniques we use in more complex systems.
Given a first order system with transfer function \(H(s) = \frac{C(s)}{R(s)}\) its output when given input \(R(s)\) must be: $$C(s) = H(s)R(s)$$
This is just a simple bit of algebra but since systems are usually given by their transfer function it pays to keep it in mind.

Find the step response of the first order system with transfer function \(H(s) = \frac{1}{Ts + 1}\)

The Laplace transform of the step function is \(\frac{1}{s}\) so our output is given by: \(C(s) = H(s)\cdot \frac{1}{s} = \frac{1}{Ts + 1}\cdot \frac{1}{s}\) Expand by partial fractions and simplify to get: \(C(s) = \frac{1}{s} - \frac{1}{s + (1/T)}\) Inverse Laplace to find: \(c(t) = 1 - e^{-\frac{t}{T}}\) You might recognise this form from capacitors and inductors charging and dischargin. In fact the same rules apply with \(T\) as the time constant (63.2% of final value in one time consant, 99.3% of final value after 5 time constants etc).
The process of getting a system's output given its transfer function and input can seem a little daunting at first but it's nothing you haven't done before. If you're having trouble go back and study the laplace transform and partial fractions, then you'll have no trouble at all.

Find the unit ramp response of the first order system with transfer function \(\frac{C(s)}{R(s)} = \frac{1}{Ts + 1}\)

The unit ramp is a \(45^\circ\) straight line, its Laplace transform is \(\frac{1}{s^2}\) so: $$ C(s) = H(s)\cdot \frac{1}{s^2} $$ $$ C(s) = \frac{1}{Ts + 1}\cdot \frac{1}{s^2} $$ Partial fractions gives us: $$ C(s) = \frac{1}{s^2} - \frac{T}{s} + \frac{T^2}{Ts + 1} $$ And an inverse Laplace transform gives us: $$ c(t) = t - T + Te^{-t/T} $$ Now the error at any point in time is: $$ e(t) = r(t) - c(t) $$ $$ e(t) = T(1 - e^{-t/T}) $$ As \(t \to \infty, \quad e(t) = T\) So our steady state error is \(T\).
If you were paying very careful attention you may have noticed something interesting about the above results. The derivative of a ramp is a step and the output of the step was the derivative of the output of the ramp. Likewise an impulse is the derivative of a step and the output to the impulse was the derivative of the output to the step.
The response to the derivative of an input signal is the differential of the response of the system to the original signal

The output of a system to a unit step input is \(c(t) = \ln(t)\). Find the output of the system to an impulse input.

We don't have the transfer function so we can't do this like the above examples. But using our last fact we know that the output to the impulse function will be the derivative of the output when given the step function. So \(c(t) = \frac{d}{dt}\ln(t)\) \(c(t) = \frac{1}{t}\) Look how easy that was.
practice problems