It is often quite easy to develop a state space model from a system diagram (see here for examples). However, there are several situations in which it is not entirely straightforward to develop a state space model from a system diagram. A few examples are given below. For the first case both a mechanical and an electrical example is given. For the other cases, only mechanical examples are given (though they are easily generalized to other systems, e.g. electrical).
The system below has input y, and output x. Find a state space representation.
We start with a free body diagram
Energy is stored in the spring and the mass, so we try using the position "x" and the velocity "v" (v = derivative of x) as our state variables. From this we try to develop our state space model
We fail because the last equation has a derivative of the input on the right hand side. There are systematic ways to choose state variables to avoid this problem (an example of rewriting a differential equation with input (i.e., right hand side) derivatives in state space form is given here).
Often, an easier way to do this is to find a transfer function and then convert the transfer function to state space (described here). In this case the transfer function is
and a state space system (observable canonic form) is
with
The system below has input e_{i}, and output e_{o}. Find a state space representation.
Let's try choosing eo and iL as our state variables.
Energy is stored in the capacitor (voltage) and the inductor (current) so we try using the capacitor voltage (e_{o}) and the inductor current (i_{L}) as our state variables.
At this point we can see that we have failed because this equation has the derivative of the input on the right hand side (and we can only have state variables and inputs in a valid state space representation). There are systematic ways to choose state variables to avoid this problem (an example of rewriting a differential equation with input (i.e., right hand side) derivatives in state space form is given here). (Note: we could succeed if we defined our state variables as q_{1}=e_{o}-e_{i}, q_{2}=i_{L}).
Often, an easier way to do this is to find a transfer function and then convert the transfer function to state space (described here). In this case the transfer function is
and a state space system (observable canonic form) is
with
Develop as state space representation for the system shown. The input is f_{a}.
At first glance it might seem that the system is third order because there are three energy storage elements (two springs and a mass). However, because the springs are in series they can be treated as a single element. Such simplifications are not always so obvious, so let us proceed to develop a third order state space model.
First we draw free body diagrams
Freebody at x | Freebody at z | |
We choose as our state variables:
so
or
While this is a valid state space system, it is, in fact, only a second order system (the state equations are not linearly independent; this is shown by the first and third rows being multiples of each other). The characteristic equation |sI-A| is only a second order polynomial.
Consider the system
It has been shown that the system shown can be represented in state space by:
Problem: Find a state space representation if we want the output to be theta;_{2}.
Solution: Since is not one of the state variables (or an algebraic sum of the state variables), we need to create another state variable. Choose as state variables
We now need to write the fourth state variable equation (which is trivial)
so our state variable system must be expanded to include this variable
Derive a state space model for the system shown. The input is f_{a} and the output is z. Note, this system was solved previously, with b and k_{1} switched.
We can write free body equations for the system at x and at y.
Freebody Diagram | Equation |
There are three energy storage elements, so we expect three state equations. The energy storage elements are the spring, k_{2}, the mass, m, and the spring, k_{1}. Therefore we choose as our state variables x (the energy in spring k_{2} is ½k_{2}x²), the velocity at x (the energy in the mass m is ½mv², where v is the first derivative of x), and y (the energy in spring k_{1} is ½k_{1}y²). Our state variables become:
Now we want equations for their derivatives. The equations of motion from the free body diagrams yield
Clearly there is a problem because the equation for the derivative of q_{2} has a derivative of a state variable on the RHS. However, we can remove this by substitution in the last equation:
so
with the input u=f_{a}.
© Copyright 2005 to 2015 Erik Cheever This page may be freely used for educational purposes.
Erik Cheever Department of Engineering Swarthmore College