Laplace Transform State Variables Circuit Analysis

Laplace Transform State Variables

Laplace transform state variables are very useful when we are analyzing a circuit with multiple inputs and outputs.

Thus far we have considered techniques for analyzing systems with only one input and only one output.

Laplace Transform State Variables

Many engineering systems have many inputs and many outputs, as shown in Figure.(1).

laplace transform state variables
Figure 1. A linear system with m inputs and p outputs.

The state variable method is a very important tool in analyzing systems and understanding such highly complex systems. Thus, the state variable model is more general than the single-input, single-output model, such as a transfer function.

Although the topic cannot be adequately covered in one post, we will cover it briefly at this point.

In the state variable model, we specify a collection of variables that describe the internal behaviour of the system. These variables are known as the state variables of the system.

They are the variables that determine the future behaviour of a system when the present state of the system and the input signals are known.

In other words, they are those variables which, if known, allow all other system parameters to be determined by using only algebraic equations.

A state variable is a physical property that characterizes the state of a system, regardless of how the system got to that state.

Common examples of state variables are pressure, volume, and temperature. In an electric circuit, the state variables are the inductor current and capacitor voltage since they collectively describe the energy state of the system.

The standard way to represent the state equations is to arrange them as a set of first-order differential equations:

Laplace Transform State Variables
(1)

where

Laplace Transform State Variables

and the dot represents the first derivative with respect to time, i.e.,

Laplace Transform State Variables

and

Laplace Transform State Variables

A and B are respectively n × n and n × m matrices. In addition to the state equation in Equation.(1), we need the output equation. The complete state model or state space is

Laplace Transform State Variables
(2a)
Laplace Transform State Variables
(2b)

where

Laplace Transform State Variables

and C and D are, respectively, p × n and p × m matrices. For the special case of single-input single-output, n = m = p = 1.

Assuming zero initial conditions, the transfer function of the system is found by taking the Laplace transform of Equation.(2a); we obtain

Laplace Transform State Variables

or

Laplace Transform State Variables
(3)

where I is the identity matrix. Taking the Laplace transform of Equation.(2b) yields

Laplace Transform State Variables
(4)

Substituting Equation.(3) into (4) and dividing by Z(s) gives the transfer function as

Laplace Transform State Variables
(5)

where:
A = system matrix
B = input coupling matrix

C = output matrix
D = feedforward matrix

In most cases, D = 0, so the degree of the numerator of H(s) in Equation.(5) is less than that of the denominator. Thus,

Laplace Transform State Variables
(6)

Because of the matrix computation involved, MATLAB can be used to find the transfer function.

To apply state variable analysis to a circuit, we follow the following three steps.

Steps to Apply the State Variable Method to Circuit Analysis:
1. Select the inductor current i and capacitor voltage v as the state variables, making sure they are consistent with the passive sign convention.
2. Apply KCL and KVL to the circuit and obtain circuit variables
(voltages and currents) in terms of the state variables. This should lead to a set of first-order differential equations necessary and sufficient to determine all state variables.
3. Obtain the output equation and put the final result in state-space representation.

Steps 1 and 3 are usually straightforward; the major task is in step  2. We will illustrate this with examples.

Laplace Transform State Variables Examples

1. Find the state-space representation of the circuit in Figure.(2). Determine the transfer function of the circuit when vs is the input and ix is the output.
Take R = 1 Ω, C = 0.25 F, and L = 0.5 H.

laplace transform state variables
Figure 2

Solution:
We select the inductor current i and capacitor voltage v as the state variables.

Laplace Transform State Variables
(1.1)
Laplace Transform State Variables
(1.2)

Applying KCL at node 1 gives

Laplace Transform State Variables
(1.3)

since the same voltage v is across both R and C. Applying KVL around the outer loop yields

Laplace Transform State Variables
(1.4)

Equations.(1.3) and (1.4) constitute the state equations. If we regard ix as the output,

Laplace Transform State Variables
(1.5)

Putting Equations.(1.3), (1.4), and (1.5) in the standard form leads to

Laplace Transform State Variables
(1.6a)
Laplace Transform State Variables
(6b)

If R = 1, C = ¼, and L = ½, we obtain from Equation.(1.6) matrices

Laplace Transform State Variables

Taking the inverse of this gives

Laplace Transform State Variables

Thus, the transfer function is given by

Laplace Transform State Variables

which is the same thing we would get by directly Laplace transforming the circuit and obtaining H(s) = Ix(s)∕Vs(s). The real advantage of the state variable approach comes with multiple inputs and multiple outputs.

In this case, we have one input vs and one output ix. In the next example, we will have two inputs and two outputs.

 

2. Consider the circuit in Figure.(3), which may be regarded as a two-input, two-output system. Determine the state variable model and fid the transfer function of the system.

laplace transform state variables
Figure 3

Solution:
In this case, we have two inputs vs and vi and two outputs vo and io. Again, we select the inductor current i and capacitor voltage v as the state variables. Applying KVL around the left-hand loop gives

Laplace Transform State Variables
(2.1)

We need to eliminate i1. Applying KVL around the loop containing vs, 1-Ω resistor, 2-Ω resistor, and 1/3-F capacitor yields

Laplace Transform State Variables
(2.2)

But at node 1, KCL gives

Laplace Transform State Variables
(2.3)

Substituting this in Equation.(2.2),

Laplace Transform State Variables
(2.4)

Substituting this in Equation.(2.1) gives

Laplace Transform State Variables
(2.5)

which is one state equation. To obtain the second one, we apply KCL at
node 2.

Laplace Transform State Variables
(2.6)

We need to eliminate vo and io. From the right-hand loop, it is evident that

Laplace Transform State Variables
(2.7)

Substituting Equation.(2.4) into (2.3) gives

Laplace Transform State Variables
(2.8)

Substituting Equations.(2.7) and (2.8) into Equation.(2.6) yields the second state equation as

Laplace Transform State Variables
(2.9)

The two output equations are already obtained in Equations.(2.7) and (2.8). Putting Equations.(2.5) and (2.7) to (2.9) together in the standard form leads to the state model for the circuit, namely,

Laplace Transform State Variables
(2.10a)
Laplace Transform State Variables
(2.10b)

 

3. Assume we have a system where the output is y(t) and the input is z(t). Let the following differential equation describe the relationship between the input and the output.

Laplace Transform State Variables
(3.1)

Obtain the state model and the transfer function of the system.

Solution:
First, we select the state variables. Let x1 = y(t), therefore

Laplace Transform State Variables
(3.2)

Now let

Laplace Transform State Variables
(3.3)

Note that at this time we are looking at a second-order system that would normally have two first-order terms in the solution.

Now we have x2 = y(t), where we can find the value x2 from Equation.(3.1), i.e.,

Laplace Transform State Variables
(3.4)

From Equations.(3.2) to (3.4), we can now write the following matrix equations:

Laplace Transform State Variables
(3.5)
Laplace Transform State Variables
(3.6)

We now obtain the transfer function.

Laplace Transform State VariablesThe inverse is

Laplace Transform State VariablesThe transfer function is

Laplace Transform State Variables

To check this, we directly apply the Laplace transfer to each term in Equation.(3.1). Given that initial conditions are zero, we get

Laplace Transform State Variables

which is in agreement with what we got previously.