Laplace Transform State Variables Circuit Analysis

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).

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:

(1)

where

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

and

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

(2a)

(2b)

where

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

(3)

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

(4)

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

(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,

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 v_sis the input and i_xis the output.
Take R = 1 Ω, C = 0.25 F, and L = 0.5 H.

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

(1.1)

(1.2)

Applying KCL at node 1 gives

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

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

(1.5)

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

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

Taking the inverse of this gives

Thus, the transfer function is given by

which is the same thing we would get by directly Laplace transforming the circuit and obtaining H(s) = I_x(s)∕V_s(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 i_x. 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.

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

(2.1)

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

(2.2)

But at node 1, KCL gives

(2.3)

Substituting this in Equation.(2.2),

(2.4)

Substituting this in Equation.(2.1) gives

(2.5)

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

We need to eliminate v_oand i_o. From the right-hand loop, it is evident that

(2.7)

Substituting Equation.(2.4) into (2.3) gives

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

(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,

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.