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Lecture 28 - Solution to Discrete State Equation, Control Systems

 

1 Solution to Discrete State Equation

Consider the following state model of a discrete time system: x(k + 1) = Ax(k) + Bu(k) where the initial conditions are x(0) and u(0). Putting k = 0 in the above equation, we get

x(1) = Ax(0) + Bu(0)

Similarly if we put k = 1, we would get

x(2) = Ax(1) + Bu(1)

Putting the expression of x(1) ⇒ x(2) = A2x(0) + AB u(0) + B u(1)

For k = 2,

x(3) = Ax(2) + Bu(2)

= A3x(0) + A2Bu(0) + ABu(1) + Bu(2) and so on.

If we combine all these equations, we would get the following expression as a general solution:

Lecture 28 - Solution to Discrete State Equation - Electrical Engineering (EE)

As seen in the above expression, x(k) has two parts. One is the contribution due to the initial state x(0) and the other one is the contribution of the external input u(i) for i = 0, 1, 2, · · · , k −1.

When the input is zero, solution of the homogeneous state equation x(k + 1) = Ax(k) can be written as

x(k) = Ak x(0)

where A= φ(k) is the state transition matrix.
 

2 Evaluation of φ(k) 

Similar to the continuous time systems, the state transition matrix of a discrete state model can be evaluated using the following different techniques.

1. Using Inverse Z-transform:

Lecture 28 - Solution to Discrete State Equation - Electrical Engineering (EE)

2. Using Similarity Transformation If Λ is the diagonal representation of the matrix A, then Λ = P −1AP . When a matrix is in diagonal form, computation of state transition matrix is straight forward:

Lecture 28 - Solution to Discrete State Equation - Electrical Engineering (EE)

Given Λk , we can compute Ak = P Λk P −1 

 

3. Using Caley Hamilton Theorem

Example Compute Ak for the following system using three different techniques and hence find y(k) for k ≥ 0.

Lecture 28 - Solution to Discrete State Equation - Electrical Engineering (EE)

Solution: A = Lecture 28 - Solution to Discrete State Equation - Electrical Engineering (EE) and eigenvalues of A are −0.3 and −0.7.


Method 1

Lecture 28 - Solution to Discrete State Equation - Electrical Engineering (EE)

Lecture 28 - Solution to Discrete State Equation - Electrical Engineering (EE)

Lecture 28 - Solution to Discrete State Equation - Electrical Engineering (EE)


Method 2

A= P Λk P −1 where Λk =Lecture 28 - Solution to Discrete State Equation - Electrical Engineering (EE)   Eigen values are −0.3 and −0.7. The corresponding eigenvectors are found, by using equation Avi = λivi, as 

Lecture 28 - Solution to Discrete State Equation - Electrical Engineering (EE) respectively. The transformation matrix is given by

Lecture 28 - Solution to Discrete State Equation - Electrical Engineering (EE)

Thus,

Ak = P ΛkP −1

Lecture 28 - Solution to Discrete State Equation - Electrical Engineering (EE)


Method 3: Caley Hamilton Theorem 

The eigenvalues are −0.3 and −0.7

(−0.3)k = β0 − 0.3β1
(−0.7)k = β0 − 0.7β1

Solving,

β0 = 1.75(−0.3)k − 0.75(−0.7)k
β1 = 2.5(−0.3)− 2.5(−0.7)k

Hence,

Lecture 28 - Solution to Discrete State Equation - Electrical Engineering (EE)

The solution x(k) is

Lecture 28 - Solution to Discrete State Equation - Electrical Engineering (EE)

Since y(k) = x2(k), we can write

Lecture 28 - Solution to Discrete State Equation - Electrical Engineering (EE)

Now,

Lecture 28 - Solution to Discrete State Equation - Electrical Engineering (EE)

Putting the above expression in y(k)

y(k) = 0.475(−0.3)k − 5.3(−0.7)k + (−0.3)k (3.33)k + 5.825(−0.7)k−1(1.43)k

 

3 State Diagram

Conventional signal flow graph method was meant for only algebraic equation, thus these are generally used for the derivation of input output relation in a transformed domain.
State diagram or state transition signal flow graph is an extension of conventional signal flow graph which can be applied to represent differential and difference equations as well.

Example 1: Draw the state diagram for the following differential equation.

Lecture 28 - Solution to Discrete State Equation - Electrical Engineering (EE)

Considering the state variables as x1(t) = y(t) and x2(t) = Lecture 28 - Solution to Discrete State Equation - Electrical Engineering (EE), we can write

Lecture 28 - Solution to Discrete State Equation - Electrical Engineering (EE)

Lecture 28 - Solution to Discrete State Equation - Electrical Engineering (EE)

Figure 1: State Diagram of Example 1

The state diagram is shown in Figure 1.


Example 2: Consider a discrete time system described by the following state difference equations.

x1(k + 1) = −x1(k) + x2(k)
x2(k + 1) = −x1(k) + u(k)
y(k) = x1(k) + x2(k)

Draw the state diagram.

The state diagram is shown in Figure 2.

Lecture 28 - Solution to Discrete State Equation - Electrical Engineering (EE)

Figure 2: State Diagram of Example 2

 

3.1 State Diagram of Zero Order Hold 

State diagram of zero order hold is important for sampled date control systems. Let the input to and output of a ZOH is e(t) and h(t) respectively. Then, for the inetrval kT ≤ t ≤

(k + 1)T , h(t)e(kT )

Or,

Lecture 28 - Solution to Discrete State Equation - Electrical Engineering (EE)

Therefore, the state diagram, as shown in Figure 3, consists of a single branch with gain s−1.

Lecture 28 - Solution to Discrete State Equation - Electrical Engineering (EE)

Figure 3: State Diagram of Zero Order Hold

 
4 System Response between Sampling Instants 

State variable method is a convenient way to evaluate the system response between the sampling instants of a sampled data system. State transition equation is given as:

Lecture 28 - Solution to Discrete State Equation - Electrical Engineering (EE)

where x(t0) is the initial state of the system and u(t) is the external input.

when t0 = kT , x(t) = φ(t − kT )x(kT ) + u(kT )Lecture 28 - Solution to Discrete State Equation - Electrical Engineering (EE)

Since we are interested in response between the sampling instants, let us consider t = (k + ∆)T where k = 0, 1, 2, · · · and 0 ≤ ∆ ≤ 1. This implies

x((k + ∆)T ) = φ(∆T )x(kT ) + θ(∆T )u(kT )

where Lecture 28 - Solution to Discrete State Equation - Electrical Engineering (EE) By varying the value of ∆ between 0 and 1 all information on x(t) for all t can be obtained.


Example 3: Consider the following state model of a continuous time system.

Lecture 28 - Solution to Discrete State Equation - Electrical Engineering (EE)

which undergoes through a sampling process with period T . To derive the discrete state space model, let us first compute the state transition matrix of the continuous time system using Caley Hamilton Theorem.

Lecture 28 - Solution to Discrete State Equation - Electrical Engineering (EE)

Let f (λ) = eλt

This implies

e−t = β0 − β11 = −1)
e−2t = β0 − 2β2 = −2)

Solving the above equations

β= e−t − e−2t β0 = 2e−t − e−2t

Then

Lecture 28 - Solution to Discrete State Equation - Electrical Engineering (EE)

Thus the discrete state matrix A is given as

Lecture 28 - Solution to Discrete State Equation - Electrical Engineering (EE)

The discrete input matrix B can be computed as

Lecture 28 - Solution to Discrete State Equation - Electrical Engineering (EE)

When t = (k + 1)T , the discrete state equation is described by

Lecture 28 - Solution to Discrete State Equation - Electrical Engineering (EE)

When t = (k + ∆)T ,

Lecture 28 - Solution to Discrete State Equation - Electrical Engineering (EE)

If the sampling period T = 1,

Lecture 28 - Solution to Discrete State Equation - Electrical Engineering (EE)

At the sampling instants we can find x(k) by putting k = 0, 1, 2 · · · . If ∆ = 0.5, then between the sampling instants,

Lecture 28 - Solution to Discrete State Equation - Electrical Engineering (EE)

The responses in between the sampling instants, i.e., x(0.5), x(1.5), x(2.5) etc., can be found by putting k = 0, 1, 2 · · · .

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FAQs on Lecture 28 - Solution to Discrete State Equation - Electrical Engineering (EE)

1. What is a discrete state equation?
Ans. A discrete state equation is a mathematical representation used to describe the evolution of a system with distinct, separate states over time. It is commonly used in various fields such as physics, engineering, and computer science to model and analyze systems that can only exist in specific states.
2. How is a discrete state equation solved?
Ans. The solution to a discrete state equation involves finding the values of the system's states at different time points. This can be done through various methods such as iteration, recursion, or using matrix operations. The specific approach depends on the complexity of the equation and the characteristics of the system being modeled.
3. What are the applications of discrete state equations?
Ans. Discrete state equations have numerous applications in different domains. For example, in physics, they can be used to model the behavior of particles in quantum mechanics or the motion of objects in classical mechanics. In computer science, they are essential for algorithms, simulations, and machine learning models. Additionally, discrete state equations find applications in financial modeling, population dynamics, and control systems.
4. Can discrete state equations be used to model real-world systems accurately?
Ans. Discrete state equations can provide accurate models for many real-world systems, but their accuracy depends on the assumptions made and the level of detail included in the model. The more accurately the states and dynamics of the system are captured, the more accurate the model will be. However, it is important to note that no model can perfectly represent the complexity of every real-world system, and there will always be some degree of approximation involved.
5. Are there any limitations or challenges in solving discrete state equations?
Ans. Yes, there are certain limitations and challenges in solving discrete state equations. One challenge is choosing appropriate initial conditions and boundary conditions, which can significantly affect the accuracy of the solution. Another challenge is dealing with nonlinearities or complex interactions between states, as they can make the equations more difficult to solve. Additionally, for systems with a large number of states, the computational complexity of finding the solution may increase, requiring advanced numerical techniques or approximations.
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