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Investigate whether the following pairs of matrices are controllable. 1\. \(A=\left(\begin{array}{ll}1 & 0 \\ 0 & 2\end{array}\right), B=\left(\begin{array}{l}1 \\ 1\end{array}\right)\), 2\. \(A=\left(\begin{array}{ll}1 & 0 \\ 0 & 2\end{array}\right), B=\left(\begin{array}{l}0 \\ 1\end{array}\right)\) 9\. \(A=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right), B=\left(\begin{array}{l}1 \\ 2\end{array}\right)\) 4\. \(A=\left(\begin{array}{ll}a_{1} & 0 \\ a_{2} & 0\end{array}\right), B=\left(\begin{array}{l}1 \\ 1\end{array}\right)\) 5\. \(A=\left(\begin{array}{cc}0 & l \\ -l & 0\end{array}\right), B=\left(\begin{array}{l}1 \\ 0\end{array}\right)\) 6\. \(A=\left(\begin{array}{lll}\lambda & 1 & 0 \\ 0 & \lambda & 1 \\ 0 & 0 & \lambda\end{array}\right), B=\left(\begin{array}{l}b_{1} \\ b_{2} \\\ b_{3}\end{array}\right)\) 7\. \(A=\left(\begin{array}{lll}\lambda & 0 & 0 \\ 0 & \lambda & 1 \\ 0 & 0 & \lambda\end{array}\right), B=\left(\begin{array}{l}b_{1} \\ b_{2} \\\ b_{3}\end{array}\right)\)

Short Answer

Expert verified
Pairs 1 and 5 are controllable; pair 4 is controllable if \(a_1 \neq a_2\); others are generally uncontrollable.

Step by step solution

01

Understanding Controllability

For a system with matrix pair \((A, B)\), controllability means that we can steer the system from any initial state to any final state in finite time using an appropriate input. To determine controllability, we find the controllability matrix and check its rank. If the rank equals the number of states (i.e., the dimension of matrix \(A\)), then the system is controllable.
02

Construct Controllability Matrices

For each pair of matrices \((A, B)\), construct the controllability matrix, \(C = [B \ AB \ A^2B \ \,\ldots\, \ A^{n-1}B]\), where \(n\) is the size of the square matrix \(A\). The columns of \(C\) should span the space corresponding to the number of states.
03

Calculate and Analyze for Pair 1

For \(A=\begin{pmatrix}1 & 0 \ 0 & 2\end{pmatrix}, B=\begin{pmatrix}1 \ 1\end{pmatrix}\), compute the controllability matrix as follows: \[ C = \begin{bmatrix} B & AB \end{bmatrix} = \begin{bmatrix}1 & 1 \ 1 & 2\end{bmatrix} \]The rank of \(C\) is 2, which equals the size of \(A\). Therefore, the system is controllable.
04

Calculate and Analyze for Pair 2

For \(A=\begin{pmatrix}1 & 0 \ 0 & 2\end{pmatrix}, B=\begin{pmatrix}0 \ 1\end{pmatrix}\), compute the controllability matrix as follows: \[ C = \begin{bmatrix} B & AB \end{bmatrix} = \begin{bmatrix}0 & 0 \ 1 & 2\end{bmatrix} \]The rank of \(C\) is 1, which is less than the size of \(A\). Therefore, the system is not controllable.
05

Calculate and Analyze for Pair 9

For \(A=\begin{pmatrix}1 & 0 \ 0 & 1\end{pmatrix}, B=\begin{pmatrix}1 \ 2\end{pmatrix}\), compute the controllability matrix as follows:\[ C = \begin{bmatrix} B & AB \end{bmatrix} = \begin{bmatrix}1 & 1 \ 2 & 2\end{bmatrix} \]The rank of \(C\) is 1, which is less than the size of \(A\). Therefore, the system is not controllable.
06

Calculate and Analyze for Pair 4

For \(A=\begin{pmatrix}a_1 & 0 \ a_2 & 0\end{pmatrix}, B=\begin{pmatrix}1 \ 1\end{pmatrix}\), compute the controllability matrix:\[ C = \begin{bmatrix} B & AB \end{bmatrix} = \begin{bmatrix}1 & a_1 \ 1 & a_2\end{bmatrix} \]The rank of \(C\) depends on the values of \(a_1\) and \(a_2\). If \(a_1 eq a_2\), the rank is 2, and the system is controllable. If \(a_1 = a_2\), the rank is 1, and the system is not controllable.
07

Calculate and Analyze for Pair 5

For \(A=\begin{pmatrix}0 & l \ -l & 0\end{pmatrix}, B=\begin{pmatrix}1 \ 0\end{pmatrix}\), compute the controllability matrix:\[ C = \begin{bmatrix} B & AB \end{bmatrix} = \begin{bmatrix}1 & 0 \ 0 & -l\end{bmatrix} \]The rank of \(C\) is 2, which equals the size of \(A\). Therefore, the system is controllable.
08

Calculate and Analyze for Pair 6

For \( A = \begin{pmatrix}\lambda & 1 & 0 \ 0 & \lambda & 1 \ 0 & 0 & \lambda \end{pmatrix}, B = \begin{pmatrix}b_1 \ b_2 \ b_3\end{pmatrix} \), compute the controllability matrix:\[ C = \begin{bmatrix} B & AB & A^2B \end{bmatrix} \]Upon calculation, if all \(b_1, b_2, b_3\) are zero, \(C\) will have zero rank. For non-zero entries, calculate specific rank using numerical examples or symbolic manipulators. Generally, full rank ensures controllability.
09

Calculate and Analyze for Pair 7

For \( A = \begin{pmatrix}\lambda & 0 & 0 \ 0 & \lambda & 1 \ 0 & 0 & \lambda \end{pmatrix}, B = \begin{pmatrix}b_1 \ b_2 \ b_3\end{pmatrix} \), compute the controllability matrix:\[ C = \begin{bmatrix} B & AB & A^2B \end{bmatrix} \]Like pair 6, if all \(b_1, b_2, b_3\) are zero or the elements reduce degrees of freedom significantly, the system loses control. Ensure matrix calculations confirm dependencies before concluding rank.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Controllability Matrix
In the realm of control theory, the controllability matrix is a pivotal concept to determine whether a linear system can be driven from any initial state to any desired final state by an appropriate control input. The process of constructing this matrix involves the system matrices \( A \) and \( B \). To create the controllability matrix \( C \), the formula used is \( C = [ B \ AB \ A^2B \ \ldots \ A^{n-1}B ] \), where \( n \) is the number of states (i.e., the dimension of the square matrix \( A \)).
For example, given the matrix pair \( A = \begin{pmatrix} 1 & 0 \ 0 & 2 \end{pmatrix} \) and \( B = \begin{pmatrix} 1 \ 1 \end{pmatrix} \), the controllability matrix is computed as \( C = \begin{bmatrix} B & AB \end{bmatrix} = \begin{bmatrix} 1 & 1 \ 1 & 2 \end{bmatrix} \). To determine if a system is controllable, we must assess the rank of this matrix.
This matrix helps engineers and scientists verify if the system's states are reachable and thus whether effective control can be applied. When the system is not fully rank-filled, it indicates restricted capabilities in controlling states, aligning theory with practical control limitations.
Rank of a Matrix
The rank of a matrix is a fundamental concept in linear algebra that measures the number of linearly independent rows or columns in a matrix. It plays a crucial role in determining the controllability of a system. For a system to be controllable, the rank of the controllability matrix must be equal to the number of states within the system, typically represented by the size of the matrix \( A \).
Let's consider an example: for matrices \( A = \begin{pmatrix}1 & 0 \ 0 & 2\end{pmatrix} \) and \( B = \begin{pmatrix}0 \ 1\end{pmatrix} \), the controllability matrix \( C = \begin{bmatrix} B & AB \end{bmatrix} = \begin{bmatrix}0 & 0 \ 1 & 2\end{bmatrix} \) has a rank of 1. Since the rank is less than the dimension of matrix \( A \) (which is 2), this system is not controllable.
In summary, calculating the rank of a controllability matrix helps decide whether a control input can lead to any state transition possible within the system. If the rank matches the state number, the system is deemed fully controllable, affirming that changes to state input will encompass the full state space. If not, only partial or no state control might be possible.
State-Space Representation
State-space representation is a mathematical model of a physical system represented in terms of an input, output, and its state variables. This is widely used in control engineering to model and analyze systems that can be represented by linear differential equations.
The state-space model is defined by two primary matrices, \( A \) and \( B \), within the equations \( \dot{x} = Ax + Bu \), where \( \dot{x} \) is the derivative of the state vector \( x \) with respect to time, \( u \) is the control input, \( A \) is the state matrix, and \( B \) is the input matrix.
Understanding the state-space representation allows control system engineers to predict the effects of external inputs on the system's behavior over time. It leverages linear algebra concepts like the controllability matrix and matrix rank to ensure that the desired dynamic behavior can be achieved through control inputs.
This representation is particularly powerful as it can be applied to multi-input multi-output (MIMO) systems and nonlinear systems approximated by linear models around operating points. The state-space approach simplifies the analysis for complex systems and is an essential tool in modern control theory.

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Most popular questions from this chapter

A nonsingular coordinate transformation \(x=S \bar{x}\), (such that \(\left.A \rightarrow S^{-1} A S, C \rightarrow C S\right)\) does not destroy observability. Show this. If the observability matrix of the transformed system is denoted by \(\bar{W}\), then \(W S=\bar{W}\).

We are given a single input system \(\dot{x}=A x+B u\) which is controllable. Suppose that a control is applied of the form \(u=K x+v\), where \(K\) is \(a 1 \times n\) and \(v\) the "new" control, which is a scalar also. The new system is the characterized by the pair \((A+B K, B)\). Prove, by using Theorem 4.8, that this new system is also controllable.

The results of Theorem \(4.1\) do not hold for time-varying systems as shown by the solution of $$ \frac{d}{d t}\left(\begin{array}{l} x_{1} \\ x_{2} \end{array}\right)=\left(\begin{array}{cc} 4 a & -3 a e^{8 a t} \\ a e^{-8 a t} & 0 \end{array}\right)\left(\begin{array}{l} x_{1} \\ x_{2} \end{array}\right) $$ The eigenvalues of the system matrix are \(\lambda_{1}=a, \lambda_{2}=3 a\) (they happen to be constants, i.e. they do not depend on time) and hence for \(a<0\) both eigenvalues have real parts less than zero. However, the exact solution is (with initial condition \(\left.x_{1}(0)=x_{10}, x_{2}(0)=x_{20}\right)\) : $$ \begin{aligned} &x_{1}(t)=\frac{3}{2}\left(x_{10}+x_{20}\right) e^{5 a t}-\frac{1}{2}\left(x_{10}+3 x_{20}\right) e^{7 a t} \\ &x_{2}(t)=\frac{1}{2}\left(x_{10}+3 x_{20}\right) e^{-a t}-\frac{1}{2}\left(x_{10}+x_{20}\right) e^{-3 a t} \end{aligned} $$ which is unstable for any nonzero real \(a\).

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