Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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.

Short Answer

Expert verified
The new system \((A + BK, B)\) is controllable because the controllability condition is unaffected by the transformation.

Step by step solution

01

Recall Controllability Definition

A system is controllable if, for any initial state \(x_0\) and any final state \(x_f\), there exists a finite time \(t\) and an input \(u(t)\) such that the system state reaches \(x_f\) from \(x_0\). For the system \(\dot{x} = Ax + Bu\), the matrix \([B, AB, A^2B, \ldots, A^{n-1}B]\) must have full rank (i.e., rank \(n\)).
02

Apply the Control Transformation

Consider the control law \(u = Kx + v\). Substituting this into the original system yields \(\dot{x} = Ax + B(Kx + v) = (A + BK)x + Bv\). The new system is characterized by the matrices \((A + BK)\) and \(B\).
03

Apply Theorem 4.8

Theorem 4.8 states that if a system \((A, B)\) is controllable, any system \((A + BK, B)\) is also controllable. This is because the transformation induced by \(K\) does not affect the rank condition of the controllability matrix, as it only affects the stability or pole placement of the system, not the controllability.
04

Verify Controllability Matrix

For the new system \((A + BK, B)\), the controllability matrix is \([B, (A+BK)B, (A+BK)^2B, \ldots, (A+BK)^{n-1}B]\). Since the original system \((A, B)\) is controllable, the controllability matrix \([B, AB, A^2B, \ldots, A^{n-1}B]\) is full rank. The term \(BK\) does not change the rank, confirming that the new system's controllability matrix also has full rank.
05

Conclusion

Since the new system's controllability matrix \([B, (A+BK)B, (A+BK)^2B, \ldots, (A+BK)^{n-1}B]\) maintains full rank, the system \((A + BK, B)\) is controllable as per Theorem 4.8 and the controllability condition.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Controllability Theorem
The Controllability Theorem is a fundamental concept in control theory which tells us when a system can be fully controlled. Specifically, it ensures that we can move a system from any initial state to any desired final state within a finite time.
By applying certain inputs, we adjust the system's behavior as needed.This theorem applies to linear systems generally expressed in state-space form as: - \( \dot{x} = Ax + Bu \), where \( x \) represents the state vector, \( u \) the input vector, \( A \) is the state matrix, and \( B \) the input matrix. The theorem requires the implementation of the controllability matrix for proper assessment. It can prove that even transformed or altered systems remain controllable under specific conditions.For instance, if a new control is implemented as \( u = Kx + v \), the system is not fundamentally altered in terms of controllability. It remains controllable as long as the original system is controllable and the conditions of the theorem are met, supporting stability and effective pole placement.
Controllability Matrix
The controllability matrix is a crucial tool utilized to determine if a system is controllable. For a given system described by the matrices \( A \) and \( B \), this matrix takes the form:- \( [B, AB, A^2B, \ldots, A^{n-1}B] \).This matrix allows us to check if the system can reach any state from any initial state. The central requirement is that the controllability matrix must have full rank.
Specifically, it should equal the dimension \( n \) of the state matrix \( A \). If the matrix has full rank, it indicates that the system is controllable.For a modified system \((A + BK, B)\), the corresponding controllability matrix is:- \( [B, (A+BK)B, (A+BK)^2B, \ldots, (A+BK)^{n-1}B] \).Importantly, the addition of the term \( BK \) doesn't affect the full rank condition of the matrix, thereby maintaining the controllability. It confirms that any state within the system can still be achieved with appropriate inputs.
State Space Representation
State Space Representation refers to a mathematical model used to describe linear systems comprehensively. It provides a big-picture viewpoint by capturing all aspects of a system with matrices. Systems are represented in a form like:- \( \dot{x} = Ax + Bu \) (for continuous systems), where: - \( \dot{x} \) is the derivative of the state vector \( x \), - \( A \) is the system matrix that dictates how the current state affects changes in the state, - \( B \) is the input matrix showing how control inputs influence the state, and - \( u \) is the input vector.- An output equation might also be included as \( y = Cx + Du \), where \( C \) and \( D \) relate the state and output directly.This representation captures the dynamic behavior of systems, making it easier to apply various analyses, including controllability checks. Whether you transform the system by introducing new control laws or keeping it as-is, state space representation helps in comprehending the interactions between various state variables and inputs. It is instrumental in ensuring that any conditions, such as controllability and reachability, are fulfilled efficiently. By using this model, engineers can design systems to achieve desirable performance, ensuring the system behaves predictably under various conditions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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 \(a\).

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 \(W S=\bar{W}\).

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\. \(A=\left(\begin{array}{lll}\lambda & 0 & 0 \\ 0 & \lambda & 1 \\ 0 & 0 & \lambda\end{array}\right), B=\left(\begin{array}{l}b_{1} \\ b_{2} \\\ b_{3}\end{array}\right)\)

See all solutions

Recommended explanations on Computer Science Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free