Logo

AskSia

Plus

For the following discrete-time state equation, COMPUTE the state response $x(k...
Oct 2, 2024
Solution by Steps
step 1
We start with the initial condition x(0)=[21] x(0) = \begin{bmatrix} 2 \\ 1 \end{bmatrix} and the state equation x(k+1)=[−1amp;20amp;1]x(k) x(k+1) = \begin{bmatrix} -1 & 2 \\ 0 & 1 \end{bmatrix} x(k)
step 2
To find x(1) x(1) , we compute x(1)=[−1amp;20amp;1][21]=[−1⋅2+2⋅10⋅2+1⋅1]=[01] x(1) = \begin{bmatrix} -1 & 2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} -1 \cdot 2 + 2 \cdot 1 \\ 0 \cdot 2 + 1 \cdot 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}
step 3
Next, we find x(2) x(2) using x(1) x(1) : x(2)=[−1amp;20amp;1][01]=[−1⋅0+2⋅10⋅0+1⋅1]=[21] x(2) = \begin{bmatrix} -1 & 2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} -1 \cdot 0 + 2 \cdot 1 \\ 0 \cdot 0 + 1 \cdot 1 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}
step 4
We observe that x(2)=x(0) x(2) = x(0) , indicating a periodic behavior. Thus, x(k) x(k) will repeat every 2 steps: x(0)=[21] x(0) = \begin{bmatrix} 2 \\ 1 \end{bmatrix} and x(1)=[01] x(1) = \begin{bmatrix} 0 \\ 1 \end{bmatrix}
Answer
The state response x(k) x(k) is [21] \begin{bmatrix} 2 \\ 1 \end{bmatrix} for even k k and [01] \begin{bmatrix} 0 \\ 1 \end{bmatrix} for odd k k .
Key Concept
The periodicity in discrete-time state equations can lead to repeating state responses.
Explanation
The computed state responses alternate between two vectors, demonstrating the periodic nature of the system defined by the state equation.
Try using the C.H. therorem
To compute the state response x(k) x(k) using the initial condition x(0)=[21] x(0) = \begin{bmatrix} 2 \\ 1 \end{bmatrix} and the state equation x(k+1)=[−1amp;20amp;1]x(k), x(k+1) = \begin{bmatrix} -1 & 2 \\ 0 & 1 \end{bmatrix} x(k), we can apply the Cayley-Hamilton theorem (C.H. theorem). This theorem states that every square matrix satisfies its own characteristic polynomial. Step 1: Find the characteristic polynomial First, we need to find the characteristic polynomial of the matrix A=[−1amp;20amp;1]. A = \begin{bmatrix} -1 & 2 \\ 0 & 1 \end{bmatrix}. The characteristic polynomial p(λ) p(\lambda) is given by p(λ)=det⁡(A−λI, p(\lambda) = \det(A - \lambda I, where I I is the identity matrix. Thus, A−λI=[−1−λamp;20amp;1−λ]. A - \lambda I = \begin{bmatrix} -1 - \lambda & 2 \\ 0 & 1 - \lambda \end{bmatrix}. Calculating the determinant: det⁡(A−λI)=(−1−λ)(1−λ)−(0⋅2)=λ2+1. \det(A - \lambda I) = (-1 - \lambda)(1 - \lambda) - (0 \cdot 2) = \lambda^2 + 1. Step 2: Apply the Cayley-Hamilton theorem According to the C.H. theorem, we have: A2+I=0, A^2 + I = 0, which implies A2=−I. A^2 = -I. Step 3: Compute the state response Now, we can express x(k) x(k) in terms of A A and the initial condition x(0) x(0) : x(k)=Akx(0). x(k) = A^k x(0). We can compute a few terms to identify a pattern: 1. For \( k = 0 \): x(0)=[21]. x(0) = \begin{bmatrix} 2 \\ 1 \end{bmatrix}. 2. For \( k = 1 \): x(1)=Ax(0)=[−1amp;20amp;1][21]=[−2+21]=[01]. x(1) = A x(0) = \begin{bmatrix} -1 & 2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} -2 + 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}. 3. For \( k = 2 \): x(2)=Ax(1)=[−1amp;20amp;1][01]=[21]. x(2) = A x(1) = \begin{bmatrix} -1 & 2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}. 4. For \( k = 3 \): x(3)=Ax(2)=[−1amp;20amp;1][21]=[−2+21]=[01]. x(3) = A x(2) = \begin{bmatrix} -1 & 2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} -2 + 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}. Conclusion From the calculations, we can see that the state response alternates between two states: - For even k k : x(k)=[21] x(k) = \begin{bmatrix} 2 \\ 1 \end{bmatrix} - For odd k k : x(k)=[01] x(k) = \begin{bmatrix} 0 \\ 1 \end{bmatrix} Thus, the state response x(k) x(k) can be summarized as: x(k)={[21]amp;if k is even[01]amp;if k is odd. x(k) = \begin{cases} \begin{bmatrix} 2 \\ 1 \end{bmatrix} & \text{if } k \text{ is even} \\ \begin{bmatrix} 0 \\ 1 \end{bmatrix} & \text{if } k \text{ is odd} \end{cases}.
© 2023 AskSia.AI all rights reserved