A Problem about State Variables

Calculus Level 5

A discrete-time system is described by the state equation

V ( k + 1 ) = A V ( k ) + B u V(k+1) = A V(k) + B u

with V ( k ) = [ x ( k ) , y ( k ) ] T V(k) = [ x(k) , y(k) ]^T being the state vector, and

A = [ 2 − 3 0.5 − 0.5 ] A = \begin{bmatrix} 2 && -3 \\ 0.5 && -0.5 \end{bmatrix}

B = [ 0 1 ] B = \begin{bmatrix} 0 \\ 1 \end{bmatrix}

u = 1 u = 1

If the system is initially relaxed, i.e. V ( 0 ) = [ 0 , 0 ] T V(0) = [ 0 , 0 ]^T , then what is the value of the following limit

lim ⁡ k → ∞ x ( k ) y ( k ) \lim_{k \to\infty} \frac{x(k)}{y(k)}


The answer is 3.

Hosam Hajjir by Hosam Hajjir , 56, Canada

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1 solution

Aareyan Manzoor
Jun 4, 2019

lemma: V ( n ) = ( ∑ k = 0 n − 1 A k ) B V(n) = \left(\sum_{k=0}^{n-1} A^k\right) B proof(by induction): the base case V ( 1 ) = B V(1) = B holds. then V ( n + 1 ) = A V ( n ) + B = A ( ∑ k = 0 n − 1 A k ) B + B = ( 1 + ∑ k = 0 n − 1 A k + 1 ) B = ( ∑ k = 0 n A k ) B V(n+1) = A V(n)+B= A \left(\sum_{k=0}^{n-1} A^k\right) B +B = \left(1+\sum_{k=0}^{n-1} A^{k+1}\right)B = \left(\sum_{k=0}^{n} A^k\right)B hence proven by induction.

now diagonalize the matrix A: A = S − 1 Λ S = [ 2 3 1 1 ] [ 1 / 2 0 0 1 ] [ − 1 3 1 − 2 ] A =S^{-1} \Lambda S= \begin{bmatrix} 2 && 3 \\ 1 && 1 \end{bmatrix} \begin{bmatrix} 1/2 && 0 \\ 0 && 1 \end{bmatrix} \begin{bmatrix} -1 && 3 \\ 1 && -2 \end{bmatrix} hence V ( n ) = ( ∑ k = 0 n − 1 A k ) B = S − 1 [ 2 ( 1 − 2 − n ) 0 0 n ] S B = [ 2 3 1 1 ] [ 2 ( 1 − 2 − n ) 0 0 n ] [ − 1 3 1 − 2 ] [ 0 1 ] = [ 4 ( 1 − 2 − n ) 3 n 2 ( 1 − 2 − n ) n ] [ 3 − 2 ] = [ 12 ( 1 − 2 − n ) − 6 n 6 ( 1 − 2 − n ) − 2 n ] V(n)=\left(\sum_{k=0}^{n-1} A^k\right) B = S^{-1} \begin{bmatrix} 2(1-2^{-n}) && 0 \\ 0 && n \end{bmatrix} S B \\ =\begin{bmatrix} 2 && 3 \\ 1 && 1 \end{bmatrix} \begin{bmatrix} 2(1-2^{-n}) && 0 \\ 0 && n \end{bmatrix} \begin{bmatrix} -1 && 3 \\ 1 && -2 \end{bmatrix}\begin{bmatrix} 0 \\ 1 \end{bmatrix} \\=\begin{bmatrix} 4(1-2^{-n}) && 3n \\ 2(1-2^{-n}) && n \end{bmatrix}\begin{bmatrix} 3 \\ -2 \end{bmatrix} =\begin{bmatrix} 12(1-2^{-n}) - 6n \\ 6(1-2^{-n}) - 2n \end{bmatrix} from this, we can plug the expressions into the limit to get lim ⁡ n → ∞ 12 ( 1 − 2 − n ) − 6 n 6 ( 1 − 2 − n ) − 2 n = 3 \lim_{n\to \infty} \dfrac{12(1-2^{-n}) - 6n}{ 6(1-2^{-n}) - 2n } = \boxed{3}

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