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Given is the following discrete system

$$\begin{align*} &x(k + 1) = Ax(k) + Bu(k)\\ &x(0) = x_0\;. \end{align*}$$

How do we prove that the explicit solution formula for $x(k)$ (analogously to the variation of constants formula in the continuous time case) is

$$A^kx_0+\sum_{j=1}^kA^{j-k}Bu(j-1)\;?$$

Thanks a lot!

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The power of $A$ inside the summation should be $k-j$. It is a convolution. –  copper.hat Nov 19 '12 at 0:35

1 Answer 1

Your formula is incorrect. The correct formula is $\phi_k = A^kx_0+\sum_{j=0}^{k-1} A^{k-j-1}Bu_{j}$, where the summation is taken to be $0$ when $k=0$.

With this convention, you have $\phi_0 = x_0$, and the induction step gives \begin{eqnarray} \phi_{k+1} & = & A^{k+1} x_0+\sum_{j=0}^{k} A^{k-j}Bu_{j}\\ &=& A(A^k x_0+\sum_{j=0}^{k-1} A^{k-j-1}Bu_{j}) + Bu_k\\ & = & A \phi_k+B u_k \end{eqnarray} Hence $\phi$ satisfies the same difference equation as $x$ with the same initial condition, hence $\phi_k = x(k)$ for all $k \geq 0$, that is $\phi$ is the solution.

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How do we derive the expression without knowing it in advance, btw? That is, if we are simply given the task to derive the explicit solution given the initial problem using discrete time) –  user49909 Nov 19 '12 at 0:27
    
Write out a few terms of the solution and guess the pattern $x_1 Ax_0+B u_0$, $x_2 = Ax_1+B u_1 = A^2 x_0 + A B u_0 + Bu_1$, etc... Alternatively, you could use the $z$-transform, but guess & prove is fairly straightforward. –  copper.hat Nov 19 '12 at 0:32

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