Difference equation with given input sequence

The output is found by solving the equation recursively, then confirmed with a closed form from the z-transform.

Step 1: Initial values

$x(k) = 0$ for $k \le 0$, so $x(0) = 0$. Put $k = -1$ in the equation, with $x(-1) = 0$, $u(-1) = 0$, $u(0) = 1$:

$$x(1) - 1.3679x(0) + 0.3679x(-1) = 0.3679u(0) + 0.2642u(-1) \;\Rightarrow\; x(1) = 0.3679$$

Step 2: Pulse transfer function

With zero initial conditions (the $x(1)$ and $u(0)$ terms cancel), taking z-transforms gives:

$$G(z) = \frac{X(z)}{U(z)} = \frac{0.3679z + 0.2642}{z^2 - 1.3679z + 0.3679} = \frac{0.3679z + 0.2642}{(z-1)(z-0.3679)}$$

The input is a finite sequence:

$$U(z) = 1 + 0.2142z^{-1} + 0.2142z^{-2} = \frac{z^2 + 0.2142z + 0.2142}{z^2}$$

Step 3: Recursive solution

$$x(k+2) = 1.3679x(k+1) - 0.3679x(k) + 0.3679u(k+1) + 0.2642u(k)$$

Step 4: Closed form for $k \ge 3$

$X(z) = G(z)U(z) = \dfrac{(0.3679z+0.2642)(z^2+0.2142z+0.2142)}{z^2(z-1)(z-0.3679)}$. For $k \ge 3$ only the poles at $z = 1$ and $z = 0.3679$ contribute to the residues of $X(z)z^{k-1}$:

$$\begin{aligned} \text{Res}_{z=1} &= \frac{(0.6321)(1.4284)}{1\,(0.6321)} = 1.4284 \\ \text{Res}_{z=0.3679} &= \frac{(0.3995)(0.4284)}{(0.3679)^2(-0.6321)}(0.3679)^{k-1} = -5.4375(0.3679)^k \end{aligned}$$ $$x(k) = 1.4284 - 5.4375(0.3679)^k, \quad k \ge 3$$

Check: $k = 3$ gives $1.4284 - 5.4375(0.04979) = 1.1576$, which matches.

Result

The final value from the theorem is $\lim_{z\to1}(1-z^{-1})X(z) = \dfrac{0.6321(1 + 0.2142 + 0.2142)}{0.6321} = 1.4284$.

Answer: $x(0) = 0$, $x(1) = 0.3679$, $x(2) = 0.8463$, and $x(k) = 1.4284 - 5.4375(0.3679)^k$ for $k \ge 3$. The output settles at 1.4284.

(With $u(2) = -0.2142$, the textbook deadbeat case, the output is exactly 1 for all $k \ge 3$.)