Difference equation with input u(2) = −0.2142

The output is found by recursion, and the z-transform then shows why it becomes constant.

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: Why the output stays at 1 (z-transform)

$$X(z) = \frac{(0.3679z + 0.2642)(z^2 + 0.2142z - 0.2142)}{z^2(z-1)(z-0.3679)}$$

The input numerator has a root at $z = 0.3679$:

$$(0.3679)^2 + 0.2142(0.3679) - 0.2142 = 0.13535 + 0.07880 - 0.2142 \approx 0$$

So $z^2 + 0.2142z - 0.2142 \approx (z - 0.3679)(z + 0.5821)$. The plant pole at $0.3679$ is cancelled:

$$X(z) = \frac{(0.3679z + 0.2642)(z + 0.5821)}{z^2(z-1)}$$

Only the pole at $z = 1$ remains, apart from poles at the origin, which give a finite transient. Expanding:

$$X(z) = 0.3679z^{-1} + 0.8463z^{-2} + z^{-3} + z^{-4} + \dots$$

Result

This is a deadbeat response: the output reaches its final value 1 in three sampling periods and stays there with no ripple. The chosen input sequence cancels the slow plant mode.

Answer: $x(0) = 0$, $x(1) = 0.3679$, $x(2) = 0.8463$, and $x(k) = 1$ for $k \ge 3$.