Difference equation with step input, x(1)=0.3679

Given (as printed) $x(k+2) + 1.379x(k+1) + 0.3679x(k) = 0.3679u(k)$, $u(k) = 1$ for $k \ge 0$, $x(0) = 0$, $x(1) = 0.3679$.

Step 1: z-transform

$$[z^2X - z^2x(0) - zx(1)] + 1.379[zX - zx(0)] + 0.3679X = 0.3679\frac{z}{z-1}$$ $$\begin{aligned} (z^2 + 1.379z + 0.3679)X(z) &= 0.3679z + \frac{0.3679z}{z-1} = \frac{0.3679z^2}{z-1} \\ X(z) &= \frac{0.3679z^2}{(z-1)(z^2+1.379z+0.3679)} \end{aligned}$$

Step 2: roots of the characteristic equation

$$z = \frac{-1.379 \pm \sqrt{1.379^2 - 4(0.3679)}}{2} = \frac{-1.379 \pm \sqrt{0.4300}}{2} = \frac{-1.379 \pm 0.6558}{2}$$

$p_1 = -0.3616$, $\quad p_2 = -1.0174$

Step 3: partial fractions of $X(z)/z = \frac{0.3679z}{(z-1)(z-p_1)(z-p_2)}$

$$\begin{aligned} A &= \frac{0.3679(1)}{(1-p_1)(1-p_2)} = \frac{0.3679}{2.7469} = 0.1339 \\ B &= \frac{0.3679p_1}{(p_1-1)(p_1-p_2)} = \frac{0.3679(-0.3616)}{(-1.3616)(0.6558)} = 0.1490 \\ C &= \frac{0.3679p_2}{(p_2-1)(p_2-p_1)} = \frac{0.3679(-1.0174)}{(-2.0174)(-0.6558)} = -0.2829 \end{aligned}$$ $$X(z) = 0.1339\frac{z}{z-1} + 0.1490\frac{z}{z+0.3616} - 0.2829\frac{z}{z+1.0174}$$

Result

$$x(k) = 0.1339 + 0.1490(-0.3616)^k - 0.2829(-1.0174)^k$$

Check

These agree with the recursion $x(k+2) = -1.379x(k+1) - 0.3679x(k) + 0.3679$. Since $|p_2| > 1$, the response oscillates with slowly growing amplitude, so the final value theorem cannot be applied.

Answer: $x(k) = 0.1339 + 0.1490(-0.3616)^k - 0.2829(-1.0174)^k$.

Note: this equation is usually printed in textbooks with $-1.3679$ (poles at $1$ and $0.3679$). With $x(k+2) - 1.3679x(k+1) + 0.3679x(k) = 0.3679u(k)$ and the same data, $X(z) = \frac{0.3679z^2}{(z-1)^2(z-0.3679)}$ and $x(k) = 0.5820k - 0.3388 + 0.3388(0.3679)^k$, i.e. $0, 0.3679, 0.8712, 1.4242, \dots$ (a ramp-like rise).