Difference equation with u(K+1) and u(K) step inputs

Assumption: the coefficient "1.379" is read as 1.3679. This is the standard textbook equation, where $1.3679 = 1 + e^{-1}$ and $0.3679 = e^{-1}$, and it agrees with the given $x(1) = 0.3679$. A note at the end gives the values for 1.379 as printed.

Step 1: Z-transform with initial conditions

Use $\mathcal{Z}[x(k+2)] = z^2X - z^2x(0) - zx(1)$, $\mathcal{Z}[x(k+1)] = zX - zx(0)$, and $\mathcal{Z}[u(k+1)] = zU - zu(0)$. Substitute $x(0) = 0$, $x(1) = 0.3679$, $u(0) = 1$:

$$z^2X - 0.3679z - 1.3679zX + 0.3679X = 0.3679zU - 0.3679z + 0.2642U$$

The $-0.3679z$ terms on both sides cancel:

$$X(z)\,(z^2 - 1.3679z + 0.3679) = (0.3679z + 0.2642)\,U(z)$$

Step 2: Substitute the step input $U(z) = \dfrac{z}{z-1}$

$z^2 - 1.3679z + 0.3679 = (z-1)(z-0.3679)$, so

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

Step 3: Partial fractions of $X(z)/z$

$$\frac{X(z)}{z} = \frac{0.3679z + 0.2642}{(z-1)^2(z-0.3679)} = \frac{A}{(z-1)^2} + \frac{B}{z-1} + \frac{C}{z-0.3679}$$ $$\begin{aligned} A &= \left[\frac{0.3679z + 0.2642}{z - 0.3679}\right]_{z=1} = \frac{0.6321}{0.6321} = 1 \\ C &= \left[\frac{0.3679z + 0.2642}{(z-1)^2}\right]_{z=0.3679} = \frac{0.1354 + 0.2642}{(0.6321)^2} = \frac{0.3996}{0.3996} = 1 \\ B &= \frac{d}{dz}\left[\frac{0.3679z + 0.2642}{z - 0.3679}\right]_{z=1} = \left[\frac{0.3679(z-0.3679) - (0.3679z+0.2642)}{(z-0.3679)^2}\right]_{z=1} = \frac{-0.3996}{0.3996} = -1 \end{aligned}$$ $$X(z) = \frac{z}{(z-1)^2} - \frac{z}{z-1} + \frac{z}{z - 0.3679}$$

Step 4: Inverse transform

Using $\dfrac{z}{(z-1)^2} \to k$, $\dfrac{z}{z-1} \to 1$, $\dfrac{z}{z-e^{-1}} \to e^{-k}$:

$$x(k) = k - 1 + (0.3679)^k = k - 1 + e^{-k}, \quad k = 0, 1, 2, \dots$$

Step 5: Check

The recursion is $x(k+2) = 1.3679x(k+1) - 0.3679x(k) + 0.6321$. The two agree to rounding. The double pole at $z = 1$ makes the output grow like a ramp ($x(k) \approx k - 1$ for large $k$). This is the step response of an open-loop system containing an integrator.

Note: with 1.379 as printed

The recursion $x(k+2) = 1.379x(k+1) - 0.3679x(k) + 0.6321$ gives $0, 0.3679, 1.1394, 2.0680, 3.0647, 4.0975, \dots$. The roots of $z^2 - 1.379z + 0.3679$ are then $z = 1.0174$ and $0.3616$, so the output grows slightly faster than a ramp.

Answer (1.3679): $x(k) = k - 1 + e^{-k} = k - 1 + (0.3679)^k$ for $k \ge 0$.