Backward difference equation with step input

Solve by z-transform with $x(k) = 0$ for $k < 0$ and $U(z) = \dfrac{1}{1-z^{-1}}$.

Step 1: Z-transform

Since $x(-1) = x(-2) = 0$, the delay theorem gives $\mathcal{Z}[x(k-n)] = z^{-n}X(z)$:

$$(2 - 2z^{-1} + z^{-2})X(z) = \frac{1}{1 - z^{-1}}$$ $$X(z) = \frac{1}{(1-z^{-1})(2 - 2z^{-1} + z^{-2})} = \frac{z^3}{(z-1)(2z^2 - 2z + 1)}$$

Step 2: Poles

$$2z^2 - 2z + 1 = 0 \;\Rightarrow\; z = \frac{2 \pm \sqrt{4 - 8}}{4} = 0.5 \pm j0.5 = \frac{1}{\sqrt2}e^{\pm j\pi/4}$$

So $r = \dfrac{1}{\sqrt2} = 0.7071$ and $\Omega = \dfrac{\pi}{4}$. Both complex poles are inside the unit circle.

Step 3: Form of the solution

The pole at $z = 1$ gives a constant, and the complex pair gives a damped sinusoid:

$$x(k) = A_0 + r^k\left(A\cos\frac{k\pi}{4} + B\sin\frac{k\pi}{4}\right)$$

• Steady state: $A_0 = \lim_{z\to1}(1-z^{-1})X(z) = \dfrac{1}{2 - 2 + 1} = 1$
• Initial values from the equation: $k = 0$ gives $2x(0) = 1$, so $x(0) = 0.5$. $k = 1$ gives $2x(1) - 2(0.5) = 1$, so $x(1) = 1$.

$$\begin{aligned} k = 0:&\quad 1 + A = 0.5 \;\Rightarrow\; A = -0.5 \\ k = 1:&\quad 1 + 0.7071\left(-0.5\times0.7071 + B\times0.7071\right) = 1 \;\Rightarrow\; B = 0.5 \end{aligned}$$

Step 4: Closed-form solution

$$x(k) = 1 + \left(\frac{1}{\sqrt2}\right)^k\left[-0.5\cos\frac{k\pi}{4} + 0.5\sin\frac{k\pi}{4}\right], \quad k \ge 0$$

Since $-0.5\cos\theta + 0.5\sin\theta = \frac{1}{\sqrt2}\sin\left(\theta - \frac{\pi}{4}\right)$, this simplifies to

$$x(k) = 1 + \left(\frac{1}{\sqrt2}\right)^{k+1}\sin\frac{(k-1)\pi}{4}$$

Step 5: Check with the recursion $x(k) = \frac12\left[1 + 2x(k-1) - x(k-2)\right]$

The formula gives the same values. For example, $k = 2$: $1 + (0.7071)^3\sin(\pi/4) = 1 + 0.3536\times0.7071 = 1.25$.

The output overshoots to 1.25 and settles at 1 with a decaying oscillation.

Answer: $x(k) = 1 + \left(\dfrac{1}{\sqrt2}\right)^{k+1}\sin\dfrac{(k-1)\pi}{4}$, with $x(\infty) = 1$.