--- title: "Backward difference equation with step input" subject: "Digital Control System" chapter: 2 key: "semester_6/digital_control_system:difference-equation-backward-step" source: https://www.bctengineer.com/q/semester_6/digital_control_system/difference-equation-backward-step site: BCT Engineering --- # Backward difference equation with step input Digital Control System · Chapter 2: The Z-Transform - Asked in: 2071 Bhadra, 6 marks - Video: [SOLUTION OF DIFFERENCE EQUATION](https://www.youtube.com/watch?v=1KuWLNN2SKw) — Department of EC SIRT Bhopal - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/digital_control_system/difference-equation-backward-step - Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-2 ## Question Solve the following difference equation: 2x(k)-2x(k-1)+x(k-2)=u(k); where x(k)=0 for k<0 and u(k) is unit step function. ## Short answer Take z-transform, solve for X(z). Poles at z=1 and 0.5±j0.5. Solution: x(k) = 1 + (1/√2)^{k+1} sin((k−1)π/4), k ≥ 0; overshoot 1.25, settles to 1. ## Answer 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 $x(k) = \frac12\left[1 + 2x(k-1) - x(k-2)\right]$ | $k$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | |---|---|---|---|---|---|---|---|---|---|---| | $x(k)$ | 0.5 | 1 | 1.25 | 1.25 | 1.125 | 1 | 0.9375 | 0.9375 | 0.9688 | 1 | 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 $x(\infty) = 1$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With x(k)=0 for k<0, Z[x(k-n)] equals: A. z^{-n}X(z) B. z^{n}X(z) C. z^{-n}X(z) - x(0) D. X(z)/n Answer: A. No initial terms remain, so a delay is simply multiplication by z^{-n}. (Key point: Delay theorem) 2. The roots of 2z^2 - 2z + 1 = 0 are: A. 0.5 ± j0.5 B. 1 ± j1 C. 0.5 and 1 D. ±j0.5 Answer: A. z = (2 ± sqrt(-4))/4 = 0.5 ± j0.5, magnitude 0.7071 and angle pi/4. (Key point: Poles) 3. The final value of x(k) is: A. 1 B. 0.5 C. 2 D. 0 Answer: A. At z=1, 2 - 2 + 1 = 1, so (1-z^{-1})X(z) tends to 1. (Key point: Steady-state value) 4. From 2x(k) - 2x(k-1) + x(k-2) = 1, what are x(0) and x(1)? A. 0.5 and 1 B. 1 and 1 C. 0.5 and 0.5 D. 0 and 1 Answer: A. k=0: 2x(0)=1; k=1: 2x(1) - 2(0.5) = 1, so x(1) = 1. 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