--- title: "Inverse Z-transform by inversion integral (cubic)" subject: "Digital Control System" chapter: 2 key: "semester_6/digital_control_system:inverse-integral-z-1-cubic" source: https://www.bctengineer.com/q/semester_6/digital_control_system/inverse-integral-z-1-cubic site: BCT Engineering --- # Inverse Z-transform by inversion integral (cubic) Digital Control System · Chapter 2: The Z-Transform - Asked in: 2068 Jestha, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/digital_control_system/inverse-integral-z-1-cubic - Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-2 ## Question Find the inverse Z-transform of X(z)=z⁻¹/((1-z⁻¹)(1+1.6z⁻¹+0.64z⁻²)) by inversion integral method. ## Short answer Rewrite X(z)=z^2/((z−1)(z+0.8)^2). Sum residues of X(z)z^{k−1} at z=1 and double pole z=−0.8 gives x(k) = (1/3.24)[1 − (1+1.8k)(−0.8)^k] for k ≥ 0. ## Answer In the **inversion integral method**, $x(k)$ is the sum of the residues of $X(z)z^{k-1}$ at its poles: $$ x(k) = \frac{1}{2\pi j}\oint_C X(z)z^{k-1}\,dz = \sum_i\text{Res}\left[X(z)z^{k-1}\right]_{z=z_i} $$ #### Step 1: Rewrite $X(z)$ in powers of $z$ $1 + 1.6z^{-1} + 0.64z^{-2} = (1 + 0.8z^{-1})^2$. Multiply numerator and denominator by $z^3$: $$ X(z) = \frac{z^{-1}}{(1 - z^{-1})(1 + 0.8z^{-1})^2} = \frac{z^2}{(z-1)(z+0.8)^2} $$ $$ X(z)z^{k-1} = \frac{z^{k+1}}{(z-1)(z+0.8)^2} $$ For $k \ge 0$ there is no pole at the origin. The poles are a simple pole at $z = 1$ and a double pole at $z = -0.8$. #### Step 2: Residue at $z = 1$ $$ R_1 = \left[\frac{z^{k+1}}{(z+0.8)^2}\right]_{z=1} = \frac{1}{(1.8)^2} = \frac{1}{3.24} = 0.3086 $$ #### Step 3: Residue at the double pole $z = -0.8$ $$ \begin{aligned} R_2 &= \frac{d}{dz}\left[\frac{z^{k+1}}{z-1}\right]_{z=-0.8} = \left[\frac{(k+1)z^{k}(z-1) - z^{k+1}}{(z-1)^2}\right]_{z=-0.8} \\ &= \frac{(-0.8)^k\left[(k+1)(-1.8) - (-0.8)\right]}{(-1.8)^2} = \frac{(-0.8)^k\left(-1.8k - 1\right)}{3.24} \end{aligned} $$ #### Step 4: Result $$ x(k) = R_1 + R_2 = \frac{1}{3.24}\left[1 - (1 + 1.8k)(-0.8)^k\right], \quad k = 0, 1, 2, \dots $$ Equivalently $x(k) = 0.3086 - (0.3086 + 0.5556k)(-0.8)^k$. #### Step 5: Check | $k$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | $\infty$ | |---|---|---|---|---|---|---|---|---| | $x(k)$ | 0 | 1 | −0.6 | 1.32 | −0.728 | 1.32 | −0.646 | 0.3086 | Expanding the denominator gives $(1 - z^{-1})(1 + 1.6z^{-1} + 0.64z^{-2}) = 1 + 0.6z^{-1} - 0.96z^{-2} - 0.64z^{-3}$. Direct division of $\dfrac{z^{-1}}{1 + 0.6z^{-1} - 0.96z^{-2} - 0.64z^{-3}}$ gives $z^{-1} - 0.6z^{-2} + 1.32z^{-3} - 0.728z^{-4} + \dots$, which matches. The final value theorem gives $\lim_{z\to1}(1 - z^{-1})X(z) = \dfrac{1}{(1.8)^2} = 0.3086$, which also matches. **Answer:** $x(k) = \dfrac{1}{3.24}\left[1 - (1 + 1.8k)(-0.8)^k\right]$ for $k \ge 0$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. X(z) = z^{-1}/((1 - z^{-1})(1 + 1.6z^{-1} + 0.64z^{-2})) becomes: A. z^2/((z-1)(z+0.8)^2) B. z/((z-1)(z+0.8)^2) C. z^2/((z-1)(z-0.8)^2) D. z^3/((z-1)(z+0.8)^2) Answer: A. 1 + 1.6z^{-1} + 0.64z^{-2} = (1 + 0.8z^{-1})^2; multiplying by z^3 gives z^2 over (z-1)(z+0.8)^2. (Key point: Rewriting in powers of z) 2. The poles of X(z)z^{k-1} for k >= 0 are: A. Simple at z = 1 and double at z = -0.8 B. Simple at z = 1 and z = -0.8 and 0 C. Double at z = 1 D. Double at z = 0.8 Answer: A. There is no pole at the origin for k >= 0. (Key point: Pole structure) 3. The residue at z = 1 equals: A. 1/(1.8)^2 = 0.3086 B. 1/(0.2)^2 C. -0.3086 D. 1/1.8 Answer: A. z^{k+1}/(z+0.8)^2 at z=1 is 1/3.24; it also equals the final value. (Key point: Residue at z = 1) 4. The solution is: A. (1/3.24)[1 - (1+1.8k)(-0.8)^k] B. (1/3.24)[1 + (1+1.8k)(-0.8)^k] C. (1/3.24)[1 - (1+1.8k)(0.8)^k] D. (1/1.8)[1 - (-0.8)^k] Answer: A. Adding both residues gives this; x(0)=0 and x(1)=1 check. 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