--- title: "Impulse response of y[n] − (3/4)y[n−1] + (1/8)y[n−2] = 2x[n]" subject: "Signals and Systems" chapter: 6 key: "semester_4/signals_and_systems:second-order-difference-equation" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/second-order-difference-equation site: BCT Engineering --- # Impulse response of y[n] − (3/4)y[n−1] + (1/8)y[n−2] = 2x[n] Signals and Systems · Chapter 6: Discrete-Time Systems - Asked in: 2082 Chaitra, 6 marks; 2080 Chaitra, 8 marks (asked 2 times) - Video: [Find the impulse response of the system for difference equation y(n)-3y(n-1)-4y(n-2)=x(n)+2x(n-1)](https://www.youtube.com/watch?v=xoWNFCkFnJI) — Dr.Shakunthala Masi - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/second-order-difference-equation - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-6 ## Question Find impulse response for the DT system given by the following linear constant coefficient difference equation. y[n] − (3/4)y[n−1] + (1/8)y[n−2] = 2x[n] ## Short answer Factor the denominator into (1−0.5e⁻ʲω)(1−0.25e⁻ʲω), use partial fractions, and apply inverse DTFT: h[n] = [4(0.5)ⁿ − 2(0.25)ⁿ] u[n]. ## Answer **Given:** $y[n] - \frac{3}{4}y[n-1] + \frac{1}{8}y[n-2] = 2x[n]$, system initially at rest. #### Step 1: Frequency response Take the DTFT ($x[n-k] \leftrightarrow e^{-j\omega k}X(e^{j\omega})$): $$ \left(1 - \tfrac{3}{4}e^{-j\omega} + \tfrac{1}{8}e^{-j2\omega}\right)Y(e^{j\omega}) = 2X(e^{j\omega}) $$ $$ H(e^{j\omega}) = \frac{2}{1 - \frac{3}{4}e^{-j\omega} + \frac{1}{8}e^{-j2\omega}} $$ #### Step 2: Factorise the denominator Let $v = e^{-j\omega}$. The roots of $1 - \frac{3}{4}v + \frac{1}{8}v^2$ come from $p^2 - \frac{3}{4}p + \frac{1}{8} = 0$, giving $p = \frac{1}{2}, \frac{1}{4}$. So $$ H(e^{j\omega}) = \frac{2}{\left(1 - \frac{1}{2}e^{-j\omega}\right)\left(1 - \frac{1}{4}e^{-j\omega}\right)} $$ #### Step 3: Partial fractions $$ H = \frac{A}{1 - \frac{1}{2}v} + \frac{B}{1 - \frac{1}{4}v} $$ $$ \begin{aligned} A &= \left.\frac{2}{1 - \frac{1}{4}v}\right|_{v=2} = \frac{2}{1 - \frac{1}{2}} = 4 \\ B &= \left.\frac{2}{1 - \frac{1}{2}v}\right|_{v=4} = \frac{2}{1 - 2} = -2 \end{aligned} $$ $$ H(e^{j\omega}) = \frac{4}{1 - \frac{1}{2}e^{-j\omega}} - \frac{2}{1 - \frac{1}{4}e^{-j\omega}} $$ #### Step 4: Inverse DTFT Using $a^n u[n] \leftrightarrow \frac{1}{1 - a e^{-j\omega}}$ ($|a| < 1$): $$ h[n] = \left[4\left(\tfrac{1}{2}\right)^n - 2\left(\tfrac{1}{4}\right)^n\right]u[n] $$ #### Check by recursion $h[n] = 2\delta[n] + \frac{3}{4}h[n-1] - \frac{1}{8}h[n-2]$, with $h[-1] = h[-2] = 0$: | $n$ | Recursion | Formula $4(0.5)^n - 2(0.25)^n$ | |---|---|---| | 0 | 2 | $4 - 2 = 2$ | | 1 | $0.75 \times 2 = 1.5$ | $2 - 0.5 = 1.5$ | | 2 | $0.75(1.5) - 0.125(2) = 0.875$ | $1 - 0.125 = 0.875$ | | 3 | $0.75(0.875) - 0.125(1.5) = 0.46875$ | $0.5 - 0.03125 = 0.46875$ | Both agree. Both poles ($0.5$, $0.25$) are inside the unit circle, so the system is causal and stable; $h[n]$ decays to zero. Also $H(e^{j0}) = \frac{2}{1 - 0.75 + 0.125} = 5.333 = \sum h[n] = \frac{4}{0.5} - \frac{2}{0.75}$. **Answer:** $h[n] = \left[4(0.5)^n - 2(0.25)^n\right]u[n]$ ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is H for y[n] - 3/4 y[n-1] + 1/8 y[n-2] = 2x[n]? A. 2/(1 - (3/4)e^(-jw) + (1/8)e^(-j2w)) B. 2/(1 + (3/4)e^(-jw) + (1/8)e^(-j2w)) C. 1/(2 - (3/4)e^(-jw) + (1/8)e^(-j2w)) D. 2(1 - (3/4)e^(-jw) + (1/8)e^(-j2w)) Answer: A. Each y[n-k] term becomes e^(-jwk) Y. (Key point: Frequency response) 2. Into what does the denominator factorise? A. (1 - 0.5 e^(-jw))(1 - 0.25 e^(-jw)) B. (1 - 0.75 e^(-jw))(1 - 0.125 e^(-jw)) C. (1 + 0.5 e^(-jw))(1 + 0.25 e^(-jw)) D. (1 - 2 e^(-jw))(1 - 4 e^(-jw)) Answer: A. p^2 - 3/4 p + 1/8 = 0 has roots 1/2 and 1/4. (Key point: Factorisation) 3. What are the partial fraction coefficients? A. A = 2, B = 2 B. A = 4, B = -2 C. A = -2, B = 4 D. A = 1, B = 1 Answer: B. A = 2/(1 - 1/2) = 4 and B = 2/(1 - 2) = -2. (Key point: Partial fractions) 4. What is h[n]? A. [4(0.5)^n + 2(0.25)^n] u[n] B. [2(0.5)^n - 4(0.25)^n] u[n] C. [4(0.5)^n - 2(0.25)^n] u[n] D. [4(2)^n - 2(4)^n] u[n] Answer: C. Check: h[0] = 4 - 2 = 2 and h[1] = 2 - 0.5 = 1.5, matching the recursion. 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