--- title: "Impulse and frequency response of y[n] + 0.55y[n−1] = x[n]" subject: "Signals and Systems" chapter: 6 key: "semester_4/signals_and_systems:difference-equation-0-55" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/difference-equation-0-55 site: BCT Engineering --- # Impulse and frequency response of y[n] + 0.55y[n−1] = x[n] Signals and Systems · Chapter 6: Discrete-Time Systems - Asked in: 2078 Baisakh, 6 marks - Video: [Frequency Response Descriptions for LTI Systems](https://www.youtube.com/watch?v=E3QfYXscsCc) — Barry Van Veen - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/difference-equation-0-55 - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-6 ## Question Derive and plot the impulse response and frequency response of a system characterized by the difference equation y[n] + 0.55y[n−1] = x[n] ## Short answer Impulse response: h[n] = (−0.55)^n u[n], alternating sign and decaying. Frequency response H(e^jω) = 1/(1+0.55e^{−jω}) is high-pass, with minimum gain at ω=0 and maximum at ω=π. ## Answer **Given:** $y[n] + 0.55\,y[n-1] = x[n]$, initially at rest. #### Impulse response Put $x[n] = \delta[n]$, so $h[n] = \delta[n] - 0.55\,h[n-1]$ with $h[-1] = 0$: - $h[0] = 1$ - $h[1] = -0.55$ - $h[2] = (-0.55)^2 = 0.3025$, and so on. So $$ h[n] = (-0.55)^n\,u[n] $$ | $n$ | 0 | 1 | 2 | 3 | 4 | 5 | |---|---|---|---|---|---|---| | $h[n]$ | 1 | −0.55 | 0.3025 | −0.1664 | 0.0915 | −0.0503 | Sketch (not to scale; the magnitudes shrink by a factor of 0.55 at each step): ```text h[n] 1.0 | o | | o o o 0 +--+--+--+--+--+--+--+--> n | 0 | 2 | 4 | 6 -0.55| o o o ``` The samples alternate in sign and decay (since $|-0.55| < 1$): causal and stable, $\sum|h[n]| = 1/0.45 = 2.222$. #### Frequency response DTFT of the difference equation: $$ \begin{aligned} Y(e^{j\omega})\left(1 + 0.55e^{-j\omega}\right) &= X(e^{j\omega}) \\ H(e^{j\omega}) &= \frac{1}{1 + 0.55e^{-j\omega}} = \frac{1}{(1 + 0.55\cos\omega) - j0.55\sin\omega} \end{aligned} $$ (This also follows from $a^n u[n] \leftrightarrow \frac{1}{1 - ae^{-j\omega}}$ with $a = -0.55$.) **Magnitude:** $$ |H(e^{j\omega})| = \frac{1}{\sqrt{(1 + 0.55\cos\omega)^2 + (0.55\sin\omega)^2}} = \frac{1}{\sqrt{1.3025 + 1.1\cos\omega}} $$ **Phase:** $$ \theta(\omega) = \tan^{-1}\frac{0.55\sin\omega}{1 + 0.55\cos\omega} $$ | $\omega$ | 0 | $\pi/4$ | $\pi/2$ | $3\pi/4$ | $\pi$ | |---|---|---|---|---|---| | $\lvert H\rvert$ | 0.645 | 0.693 | 0.876 | 1.381 | 2.222 | | $\theta$ | 0° | 15.6° | 28.8° | 32.5° | 0° | ```text |H| 2.22|* * | '. .' 1.38| * * 0.88| * * 0.65| '-*-' +--+-------+-------+--> w -pi 0 pi ``` The gain is smallest at $\omega = 0$ ($1/1.55 = 0.645$) and largest at $\omega = \pi$ ($1/0.45 = 2.222$), so this system is a **high pass filter**: the pole at $z = -0.55$ (near $\omega = \pi$) makes $h[n]$ alternate in sign, so it emphasises rapid sample-to-sample changes. $|H|$ is even and $\theta(\omega)$ is odd, and both are periodic with period $2\pi$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For y[n] + 0.55y[n-1] = x[n], what is h[n]? A. (0.55)^n u[n] B. (-0.55)^n u[n] C. -0.55 delta[n] D. (0.55)^n u[-n] Answer: B. Recursion h[n] = delta[n] - 0.55h[n-1] gives 1, -0.55, 0.3025, alternating in sign. (Key point: Impulse response) 2. H(e^jw) is: A. 1/(1 - 0.55e^(-jw)) B. 1/(1 + 0.55e^(-jw)) C. 1 + 0.55e^(-jw) D. 0.55/(1 + e^(-jw)) Answer: B. DTFT gives Y(1 + 0.55e^(-jw)) = X. (Key point: Frequency response) 3. |H(0)| = 0.645 and |H(pi)| = 2.222. The system is: A. Low pass B. High pass C. Band stop D. All pass Answer: B. Gain rises with frequency, so it emphasises rapid changes. (Key point: Filter type) 4. The system is stable because: A. The samples alternate in sign B. |-0.55| < 1 so the sum of |h[n]| = 1/0.45 is finite C. It is an FIR system D. h[0] = 1 Answer: B. A decaying geometric h[n] is absolutely summable. 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