--- title: "H(e^(jω)), magnitude plot and h[n] for y[n] = 0.5y[n−1] + x[n]" subject: "Signals and Systems" chapter: 6 key: "semester_4/signals_and_systems:first-order-difference-plot" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/first-order-difference-plot site: BCT Engineering --- # H(e^(jω)), magnitude plot and h[n] for y[n] = 0.5y[n−1] + x[n] Signals and Systems · Chapter 6: Discrete-Time Systems - Asked in: 2081 Chaitra, 6 marks; 2069 Bhadra, 8 marks (asked 2 times) - Video: [Magnitude and phase spectrum graphs - Frequency response in signal and system / DSP](https://www.youtube.com/watch?v=LXw1uTE6Zv4) — Shrenik Jain - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/first-order-difference-plot - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-6 ## Question Find the frequency response H(e^(jω)), plot the magnitude and impulse response h[n] for a system characterized by linear constant coefficient difference equation y[n] = 0.5y[n−1] + x[n] ## Short answer H(eʲω) = 1/(1−0.5e⁻ʲω), magnitude |H| = 1/√(1.25−cos ω), with max 2 at ω=0 and min 2/3 at ω=π (lowpass). Impulse response: h[n] = (0.5)ⁿ u[n]. ## Answer **Given:** $y[n] = 0.5\,y[n-1] + x[n]$, initially at rest. #### Frequency response DTFT of both sides: $$ \begin{aligned} Y(e^{j\omega})\left(1 - 0.5e^{-j\omega}\right) &= X(e^{j\omega}) \\ H(e^{j\omega}) &= \frac{1}{1 - 0.5e^{-j\omega}} = \frac{1}{(1 - 0.5\cos\omega) + j0.5\sin\omega} \end{aligned} $$ **Magnitude:** $$ |H(e^{j\omega})| = \frac{1}{\sqrt{(1 - 0.5\cos\omega)^2 + 0.25\sin^2\omega}} = \frac{1}{\sqrt{1.25 - \cos\omega}} $$ **Phase:** $$ \theta(\omega) = -\tan^{-1}\frac{0.5\sin\omega}{1 - 0.5\cos\omega} $$ #### Magnitude plot | $\omega$ | 0 | $\pm\pi/4$ | $\pm\pi/2$ | $\pm3\pi/4$ | $\pm\pi$ | |---|---|---|---|---|---| | $\lvert H\rvert$ | 2.000 | 1.357 | 0.894 | 0.715 | 0.667 | | $\theta$ (for $+\omega$) | 0° | −28.7° | −26.6° | −14.6° | 0° | ```text |H| 2.0 | * | .' '. 1.36| * * | .' '. 0.89| * * 0.67|* * +--+---+---+---+---+--> w -pi -pi/2 0 pi/2 pi ``` The magnitude is even in $\omega$, has its maximum 2 at $\omega = 0$ and its minimum $2/3$ at $\omega = \pm\pi$, and repeats every $2\pi$. So the system is a **low pass filter**. The phase is odd in $\omega$ (maximum lag 30° at $\omega = \pm\pi/3$). #### Impulse response Using $a^n u[n] \leftrightarrow \frac{1}{1 - ae^{-j\omega}}$ with $a = 0.5$: $$ h[n] = (0.5)^n\,u[n] $$ | $n$ | <0 | 0 | 1 | 2 | 3 | 4 | 5 | |---|---|---|---|---|---|---|---| | $h[n]$ | 0 | 1 | 0.5 | 0.25 | 0.125 | 0.0625 | 0.03125 | ```text h[n] 1.0 | o | | 0.5 | | o 0.25 | | | o | | | | o o . 0 +-o-+--+--+--+--+--+--> n -1 0 1 2 3 4 5 ``` The system is causal, stable ($\sum h[n] = 2$) and IIR. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is H(e^(jw)) for y[n] = 0.5 y[n-1] + x[n]? A. 1/(1 + 0.5 e^(-jw)) B. 0.5/(1 - e^(-jw)) C. 1/(1 - 0.5 e^(-jw)) D. 1 - 0.5 e^(-jw) Answer: C. Y(1 - 0.5e^(-jw)) = X. (Key point: Frequency response) 2. What are |H| at w = 0 and w = pi? A. 2 and 2/3 B. 1 and 0 C. 1/2 and 2 D. 2/3 and 2 Answer: A. |H| = 1/sqrt(1.25 - cos w), giving 1/0.5 = 2 and 1/1.5 = 0.667. (Key point: Magnitude values) 3. What type of filter and which symmetry/periodicity? A. High pass; odd magnitude B. Low pass; even magnitude, period 2 pi C. Band pass; not periodic D. All pass; period pi Answer: B. Gain is largest at w = 0 and smallest at pi; |H| is even and repeats every 2 pi. (Key point: Filter type) 4. How does h[n] look? A. Alternating in sign B. Constant at 1 for n >= 0 C. Zero for n > 0 D. Samples 1, 0.5, 0.25, ... for n >= 0 (decaying, causal) Answer: D. h[n] = (0.5)^n u[n] decays geometrically, sum = 2. 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