--- title: "Frequency response of y[n] = 0.5x[n] − 0.5x[n−1]" subject: "Signals and Systems" chapter: 6 key: "semester_4/signals_and_systems:fir-difference-frequency-response" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/fir-difference-frequency-response site: BCT Engineering --- # Frequency response of y[n] = 0.5x[n] − 0.5x[n−1] Signals and Systems · Chapter 6: Discrete-Time Systems - Asked in: 2082 Kartik, 5 marks; 2083 Baisakh (new course), 4 marks (asked 2 times) - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/fir-difference-frequency-response - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-6 ## Question Find out the frequency response of the following signal y[n] = 0.5x[n] − 0.5x[n−1]; h[n] = 1/2 for n = 0 and −1/2 for n = 1 ## Short answer H(eʲω) = 0.5 − 0.5e⁻ʲω = j sin(ω/2) e⁻ʲω/². Magnitude is |sin(ω/2)|, phase is π/2 − ω/2 (linear). It is a highpass FIR filter that blocks DC. ## Answer **Given:** $y[n] = 0.5\,x[n] - 0.5\,x[n-1]$, so $h[n] = 0.5\,\delta[n] - 0.5\,\delta[n-1] = \{\underset{\uparrow}{0.5}, -0.5\}$ (a two-tap FIR system: a scaled first difference). #### Frequency response $$ H(e^{j\omega}) = \sum_n h[n]e^{-j\omega n} = 0.5 - 0.5e^{-j\omega} $$ Factor out $e^{-j\omega/2}$: $$ \begin{aligned} H(e^{j\omega}) &= 0.5\,e^{-j\omega/2}\left(e^{j\omega/2} - e^{-j\omega/2}\right) \\ &= 0.5\,e^{-j\omega/2}\,(2j\sin\tfrac{\omega}{2}) \\ &= j\sin\left(\tfrac{\omega}{2}\right)e^{-j\omega/2} = \sin\left(\tfrac{\omega}{2}\right)e^{j\left(\frac{\pi}{2} - \frac{\omega}{2}\right)} \end{aligned} $$ **Magnitude:** $|H(e^{j\omega})| = \left|\sin\frac{\omega}{2}\right|$ **Phase:** $\theta(\omega) = \frac{\pi}{2} - \frac{\omega}{2}$ for $0 < \omega < \pi$ (and $-\frac{\pi}{2} - \frac{\omega}{2}$ for $-\pi < \omega < 0$): linear phase, apart from a jump of $\pi$ at $\omega = 0$, where $\sin(\omega/2)$ changes sign. | $\omega$ | 0 | $\pi/4$ | $\pi/2$ | $3\pi/4$ | $\pi$ | |---|---|---|---|---|---| | $\lvert H\rvert$ | 0 | 0.383 | 0.707 | 0.924 | 1 | | $\theta$ | – | 67.5° | 45° | 22.5° | 0° | ```text |H| 1 |* * | '. .' 0.7 | * * | '. .' 0 +--------'-*-'--------> w -pi 0 pi ``` **Interpretation:** DC ($\omega = 0$) is blocked completely and the highest frequency ($\omega = \pi$) passes with gain 1, so this is a simple **high pass filter** (a difference/differentiator-like system). Being FIR, it is always stable. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For y[n] = 0.5x[n] - 0.5x[n-1], what is h[n]? A. h[n] = 0.5 for n = 0 and n = 1 B. h[n] = 0.5 for n = 0 and -0.5 for n = 1 C. h[n] = 0.5 for n = 0 and -0.5 for n = -1 D. h[n] = 1 for n = 0 and -1 for n = 1 Answer: B. Putting x[n] = delta[n] gives 0.5 delta[n] - 0.5 delta[n-1], so h = {0.5, -0.5}. (Key point: Impulse response) 2. Why is e^(-jw/2) factored out of 0.5 - 0.5e^(-jw)? A. So the remaining bracket becomes e^(jw/2) - e^(-jw/2) = 2j sin(w/2) B. So the system becomes causal C. So the magnitude becomes 1 D. To remove the pi/2 phase term Answer: A. The bracket is then a difference of complex exponentials, which equals 2j sin(w/2), giving a clean magnitude and phase. (Key point: Factoring step) 3. What is the magnitude response of this system? A. |H| = cos(w/2) B. |H| = |sin(w/2)| C. |H| = 0.5 D. |H| = |sin(w)| Answer: B. H = j sin(w/2) e^(-jw/2), and the factors j and e^(-jw/2) have unit magnitude, so |H| = |sin(w/2)|. (Key point: Magnitude) 4. Since |H| = 0 at w = 0 and |H| = 1 at w = pi, the system is a: A. Low pass filter B. High pass filter C. Band stop filter D. All pass filter Answer: B. DC is blocked completely while the highest frequency passes with gain 1, which is a high pass (difference) filter. (Key point: Filter type) ## Related in Signals and Systems - [Impulse and frequency response of y[n] = 0.5y[n−1] + x[n] + x[n+1]](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Adifference-equation-noncausal) — Signals and Systems, chapter 6 - [Frequency response of a CT LTI system](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Act-frequency-response) — Signals and Systems, chapter 5, asked 3× - [Frequency response; ideal LPF impulse response](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aideal-lpf-cutoff-impulse) — Signals and Systems, chapter 5, asked 2× - [Impulse and frequency response of y[n] + 0.55y[n−1] = x[n]](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Adifference-equation-0-55) — Signals and Systems, chapter 6 - [Frequency of a signal sampled at 300 Hz](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Asampled-signal-frequency-300) — Signals and Systems, chapter 4 - [Impulse response of ideal low pass filter](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aideal-lpf-impulse-response) — Signals and Systems, chapter 5, asked 5× ## Related in other subjects - [Frequency response of networks and complex frequency](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Afrequency-response-complex-frequency-role) — Electrical Circuit Theory, chapter 5 - [Frequency response design procedure](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Afrequency-response-design-procedure) — Digital Control System, chapter 4 - [Second-order transducer frequency and step response](https://www.bctengineer.com/question.md?key=semester_4%2Finstrumentation_and_measurement%3Asecond-order-transducer-numerical) — Instrumentation and Measurement, chapter 4 - [Pole-zero and magnitude response, third order](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Amagnitude-response-0-4-0-2-0-5) — Digital Signal Analysis and Processing, chapter 3, asked 3×