--- title: "CTFT convolution; DTFT of 2ⁿu[n]" subject: "Signals and Systems" chapter: 3 key: "semester_4/signals_and_systems:ctft-convolution-dtft-2n" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/ctft-convolution-dtft-2n site: BCT Engineering --- # CTFT convolution; DTFT of 2ⁿu[n] Signals and Systems · Chapter 3: Fourier Transform - Asked in: 2074 Bhadra, 5+5 marks - Video: [Determine DTFT of given sequences](https://www.youtube.com/watch?v=DnhOxaJSV3E) — Engg-Course-Made-Easy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/ctft-convolution-dtft-2n - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-3 ## Question Show that convolution in time domain results multiplication in frequency domain using continuous time Fourier transform. Determine discrete time Fourier transform of the discrete time signal x[n] = 2ⁿ u[n] and also plot magnitude and phase spectrum. ## Short answer Convolution in time becomes multiplication in frequency: x(t) * h(t) ↔ X(jω)H(jω). For x[n] = 2^n u[n], the DTFT does not exist because the series diverges; if intended as (1/2)^n u[n], X(e^(jω)) = 1/(1 - 0.5e^(-jω)). ## Answer #### Convolution property of CTFT **Statement:** If $x(t) \leftrightarrow X(j\omega)$ and $h(t) \leftrightarrow H(j\omega)$, then $$ y(t) = x(t) * h(t) \;\longleftrightarrow\; Y(j\omega) = X(j\omega)\,H(j\omega) $$ **Proof:** The convolution is $y(t) = \int_{-\infty}^{\infty} x(\tau)h(t-\tau)d\tau$. Its FT is $$ \begin{aligned} Y(j\omega) &= \int_{-\infty}^{\infty}\left[\int_{-\infty}^{\infty} x(\tau)h(t-\tau)d\tau\right]e^{-j\omega t}dt \\ &= \int_{-\infty}^{\infty} x(\tau)\left[\int_{-\infty}^{\infty} h(t-\tau)e^{-j\omega t}dt\right]d\tau \end{aligned} $$ The inner integral is the FT of $h(t)$ shifted by $\tau$, which by the time-shifting property is $e^{-j\omega\tau}H(j\omega)$: $$ \begin{aligned} Y(j\omega) &= \int_{-\infty}^{\infty} x(\tau)\,e^{-j\omega\tau}H(j\omega)\,d\tau \\ &= H(j\omega)\int_{-\infty}^{\infty} x(\tau)e^{-j\omega\tau}d\tau = X(j\omega)H(j\omega) \end{aligned} $$ Hence convolution in time becomes multiplication in frequency. This is why the output of an LTI system is found simply as $Y = XH$, where $H(j\omega)$ is the frequency response. #### DTFT of $x[n] = 2^n u[n]$ $$ X(e^{j\omega}) = \sum_{n=0}^{\infty} 2^n e^{-j\omega n} = \sum_{n=0}^{\infty}\left(2e^{-j\omega}\right)^n $$ This geometric series converges only if $|2e^{-j\omega}| < 1$, i.e. $2 < 1$, which is false. Also $\sum|x[n]| = \sum 2^n = \infty$, so the sequence is not absolutely summable. **The DTFT of $2^n u[n]$ does not exist** (its z-transform $\frac{1}{1-2z^{-1}}$, ROC $|z|>2$, does not include the unit circle). The question is usually intended as $x[n] = \left(\frac{1}{2}\right)^n u[n]$, which is solved below. $$ X(e^{j\omega}) = \sum_{n=0}^{\infty}\left(\tfrac12 e^{-j\omega}\right)^n = \frac{1}{1 - 0.5\,e^{-j\omega}} $$ $$ |X(e^{j\omega})| = \frac{1}{\sqrt{1.25 - \cos\omega}}, \qquad \angle X(e^{j\omega}) = -\tan^{-1}\frac{0.5\sin\omega}{1 - 0.5\cos\omega} $$ | $\omega$ | $\lvert X\rvert$ | Phase | |---|---|---| | $0$ | 2 | $0^\circ$ | | $\pi/3$ | 1.155 | $-30^\circ$ (minimum) | | $\pi/2$ | 0.894 | $-26.57^\circ$ | | $\pi$ | 0.667 | $0^\circ$ | ```text |X| 2 | * | * * | | * * | * * | * * * * 0.67 |* * * * * * +-------------------------------------------------> w -pi 0 pi phase (degrees) +30 | * * * | * * | * | * 0 |* * * | * | * | * * -30 | * * * +-------------------------------------------------> w -pi 0 pi ``` The magnitude is even (low-pass, peak 2 at $\omega = 0$), the phase is odd (extremes $\pm30^\circ$ at $\omega = \mp\pi/3$), and both repeat every $2\pi$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the FT of x(t)*h(t)? A. X(jω) * H(jω) B. X(jω) + H(jω) C. X(jω)/H(jω) D. X(jω) H(jω) Answer: D. Convolution in time becomes multiplication in frequency. (Key point: Convolution property) 2. In the proof, how is the inner integral ∫h(t-τ)e^(-jωt)dt evaluated? A. By the time-shifting property: e^(-jωτ)H(jω) B. By the frequency-shifting property C. By integration by parts D. It is zero Answer: A. h(t-τ) is h delayed by τ, so its FT is e^(-jωτ)H(jω). (Key point: Proof step) 3. Does the DTFT of 2^n u[n] exist? A. Yes, it equals 1/(1 - 2e^(-jω)) B. No, it is not absolutely summable and the ROC |z|>2 excludes the unit circle C. Yes, but only for ω = 0 D. Yes, because it is causal Answer: B. The series sum of (2e^(-jω))^n diverges since |2e^(-jω)| = 2 > 1. (Key point: Existence for 2^n u[n]) 4. For (1/2)^n u[n], what is |X| at ω = 0? A. 1 B. 0.5 C. 2 D. 0.667 Answer: C. X = 1/(1 - 0.5e^(-jω)), which at ω = 0 is 1/0.5 = 2. (Key point: Intended signal) ## Related in Signals and Systems - [Magnitude and phase of DTFT of aⁿu[n]](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Adtft-geometric-magnitude-phase) — Signals and Systems, chapter 3 - [DTFT of aⁿu[n] with spectrum plot](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Adtft-geometric-plot) — Signals and Systems, chapter 3 - [CTFT convolution property](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Actft-convolution-property) — Signals and Systems, chapter 3, asked 12× - [DTFT convolution property](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Adtft-convolution-property) — Signals and Systems, chapter 3, asked 12× - [Fourier series vs transform; DTFT of aⁿ](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Afs-vs-ft-dtft-geometric) — Signals and Systems, chapter 3 - [Derive the discrete-time Fourier series](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Adtfs-derivation) — Signals and Systems, chapter 2, asked 15× ## Related in other subjects - [Fourier transform of a DC signal](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Aft-dc-signal) — Data Communication, chapter 3 - [Single tone DSB-FC AM in time and frequency domain](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Asingle-tone-dsbfc-time-freq) — Communication Systems, chapter 2 - [Discrete vs continuous signal; convert (25.25)₁₀](https://www.bctengineer.com/question.md?key=semester_3%2Fdigital_logic%3Adiscrete-continuous-convert-25-25) — Digital Logic, chapter 1 - [Angle modulation; single tone FM time and frequency domain](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aangle-mod-single-tone-fm-time-freq) — Communication Systems, chapter 3