--- title: "DTFT of sgn[n] and u[n]" subject: "Signals and Systems" chapter: 3 key: "semester_4/signals_and_systems:dtft-signum-unit-step" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/dtft-signum-unit-step site: BCT Engineering --- # DTFT of sgn[n] and u[n] Signals and Systems · Chapter 3: Fourier Transform - Asked in: 2081 Chaitra, 6+6 marks; 2079 Asoj, 4+8 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/dtft-signum-unit-step - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-3 ## Question Determine the discrete time fourier transform of signum function (sgn[n]) and use it to find the DTFT of unit-step signal u[n]. ## Short answer Define sgn[n] as limit of decaying sequences, giving DTFT 2/(1 - e^{-jw}) for w ≠ 2pi k. Then u[n] = 1/2 + (1/2)sgn[n] yields U(e^{jw}) = 1/(1 - e^{-jw}) + pi sum delta(w - 2pi k). ## Answer The definition used here is $\mathrm{sgn}[n] = 1$ for $n \ge 0$ and $-1$ for $n < 0$, so that $u[n] = \frac{1}{2}(1 + \mathrm{sgn}[n])$. (A note at the end covers the version with $\mathrm{sgn}[0] = 0$.) ```text sgn[n] 1 * * * * | | | | ----+--+--+--+--+--+--+--+--- n | | | | -1 * * * * -4 -3 -2 -1 0 1 2 3 ``` #### DTFT of sgn[n] $\mathrm{sgn}[n]$ is not absolutely summable, so take it as a limit of decaying sequences ($0 < a < 1$): $$ \mathrm{sgn}[n] = \lim_{a\to 1}\left[a^n u[n] - a^{-n}u[-n-1]\right] $$ First part: $$ \sum_{n=0}^{\infty}a^n e^{-j\omega n} = \frac{1}{1 - ae^{-j\omega}} $$ Second part, with $m = -n$: $$ \sum_{n=-\infty}^{-1}a^{-n}e^{-j\omega n} = \sum_{m=1}^{\infty}\left(ae^{j\omega}\right)^m = \frac{ae^{j\omega}}{1 - ae^{j\omega}} $$ Hence $$ S_a(e^{j\omega}) = \frac{1}{1 - ae^{-j\omega}} - \frac{ae^{j\omega}}{1 - ae^{j\omega}} $$ Let $a \to 1$. In the second term, divide the numerator and denominator by $e^{j\omega}$: $$ \frac{e^{j\omega}}{1 - e^{j\omega}} = \frac{1}{e^{-j\omega} - 1} = -\frac{1}{1 - e^{-j\omega}} $$ Therefore $$ \boxed{\mathrm{sgn}[n] \xleftrightarrow{DTFT} \frac{2}{1 - e^{-j\omega}}, \quad \omega \ne 2\pi k} $$ Since $\mathrm{sgn}[n]$ has zero average value, there is no impulse at $\omega = 0$. **Magnitude and phase:** $1 - e^{-j\omega} = e^{-j\omega/2}\cdot 2j\sin(\omega/2)$, so $$ \frac{2}{1 - e^{-j\omega}} = \frac{e^{j\omega/2}}{j\sin(\omega/2)}, \qquad |S(e^{j\omega})| = \frac{1}{|\sin(\omega/2)|} $$ The phase is $\frac{\omega}{2} - \frac{\pi}{2}$ for $0 < \omega < \pi$ (and $\frac{\omega}{2} + \frac{\pi}{2}$ for $-\pi < \omega < 0$). #### DTFT of u[n] $$ u[n] = \frac{1}{2} + \frac{1}{2}\mathrm{sgn}[n] $$ The constant $1$ has DTFT $2\pi\sum_k\delta(\omega - 2\pi k)$. (Its inverse DTFT over one period is $\frac{1}{2\pi}\int 2\pi\delta(\omega)d\omega = 1$.) So $$ \frac{1}{2} \xleftrightarrow{DTFT} \pi\sum_{k=-\infty}^{\infty}\delta(\omega - 2\pi k) $$ By linearity: $$ \boxed{U(e^{j\omega}) = \frac{1}{1 - e^{-j\omega}} + \pi\sum_{k=-\infty}^{\infty}\delta(\omega - 2\pi k)} $$ #### Check - As $a \to 1$, $a^n u[n] \to u[n]$ and $\frac{1}{1 - ae^{-j\omega}} \to \frac{1}{1 - e^{-j\omega}}$. The impulses account for the DC value of $u[n]$. - Accumulation property: $u[n] = \sum_{k=-\infty}^{n}\delta[k]$, which gives $\frac{1}{1 - e^{-j\omega}}\cdot 1 + \pi\cdot 1\cdot\sum_k\delta(\omega - 2\pi k)$, the same result. - This is the DT counterpart of $u(t) \leftrightarrow \frac{1}{j\omega} + \pi\delta(\omega)$. **Note:** if $\mathrm{sgn}[n]$ is defined with $\mathrm{sgn}[0] = 0$, it equals the version above minus $\delta[n]$. Its DTFT is then $\frac{2}{1 - e^{-j\omega}} - 1 = \frac{1 + e^{-j\omega}}{1 - e^{-j\omega}}$. Using $u[n] = \frac{1}{2}(1 + \mathrm{sgn}[n] + \delta[n])$ gives the same $U(e^{j\omega})$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is sgn[n] written as a limit of a^n u[n] - a^(-n) u[-n-1] as a -> 1 before taking its DTFT? A. Because sgn[n] is not absolutely summable, so the DTFT sum does not converge directly B. Because sgn[n] is complex valued C. Because sgn[n] is periodic in n D. Because the DTFT is only defined for causal signals Answer: A. sgn[n] does not decay, so the sum diverges; using decaying sequences with 0 1 is taken. (Key point: Limit of decaying sequences) 2. What is the DTFT of sgn[n] (with sgn[0] = 1) for omega not equal to 2*pi*k? A. 1/(1 - e^(-jω)) B. 2/(1 - e^(-jω)) C. 2/(1 + e^(-jω)) D. 1/(1 - e^(-jω)) + π δ(ω) Answer: B. The two parts become 1/(1 - e^(-jω)) each after a -> 1 (the second term is -(-1/(1 - e^(-jω)))), adding to 2/(1 - e^(-jω)). (Key point: DTFT of sgn[n]) 3. Why does the DTFT of sgn[n] have no impulse at omega = 0? A. Because sgn[n] is causal B. Because sgn[n] has zero average value C. Because sgn[n] is real D. Because sgn[n] is bounded Answer: B. An impulse at ω = 0 represents a DC component; sgn[n] has zero average value, so no impulse appears. (Key point: No impulse in sgn spectrum) 4. Using u[n] = 1/2 + (1/2) sgn[n], what is the DTFT of u[n]? A. 2/(1 - e^(-jω)) + 2π Σ δ(ω - 2πk) B. 1/(1 - e^(-jω)) only C. 1/(1 - e^(-jω)) + π Σ δ(ω - 2πk) D. 1/(1 - e^(-jω)) - π Σ δ(ω - 2πk) Answer: C. The 1/2 gives π Σ δ(ω - 2πk) and the sgn part gives half of 2/(1 - e^(-jω)), i.e. 1/(1 - e^(-jω)). 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