--- title: "Fourier transform of periodic sequences" subject: "Signals and Systems" chapter: 3 key: "semester_4/signals_and_systems:ft-of-periodic-sequence" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/ft-of-periodic-sequence site: BCT Engineering --- # Fourier transform of periodic sequences Signals and Systems · Chapter 3: Fourier Transform - Asked in: 2078 Poush, 5 marks; 2080 Chaitra, 7 marks; 2070 Magh, 3 marks (asked 3 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/ft-of-periodic-sequence - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-3 ## Question Derive the expression of Discrete Time Fourier Transform for periodic sequence (how do you calculate the Fourier transform of a periodic sequence?). ## Short answer A periodic sequence is not absolutely summable, so its DTFT consists of impulses. Compute the DTFS coefficients a_k, then X(e^{jw}) = sum 2pi a_k delta(w - 2pi k/N). ## Answer A periodic sequence is not absolutely summable, so its DTFT does not exist in the ordinary sense. It is found by allowing impulses in the frequency domain: each Fourier series term becomes a train of impulses. #### Step 1: DTFT of a single complex exponential Consider the spectrum consisting of impulses of area $2\pi$ at $\omega_0$, repeated every $2\pi$: $$ X(e^{j\omega}) = \sum_{l=-\infty}^{\infty}2\pi\,\delta(\omega - \omega_0 - 2\pi l) $$ Its inverse DTFT, taken over one period of $2\pi$ that contains only the impulse at $\omega_0$, is $$ x[n] = \frac{1}{2\pi}\int_{2\pi}2\pi\,\delta(\omega - \omega_0)e^{j\omega n}d\omega = e^{j\omega_0 n} $$ Hence $$ e^{j\omega_0 n} \xleftrightarrow{DTFT} \sum_{l=-\infty}^{\infty}2\pi\,\delta(\omega - \omega_0 - 2\pi l) $$ #### Step 2: Periodic sequence as a sum of exponentials A periodic $x[n]$ with period $N$ has the DTFS $$ x[n] = \sum_{k=\langle N\rangle}a_k e^{jk(2\pi/N)n}, \qquad a_k = \frac{1}{N}\sum_{n=\langle N\rangle}x[n]e^{-jk(2\pi/N)n} $$ #### Step 3: Apply linearity Each term $a_k e^{jk(2\pi/N)n}$ gives impulses of area $2\pi a_k$ at $\omega = 2\pi k/N + 2\pi l$. Since $a_k$ is periodic in $k$ with period $N$, the impulses for all $k$ and $l$ can be combined into one sum over all integers $k$: $$ \boxed{X(e^{j\omega}) = \sum_{k=-\infty}^{\infty}2\pi\,a_k\,\delta\!\left(\omega - \frac{2\pi k}{N}\right)} $$ **Procedure:** find the DTFS coefficients $a_k$, then place an impulse of area $2\pi a_k$ at each frequency $2\pi k/N$. ```text X(e^jw) 2pi a0 2pi a1 2pi a0 (repeat) ^ ^ ^ ^ | ^ | | | ------+---+----+------+------+---- w 0 2pi/N 4pi/N ... 2pi ``` #### Example: periodic impulse train $x[n] = \sum_k\delta[n - kN]$ has $a_k = 1/N$ for all $k$, so $$ X(e^{j\omega}) = \frac{2\pi}{N}\sum_{k=-\infty}^{\infty}\delta\!\left(\omega - \frac{2\pi k}{N}\right) $$ An impulse train in time gives an impulse train in frequency. **Example 2:** $\cos\omega_0 n = \frac{1}{2}(e^{j\omega_0 n} + e^{-j\omega_0 n})$ gives $X(e^{j\omega}) = \sum_l \pi[\delta(\omega - \omega_0 - 2\pi l) + \delta(\omega + \omega_0 - 2\pi l)]$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why can't the DTFT of a periodic sequence be found by the ordinary summation? A. It is not absolutely summable, so impulses in frequency are used B. It has no Fourier series C. It is always zero D. It is always complex Answer: A. A periodic sequence is not absolutely summable, so its DTFT consists of impulses. (Key point: Why special treatment) 2. The DTFT of e^{jω0 n} is: A. δ(ω - ω0) B. Σ 2π δ(ω - ω0 - 2πl) C. Σ π δ(ω - ω0 - 2πl) D. 2π δ(ω) only Answer: B. Impulses of area 2π at ω0, repeated every 2π (the DTFT is periodic). (Key point: DTFT of an exponential) 3. For periodic x[n] with DTFS coefficients a_k and period N, X(e^{jω}) is: A. Σ a_k δ(ω - 2πk/N) B. Σ 2π a_k δ(ω - 2πk/N) C. Σ (2π/N) a_k δ(ω - k) D. Σ a_k e^{-jωk} Answer: B. Each term a_k e^{jk(2π/N)n} gives an impulse of area 2π a_k at 2πk/N. (Key point: Final result) 4. For x[n] = Σ δ[n - kN], the transform is: A. (2π/N) Σ δ(ω - 2πk/N) B. N Σ δ(ω - 2πk/N) C. Σ δ(ω - 2πk) D. 2π δ(ω) Answer: A. a_k = 1/N, so the impulse areas are 2π/N. 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