--- title: "DTFS coefficients are periodic" subject: "Signals and Systems" chapter: 2 key: "semester_4/signals_and_systems:dtfs-coefficients-periodic" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/dtfs-coefficients-periodic site: BCT Engineering --- # DTFS coefficients are periodic Signals and Systems · Chapter 2: Fourier Series - Asked in: 2075 Bhadra, 6 marks - Video: [Introduction to DT Fourier Series](https://www.youtube.com/watch?v=OMdUYrKMZjE) — Darryl Morrell - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/dtfs-coefficients-periodic - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-2 ## Question Show that fourier series coefficient of discrete time signal is periodic in nature. ## Short answer The DTFS coefficient is a_k = (1/N) Σ_n x[n] e^{-j k (2π/N) n}. Replacing k by k+N gives a_{k+N} = a_k because e^{-j 2π n}=1 for integer n. Thus coefficients are periodic with period N, since only N distinct complex exponentials exist. ## Answer For a discrete-time periodic signal $x[n]$ with period $N$, the DTFS coefficients are $$ a_k = \frac{1}{N}\sum_{n=\langle N\rangle}x[n]\,e^{-jk(2\pi/N)n} $$ #### To show: $a_{k+N} = a_k$ Replace $k$ by $k + N$: $$ \begin{aligned} a_{k+N} &= \frac{1}{N}\sum_{n=\langle N\rangle}x[n]\,e^{-j(k+N)(2\pi/N)n} \\ &= \frac{1}{N}\sum_{n=\langle N\rangle}x[n]\,e^{-jk(2\pi/N)n}\,e^{-j2\pi n} \end{aligned} $$ Since $n$ is an integer, $e^{-j2\pi n} = \cos(2\pi n) - j\sin(2\pi n) = 1$. Hence $$ a_{k+N} = \frac{1}{N}\sum_{n=\langle N\rangle}x[n]\,e^{-jk(2\pi/N)n} = a_k $$ Repeating the argument, $a_{k+mN} = a_k$ for any integer $m$. So the sequence of DTFS coefficients is **periodic in $k$ with period $N$**. #### Reason The root cause is that discrete-time complex exponentials whose frequencies differ by $2\pi$ are identical: $$ \phi_{k+N}[n] = e^{j(k+N)(2\pi/N)n} = e^{jk(2\pi/N)n} = \phi_k[n] $$ So there are only $N$ distinct harmonics, and the "$(k+N)$th harmonic" is the same signal as the $k$th; its coefficient must be the same. This is why the synthesis equation sums over only $N$ consecutive $k$: $$ x[n] = \sum_{k=\langle N\rangle}a_k e^{jk(2\pi/N)n} $$ This is different from the CT Fourier series, where every $e^{jk\omega_0 t}$ is distinct and the coefficients are not periodic. #### Example $x[n] = \cos(\pi n/2)$, $N = 4$: $x[n] = \frac{1}{2}e^{j(2\pi/4)n} + \frac{1}{2}e^{-j(2\pi/4)n}$, so $a_1 = a_{-1} = \frac12$. Computing from the analysis equation with samples $\{1, 0, -1, 0\}$: | $k$ | $-3$ | $-2$ | $-1$ | 0 | 1 | 2 | 3 | 4 | 5 | |---|---|---|---|---|---|---|---|---|---| | $a_k$ | 0.5 | 0 | 0.5 | 0 | 0.5 | 0 | 0.5 | 0 | 0.5 | The pattern $\{0, 0.5, 0, 0.5\}$ repeats every 4 values of $k$, e.g. $a_5 = a_1$ and $a_3 = a_{-1}$. ```text a_k 0.5 | | | | | -+--o--+--o--+--o--+--o--+--> k -3 -2 -1 0 1 2 3 4 5 ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In showing a_{k+N} = a_k, which fact is used? A. x[n] is even B. e^{-j2 pi n} = 1 because n is an integer C. The sum is over infinitely many terms D. N is odd Answer: B. Replacing k by k+N adds the factor e^{-j2 pi n}, which equals 1 for integer n. (Key point: Key step of the proof) 2. Why are the DTFS coefficients periodic in k? A. Because x[n] is real B. Because the DT exponentials with frequencies differing by 2 pi are identical, so only N distinct harmonics exist C. Because the CT series is periodic D. Because the coefficients are always zero beyond N Answer: B. phi_{k+N}[n] = phi_k[n], so the (k+N)th harmonic is the same signal as the kth and has the same coefficient. (Key point: Reason for periodicity) 3. As a consequence, the synthesis sum needs: A. Infinitely many terms B. Only N consecutive values of k C. Only k = 0 D. Only positive k Answer: B. Only N harmonics are distinct, so any N consecutive k suffice. (Key point: Consequence for synthesis) 4. For x[n] = cos(pi n/2), N = 4, which coefficients are nonzero over one period? A. a_0 and a_2 B. a_1 = a_3 = 1/2 (a_3 = a_{-1}) C. a_1 = 1 only D. a_0 = 1/2 Answer: B. x = (1/2)e^{j(2 pi/4)n} + (1/2)e^{-j(2 pi/4)n}, so a_1 = a_{-1} = 1/2, and a_{-1} = a_3. (Key point: Example check) ## Related in Signals and Systems - [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× - [Periodic signal whose Fourier series fails](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Afourier-series-existence-examples) — Signals and Systems, chapter 2 - [Fourier series coefficients and plot](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Afourier-series-coefficients-plot) — Signals and Systems, chapter 2 - [FS coefficients of periodic square wave](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Asquare-wave-fourier-coefficients) — Signals and Systems, chapter 2 - [Fourier transform of periodic sequences](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aft-of-periodic-sequence) — Signals and Systems, chapter 3, asked 3× - [FS coefficients; time scaling and conjugation](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Actfs-significance-scaling-conjugation) — Signals and Systems, chapter 2 ## Related in other subjects - [Fourier series coefficients of a square wave](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Afs-coefficients-square-wave) — Data Communication, chapter 3 - [How Fourier series coefficients are calculated](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Afourier-series-coefficients) — Digital Signal Analysis and Processing, chapter 3, asked 2× - [Properties of continuous-time Fourier series](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Afourier-series-properties) — Data Communication, chapter 3 - [Rayleigh energy theorem; power from Fourier series](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Arayleigh-periodic-power-fourier) — Communication Systems, chapter 1