--- title: "CTFS analysis and synthesis equations" subject: "Signals and Systems" chapter: 2 key: "semester_4/signals_and_systems:ctfs-analysis-synthesis-equations" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/ctfs-analysis-synthesis-equations site: BCT Engineering --- # CTFS analysis and synthesis equations Signals and Systems · Chapter 2: Fourier Series - Asked in: 2081 Asoj, 6 marks - Video: [Continuous Time Fourier Series (CTFS) Equation](https://www.youtube.com/watch?v=vDY9AtNl5q8) — SAJAN P PHILIP BIT - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/ctfs-analysis-synthesis-equations - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-2 ## Question What is Fourier series representation of a signal? Derive an expression for analysis and synthesis equation of the Fourier Series for a CT periodic signal x(t). ## Short answer Fourier series represents a periodic signal as sum of harmonically related complex exponentials. The synthesis equation is x(t)=∑ a_k e^{jkω0t}; the analysis equation extracts coefficients via a_k = (1/T)∫_T x(t)e^{-jkω0t} dt. ## Answer The **Fourier series representation** of a periodic signal expresses it as a sum (linear combination) of harmonically related sinusoids or complex exponentials, whose frequencies are integer multiples of the fundamental frequency $\omega_0 = 2\pi/T$. It shows which frequencies the signal contains and how strongly. #### Synthesis equation Each $e^{jk\omega_0 t}$, $k = 0, \pm1, \pm2, \dots$, is periodic with period $T$, so their combination is also periodic with $T$: $$ x(t) = \sum_{k=-\infty}^{\infty}a_k\,e^{jk\omega_0 t} $$ This builds (synthesises) $x(t)$ from its harmonics; $a_k$ are the Fourier series coefficients. #### Analysis equation Multiply both sides by $e^{-jn\omega_0 t}$ and integrate over one period: $$ \int_0^T x(t)e^{-jn\omega_0 t}dt = \sum_{k=-\infty}^{\infty}a_k\int_0^T e^{j(k-n)\omega_0 t}dt $$ For $k \ne n$, $e^{j(k-n)\omega_0 t}$ completes $(k - n)$ whole cycles in $T$, so $$ \int_0^T e^{j(k-n)\omega_0 t}dt = \int_0^T\cos((k-n)\omega_0 t)dt + j\int_0^T\sin((k-n)\omega_0 t)dt = 0 $$ For $k = n$ the integrand is 1 and the integral is $T$. So only $k = n$ survives: $$ \int_0^T x(t)e^{-jn\omega_0 t}dt = T a_n $$ $$ a_k = \frac{1}{T}\int_T x(t)\,e^{-jk\omega_0 t}\,dt $$ This extracts (analyses) the amount of the $k$th harmonic in $x(t)$; the integral may be taken over any interval of length $T$. #### Summary | Name | Equation | |---|---| | Synthesis | $x(t) = \sum_{k=-\infty}^{\infty}a_k e^{jk\omega_0 t}$ | | Analysis | $a_k = \frac{1}{T}\int_T x(t)e^{-jk\omega_0 t}dt$ | | DC term | $a_0 = \frac{1}{T}\int_T x(t)dt$ | Example: $x(t) = \sin\omega_0 t = \frac{1}{2j}e^{j\omega_0 t} - \frac{1}{2j}e^{-j\omega_0 t}$, so $a_1 = \frac{1}{2j}$, $a_{-1} = -\frac{1}{2j}$, all others 0. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the Fourier series representation? A. A sum of harmonically related sinusoids/exponentials at multiples of w0 B. A single sinusoid at the average frequency C. An integral transform of a non-periodic signal D. A Taylor expansion Answer: A. The signal is a linear combination of harmonics of w0 = 2pi/T. (Key point: Definition) 2. Which is the synthesis equation? A. x(t) = sum a_k e^(jk w0 t) B. x(t) = sum a_k e^(-jk w0 t) / T C. x(t) = (1/T) integral a_k dk D. x(t) = a_0 Answer: A. It builds x(t) from its harmonics. (Key point: Synthesis) 3. Why do all terms except k = n vanish? A. They complete whole cycles over T, so integrate to 0 B. They are multiplied by 0 C. The integral is infinite D. They are dc Answer: A. cos and sin over whole cycles integrate to zero. (Key point: Why orthogonality helps) 4. What is a_k? A. (1/T) integral over T of x(t) e^(-jk w0 t) dt B. integral over T of x(t) e^(jk w0 t) dt C. T integral of x(t) e^(-jk w0 t) dt D. (2/T) integral of x(t) cos(k w0 t) dt Answer: A. Only k = n term survives, equal to T a_n. (Key point: Analysis) ## Related in Signals and Systems - [Derive exponential Fourier series (CT)](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Actfs-exponential-derivation) — Signals and Systems, chapter 2, asked 15× - [Derive trigonometric Fourier series (CT)](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Actfs-trigonometric-derivation) — Signals and Systems, chapter 2, asked 3× - [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 - [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× - [Fourier transform of CT periodic signals](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aft-periodic-signal-derivation) — Signals and Systems, chapter 3, asked 3× - [Derive the CT Fourier transform pair](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Actft-derivation) — Signals and Systems, chapter 3, asked 3× ## Related in other subjects - [Image synthesis, analysis and transmission](https://www.bctengineer.com/question.md?key=semester_8%2Fmultimedia_system%3Aimage-synthesis-analysis-transmission) — Multimedia System, 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 - [Fourier series of half-wave rectified sine](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Afs-half-wave-rectified) — Data Communication, chapter 3 - [Fourier series of full-wave rectified sine](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Afs-full-wave-rectified) — Data Communication, chapter 3