--- title: "Exponential Fourier series of A sin ω₀n" subject: "Signals and Systems" chapter: 2 key: "semester_4/signals_and_systems:exponential-fs-sine" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/exponential-fs-sine site: BCT Engineering --- # Exponential Fourier series of A sin ω₀n Signals and Systems · Chapter 2: Fourier Series - Asked in: 2073 Magh, 6 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/exponential-fs-sine - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-2 ## Question Derive the expression of continuous time exponential Fourier series x[n] = A sin ω₀n ## Short answer Express A sin ω₀t as (A/(2j))e^{jω₀t} - (A/(2j))e^{-jω₀t}. Thus exponential Fourier coefficients: a1 = A/(2j) = -jA/2, a-1 = -A/(2j) = jA/2; all other ak = 0. Magnitude A/2, phase ∓π/2. ## Answer The signal is read as the continuous-time sinusoid $x(t) = A\sin\omega_0 t$ (the question writes $x[n]$, but asks for the continuous-time series). The derivation of the general coefficient formula comes first, then the formula is applied. #### Exponential Fourier series and its coefficients A periodic signal with period $T$ and $\omega_0 = 2\pi/T$ can be written as a sum of harmonically related complex exponentials: $$ x(t) = \sum_{k=-\infty}^{\infty} a_k e^{jk\omega_0 t} $$ To find $a_n$, multiply both sides by $e^{-jn\omega_0 t}$ and integrate over one period: $$ \int_0^T x(t)e^{-jn\omega_0 t}dt = \sum_{k=-\infty}^{\infty} a_k \int_0^T e^{j(k-n)\omega_0 t}dt $$ Orthogonality of the exponentials gives $$ \int_0^T e^{j(k-n)\omega_0 t}dt = \begin{cases} T, & k = n \\ 0, & k \ne n \end{cases} $$ so only the $k = n$ term remains, and $$ a_k = \frac{1}{T}\int_T x(t)e^{-jk\omega_0 t}dt $$ #### Applying it to x(t) = A sin ω₀t Using Euler's identity: $$ x(t) = A\,\frac{e^{j\omega_0 t} - e^{-j\omega_0 t}}{2j} = \frac{A}{2j}e^{j\omega_0 t} - \frac{A}{2j}e^{-j\omega_0 t} $$ Comparing with $\sum a_k e^{jk\omega_0 t}$: $$ a_1 = \frac{A}{2j} = -j\frac{A}{2}, \qquad a_{-1} = -\frac{A}{2j} = j\frac{A}{2}, \qquad a_k = 0 \ (k \ne \pm 1) $$ The integral formula gives the same result. For example, $$ a_1 = \frac{1}{T}\int_0^T A\,\frac{e^{j\omega_0 t}-e^{-j\omega_0 t}}{2j}e^{-j\omega_0 t}dt = \frac{A}{2j}\cdot\frac{1}{T}\left[T - 0\right] = \frac{A}{2j} $$ #### Spectrum - Magnitude: $|a_1| = |a_{-1}| = A/2$ - Phase: $\angle a_1 = -\pi/2$, $\angle a_{-1} = +\pi/2$ ```text |a_k| A/2 | | -+--o--+--> k -1 0 1 angle(a_k) +pi/2 | -+--o--+--> k -pi/2 | -1 0 1 ``` The spectrum has only two lines, at $\pm\omega_0$. The DT signal $A\sin\omega_0 n$ with $\omega_0 = 2\pi/N$ gives the same coefficients $a_{\pm1} = \pm A/(2j)$, repeated every $N$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. A sin(w0 t) written with complex exponentials is: A. A (e^{j w0 t} + e^{-j w0 t})/2 B. A (e^{j w0 t} - e^{-j w0 t})/(2j) C. A (e^{j w0 t} - e^{-j w0 t})/2 D. A (e^{j w0 t} + e^{-j w0 t})/(2j) Answer: B. sin(theta) = (e^{j theta} - e^{-j theta})/(2j). (Key point: Euler form) 2. What are a_1 and a_{-1}? A. A/2 and A/2 B. A/(2j) = -jA/2 and -A/(2j) = jA/2 C. jA/2 and -jA/2 D. A and -A Answer: B. Comparing with sum a_k e^{jk w0 t} gives a_1 = A/(2j) and a_{-1} = -A/(2j). (Key point: Coefficients) 3. What do the magnitude and phase spectra look like? A. Infinite lines with 1/k decay B. Single line at k = 0 C. Lines at k = +-1 with magnitude A/2 and phases -pi/2, +pi/2 D. Lines at k = +-1 with magnitude A and phase 0 Answer: C. Only two lines exist at +-w0, each of magnitude A/2, with phases -pi/2 at k = 1 and +pi/2 at k = -1. (Key point: Spectrum) 4. In the integral check of a_1, why does the integral reduce to T? A. The term e^{j w0 t} e^{-j w0 t} = 1 integrates to T while the other term integrates to 0 B. Because sin is odd C. Because A cancels D. Because the period is 1 Answer: A. The first term gives 1 over the period (integral T) and the second gives e^{-j2 w0 t}, which integrates to zero. 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