--- title: "FT of periodic signals; e^(−2|t|)sin t" subject: "Signals and Systems" chapter: 3 key: "semester_4/signals_and_systems:ft-periodic-damped-sine" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/ft-periodic-damped-sine site: BCT Engineering --- # FT of periodic signals; e^(−2|t|)sin t Signals and Systems · Chapter 3: Fourier Transform - Asked in: 2078 Baisakh, 5+5 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/ft-periodic-damped-sine - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-3 ## Question Derive expression for Fourier transform of continuous time periodic signal. Find Fourier transform of signal x(t) = e^(−2|t|) sin(t). ## Short answer A periodic signal's FT is an impulse train: X(ω) = ∑ 2π a_k δ(ω−kω₀). For x(t)=e^{−2|t|} sin t, use e^{−2|t|} ↔ 4/(4+ω²) and frequency shift to get X(ω) = −j8ω/(ω⁴+6ω²+25). ## Answer #### Fourier transform of a CT periodic signal A periodic signal is not absolutely integrable, so its Fourier transform is built from impulses. **Step 1:** consider the spectrum $X(\omega) = 2\pi\delta(\omega - k\omega_0)$. Its inverse transform is $$ x(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty}2\pi\delta(\omega - k\omega_0)e^{j\omega t}d\omega = e^{jk\omega_0 t} $$ So $$ e^{jk\omega_0 t} \xleftrightarrow{F} 2\pi\delta(\omega - k\omega_0) $$ **Step 2:** a periodic signal with period $T$ ($\omega_0 = 2\pi/T$) has the Fourier series $$ x(t) = \sum_{k=-\infty}^{\infty}a_k e^{jk\omega_0 t}, \qquad a_k = \frac{1}{T}\int_T x(t)e^{-jk\omega_0 t}dt $$ **Step 3:** apply linearity term by term: $$ \boxed{X(\omega) = \sum_{k=-\infty}^{\infty}2\pi\,a_k\,\delta(\omega - k\omega_0)} $$ The FT of a periodic signal is a train of impulses at the harmonic frequencies $k\omega_0$. The area of each impulse is $2\pi$ times the Fourier series coefficient. ```text X(w) 2pi a0 2pi a-1 ^ 2pi a1 ^ | ^ ^ | | | ^ -+---+-----+-----+---+-- w -2w0 -w0 0 w0 2w0 ``` **Examples:** - $\cos\omega_0 t \leftrightarrow \pi[\delta(\omega - \omega_0) + \delta(\omega + \omega_0)]$ - $\sum_n\delta(t - nT) \leftrightarrow \frac{2\pi}{T}\sum_k\delta\!\left(\omega - \frac{2\pi k}{T}\right)$ #### FT of x(t) = e^{−2|t|} sin(t) This signal is aperiodic (a decaying envelope times a sine). Use a known pair and the frequency-shift property. **Known pair:** with $a = 2$, $$ e^{-a|t|} \leftrightarrow \frac{2a}{a^2 + \omega^2} \quad\Rightarrow\quad e^{-2|t|} \leftrightarrow G(\omega) = \frac{4}{4 + \omega^2} $$ **Euler's formula:** $\sin t = \frac{e^{jt} - e^{-jt}}{2j}$, so $$ x(t) = \frac{1}{2j}\left[e^{-2|t|}e^{jt} - e^{-2|t|}e^{-jt}\right] $$ **Frequency shift** ($e^{j\omega_0 t}g(t) \leftrightarrow G(\omega - \omega_0)$, here $\omega_0 = 1$): $$ X(\omega) = \frac{1}{2j}\left[G(\omega - 1) - G(\omega + 1)\right] = \frac{1}{2j}\left[\frac{4}{4 + (\omega - 1)^2} - \frac{4}{4 + (\omega + 1)^2}\right] $$ **Simplify:** over a common denominator, the numerator is $$ 4\left[(4 + (\omega + 1)^2) - (4 + (\omega - 1)^2)\right] = 4\cdot 4\omega = 16\omega $$ So $$ X(\omega) = \frac{16\omega}{2j\,[4 + (\omega - 1)^2][4 + (\omega + 1)^2]} = \frac{-j\,8\omega}{[4 + (\omega - 1)^2][4 + (\omega + 1)^2]} $$ Expanding the denominator, $(\omega^2 - 2\omega + 5)(\omega^2 + 2\omega + 5) = \omega^4 + 6\omega^2 + 25$: $$ \boxed{X(\omega) = \frac{-j\,8\omega}{\omega^4 + 6\omega^2 + 25}} $$ **Checks:** - $x(t)$ is real and odd (an even envelope times an odd sine), so $X(\omega)$ must be purely imaginary and odd. It is. - $X(0) = 0$, which agrees with $\int x(t)dt = 0$ for an odd signal. - A numerical integration at $\omega = 0.7$ gives $X = -j0.1987$, matching the formula: $-j\frac{5.6}{0.2401 + 2.94 + 25} = -j0.1987$. **Spectrum:** $|X(\omega)| = \frac{8|\omega|}{\omega^4 + 6\omega^2 + 25}$, which is zero at $\omega = 0$, peaks where $3\omega^4 + 6\omega^2 - 25 = 0$, i.e. at $\omega = \pm1.434$ rad/s (where $|X| \approx 0.276$), and decays as $8/|\omega|^3$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which pair is the basis for the FT of a periodic signal? A. e^(jkω0 t) <-> 2π δ(ω - kω0) B. e^(jkω0 t) <-> δ(ω - kω0) C. e^(jkω0 t) <-> 1/(jω - jkω0) D. δ(t) <-> 1 Answer: A. The inverse FT of 2π δ(ω - kω0) is e^(jkω0 t). (Key point: Pair for exponentials) 2. The FT of a periodic signal with Fourier series coefficients a_k is: A. Σ a_k δ(ω - kω0) B. Σ 2π a_k δ(ω - kω0) C. Σ a_k/(2π) δ(ω - kω0) D. Σ 2π a_k δ(t - kT) Answer: B. Linearity applied to each series term gives impulses at kω0 of area 2π a_k. (Key point: Periodic FT) 3. How is e^(-2|t|) sin(t) transformed? A. Fourier series B. Known pair e^(-2|t|) <-> 4/(4+ω^2), Euler's formula and frequency shift by ±1 C. Parseval's relation D. Duality alone Answer: B. sin t = (e^(jt) - e^(-jt))/2j, so X = (1/2j)[G(ω-1) - G(ω+1)]. (Key point: Method for damped sine) 4. Why must the FT of e^(-2|t|) sin(t) be purely imaginary and odd? A. x(t) is real and odd B. x(t) is real and even C. x(t) is periodic D. x(t) is causal Answer: A. An even envelope times an odd sine is real and odd, so its FT is imaginary and odd; this matches -j8ω/(ω^4+6ω^2+25). 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