--- title: "Parseval's theorem for CT aperiodic signals" subject: "Signals and Systems" chapter: 3 key: "semester_4/signals_and_systems:ct-parseval-aperiodic" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/ct-parseval-aperiodic site: BCT Engineering --- # Parseval's theorem for CT aperiodic signals Signals and Systems · Chapter 3: Fourier Transform - Asked in: 2080 Asoj, 2+6 marks; 2073 Magh, 2+4 marks; 2071 Magh, 2+4 marks; 2069 Bhadra, 4 marks; 2083 Baisakh (new course), 4 marks (asked 5 times) - Video: [Parseval's Energy Theorem](https://www.youtube.com/watch?v=IrKSx4mMK5M) — Neso Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/ct-parseval-aperiodic - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-3 ## Question State and prove Parseval's theorem for continuous time aperiodic (energy) signal. ## Short answer Energy is conserved: ∫|x(t)|² dt = (1/2π)∫|X(ω)|² dω. Proof: write |x(t)|² = x(t)x*(t), substitute x*(t) via inverse FT conjugate, swap integration order, identify X(ω) to obtain the relation. ## Answer #### Statement For a continuous-time aperiodic (energy) signal $x(t)$ with Fourier transform $X(\omega)$, the total energy found in the time domain equals the total energy found in the frequency domain: $$ E = \int_{-\infty}^{\infty}|x(t)|^2dt = \frac{1}{2\pi}\int_{-\infty}^{\infty}|X(\omega)|^2d\omega $$ In terms of hertz ($\omega = 2\pi f$): $\int|x(t)|^2dt = \int|X(f)|^2df$. #### Proof Write $|x(t)|^2 = x(t)x^*(t)$: $$ E = \int_{-\infty}^{\infty} x(t)x^*(t)\,dt $$ Replace $x^*(t)$ using the conjugate of the inverse Fourier transform: $$ x(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty}X(\omega)e^{j\omega t}d\omega \;\Rightarrow\; x^*(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty}X^*(\omega)e^{-j\omega t}d\omega $$ Then $$ E = \int_{-\infty}^{\infty} x(t)\left[\frac{1}{2\pi}\int_{-\infty}^{\infty}X^*(\omega)e^{-j\omega t}d\omega\right]dt $$ Interchange the order of integration: $$ E = \frac{1}{2\pi}\int_{-\infty}^{\infty}X^*(\omega)\left[\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt\right]d\omega $$ The bracket is $X(\omega)$ by definition, so $$ E = \frac{1}{2\pi}\int_{-\infty}^{\infty}X^*(\omega)X(\omega)\,d\omega = \frac{1}{2\pi}\int_{-\infty}^{\infty}|X(\omega)|^2d\omega $$ #### Meaning - $|X(\omega)|^2$ is the **energy spectral density**. The energy in a band $\omega_1$ to $\omega_2$ is $\frac{1}{2\pi}\int_{\omega_1}^{\omega_2}|X(\omega)|^2d\omega$. - Energy can be computed in whichever domain is easier. - The Fourier transform preserves energy (apart from the $1/2\pi$ factor), so the phase of $X(\omega)$ has no effect on energy. #### Example For $x(t) = e^{-at}u(t)$ with $a > 0$, $X(\omega) = \frac{1}{a+j\omega}$. - Time domain: $E = \int_0^\infty e^{-2at}dt = \frac{1}{2a}$. - Frequency domain: $E = \frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{d\omega}{a^2+\omega^2} = \frac{1}{2\pi}\cdot\frac{\pi}{a} = \frac{1}{2a}$. The two results agree. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Parseval's theorem for a CT aperiodic signal states that the energy E equals: A. ∫|X(ω)|^2 dω B. (1/2π)∫|X(ω)|^2 dω C. 2π∫|X(ω)|^2 dω D. (1/2π)∫|X(ω)| dω Answer: B. E = ∫|x(t)|^2 dt = (1/2π)∫|X(ω)|^2 dω. (Key point: Statement) 2. What is the key step of the proof? A. Write |x|^2 = x x*, replace x*(t) using the conjugate of the inverse FT, and swap integrals to get ∫x(t)e^{-jωt}dt = X(ω) B. Differentiate x(t) C. Use the convolution theorem on x(t) D. Assume x(t) is periodic Answer: A. After interchanging the integrals, the inner bracket is exactly X(ω), leaving X* X = |X|^2. (Key point: Proof idea) 3. How does the phase of X(ω) affect the energy? A. It changes the energy B. It has no effect on the energy C. It doubles the energy D. It only matters in the time domain Answer: B. Energy depends only on |X(ω)|^2, so phase is irrelevant. (Key point: Role of phase) 4. For x(t) = e^{-at}u(t), a > 0, the energy is: A. 1/a B. 1/(2a) C. 2a D. 1/a^2 Answer: B. E = ∫0^∞ e^{-2at} dt = 1/(2a), which matches the frequency-domain result. 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