--- title: "Energy density spectrum of CT signals" subject: "Signals and Systems" chapter: 3 key: "semester_4/signals_and_systems:energy-density-spectrum" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/energy-density-spectrum site: BCT Engineering --- # Energy density spectrum of CT signals Signals and Systems · Chapter 3: Fourier Transform - Asked in: 2082 Kartik, 6 marks; 2077 Chaitra, 5 marks; 2075 Baisakh, 5 marks (asked 3 times) - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/energy-density-spectrum - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-3 ## Question Derive and explain the expression of energy density spectrum for a continuous time aperiodic signal. ## Short answer Energy density spectrum is S_{xx}(ω)=|X(ω)|², derived from Parseval: E=∫|x(t)|²dt = (1/2π)∫|X(ω)|²dω. The energy in a band dω is (1/2π)|X(ω)|² dω. It is real, nonnegative, even for real signals, and equals the FT of autocorrelation. ## Answer The **energy density spectrum** (energy spectral density, ESD) of an aperiodic signal shows how the signal's total energy is spread over frequency. It is defined as $|X(\omega)|^2$. #### Derivation The energy of a CT aperiodic signal is $$ E = \int_{-\infty}^{\infty}|x(t)|^2dt = \int_{-\infty}^{\infty}x(t)x^*(t)\,dt $$ Express $x^*(t)$ through the inverse Fourier transform: $$ x^*(t) = \left[\frac{1}{2\pi}\int_{-\infty}^{\infty}X(\omega)e^{j\omega t}d\omega\right]^* = \frac{1}{2\pi}\int_{-\infty}^{\infty}X^*(\omega)e^{-j\omega t}d\omega $$ Substitute and change the order of integration: $$ \begin{aligned} E &= \frac{1}{2\pi}\int_{-\infty}^{\infty}X^*(\omega)\left[\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt\right]d\omega \\ &= \frac{1}{2\pi}\int_{-\infty}^{\infty}X^*(\omega)X(\omega)\,d\omega = \frac{1}{2\pi}\int_{-\infty}^{\infty}|X(\omega)|^2d\omega \end{aligned} $$ This is Parseval's relation. Write it as $$ E = \frac{1}{2\pi}\int_{-\infty}^{\infty}S_{xx}(\omega)\,d\omega, \qquad S_{xx}(\omega) = |X(\omega)|^2 $$ In hertz: $E = \int_{-\infty}^{\infty}|X(f)|^2df$, with ESD $|X(f)|^2$ in J/Hz. #### Explanation - $\frac{1}{2\pi}|X(\omega)|^2 d\omega$ is the energy carried by the frequency components in a small band $d\omega$ around $\omega$. The energy between $\omega_1$ and $\omega_2$ is therefore $\frac{1}{\pi}\int_{\omega_1}^{\omega_2}|X(\omega)|^2d\omega$ for a real signal, counting both the positive and the negative band. - The ESD is real, non-negative, and holds no phase information. Signals with different phase but the same $|X(\omega)|$ have the same ESD. - For real $x(t)$, $|X(-\omega)| = |X(\omega)|$, so the ESD is even. - The ESD is the Fourier transform of the autocorrelation $R_{xx}(\tau) = \int x(t)x^*(t-\tau)dt$ (Wiener–Khinchin relation for energy signals). - **LTI systems:** if $Y(\omega) = H(\omega)X(\omega)$, then $S_{yy}(\omega) = |H(\omega)|^2 S_{xx}(\omega)$. - It applies only to energy signals (finite $E$). Power signals use the power spectral density instead. #### Example For $x(t) = e^{-at}u(t)$: $S_{xx}(\omega) = \frac{1}{a^2+\omega^2}$, and $$ E = \frac{1}{2\pi}\cdot\frac{\pi}{a} = \frac{1}{2a} $$ Half of this energy lies in the band $|\omega| < a$: $$ \frac{1}{2\pi}\int_{-a}^{a}\frac{d\omega}{a^2+\omega^2} = \frac{1}{2\pi}\cdot\frac{2}{a}\tan^{-1}(1) = \frac{1}{2\pi}\cdot\frac{2}{a}\cdot\frac{\pi}{4} = \frac{1}{4a} $$ ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The energy density spectrum of an aperiodic signal is defined as: A. X(ω) B. |X(ω)|^2 C. angle X(ω) D. |x(t)|^2 Answer: B. S_xx(ω) = |X(ω)|^2 shows how the energy is distributed over frequency. (Key point: Definition of ESD) 2. E = (1/2π)∫|X(ω)|^2 dω is derived by: A. Writing x*(t) via the inverse FT, interchanging the integrals and recognising ∫x(t)e^{-jωt}dt = X(ω) B. Differentiating X(ω) C. Using the sampling theorem D. Assuming x(t) is periodic Answer: A. This is Parseval's relation: E = ∫x x* dt becomes (1/2π)∫X* X dω. (Key point: Derivation) 3. If Y(ω) = H(ω)X(ω), then S_yy(ω) is: A. |H(ω)| S_xx(ω) B. |H(ω)|^2 S_xx(ω) C. H(ω) S_xx(ω) D. S_xx(ω)/|H(ω)|^2 Answer: B. |Y|^2 = |H|^2 |X|^2. (Key point: Effect of an LTI system) 4. For x(t) = e^{-at}u(t), what fraction of the total energy 1/(2a) lies in |ω| < a? A. One quarter B. One half C. All of it D. One third Answer: B. The band energy is 1/(4a), which is half of 1/(2a). 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