--- title: "ESD of an energy signal (ideal BPF example)" subject: "Communication Systems" chapter: 1 key: "semester_6/communication_systems:esd-energy-signal-ideal-bpf" source: https://www.bctengineer.com/q/semester_6/communication_systems/esd-energy-signal-ideal-bpf site: BCT Engineering --- # ESD of an energy signal (ideal BPF example) Communication Systems · Chapter 1: Introduction - Asked in: 2064 Shrawan (CS I), 5+3 marks - Video: [Energy density spectrum](https://www.youtube.com/watch?v=Rt_eWkz_MF4) — Saritha's Tutorials - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/esd-energy-signal-ideal-bpf - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-1 ## Question Derive a general expression for an energy spectral density of an energy signal with an example of an ideal BPF. Mention basic properties of the energy spectral density. ## Short answer Energy spectral density Ψ_g(f) = |G(f)|² (J/Hz) describes energy distribution of an energy signal. Derived by passing g(t) through an ideal narrow BPF of bandwidth Δf at f₀; output energy ≈ 2|G(f₀)|²Δf, giving energy per unit bandwidth as |G(f)|². Properties: area equals total energy, real and non-negative, even for real signals, no phase information, Fourier pair with energy autocorrelation, and Ψ_y(f)=|H(f)|²Ψ_x(f) through LTI systems. ## Answer The **energy spectral density (ESD)** $\Psi_g(f)$ of an energy signal tells how its energy is distributed with frequency; $\Psi_g(f)\,df$ is the energy in a small band $df$. #### Derivation using an ideal BPF Let $g(t)$ be an energy signal with Fourier transform $G(f)$ and energy $E=\int\lvert g(t)\rvert^2dt$. To find how much energy lies near a frequency $f_0$, pass $g(t)$ through an **ideal narrow band-pass filter** centred at $f_0$ with bandwidth $\Delta f$: $$ H(f)=\begin{cases}1, & f_0-\frac{\Delta f}{2}\le\lvert f\rvert\le f_0+\frac{\Delta f}{2}\\0, & \text{otherwise}\end{cases} $$ ```text |H(f)| +--+ | +--+ | | | | | <- width df each ----+--+-----+-----+--+----> f -f0 0 f0 ``` **Step 1 – output spectrum:** $Y(f)=H(f)G(f)$. **Step 2 – output energy** by Rayleigh's theorem: $$ E_y=\int_{-\infty}^{\infty}\lvert Y(f)\rvert^2df=\int_{-\infty}^{\infty}\lvert H(f)\rvert^2\lvert G(f)\rvert^2df $$ **Step 3 – narrow filter:** $\lvert H(f)\rvert^2=1$ only in the two narrow bands. If $\Delta f$ is small, $\lvert G(f)\rvert^2$ is nearly constant there, and for a real signal $\lvert G(-f_0)\rvert=\lvert G(f_0)\rvert$: $$ E_y\approx\lvert G(f_0)\rvert^2\Delta f+\lvert G(-f_0)\rvert^2\Delta f=2\lvert G(f_0)\rvert^2\Delta f $$ **Step 4 – interpret:** the energy in the band of total width $2\Delta f$ (positive plus negative frequencies) is $2\lvert G(f_0)\rvert^2\Delta f$. So the energy per unit bandwidth at $f_0$ is $$ \Psi_g(f)=\lvert G(f)\rvert^2\quad(\text{J/Hz}) $$ Summing over all bands gives the total energy: $$ E=\int_{-\infty}^{\infty}\Psi_g(f)\,df=\int_{-\infty}^{\infty}\lvert G(f)\rvert^2df $$ **Example:** for $g(t)=A\,\text{rect}(t/T)$, $G(f)=AT\,\text{sinc}(fT)$ and $\Psi_g(f)=A^2T^2\text{sinc}^2(fT)$; its area is $A^2T$, equal to the time-domain energy. #### Basic properties of ESD 1. **Area gives energy:** $\int\Psi_g(f)df=E$. 2. **Real and non-negative:** $\Psi_g(f)=\lvert G(f)\rvert^2\ge0$. 3. **Even function** for real $g(t)$: $\Psi_g(-f)=\Psi_g(f)$. 4. **No phase information:** time-shifted signals have the same ESD. 5. **Wiener–Khinchin pair:** $\Psi_g(f)$ is the Fourier transform of the energy autocorrelation $R_g(\tau)=\int g(t)g^*(t-\tau)dt$. 6. **Through an LTI system:** $\Psi_y(f)=\lvert H(f)\rvert^2\Psi_x(f)$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The energy spectral density of an energy signal is A. |G(f)| B. |G(f)|^2 C. G(f)^2 including phase D. |g(t)|^2 Answer: B. Ψ_g(f) = |G(f)|^2 in J/Hz. (Key point: ESD definition) 2. For a narrow ideal BPF at f0 of width Δf, the output energy for a real signal is approximately A. |G(f0)|^2 Δf B. 2|G(f0)|^2 Δf C. |G(f0)| Δf D. |G(f0)|^2 / Δf Answer: B. Positive and negative frequency bands each contribute |G(f0)|^2 Δf. (Key point: Narrow BPF derivation) 3. Which is NOT a property of ESD? A. Area gives total energy B. Real and non-negative C. Carries phase information D. Even for real g(t) Answer: C. ESD has no phase information; time-shifted signals share the same ESD. (Key point: Properties of ESD) 4. If x(t) passes through an LTI system H(f), the output ESD is A. Ψ_x(f)H(f) B. |H(f)|^2 Ψ_x(f) C. Ψ_x(f)/|H(f)| D. |H(f)| Ψ_x(f) Answer: B. Energy is weighted by the squared magnitude response. 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