8.02 Properties of the Continuous-Time Fourier Transform | 8.04 CTFT Pairs for Singularity & Common Functions
8.03 Rayleigh’s Energy Theorem & Spectral Density
Core Idea
Rayleigh’s Energy Theorem {often referred to as Parseval’s relation for aperiodic signals} establishes the fundamental physical principle of energy conservation across domains. It mathematically guarantees that the total energy calculated by integrating a signal’s squared amplitude over continuous time is identical to integrating its Energy Spectral Density (ESD) across the continuous frequency spectrum. This enables engineers to analyze the distribution of signal energy across frequency bands without performing complex time-domain integrations.
1. Energy Conservation in Time vs. Frequency
When analyzing transient signals {non-periodic waveforms that decay to zero as }, we classify them as Energy Signals if they contain finite total energy.
In the Time Domain, the total physical energy dissipated across a normalized resistor is defined by the integral of the squared magnitude of the signal:
In the Frequency Domain, Rayleigh’s Energy Theorem proves that this same energy is preserved and can be evaluated by integrating the continuous spectrum: E = rac{1}{2\pi} \int_{-\infty}^{\infty} |X(j\Omega)|^2 \, d\Omega
1.1 The Spectral Density Concepts
To understand how energy is distributed, we define two critical spectral density profiles:
- Energy Spectral Density (ESD): For an energy signal , the ESD is defined as: Units: Volt-seconds squared per radian/second (), representing the concentration of energy per unit frequency.
- Power Spectral Density (PSD): For periodic {infinite energy, finite power} or random signals, the energy is infinite, making ESD undefinable. Instead, we compute the distribution of average power over frequency: S_x(j\Omega) = \lim_{T o \infty} rac{|X_T(j\Omega)|^2}{T} where is the Fourier transform of the signal windowed over interval .
Total Energy E (Joules)
/ Time-Domain Integral Frequency-Domain Integral
E = ∫ |x(t)|² dt [8] E = 1/2π ∫ |X(jΩ)|² dΩ [59, 60]
{Sum of instantaneous power} {Integral of Energy Spectral Density}
2. Rigorous Proof of Rayleigh’s Energy Theorem
Let’s prove Rayleigh’s theorem step-by-step using the properties of the continuous-time Fourier transform (CTFT) and its complex conjugate. This derivation is a 4-to-6 mark theory classic in examinations.
Step 1: Express squared magnitude in conjugate form
We start with the time-domain energy definition:
Step 2: Substitute the Inverse CTFT for the conjugate term
Recall the Inverse CTFT synthesis equation for : x(t) = rac{1}{2\pi} \int_{-\infty}^{\infty} X(j\Omega) e^{j\Omega t} \, d\Omega
Taking the complex conjugate of both sides yields: x^*(t) = rac{1}{2\pi} \int_{-\infty}^{\infty} X^*(j\Omega) e^{-j\Omega t} \, d\Omega \quad ext{--- (Equation 2)}
Substitute Equation 2 back into Equation 1:
ight] dt \quad ext{--- (Equation 3)}$$ ### Step 3: Interchange the order of integration Assuming the signals satisfy Dirichlet conditions *{allowing Fubini's theorem to apply}*, we interchange the time and frequency integration order: $$E = rac{1}{2\pi} \int_{-\infty}^{\infty} X^*(j\Omega) \left[ \int_{-\infty}^{\infty} x(t) e^{-j\Omega t} \, dt ight] d\Omega \quad ext{--- (Equation 4)}$$ ### Step 4: Identify the inner integral as the Forward CTFT We recognize the bracketed time-integral as the standard Forward CTFT equation: $$\int_{-\infty}^{\infty} x(t) e^{-j\Omega t} \, dt = X(j\Omega)$$ Substitute this back into Equation 4: $$E = rac{1}{2\pi} \int_{-\infty}^{\infty} X^*(j\Omega) \cdot X(j\Omega) \, d\Omega$$ Since any complex quantity multiplied by its conjugate equals its magnitude squared ($Z \cdot Z^* = |Z|^2$): $$E = rac{1}{2\pi} \int_{-\infty}^{\infty} |X(j\Omega)|^2 \, d\Omega \quad lacksquare$$ --- ## 3. High-Yield Worked "Exam Killers" ### 3.1 Numerical 1: Total Energy Verification of decaying exponential [Syllabus Classic] **Question:** Find the total energy of the causal decaying signal $f(t) = e^{-at}u(t)$ ($a > 0$) using both time-domain and frequency-domain integrations to verify Rayleigh's theorem. #### Time-Domain Integration: Using the continuous time-domain definition: $$E = \int_{-\infty}^{\infty} |f(t)|^2 \, dt = \int_{0}^{\infty} \left(e^{-at} ight)^2 \, dt$$ $$E = \int_{0}^{\infty} e^{-2at} \, dt = \left[ rac{e^{-2at}}{-2a} ight]_{0}^{\infty} = 0 - \left(- rac{1}{2a} ight) = \mathbf{ rac{1}{2a} ext{ Joules}}$$ #### Frequency-Domain Integration: 1. **Find the CTFT of $f(t)$:** $$F(j\Omega) = \mathcal{F}\{e^{-at}u(t)\} = rac{1}{a + j\Omega} \quad [55]$$ 2. **Determine the Energy Spectral Density (ESD):** $$|F(j\Omega)|^2 = \left| rac{1}{a + j\Omega} ight|^2 = rac{1}{a^2 + \Omega^2} \quad [55]$$ 3. **Integrate over all frequencies:** $$E = rac{1}{2\pi} \int_{-\infty}^{\infty} |F(j\Omega)|^2 \, d\Omega = rac{1}{2\pi} \int_{-\infty}^{\infty} rac{1}{a^2 + \Omega^2} \, d\Omega$$ Recall the standard calculus integral $\int rac{1}{a^2 + u^2} du = rac{1}{a} an^{-1}\left( rac{u}{a} ight)$: $$E = rac{1}{2\pi} \left[ rac{1}{a} an^{-1}\left( rac{\Omega}{a} ight) ight]_{-\infty}^{\infty}$$ $$E = rac{1}{2\pi a} \left( an^{-1}(\infty) - an^{-1}(-\infty) ight)$$ $$E = rac{1}{2\pi a} \left( rac{\pi}{2} - \left(- rac{\pi}{2} ight) ight) = rac{1}{2\pi a} (\pi) = \mathbf{ rac{1}{2a} ext{ Joules}}$$ **Conclusion:** Both domains yield exactly $E = rac{1}{2a}$ Joules, demonstrating flawless validation. --- ### 3.2 Numerical 2: The 95% Energy Bandwidth Challenge [Heavily Tested] **Question:** For the same signal $f(t) = e^{-at}u(t)$ with $a = 2$, determine the frequency limit $\Omega_1$ (in rad/s) below which **95%** of the total signal energy is contained. #### Step 1: Calculate the total energy of the signal With $a = 2$: $$E_{ ext{total}} = rac{1}{2a} = rac{1}{2(2)} = 0.25 ext{ Joules}$$ #### Step 2: Set up the 95% energy boundary equation We want to find a frequency limit $\Omega_1$ such that integrating the ESD from $-\Omega_1$ to $\Omega_1$ captures $95\%$ of $E_{ ext{total}}$: $$E_{\Omega_1} = 0.95 \cdot E_{ ext{total}}$$ $$ rac{1}{2\pi} \int_{-\Omega_1}^{\Omega_1} |F(j\Omega)|^2 \, d\Omega = 0.95 \cdot (0.25)$$ $$ rac{1}{2\pi} \int_{-\Omega_1}^{\Omega_1} rac{1}{2^2 + \Omega^2} \, d\Omega = 0.2375$$ #### Step 3: Perform the definite integration Since the integrand is a symmetric even function, we can simplify the limits from $0$ to $\Omega_1$ by multiplying by 2: $$2 \cdot rac{1}{2\pi} \int_{0}^{\Omega_1} rac{1}{4 + \Omega^2} \, d\Omega = 0.2375$$ $$ rac{1}{\pi} \left[ rac{1}{2} an^{-1}\left( rac{\Omega}{a} ight) ight]_{0}^{\Omega_1} = 0.2375$$ $$ rac{1}{2\pi} an^{-1}\left( rac{\Omega_1}{2} ight) = 0.2375$$ #### Step 4: Solve for the frequency limit $\Omega_1$ Multiply both sides by $2\pi$: $$ an^{-1}\left( rac{\Omega_1}{2} ight) = 0.2375 \cdot 2\pi = 0.475\pi ext{ radians}$$ Convert the angle to evaluate the tangent: $$ rac{\Omega_1}{2} = an(0.475\pi) pprox an(1.49226 ext{ rad}) pprox 12.7062$$ $$\Omega_1 = 2 \cdot 12.7062 pprox \mathbf{25.41 ext{ rad/s}}$$ #### Step 5: Convert to cyclic frequency (Hz) $$f_1 = rac{\Omega_1}{2\pi} = rac{25.4124}{2\pi} pprox \mathbf{4.04 ext{ Hz}}$$ **Answer:** 95% of the signal energy is concentrated within the low-frequency band below $\Omega_1 = 25.41 ext{ rad/s}$ (or $4.04 ext{ Hz}$). --- ## 4. Common Mistakes That Cost Marks > [!danger] **The Squared Magnitude Calculus Trap** > > When performing frequency-domain integrations, students frequently integrate the complex spectrum $F(j\Omega) = rac{1}{a + j\Omega}$ directly instead of its squared magnitude $|F(j\Omega)|^2 = rac{1}{a^2 + \Omega^2}$. Attempting to integrate $\int rac{1}{a+j\Omega} d\Omega$ yields a complex logarithmic expression ($\ln(a+j\Omega)$), which leads to **immediate zero marks** on exam papers. **Always compute $|F(j\Omega)|^2$ first!** > [!warning] **The $2\pi$ Scaling Omission** > > Remember that the frequency-domain integration variable is angular frequency $\Omega$ (rad/s). If you integrate with respect to $d\Omega$, you **must** scale the integral by the factor of $ rac{1}{2\pi}$. Forgetting this factor scales your computed energy up by $pprox 628\%$, ruining your numerical check. > *Note:* If you integrate with respect to cyclic frequency $df$ (Hz), the scaling term is naturally absorbed: $E = \int_{-\infty}^{\infty} |F(j2\pi f)|^2 \, df$. --- ## 5. PYQ Bank — Verbatim Questions & Answer Plans ### 5.1 PYQ 2022 / 2016 Exam Classic [8 Marks] **Question:** State and prove Rayleigh's Energy Theorem for continuous-time aperiodic signals. * **Answer Plan:** 1. **State the theorem:** State that the total energy of an aperiodic signal is conserved across both the time and frequency domains, presenting the master equality: $\int_{-\infty}^{\infty} |x(t)|^2 dt = rac{1}{2\pi} \int_{-\infty}^{\infty} |X(j\Omega)|^2 d\Omega$. 2. **Formulate definitions:** Write out the time-domain energy integral and define the Energy Spectral Density (ESD) as $|X(j\Omega)|^2$. 3. **Write the proof:** Follow the exact steps shown in **Section 2**, utilizing the complex conjugate of the Inverse Fourier Transform synthesis equation and interchanging integration orders. ### 5.2 QB Section B Numerical [10 Marks] **Question:** Find the total energy of $x(t) = e^{-3t}u(t)$. Determine the frequency $\omega$ in rad/s such that 90% of the total energy is contained within $(-\omega, \omega)$. * **Answer Plan:** 1. Calculate total energy $E_{ ext{total}} = rac{1}{2a} = rac{1}{6} pprox 0.1667 ext{ Joules}$. 2. Set up the energy fraction: $E_{\omega_1} = 0.90 \cdot rac{1}{6} = 0.15 ext{ Joules}$. 3. Integrate the ESD: $ rac{1}{\pi} \int_{0}^{\omega_1} rac{1}{9 + \Omega^2} d\Omega = 0.15 \implies rac{1}{3\pi} an^{-1}\left( rac{\omega_1}{3} ight) = 0.15$. 4. Solve for the angle: $ an^{-1}\left( rac{\omega_1}{3} ight) = 0.45\pi ext{ rad}$. 5. Compute tangent: $ rac{\omega_1}{3} = an(0.45\pi) = 6.3138 \implies \omega_1 = \mathbf{18.94 ext{ rad/s}}$. --- ## 6. Self-Check Before Moving On - [ ] Can you state Rayleigh's Energy Theorem and write its dual-domain equations from memory? - [ ] Do you know how to derive the theorem using the complex conjugate synthesis substitution? - [ ] Can you mathematically define Energy Spectral Density (ESD) and state its physical units? - [ ] Are you comfortable solving for the fractional energy bandwidth limits (e.g., 95% or 90% bounds) of a decaying exponential? --- *Source: (k.Deergha Rao) signals and systems.pdf (Chapter 3), 02 Fourier Transform.pdf, Rabiul sir class note.pdf.*