Parseval's Theorem and Energy Conservation

Hello! Welcome to your seventh lesson.

In our last lesson, we explored the powerful Convolution Theorem, which established that convolution in the time domain is equivalent to multiplication in the frequency domain. This gave us a profound insight into how LTI systems act as spectral filters. Today, we'll continue exploring the deep connections between the time and frequency domains by focusing on a fundamental conservation principle for signals.

This lesson is about Parseval's Theorem. For someone with your background in electrical engineering, the concept of energy is second nature. You're used to thinking about the energy dissipated in a resistor or stored in a capacitor. Parseval's theorem extends this physical idea into the abstract world of signals, showing that the total energy of a signal is the same whether you calculate it in the time domain or the frequency domain. It's a conservation law for signal analysis.

Our goal is to apply Parseval's theorem to relate the energy of a signal in the time and frequency domains. We will derive the theorem, understand its physical interpretation, and use it as a practical tool for signal analysis and problem-solving.

1. Signal Energy and Energy Spectral Density

Before stating the theorem, we must first formally define what we mean by "signal energy."

For a continuous-time signal $x (t)$ , the total energy $E_{x}$ is defined as the integral of its squared magnitude over all time:
$E_{x} = \int_{- \infty}^{\infty} ∣ x (t) ∣^{2} d t$
Physically, you can interpret this as the total energy dissipated in a 1-ohm resistor if a voltage $x (t)$ is applied across it.

Signals for which this integral is finite and non-zero ( $0 < E_{x} < \infty$ ) are called energy signals. These are typically pulses or signals of finite duration. In contrast, signals like a continuous sinusoid or a DC voltage have infinite energy but a finite average power, and are called power signals. Parseval's theorem, in the form we will study it today, applies to energy signals.

To understand these concepts more formally, please read the following brief section.

Topic 5: Energy & Power Signals, Correlation & Spectral Density

These notes from a course at the University of Oxford provide a clear and mathematically precise distinction between energy signals and power signals, which is the foundational context for Parseval's theorem.

Please read Section 5.2, titled 'Energy signals vs Power signals'. Focus on the definitions of an energy signal (Eq. 5.5) and a power signal (Eq. 5.6).

The Statement of Parseval's Theorem

Parseval's theorem states that the total energy of a signal can also be calculated by integrating its energy distribution in the frequency domain. It provides an equality between the time-domain energy and the frequency-domain energy.

The theorem has two common forms, depending on whether you use ordinary frequency $f$ (in Hz) or angular frequency $ω$ (in rad/s):

Using ordinary frequency $f$ : $\int_{- \infty}^{\infty} ∣ x (t) ∣^{2} d t = \int_{- \infty}^{\infty} ∣ X (f) ∣^{2} df$
Using angular frequency $ω$ : $\int_{- \infty}^{\infty} ∣ x (t) ∣^{2} d t = \frac{1}{2 π} \int_{- \infty}^{\infty} ∣ X (j ω) ∣^{2} d ω$ The factor of $1/ (2 π)$ arises from the definition of the Fourier transform using $ω$ . We will use the $ω$ form, as it is consistent with our previous lessons.

The term $∣ X (j ω) ∣^{2}$ has a special name and significance: it is the Energy Spectral Density (ESD). The ESD tells us how the energy of the signal $x (t)$ is distributed across different frequencies. Parseval's theorem shows that if you sum up all the energy contributions from every frequency (i.e., integrate the ESD), you get the total energy of the signal.

2. Derivation of Parseval's Theorem

Given your preference for mathematical rigor, we'll derive the theorem in two distinct ways.

Derivation 1: From the Convolution Theorem

The first derivation elegantly uses the frequency convolution property we learned in the last lesson, highlighting the interconnectedness of these Fourier properties.

Recall the theorem for multiplication in the time domain:

F {x (t) y (t)} = \frac{1}{2 π} [X (j ω) * Y (j ω)] = \frac{1}{2 π} \int_{- \infty}^{\infty} X (j λ) Y (j (ω - λ)) d λ

The full derivation is laid out nicely in the Oxford notes.

Topic 5: Energy & Power Signals, Correlation & Spectral Density

This section of the Oxford notes provides a concise and elegant proof of Parseval's Theorem. It starts with a general result from the convolution theorem and makes a clever substitution to arrive at the final identity.

Please read Section 5.3, 'Parseval’s theorem revisited'. Follow the argument starting from the general convolution result (Eq. 5.7), setting ω=0 (Eq. 5.9), and then substituting g(t) = x*(t). This shows how the energy integral emerges naturally.

In summary, the key steps are:

Start with the identity for the inverse Fourier transform at $t = 0$ : $\int_{- \infty}^{\infty} X (j ω) d ω = 2 π x (0)$ . A more general version is that the integral of a product in one domain relates to the convolution evaluated at the origin in the other. Specifically, by setting $ω = 0$ in the frequency convolution theorem, we get: $\int_{- \infty}^{\infty} x (t) y (t) d t = \frac{1}{2 π} \int_{- \infty}^{\infty} X (j λ) Y (- j λ) d λ$
Choose a specific function for $y (t)$ . Let $y (t) = x^{*} (t)$ (the complex conjugate of $x (t)$ ).
We need the Fourier transform of $y (t)$ . Using the property $F {x^{*} (t)} = X^{*} (- j ω)$ , we find that $Y (j ω) = X^{*} (- j ω)$ . Therefore, $Y (- j λ) = X^{*} (j λ)$ .
Substitute these into the identity from step 1: $\int_{- \infty}^{\infty} x (t) x^{*} (t) d t = \frac{1}{2 π} \int_{- \infty}^{\infty} X (j λ) X^{*} (j λ) d λ$
Since $z z^{*} = ∣ z ∣^{2}$ , this simplifies to the final theorem (renaming the dummy variable $λ$ to $ω$ ): $\int_{- \infty}^{\infty} ∣ x (t) ∣^{2} d t = \frac{1}{2 π} \int_{- \infty}^{\infty} ∣ X (j ω) ∣^{2} d ω$

Derivation 2: From First Principles

Another powerful derivation involves substituting the Fourier transform definition directly into the energy integral and using the integral representation of the Dirac delta function. The following video from MIT OpenCourseWare presents this approach.

Note on Notation: The video uses notation from quantum mechanics. Please make the following mental substitutions to map it to our signal processing context:

Time-domain function $ψ (x)$ is our $x (t)$ .
Frequency-domain function $ϕ (k)$ is our $X (j ω)$ .
The time variable $x$ is our $t$ .
The frequency variable $k$ is our $ω$ .

Parseval identity

This video from MIT OCW provides a detailed mathematical derivation of Parseval's identity. It starts with the energy integral and directly substitutes the inverse Fourier transform definitions, using the properties of the Dirac delta function to simplify the expression. This is a classic and fundamental proof.

Watch the first 6 minutes and 45 seconds of the video. Follow the derivation as it expands the integral of |ψ(x)|² by substituting the Fourier transforms for ψ(x) and ψ*(x), rearranges the integrals, identifies the Dirac delta function, and arrives at the final identity.

3. Applying Parseval's Theorem

Now, let's see how the theorem is applied. It serves two main purposes:

As a way to verify calculations by computing energy in both domains.
As a shortcut to solve integrals that are difficult in one domain but easy in the other.

Example 1: Calculation and Verification

This first example is a practical demonstration. We will calculate the energy of a decaying exponential signal, first in the time domain, and then verify the result using Parseval's theorem in the frequency domain.

Energy Example and Verification Using Parseval's Theorem in Signals & Systems

This video from Engineering Funda provides a clear, step-by-step calculation for the signal x(t) = e^(-4t)u(t). It is a perfect practical application of the theorem.

Please watch the entire video. It is broken down into clear segments: introduction, time-domain energy calculation, statement of Parseval's theorem, Fourier transform calculation, and finally, the frequency-domain energy calculation and verification.

As the video demonstrates, both methods yield the same result, $E_{x} = 1/8$ . The time-domain integral was straightforward, but the frequency-domain integral required knowledge of the integral of $1/ (a^{2} + ω^{2})$ , which is an arctangent function. This shows that while both paths lead to the same answer, one might be computationally easier than the other.

Example 2: The Power of a Shortcut

Now for an example where the frequency-domain approach is vastly simpler. Let's evaluate the integral:
$\int_{- \infty}^{\infty} sinc^{2} (t) d t$
where $sinc (t) = \frac{s i n ( π t )}{π t}$ . Integrating $sinc^{2} (t)$ directly is a challenging calculus problem. However, we can use Parseval's theorem.

Identify the signal: Our signal is $x (t) = sinc (t)$ . The integral we want to find is its total energy.
Find its Fourier Transform: In a previous lesson, we encountered the duality property. A key transform pair is the rectangular pulse and the sinc function: $rect (f) \Leftrightarrow sinc (t)$ where $rect (f)$ is a rectangular pulse of height 1 from $f = - 1/2$ to $f = + 1/2$ . So, $X (f) = rect (f)$ .
Apply Parseval's Theorem: We'll use the version with frequency $f$ , as it's simpler here. $\int_{- \infty}^{\infty} sinc^{2} (t) d t = \int_{- \infty}^{\infty} ∣ rect (f) ∣^{2} df$
Solve the easy integral: The integral of the squared rectangular pulse is just the area of a rectangle of width 1 and height $1^{2} = 1$ . $\int_{- \infty}^{\infty} ∣ rect (f) ∣^{2} df = \int_{- 1/2}^{1/2} 1^{2} df = [f]_{- 1/2}^{1/2} = \frac{1}{2} - (- \frac{1}{2}) = 1$

Thus, without any difficult integration, we found that $\int_{- \infty}^{\infty} sinc^{2} (t) d t = 1$ . This demonstrates the power of the theorem as a problem-solving tool.

Test your understanding!

A signal $x (t)$ has a Fourier transform $X (j ω) = \frac{1}{a + j ω}$ . What is the total energy of the signal $y (t) = x (t) * x (t)$ , where '*' denotes convolution? Express your answer in terms of $a$ . (Hint: You do not need to compute any time-domain integrals.)

Show answer

Find the spectrum of the output signal: Using the Convolution Theorem from the last lesson, the Fourier transform of $y (t) = x (t) * x (t)$ is $Y (j ω) = X (j ω) \cdot X (j ω) = (X (j ω))^{2}$ .
$Y (j ω) = (\frac{1}{a + j ω})^{2} = \frac{1}{( a + j ω ) ^{2}}$
Use Parseval's Theorem to find the energy of y(t):
$E_{y} = \frac{1}{2 π} \int_{- \infty}^{\infty} ∣ Y (j ω) ∣^{2} d ω = \frac{1}{2 π} \int_{- \infty}^{\infty} \frac{1}{( a + j ω ) ^{2}}^{2} d ω$
Calculate the magnitude:
$∣ Y (j ω) ∣^{2} = \frac{1}{( a + j ω ) ^{2}}^{2} = (\frac{1}{∣ a + j ω ∣ ^{2}})^{2} = (\frac{1}{a ^{2} + ω ^{2}})^{2} = \frac{1}{( a ^{2} + ω ^{2} ) ^{2}}$
Set up the integral:
$E_{y} = \frac{1}{2 π} \int_{- \infty}^{\infty} \frac{1}{( a ^{2} + ω ^{2} ) ^{2}} d ω$
This is a known definite integral. A standard integral table shows $\int_{- \infty}^{\infty} \frac{1}{( x ^{2} + a ^{2} ) ^{2}} d x = \frac{π}{2 a ^{3}}$ .
Solve the integral:
$E_{y} = \frac{1}{2 π} (\frac{π}{2 a ^{3}}) = \frac{1}{4 a ^{3}}$
The total energy of the convolved signal is $\frac{1}{4 a ^{3}}$ . Notice we found this by combining the convolution and Parseval's theorems, operating entirely in the frequency domain.

Conclusion

Today we formalized the concept of signal energy and established the profound connection between a signal's time-domain representation and its energy spectrum.

Key Takeaways:

Parseval's Theorem states that the total energy of a signal—the integral of its squared magnitude—is conserved across the time and frequency domains.
The quantity $∣ X (j ω) ∣^{2}$ is the Energy Spectral Density (ESD), which describes how the signal's energy is distributed over frequency. The total energy is the area under the ESD curve (scaled by $1/2 π$ ).
Historically, Marc-Antoine Parseval developed this for Fourier series, and Lord Rayleigh later extended it to Fourier transforms. It is also sometimes referred to as Plancherel's theorem.
The theorem is a powerful practical tool, allowing us to calculate signal energy or solve complex integrals by choosing the domain (time or frequency) where the calculation is simpler.

Preview of the Next Lesson:
In this lesson and the last, we've relied on knowing certain Fourier transform pairs, like the rect/sinc pair. In our next lesson, we will systematically derive the Fourier transforms of common, fundamental signals, including rectangular pulses, sinc functions, and Gaussian functions. This will solidify our library of essential transform pairs that form the building blocks for analyzing more complex signals and systems.

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