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Parseval’s Theorem – Definition, Conditions and Applications

Parseval’s theorem is an important theorem used to relate the product or square of functions using their respective Fourier series components. Theorems like Parseval’s theorem are helpful in signal processing, studying behaviors of random processes, and relating functions from one domain to another.

Parseval’s theorem states that the integral of the square of its function is equal to the square of the function’s Fourier components.

This article covers the fundamentals of Parseval’s theorem and its proof. Learn when to apply the theorem and how to apply it given a particular function.

Take a refresher on Fourier transform before trying out the examples prepared just for you, so that by the end of this discussion, you can feel confident when working with functions and the Fourier series that represent them!

What Is the Parseval’s Theorem?

Parseval’s theorem (also known as Rayleigh’s theorem or energy theorem) is a theorem stating that the energy of a signal can be expressed as its frequency components’ average energy. Think of Parseval’s theorem as a Pythagorean theorem of Fourier transform.

In terms of integrals, Parseval’s theorem states that the integral of the function’s square is equivalent to the square of the function’s Fourier transform. This means that through Parseval’s theorem, the equation shown below holds.

\begin{aligned}\color{DarkOrange} \textbf{Parsev} &\color{DarkOrange}\textbf{al’s Theorem}\\\\\int_{-\infty}^{\infty} |g(t)|^2 \phantom{x}dt &= \dfrac{1}{2\pi} \int_{-\infty}^{\infty} |G(\omega)|^2 \phantom{x}d\omega\end{aligned}

This theorem is helpful when dealing with signal processing and when observing the behavior of random processes. When signals are challenging to process with time as their domain, transforming the domain is the best course of action so that the values are easier to work with. This is where Fourier transforms and the Parseval’s theorem enters.

Taking a look at the equation of Parseval’s theorem for continuous functions, a signal’s power (or energy) will be much easier to capitalize on and will provide insight on how these quantities behave through a different domain, say frequency. When dealing with discrete quantities, Parseval’s theorem can also be expressed by the equation shown below:

\begin{aligned}\color{DarkOrange} \textbf{Parsev} &\color{DarkOrange}\textbf{al’s Theorem}\\\\\sum_{i = 0}^{n – 1} |x_i|^2 &= \dfrac{1}{n} \sum_{k = 0}^{n – 1} |x_k|^2\end{aligned}

For the equation to be true, $x_i$ and $x_k$ must be pairs of fast Fourier transform (also known as FFT) and $n$ must be the total number of terms present in the sequence. Now, to better understand how Parseval’s theorem is used to rewrite different functions in a new domain, take a look at the proof and application of Parseval’s theorem in the sections that follow.

Proof of Parseval’s Theorem

To prove Parseval’s theorem, rewrite the left-hand side of the equation and express the square of the function as the product of the function and its conjugate’s inverse Fourier transform. Use the Dirac delta function’s identity to simplify the expression and prove Parseval’s theorem.

Recall that the function’s Fourier transform and inverse Fourier transform are related to each other as shown below:

\begin{aligned}\color{DarkOrange} \textbf{Fourier } &\color{DarkOrange}\textbf{Transform}\\\\G(\omega) = \int_{-\infty}^{\infty} & g(t)e^{-i\omega t} \phantom{x}dt\\\color{DarkOrange} \textbf{Inverse Fourier } &\color{DarkOrange}\textbf{Transform}\\\\g(t) = \dfrac{1}{2\pi} \int_{-\infty}^{\infty} & G(\omega)e^{i\omega t} \phantom{x}d\omega\end{aligned}

Use these two properties to rewrite the left-hand side of Parseval’s theorem: $\int_{-\infty}^{\infty} |g(t)|^2 \phantom{x}dt$.

\begin{aligned}\int_{-\infty}^{\infty} |g(t)|^2 \phantom{x}dt &=\int_{-\infty}^{\infty} |g(t)|^2 \phantom{x}dt \\&=\int_{-\infty}^{\infty} g(t) \cdot g(t)\phantom{x}dt \\&=\int_{-\infty}^{\infty} g(t) \cdot \left[\dfrac{1}{2\pi} \int_{-\infty}^{\infty} G(\omega)e^{i\omega t} \phantom{x}d \omega\right]\phantom{x}dt  \end{aligned}

Rewrite the resulting expression by factoring out $\dfrac{1}{2\pi}$ then interchanging the order of $dt$ and $d\omega$ as shown below. Recall that the complex conjugate of $G(\omega)$ is equal to $G^{*}(\omega) = \int_{-\infty}^{\infty} g(t)e^{i\omega t} \phantom{x}dt$.

\begin{aligned}\int_{-\infty}^{\infty} |g(t)|^2 \phantom{x}dt &=\dfrac{1}{2\pi} \int_{-\infty}^{\infty} G(\omega) \cdot \left[\int_{-\infty}^{\infty} g(t)e^{i\omega t} \phantom{x}d t\right]\phantom{x}d\omega\\&= \dfrac{1}{2\pi} \int_{-\infty}^{\infty} G(\omega)G^*(\omega) \phantom{x}d\omega\end{aligned}

The integral identity of the Dirac delta function establishes that the integral of the function and its conjugate’s propduct is equal to the integral of the function’s square. This means that $\int_{-\infty}^{\infty} |g(t)|^2 \phantom{x}dt = \int_{-\infty}^{\infty} g(t)g^{*}(t) \phantom{x}dt$, so use this to simplify the resulting expression further.

  \begin{aligned}\int_{-\infty}^{\infty} |g(t)|^2 \phantom{x}dt &= \dfrac{1}{2\pi} \int_{-\infty}^{\infty} G(\omega)G^*(\omega) \phantom{x}d\omega\\&= \dfrac{1}{2\pi} \int_{-\infty}^{\infty} |G(\omega)|^2 \phantom{x}d\omega\end{aligned}

This proves Parseval’s theorem, $\int_{-\infty}^{\infty} |g(t)|^2 \phantom{x}dt = \dfrac{1}{2\pi} \int_{-\infty}^{\infty} |G(\omega)|^2 \phantom{x}d\omega$. Now that the Parseval’s theorem is established, learn how to apply it to work out different problems. When ready, head on over to the section below!

Example 1

To appreciate Parseval’s theorem, use it to find the Fourier series that represents $f(x) = 1 + x$, where $x$ is defined by the interval $x \in (-\pi, \pi)$.

Solution

This function is a periodic function for the interval $-j < x< j$. In the past, it has been shown that periodic functions such as $f(x)$ can be written as a sum of three periodic terms:

\begin{aligned}f(x) = \dfrac{a_o}{2} + \sum_{n = 1}^{\infty} a_n \cos \dfrac{n\pi x}{j} + \sum_{n = 1}^{\infty} b_n \sin  \dfrac{n\pi x}{j} \end{aligned}

Substitute $f(x) = 1 +x$ and $j = \pi$ into the equation to rewrite $f(x)$. Keep in mind that $a_o$, $a_n$, and $b_n$ are Fourier coefficients equivalent to:

\begin{aligned}a_o &= \dfrac{1}{\pi}\int_{-\pi}^{\pi} \dfrac{f(x)}{\sqrt{2}} \phantom{x}dx\\a_n &=\dfrac{1}{\pi}\int_{-\pi}^{\pi} f(x)\cos(nx) \phantom{x}dx\\b_n &=\dfrac{1}{\pi}\int_{-\pi}^{\pi} f(x)\sin(nx) \phantom{x}dx \end{aligned}

\begin{aligned}\boldsymbol{a_o}\end{aligned}

\begin{aligned}\boldsymbol{a_n}\end{aligned}

\begin{aligned}\boldsymbol{b_n}\end{aligned}

\begin{aligned}a_o &= \dfrac{1}{\pi}\int_{-\pi}^{\pi} \dfrac{(1 + x)}{\sqrt{2}} \phantom{x}dx\\&= 2 \end{aligned}

\begin{aligned}a_n &=\dfrac{1}{\pi}\int_{-\pi}^{\pi} (1 + x)\cos(nx) \phantom{x}dx \\&= 0 \end{aligned}

\begin{aligned} b_n &=\dfrac{1}{\pi}\int_{-\pi}^{\pi} (1 + x)\sin(nx) \phantom{x}dx \\&= (-1)^{n + 1} \dfrac{2}{n} \end{aligned}

When working with periodic functions, Parseval’s theorem can be applied to write $f(x)$ as shown below:

\begin{aligned}\color{DarkOrange} \textbf{Parsev} &\color{DarkOrange}\textbf{al’s Theorem}\\\\ \dfrac{1}{2j}\int_{-j}^{j} [f(x)]^2 \phantom{x}dx &= \dfrac{1}{4}a_o^2 + \dfrac{1}{2}\sum_{n = 1}^{\infty} (a_n^2 + b_n^2)\end{aligned}

Keep in mind that $f(x)$ is bounded by the interval $-j.

\begin{aligned}\dfrac{1}{2\pi}\int_{-\pi}^{\pi} [f(x)]^2 &= \dfrac{1}{2\pi}\int_{-\pi}^{\pi} (1 + x)^2  \phantom{x}dx\\ &= \dfrac{1}{4} (2)^2 + \dfrac{1}{2}\sum_{n = 1}^{\infty} \left[0 + \left((-1)^{n + 1} \dfrac{2}{n} \right)^2\right]\\&= 1 + \dfrac{1}{2} \sum_{n =1}^{\infty} \dfrac{4}{n^2}\\&= 1 + 2\sum_{n = 1}^{\infty} \dfrac{1}{n^2}\end{aligned}

This relationship is also called Parseval’s identity for the Fourier series. To find the Fourier series for $(1 + x)$, rewrite the resulting equation.

 \begin{aligned}\dfrac{1}{2\pi}\int_{-\pi}^{\pi} (1 + x)^2  \phantom{x}dx&= 1 + 2\sum_{n = 1}^{\infty} \dfrac{1}{n^2}\\-1 + \dfrac{1}{2\pi}\int_{-\pi}^{\pi} (1 + x)^2  \phantom{x}dx&= 2\sum_{n = 1}^{\infty} \dfrac{1}{n^2}\\-\dfrac{1}{2} + \dfrac{1}{4\pi}\int_{-\pi}^{\pi} (1 + x)^2  \phantom{x}dx&= \sum_{n = 1}^{\infty} \dfrac{1}{n^2}\\\sum_{n = 1}^{\infty} \dfrac{1}{n^2} &= -\dfrac{1}{2} + \dfrac{1}{4\pi}\int_{-\pi}^{\pi} (1 + x)^2  \phantom{x}dx\end{aligned}

Apply properties learned in integral calculus to evaluate the right-hand side of the equation.

\begin{aligned}-\dfrac{1}{2} + \dfrac{1}{4\pi}\int_{-\pi}^{\pi} (1 + x)^2  \phantom{x}dx &= -\dfrac{1}{2} + \dfrac{1}{4\pi} \int_{-\pi}^{\pi}(1 + 2x + x^2) \phantom{x}dx\\&= -\dfrac{1}{2}+ \dfrac{1}{4\pi}\left[x + x^2 + \dfrac{x^3}{3}\right]_{-\pi}^{\pi}\\&= -\dfrac{1}{2} + \dfrac{1}{4\pi} \left(2\pi +\frac{2\pi ^3}{3}\right)\\&= \dfrac{\pi^2}{6} \end{aligned}

This means that through the Parseval’s theorem, $\sum_{n = 1}^{\infty} \dfrac{1}{n^2} = \dfrac{\pi^2}{6}$.

Example 2

Evaluate the integral $\int_{0}^{\infty} \dfrac{1}{(t^2 + m^2)(t^2 + n^2)} \phantom{x}dt$.

Hint: Use the fact that when $f(t) =e^{-m |t|}$, the inverse Fourier transform, $F(\omega) = \sqrt{\dfrac{2}{\pi}} \dfrac{m}{m^2 + \omega^2}$.

Solution

Express the rational expression $\dfrac{1}{(x^2 + m^2)(x^2 + n^2)}$ as a product of two functions: $f(t) = \dfrac{1}{t^2 +m^2}$ and $g(t) = \dfrac{1}{t^2 + n^2}$.

Use the hint and rewrite these two functions:

\begin{aligned}f(t) &= e^{-m|t| }\\g(t) &= e^{-n|t|}\end{aligned}

Parseval’s theorem can also be extended to account for the integral of two functions’ products.

\begin{aligned}\color{DarkOrange} \textbf{Parsev} &\color{DarkOrange}\textbf{al’s Theorem}\\\\\int_{-\infty}^{\infty} f(t)g(t) \phantom{x}dt &= \int_{-\infty}^{\infty} F(\omega) G(\omega) \phantom{x}d\omega\end{aligned}

Use this equation and rewrite the left-hand side using the exponential forms of $f(t)$ and $g(t)$. Similarly, rewrite the right-hand side in terms of the inverse Fourier transform from the hint.

\begin{aligned}\int_{-\infty}^{\infty} f(t)g(t) \phantom{x}dt &= \int_{-\infty}^{\infty} F(\omega)G(\omega) \phantom{x}d\omega\\ \int_{-\infty}^{\infty} e^{-m|t|}e^{-n|t|} \phantom{x}dt &=\int_{-\infty}^{\infty} F(\omega)G(\omega) \phantom{x}d\omega\\\int_{-\infty}^{\infty} e^{-m|t|}e^{-n|t|} \phantom{x}dt &=\int_{-\infty}^{\infty} \sqrt{\dfrac{2}{\pi}} \dfrac{m}{m^2 + \omega^2} \cdot \sqrt{\dfrac{2}{\pi}} \dfrac{n}{n^2 + \omega^2} \phantom{x}d\omega\end{aligned}

Simplify both sides of the equation by applying appropriate algebraic techniques.

\begin{aligned}\int_{-\infty}^{\infty} e^{-(m + n)|t|}\phantom{x}dt &=\int_{-\infty}^{\infty} \sqrt{\dfrac{2}{\pi}} \dfrac{m^2}{m^2 + \omega^2} \cdot \sqrt{\dfrac{2}{\pi}} \dfrac{n^2}{n^2 + \omega^2} \phantom{x}d\omega\\\int_{-\infty}^{\infty} e^{-(m+n)|t|}\phantom{x}dt&= \int_{-\infty}^{\infty} \dfrac{2}{\pi}\dfrac{mn}{(m^2 + \omega^2)(n^2 + \omega^2)} \phantom{x}d\omega \\\int_{-\infty}^{\infty} e^{-(m + n)|t|}\phantom{x}dt&= \int_{-\infty}^{\infty} \dfrac{2mn}{\pi}\dfrac{d\omega}{(m^2 + \omega^2)(n^2 + \omega^2)}\end{aligned}

Focus on the upper half of the limits $[0, \pi]$, so divide both intervals by half and focus on the positive values of the domain.

\begin{aligned}\int_{0}^{\infty} e^{-(m + n)t}\phantom{x}dt&= \dfrac{2mn}{\pi}\int_{0}^{\infty} \dfrac{d\omega}{(m^2 + \omega^2)(n^2 + \omega^2)}\\\dfrac{2mn}{\pi}\int_{0}^{\infty} \dfrac{d\omega}{(m^2 + \omega^2)(n^2 + \omega^2)} &= \int_{0}^{\infty} e^{-(m + n)t}\phantom{x}dt\end{aligned}

Evaluate the integral of the expression on the right-hand side of the equation.

\begin{aligned}\dfrac{2mn}{\pi}\int_{0}^{\infty} \dfrac{d\omega}{(m^2 + \omega^2)(n^2 + \omega^2)} &= \int_{0}^{\infty} e^{-(m + n)t}\phantom{x}dt\\\dfrac{2mn}{\pi}\int_{0}^{\infty} \dfrac{d\omega}{(m^2 + \omega^2)(n^2 + \omega^2)} &= \left[\dfrac{1}{m + n}e^{-(m + n)t}\right]_{\infty}^{0}\\\dfrac{2mn}{\pi}\int_{0}^{\infty} \dfrac{d\omega}{(m^2 + \omega^2)(n^2 + \omega^2)} &= \dfrac{1}{m + n}\\\int_{0}^{\infty} \dfrac{d\omega}{(m^2 + \omega^2)(n^2 + \omega^2)} &= \dfrac{\pi}{2mn}\cdot \dfrac{1}{m + n}\\\int_{0}^{\infty} \dfrac{d\omega}{(m^2 + \omega^2)(n^2 + \omega^2)} &= \dfrac{\pi}{2mn(m + n)}\end{aligned}

Replace $\omega$ with $t$ and the conclusion will still remain. This means that through the Parseval’s theorem, $\int_{0}^{\infty} \dfrac{1}{(t^2 + m^2)(t^2 + n^2)} \phantom{x} dt$ is also equal to $\dfrac{\pi}{2mn(m + n)}$.

Practice Questions

1. Using Parseval’s theorem, which of the following shows the Fourier series for $g(x) = x^2$, where $x$ is defined by the interval $x \in (-\pi, \pi)$?A. $\sum_{n = 1}^{\infty} \dfrac{1}{n^4} = \dfrac{\pi^4}{90}$
B. $\sum_{n = 1}^{\infty} \dfrac{1}{n^4} = \dfrac{\pi^2}{40}$
C. $\sum_{n = 1}^{\infty} \dfrac{1}{n^3} = \dfrac{\pi^4}{90}$
D. $\sum_{n = 1}^{\infty} \dfrac{1}{n^3} = \dfrac{\pi^2}{40}$

2. Given that $h(x) = -\pi^2 x + x^3$ and the function has the Fourier series, $h(x) = \sum_{n = 1}^{\infty} (-1)^n \dfrac{12}{n^3} \sin (nx)$, which of the following shows the value of $\sum_{n = 1}^{\infty}\dfrac{1}{n^6}$?
A. $\sum_{n = 1}^{\infty} \dfrac{1}{n^6} = \dfrac{\pi^5}{455}$
B. $\sum_{n = 1}^{\infty} \dfrac{1}{n^6} = \dfrac{\pi^6}{455}$
C. $\sum_{n = 1}^{\infty} \dfrac{1}{n^6} = \dfrac{\pi^5}{945}$
D. $\sum_{n = 1}^{\infty} \dfrac{1}{n^6} = \dfrac{\pi^6}{945}$

Answer Key

1. A

2. D

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