Parseval's Theorem

Discover the relation between signal energy in the time and frequency domains.

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The energy of a length-NN discrete-time signal is defined as:

Ex=n=0N1x[n]2E_x = \sum_{n=0}^{N-1} |x[n]|^2

Parseval’s relation links the energy of a signal in the time domain to its energy in the frequency domain. In the equation form, it is expressed as:

n=0N1x[n]2=1Nk=0N1X[k]2\sum _{n=0} ^{N-1} |x[n]|^2 = \frac{1}{N} \sum _{k=0} ^{N-1} |X[k]|^2

In other words, the energy of a signal in the time domain is equal to its energy in the transform domain after scaling by 1/N1/N.

Derivation

From the definition of the inverse discrete Fourier transform, we have:

x[n]=1Nk=0N1X[k]e+j2πkNnx[n]=\frac{1}{N}\sum_{k=0}^{N-1}X[k]e^{+j2\pi\frac{k}{N}n}

Using this in the expression for the energy of a signal:

n=0N1x[n]2=n=0N1x[n]x[n]=n=0N1x[n](1Nk=0N1X[k]e+j2πkNn)\begin{align*} \sum _{n=0} ^{N-1} |x[n]|^2 &= \sum _{n=0} ^{N-1} x[n]\cdot x^*[n]\\ &= \sum _{n=0} ^{N-1} x[n]\cdot \left(\frac{1}{N}\sum_{k=0}^{N-1}X[k]e^{+j2\pi\frac{k}{N}n}\right)^* \end{align*}

From here, we can rearrange the terms after conjugation.

n=0N1x[n]2=n=0N1x[n]1Nk=0N1X[k]ej2πkNn=1Nk=0N1X[k]n=0N1x[n]ej2πkNnX[k]=1Nk=0N1X[k]2\begin{align*} \sum _{n=0} ^{N-1} |x[n]|^2 &= \sum _{n=0} ^{N-1} x[n]\cdot \frac{1}{N}\sum_{k=0}^{N-1}X^*[k]e^{-j2\pi\frac{k}{N}n}\\ &= \frac{1}{N}\sum_{k=0}^{N-1}X^*[k]\underbrace{\sum _{n=0} ^{N-1} x[n] e^{-j2\pi\frac{k}{N}n}}_{X[k]}\\ &= \frac{1}{N}\sum_{k=0}^{N-1}|X[k]|^2 \end{align*}

Intuition

Intuitively, going into the frequency domain doesn’t alter the energy contents of the observed signal. In mathematics, this is known as a unitary transformation.

As far as the scaling factor of 1/N1/N is concerned, recall that the DFT outputs a sum of complex sinusoids. For a constant input, this kind of sum magnifies the input by NN. This is why a reduction by 1/N1/N is necessary to balance the energy in the two domains. Due to this reason, many software routines define the DFT and inverse DFT with the same scaling factor of 1/N1/\sqrt{N}.

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