If I have a time-dependent electric field $E(t)$ and want to calculate the spectrum of this field by $$A_0\cdot|E(\omega)|^2\,$$ where $A_0$ is some constant factor.

My questions is which definition of fourier transform we should use when transforming $E(t)$ to $E(\omega)$? It seems that the absolute magnitude of $E(\omega)$ will depend on the choice of fourier transform. For example, we can have different definitions for fourier transform from mathematics: 1. \begin{align*} F(\omega) &= \int\dfrac{1}{\sqrt{2\pi}}f(t)\exp(-j\omega t)dt \\ f(t) &= \int\dfrac{1}{\sqrt{2\pi}}F(\omega)\exp(j\omega t)d\omega \end{align*}

2. \begin{align*} F(\omega) &= \int f(t)\exp(-j\omega t)dt \\ f(t) &= \int\dfrac{1}{2\pi}F(\omega)\exp(j\omega t)d\omega \end{align*}

3. \begin{align*} F(\omega) &= \int\dfrac{1}{2\pi}f(t)\exp(-j\omega t)dt \\ f(t) &= \int F(\omega)\exp(j\omega t)d\omega \end{align*}


You can use either one of those pairs. The choice will affect the constant $A_0$, but nothing else.

This is an application of the Plancherel theorem.

If you use the first definition $$F(\omega) = \frac{1}{\sqrt{2\pi}} \int_{-\infty} ^\infty f(t) \exp(-j\omega t) \ \text{d} t, \qquad f(t) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty F(\omega) \exp(j\omega t) \ \text{d} \omega$$ then you obtain $A_0 = 1$, that is $$\int_{-\infty}^\infty |E(t)|^2 \ \text{d}t = \int_{-\infty} ^\infty |\tilde{E}(\omega)|^2 \ \text{d} \omega,$$ where $\tilde{E}(\omega)$ is the Fourier transform of $E(t)$.

If instead you use definition 2 of the Fourier transform, you see that you now scale the frequency domain by a factor $\sqrt{2\pi}$ compared to definition 1. This is why you have to correct for this extra factor when transforming back to the time domain. This also affects the calculation of the power spectrum, and using definition 2, $\tilde{E}(\omega) = \int_{-\infty} ^\infty E(t) \exp(-j\omega t) \ \text{d}t$, you would now obtain $$\int_{-\infty}^\infty |E(t)|^2 \ \text{d}t = \frac{1}{2\pi} \int_{-\infty} ^\infty |\tilde{E}(\omega)|^2 \ \text{d} \omega.$$ With this definition of the Fourier transform, you would therefore instead have to use $A_0 = \frac{1}{2\pi}$ in your calculation of the power spectrum.

  • 1
    $\begingroup$ Thank you so much for your answer! $\endgroup$ – Yikai Kan Mar 10 at 13:39

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