## Frequency domain and time domain In electrodynamics, we often convert between frequency domain and time domain for convenience. The general rule for conversion follows the Fourier transform, namely $F(\omega) = \int_{-\infty}^{\infty} f(t) e^{-i\omega t} \mathrm{d}t$ and $f(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) e^{i\omega t} \, \mathrm{d}\omega$ General properties like linearity and convolution are also the same $a f(t) + b g(t) \longleftrightarrow a F(\omega) + b G(\omega)$ $(f * g)(t) \longleftrightarrow F(\omega) \cdot G(\omega)$ the convolution is define as $(f * g)(t) = \int_{-\infty}^{\infty} f(\tau) g(t-\tau) d\tau$. $f(t) \cdot g(t) \longleftrightarrow \frac{1}{2\pi} (F * G)(\omega)$ Also with time/frequency displacement we have $f(t - t_0) \longleftrightarrow F(\omega) e^{-i\omega t_0}$ $f(t) e^{i2\pi f_0 t} \longleftrightarrow F(\omega - 2\pi f_0)$ The later one is also called 'modulation'. Under Fourier transform, $\nabla$ would become $i \omega$. For time derivatives $\frac{d^n f(t)}{\mathrm{d}t^n} \longleftrightarrow (i\omega)^n F(\omega)$ People may use $f$ as time domain expression and $\tilde{f}$ as frequency domain expression, but this is not always the case. >[!Notice] >It is worth noting that, the Fourier transform of a real function satisfies the Hermitian symmetry, namely for a real function $f(t)$, its Fourier transform $F(t)$ satisfies >$F(-\omega) = F(\omega)^{*}$ >This is valid even for single side Fourier transform. >>[!Note]- >>The proof itself is easy. For $f(t)$, the Fourier transform is >>$F(\omega) = \int_{-\infty}^{\infty} f(t) e^{-i\omega t} \mathrm{d}t$ >>and the negative frequency is >>$F(-\omega) = \int_{-\infty}^{\infty} f(t) e^{i\omega t} \mathrm{d}t$ >>Consider the complex conjugate of $F(\omega)$, >>$F(\omega)^* = \left( \int_{-\infty}^{\infty} f(t) e^{-i\omega t} \mathrm{d}t \right)^*$ >>This can be written as >>$F(\omega)^* = \int_{-\infty}^{\infty} f(t)^* e^{i\omega t} \mathrm{d}t$ >>If (and only if) $f(t)$ is real, we have $f(t)^{*}=f(t)$, so >>$F(\omega)^* = \int_{-\infty}^{\infty} f(t) e^{i\omega t} \mathrm{d}t = F(-\omega)$ >>Therefore, $F$ satisfies the Hermitian symmetry, $F(-\omega) = F(\omega)^{*}$. ^7b97a0 >[!Question] >Why use Fourier transform to decompose these waves? Except for the obvious reason that with the linear combinations of sinusoidal functions the mathematical expressions are greatly simplified, which further facilitates the calculation of differentials and convolutions, etc. The most important intuition is that we wanna get the frequency domain expressions. In physics there are tons of quantities are better characterized by frequencies (also spatial frequencies) than time. >As for linearity, completeness and orthonormality, this is not special and we play the same trick in Hilbert spaces. >[!Info] >For Fourier transform in more detail, see contents in [[Index (Fourier Analysis)]]. >For real/reciprocal space conversion, see [[Real space and reciprocal space]].