Interpreting Time Domain Derivative in Frequency Domain

One of the properties of Fourier Transform is that the derivative of a signal in time domain gets translated to multiplication of the signal spectrum by $j2\pi f$ in frequency domain. This property is usually derived as follows.

For a signal $s(t)$ with Fourier Transform $S(f)$
\begin{equation*}
s(t) = \frac{1}{2\pi}\int \limits _{-\infty}^{+\infty} S(f) e^{j2\pi ft}df,
\end{equation*}
we have
\begin{align*}
\frac{d}{dt} s(t) &= \frac{1}{2\pi}\int \limits_{-\infty}^{+\infty}S(f) j2\pi f e^{j2\pi ft}df\\
&= \frac{1}{2\pi}\int \limits_{-\infty}^{+\infty}\bigg\{j2\pi f S(f)\bigg\} e^{j2\pi ft}df
\end{align*}
which is the inverse Fourier Transform of $j2\pi f S(f)$.

Now we want to understand this relation one level deeper, i.e., the reason behind the factor $j2\pi f$? There are two parts of this expression: one is $j$ and the other is $2\pi f$. We start with $2\pi f$.

Notice from the definition of Fourier Transform that this operation decomposes a signal into a sequence of complex sinusoids with frequencies ranging from $-\infty$ to $+\infty$. This is shown in Figure 1 below.

Three complex sinusoids in time IQ plane
Complex sinusoids decomposed in IQ components

Figure 1: Three complex sinusoids and their decomposition into sines and cosines

By Euler’s formula,
\begin{equation*}
e^{j2\pi ft} = \cos 2\pi ft + j\sin 2\pi ft
\end{equation*}

Naturally, the higher the frequency, the steeper the slope and hence larger the derivative. After all, a derivative is nothing but the slope of the line tangent to the curve at a point. This is where the factor $2\pi f$ comes from (simply put, the derivative of $\cos 2\pi ft$ is $-2\pi f\cdot \sin 2\pi ft$).

The term $j$ is more interesting. The derivative of $\cos 2\pi ft$ is $-2\pi f\cdot \sin 2\pi ft$ while that of $\sin 2\pi ft$ is $2\pi f\cdot \cos 2\pi ft$. So from Euler’s formula and using $j^2=-1$,
\begin{align*}
\frac{d}{dt} e^{j2\pi ft} &= 2\pi f\bigg\{-\sin 2\pi ft + j\cos 2\pi ft \bigg\}\\
&= 2\pi f\cdot j\bigg\{j\sin 2\pi ft + \cos 2\pi ft \bigg\}\\
&= j 2\pi f\bigg\{\cos 2\pi ft + j\sin 2\pi ft\bigg\}\\
\end{align*}

Remembering that $j=e^{j\pi/2}$, the factor $j$ is therefore necessary to rotate $\cos$ and $-\sin$ by their corresponding angles such that we get our basis signals $e^{j2\pi ft}$ back. This results in getting the same signal $S(f)$ at the output with multiplication by $2\pi f$.

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