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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