Technical Report: About FFT Analyzers 9

Next, let's look at the Fourier transform and inverse transform of basic waveforms. Understand what form f(t) and F(ω) = R(ω) + jI(ω) take. First, let's look at the properties of the Fourier transform and inverse transform.

Let F(ω) be the Fourier transform of f(t);

(Formula 5-17)

(Formula 5-18)

Therefore, if we swap ω ↔ t, from equations 5-17 and 5-18;

(Formula 5-19)

This relationship is derived. Equation 5-19 represents the symmetry of the Fourier transform and its inverse transform.

The Fourier transform F(ω) of f(t) shifted by τ, to f(t-τ), is given by setting t - τ = x:

(Formula 5-20)

Multiplying F(ω) by e^{- jωτ}, i.e., e^{- jωt} represents circular motion, so the waveform is shifted in phase by ωτ in the frequency domain. For example, f(t - τ) shifted by τ = 1 s will have a phase lag proportional to the frequency compared to F(ω) in the frequency domain: 2π at 1 Hz, 4π at 2 Hz, and 8π at 4 Hz.

The basic function of the Fourier transform is called the Fourier kernel.This is shown in Figure 5-3.

This waveform is tangent to f(t) = 1/t, and it is a fundamental equation that frequently appears in Fourier transforms, so it would be good to keep it in mind.

Let's try applying what we've learned so far to basic functions.

(Formula 5-21)

The Fourier transform of is:

(Formula 5-22)

This relationship can be illustrated as follows:

From equation 5-19, which expresses symmetry, we swap ω → -t and T → a in equations 5-21 and 5-22, and divide by 2π:

(Formula 5-23)

The inverse transform of this equation is:

(Formula 5-24)

You can obtain this.

This relationship can be illustrated as follows:

Consider a square waveform like the one in Figure 5-8 below. From -T to 0, if we consider the square pulse in Figure 5-4 from equations 5-21 and 5-22 to have T reduced to T/2 and its position shifted by -T/2, then the Fourier transform for this range PT/2 can be obtained by reducing T to T/2 and multiplying by e ^jωT/2 for the shift, as shown in equation 5-19:

(Formula 5-25)

Similarly, considering the range from 0 to T, the Fourier transform is:

(Formula 5-26)

Considering the period from -T to T, the Fourier transform is the sum of _F1 (ω) and _F2 (ω):

(Formula 5-27)

Up until now, since there was no 'j', only the real part was represented. This one has 'j', so it only represents the imaginary part. This can be illustrated as shown in Figure 5-9.

The triangular pulse wave qt shown in Figure 5-10 is the waveform obtained by integrating the waveform in Figure 5-8 with the Y-axis set to 1/T. Therefore, the Fourier transform can be obtained by multiplying equation 5-27 by 1/T * 1/jω. That is:

(Formula 5-28)

This can be illustrated as shown in Figure 5-11.

The data obtained by cutting cos ω ₀ t between -T and T can be considered to be cos ω ₀ t multiplied by Pt from 5-3 (1), or in other words, the cosine waveform amplitude-modulated by a square pulse.

The Fourier transform of is:

(Formula 5-29)

Figure 5-12 below shows the case where f ₀ = 10/T.

Figure 5-13-1 illustrates the first term of equation 5-29, and Figure 5-13-2 illustrates the second term of the same equation. These are the waveforms in Figure 5-4 shifted by _ω0. Figure 5-13-3 illustrates the sum of the two waveforms, i.e., equation 5-29 itself.

Similar to the previous example, the transformation of the data segmented from -T to T in sin ω ₀ t is:

(Formula 5-30)

The first and second terms of equation 5-30 are the same as those in equation 5-29, and are obtained by taking their difference. However, note that because there is a 'j' before it, it becomes a purely imaginary number.