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Technical Report on FFT Analyzer 9

5. Properties of Fourier Transforms and Fourier Series

5.2 Basic Waveform Fourier Transform and Inverse Transform

Next, let's take a look at the basic Fourier transform and inverse transform of waveforms. Please understand what the forms of f(t) and F(ω) = R(ω) + jI(ω) will be. First, let's examine the properties of the Fourier transform and inverse transform.

symmetry

If the Fourier transform of f(t) is denoted as F(ω);

Equation 5-17

Equation 5-18

By interchanging ω and t, from equations 5-17 and 5-18;

Equation 5-19

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

(2) Parallel movement (advancing/delaying)

The Fourier transform F(ω) of f(t) shifted by τ is expressed by substituting t - τ = x.

(Equation 5-20)

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

(3) A function known as the Fourier kernel

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

  • Figure 5-3

This waveform is related to the function f(t) = 1/t, which is a fundamental expression that often appears in Fourier transforms, so I think it would be good to remember it.

5.3 Fourier Transform of Basic Functions

Let's consider applying the content up to this point to basic functions.

(1) Fourier transform of the square pulse Pt

Equation 5-21

The Fourier transform of:

Equation 5-22

If we illustrate this relationship, it will look like this.

  • Figure 5-4
  • Figure 5-5

(2) Consider the inverse transformation of a square pulse in the frequency domain.

From the expression representing symmetry in equation 5-19, by swapping ω with -t and T with a in equations 5-21 and 5-22, and dividing by 2π:

Equation 5-23

The inverse transformation of this formula is;

Equation 5-24

can obtain.

If we illustrate this relationship, it will look like this.

  • Figure 5-6
  • Figure 5-7

Fourier Transform of a Square Wave

Consider a square waveform as shown in Figure 5-8. From -T to 0, if we take the square pulse in Figure 5-4 from the previous equations 5-21 and 5-22 with T set to T/2 and shifted by -T/2, the Fourier transform in this range PT/2 can be obtained by setting T to T/2 and applying the shift from equation 5-19, which means we multiply by e jωT/2:

Equation 5-25

Similarly, considering from 0 to T, its Fourier transform is,

Equation 5-26

Considering from -T to T, the Fourier transform becomes the sum of F 1 (ω) and F 2 (ω):

(Equation 5-27)

Until now, there was no 'j', so only the real part was considered. Since this has 'j', it means that this is only the imaginary part. When illustrated, it looks like Figure 5-9.

  • Figure 5-8
  • Figure 5-9

triangular pulse wave

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

(Equation 5-28)

When this is illustrated, it will look like Figure 5-11.

  • Figure 5-10
  • Figure 5-11

(5) Time - cut from T to T cos ω 0 t

The data cut from -T to T of cos ω 0 t is what you get when you multiply cos ω 0 t by the 5-3 term (1) of Pt, or in other words, it can be considered as a cosine waveform amplitude modulated by a square pulse.

The Fourier transform of:

Equation 5-29

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

  • Figure 5-12

Figure 5-13-1 illustrates the first term of equation 5-29, while Figure 5-13-2 shows the second term of the same equation, which is a shifted version of the waveform in Figure 5-4 by ω 0. Additionally, Figure 5-13-3 depicts the sum of the two waveforms, that is, equation 5-29 itself.

  • Figure 5-13-1
  • Figure 5-13-2
  • Figure 5-13-3

(6) Time - cut from T 〜 T sin ω 0 t

The transformation of the data cut from -T to T of sin ω 0 t is as follows:

Equation 5-30

The first and second terms of equation 5-30 are the same as those in the previous equation 5-29, and the difference between them is taken. However, please note that the presence of 'j' indicates that it is a pure imaginary number.

  • Figure 5-14
  • Figure 5-15