# How can I interpret the superpostion of Fourier analysis?

In summary: The textbook is telling you that either way works. You can "read out" the initial velocity in the time domain by taking the derivative then setting t=0. In the frequency (Fourier) domain, you just need to set t=0 to see the initial velocity. If you don't get the same answers, something is wrong.

I have been studied classical mechanics and quantum mechanics a little and after that,
I got a kind of feeling "I have to study Fourier analysis(FA) again".

So I have been studying FA again.

Here're my question.

1. From my viewpoint, FA is a method of expanding functions into several fraction of wave function using orthogonality. how do I put this? like using a lot of shadows of some physics phenomenon on each plains.?(For FA, wave function) so from one plain, one side of phenomenon will be seen and other sides will not be seen.
Is it right?

2. And about FA solution. For simple oscillation problem with two variables(horizontal axis "x" and time variable t) , initial displacement function called f(x) and initial velocity 0(m/s or any kind of velocity demension), assume that I got a simple solution with variable x, t which have undetermined constant.
for your clear understanding of situation, like this.
solution U=XT=(Acosx + Bsinx)*(Ccost + Dsint)

from here, there's a problem.

with two initial condition, FA textbook says,
for initial velocity condition, you have to just get a derivate of function U and set t=0.
but for initial velocity condition, you have to get a Fourier expansion of U and set t=0.

What's the reason of that? if textbook said about only one for derivate or Fourier expansion, I would have no question about that. but why?
I'm wondering how I can interpret that solution physically and mathematically.
If someone can answer this without math, I would appreciate it. because that is more beautiful.

regards and sorry for my poor English.

regards.

I'm not sure what you are asking, but I will try to answer. Fourier Analysis transforms data from normal space (the time domain) into a frequency domain. Any arbitrary (or not) function of t (time) can be represented as the sum of sine waves, and that is what FA does.

There is also phase information, which is why you see the sum of sine and cosine functions. A pure sine wave of a single frequency, which is zero at t=0 has no cosine component. if it is maximum at t=0, then it is a pure cosine function. If it is minimum at t=0, the cosine will be negative, etc.

Let's go back to your U=XT=(Acosx + Bsinx)*(Ccost + Dsint) this is wrong. For a Fourier transform you would have U=X(t)= Bx(C sin(At) (1-C) cos(At)), well I would use different names for the constants but:
A is the frequency of the sine wave or oscillation
B is the amplitude of the sine wave, and
C is the phase of the signal.

"with two initial condition, FA textbook says,
for initial velocity condition, you have to just get a derivative of function U and set t=0.
but for initial velocity condition, you have to get a Fourier expansion of U and set t=0."

What the textbook is telling you here is that either way works. You can "read out" the initial velocity in the time domain by taking the derivative then setting t=0. In the frequency (Fourier) domain, you just need to set t=0 to see the initial velocity. If you don't get the same answers, something is wrong.

Why are there three initial conditions in the frequency domain, and only two in the time domain? There aren't. The frequency of the oscillation is very visible in the Fourier domain, but it is an emergent/hidden property of the physical dimensions of the pendulum or spring you are analyzing in the time domain. That is usually why you do the transform, to make the frequency, or frequencies of a system visible.

I typically use FA (in particular the fast Fourier Transform--FFT) to analyze data sets with multiple frequency components. For example there is one star where astronomers have teased the periods of four different planets out of the very slight motions of the star toward or away from Earth.

Thanks for your answering and pointing out major errors in my question.
about solution, you are correct. thanks. There's no way wave function don't have a frequency term.
but for initial conditions, I really did big mistake.

Here's again.
1. get a first derivate of single solution and set t=0. (or get a first derivate of expanded or superposed solution.)

2. (Corrected)I necessarily have to get a superposed solution(Fourier expanded) and set t=0 and it will be function of mere "x" domain and be a function of displacement f(x).

Why is that?

regards.

## 1. What is the superposition principle in Fourier analysis?

The superposition principle in Fourier analysis states that any periodic function can be represented as a sum of sine and cosine functions with different frequencies and amplitudes. This means that a complex signal can be broken down into simpler components, making it easier to analyze and interpret.

## 2. How do I interpret the Fourier coefficients in a superposition?

The Fourier coefficients represent the amplitudes and phases of the individual sine and cosine functions that make up a signal's Fourier series. These coefficients can tell us about the frequency content of a signal and how different components contribute to the overall shape of the signal.

## 3. What is the difference between frequency and amplitude in Fourier analysis?

Frequency in Fourier analysis refers to the number of oscillations per unit of time in a signal. Amplitude, on the other hand, describes the magnitude or strength of a signal. In other words, frequency is a measure of how often a signal repeats, while amplitude is a measure of its intensity.

## 4. How can I use Fourier analysis to filter out unwanted frequencies?

By analyzing the frequency content of a signal using Fourier analysis, we can identify and isolate specific frequencies that we want to remove. This can be done by manipulating the Fourier coefficients or using filters to remove certain frequencies from the signal.

## 5. Can Fourier analysis be applied to non-periodic signals?

Yes, Fourier analysis can also be applied to non-periodic signals through a technique called the Fourier transform. This allows us to analyze the frequency content of a signal that is not repeating over time. However, the results may be different from those obtained for periodic signals due to the nature of the Fourier transform.