Sounds contain a spectrum of different pitches or frequencies which our sense of hearing can sort out to a certain degree. Your ear responds to variations in air pressure, that is pressure as a function of time. There is a relatively simple mathematical theory for how to decompose any function of time into components of different frequencies, which corresponds fairly well to the way you perceive different pitches. Frequency analysis turns out to be one of the most important mathemetical techniques ever developed. Not only acoustics, but signal processing, image processing, and the fundamental theory of quantum mechanics all rely on the technique pioneered by Fourier in the early 1800s.
Unfortunately, Fourier analysis is not introduced until roughly the second year of college calculus. There are two historical accidents that account for this neglect: First, Fourier lived a hundred and fifty years after Newton; it took a long time for people to recognize the importance of frequency decomposition. Second, the suite of mathematical techniques that bear Fourier's name did not become ubiquitous practical tools for another hundred and fifty years, with the advent of the digital age. Here we present the essentials of Fourier analysis, the decomposition of signals into their component frequencies, from an elementary point of view. You will need algebra and complex numbers, but little to no calculus.
The mathematical breakthrough that makes frequency analysis simple is the invention of complex numbers. In particular, introducing $i=\sqrt{-1}$ makes the operation of complex multiplication correspond to the physical operations of rotation and scaling in the complex plane. That is, the product of two complex numbers is the complex number whose argument (angle from the positive real axis) is the sum of the angles of the factors, and whose modulus (distance from the origin, $0$) is the product of their moduli. In particular, this means that multiplying two complex numbers which lie on the unit circle produces another point on the unit circle at the sum of their angles from the point $1$.
The $N^\text{th}$ roots of unity are the complex numbers $z$ for which $z^N=1$. Now $z=1$ is always one such point, but there are obviously $N-1$ other such points equally spaced around the unit circle forming a regular $N\text{-gon}.$ We will write $1^{1/N}$ for the $N^\text{th}$ root of unity nearest to $1$ with non-negative imaginary part. Then the $N$ $N^\text{th}$ roots of unity are $1^{n/N}$ for powers $n=0,1,2,...,N-1.$ ($1^x$ is non-standard notation written $e^{i2\pi x}$ in standard notation. We adopt $1^x$ here because it makes the equations much easier to remember.)
Raising any of these to the $N^\text{th}$ power produces $(1^{n/N})^N=1^n=1.$ That is, multiplying any of them by itself $N$ times returns to the starting vertex of the polygon at $1.$ The roots differ, however, in how many times they orbit the circle when you imagine marching up through all of their powers, $1^{mn/N}$ for $m=0,1,2,...,N-1$ and any fixed root $n.$ The root we labeled $n$ orbits the unit circle $n$ times before returning exactly to $1$ on the $N^\text{th}$ step.
What do the complex roots of unity have to do with frequencies? If we think of $m$ as time, continuing to count indefinitely, then the point $1^{mn/N}$ will continue to orbit the unit circle, completing $n$ revolutions for every increase of $m$ by $N.$ We describe this motion as repeating with a frequency of $n$ cycles every $N$ steps. Hence each $n$ represents a different frequency of this orbital motion, while $m$ represents a time. In any real application, time $t$ will not be an integer, but some number of seconds or other unit. We therefore introduce the fundamental period $T$ of the motion, which has units of time, and corresponds to $N$ increments of $m,$ $$t/T = m/N. \tag{1}$$ Similarly, frequency $\nu$ is measured in Hertz (cycles per second) or some corresponding frequency unit. Since root $n$ completes $n$ cycles in time $T,$ $$\nu = n/T. \tag{2}$$ Thus, $1^{mn/N}=1^{\nu t}$; the exponent is an angle measured in units of revolutions or cycles.
Our model for time is a succession of powers $m$, and each $N^\text{th}$ root of unity is a base $1^{n/N}$ for those powers corresponding to a succession of frequencies $n$. In this model, a function of time is just a list of complex numbers $f_m$, representing the evolution of the function over time. Each of our sequences of powers $1^{mn/N}$ corresponds to the function of time which orbits the unit circle with frequency $n$, so if we can write $f$ in the form $$f_m = \sum_{n=0}^{N-1}F_n 1^{mn/N}, \tag{3}$$ we will have expressed $f$ as a sum of frequency components. Here the complex number $F_n$ represents the amplitude of frequency $n$ in the function $f_m.$ Sometimes "amplitude" refers to just the modulus of the complex $F_n$ in which case the argument of $F_n$ is called the "phase" of frequency component $n.$
Given $f_m$, how do we find $F_n$? The brute force approach would be to recognize (3) as an $N\times N$ matrix equation and solve it. However, in this case we can explicitly write the inverse matrix. The trick is to notice that $$\delta_{n0} = \frac{1}{N} \sum_{m=0}^{N-1}1^{mn/N}, \tag{4}$$ where $\delta_{jk}$ is the Kronecker delta, defined to be $1$ when $j=k$ and $0$ otherwise. The $n=0$ case is obvious since every term in the sum is $1$ in that case and there are $N$ terms. Multiplying the sum by $1^{n/N}$ does not change its value, since each term simply becomes its succeeding term and the last term times $1^{n/N}$ is $1$, wrapping back to the first term. The only number which remains unchanged when you multiply by something other than $1$ (for $n\ne 0$) is $0.$ Formula (4) provides a way to invert the matrix to get $$F_n = \frac{1}{N} \sum_{m=0}^{N-1}f_m 1^{-mn/N}. \tag{5}$$
The symmetry between (3) and (5) is astonishing. The only differences are the leading factor of $1/N$ and the sign of the exponent. Roughly speaking, though, the values of the time sequence are the frequency components of the frequency sequence as well as vice-versa. Together, these two equations comprise the discrete Fourier transform (DFT) and its inverse. There are many variants, but all of them come back to these.
Notice that (5) demonstrates that no matter what function of time $f_{m}$ we choose, we can always decompose it into frequency coefficients $F_n.$
If we were to use equation (5) directly to compute the Fourier coefficients $F_n$ given the $f_m$, we would need to perform $N^2$ multiplies and $N(N-1)$ adds. There is a far, far more efficient algorithm used by Gauss in 1805, then forgotten until its rediscovery in 1965 by Cooley and Tukey. This fast Fourier transform (FFT) algorithm is the reason frequency analysis has become ubiquitous in the age of digital computers.
The idea is simple, but obviously subtle enough to have been forgotten: Given any two sequences of $N$ complex numbers $f_m$ and $g_m$ and their frequency amplitudes $F_n$ and $G_n$ we can immediately write down the frequency amplitudes for the sequence of length $P=2N$ built by interleaving $f$ and $g$, $f_0, g_0, f_1, g_1, ...$ until $f_{N-1}, g_{N-1}$. Calling the interleaved sequence $h_p$ and its frequency amplitudes $H_q$ where $p$ and $q$ run from $0$ to $2N-1,$ we see that: $$\begin{align*} H_q &= \frac{1}{P} \sum_{p=0}^{P-1} h_p 1^{pq/P} \\ &= \frac{1}{2N} \left(\sum_{m=0}^{N-1} f_m 1^{2mq/P} + \sum_{m=0}^{N-1} g_m 1^{(2m+1)q/P}\right) \\ &= \tfrac{1}{2} (F_q + 1^{q/P}G_q) \tag{7} \end{align*}$$ Notice that frequency coefficients are always periodic in their index $n$, so that $F_q$ and $G_q$ are the same as $F_{q-N}$ and $G_{q-N},$ so the fact that $q$ runs from $0$ to $P=2N$ is not a problem. Setting aside the leading factor of $\tfrac{1}{2}$ for a moment (although it does not affect the scaling argument we are about to make), given $F_m$ and $G_m$ of length $N$, we can compute a Fourier series of double the length using only one multiply and one add per value of $q.$
The insight is that if we apply this interleaving procedure backwards, recursively, we will use far fewer than the $N^2$ operations of the direct application of (5). Assume that $N$ is a power of $2$. We first take every other element and compute their coefficients; to compute those coefficents, we take every other element of those sequences (every fourth of the original), and so on until we reach sequences of only a single element, for which $F_0=f_0.$ There are a total of $\log_2 N$ such stages. The stage with subsequences of length $M$ will have $N/M$ such subseqences, each of which requires $M$ multiplies and adds to construct from the following stage, so that every stage involves a grand total of $N$ multiplies and adds. Since there are $\log_2 N$ stages, the total operation count for the whole procedure is $N\log_2 N.$ This recursive algorithm for computing $F_n$ given $f_m$ or vice-versa is called FFT. For values of $N$ which are not powers of two, there are variants of this algorithm that involve extracting every third element, or fifth element, etc. You can easily write down the equivalent of equation (7) for these cases.
The FFT operation count of $N\log_2 N$ is far, far less than the count of $N^2$ corresponding to direct application of (5) or (3). Thus, even though those formulas suggest a technique for computing the DFT, they should never be used for that purpose. The existence of the FFT algorithm makes it so cheap to compute decompose and recompose frequency coefficients that frequency decomposition has become a ubiquitous computing tool, shifting the Fourier transform from an important theoretical concept to an indispensible practical one.
For practical purposes, we must always sample functions at a finite rate, and collect only a finite number of samples. Hence the DFT in one form or another is the only practical way to decompose a function into its frequency components. However, for theoretical purposes, we often think about continuous functions of time and an infinite number of samples. These are important limits, because we can often choose the number of samples and their spacings, and we need to understand what might happen if we could collect many more samples. There are two distinct cases: The continuous function $f(t)$ may be precisely periodic with period $T$, but we collect an infinite number of infinitessimally spaced samples. On the other hand, the function $f(t)$ may not be periodic at all, which we can model by allowing the period $T$ to become infinite.
If we let the number of samples become infinite while the period $T$ remains finite, we immediately hit a snag. As long as $N$ was finite, we let the label $n$ for the frequency run from $0$ to $N-1$. The beginning and ending point did not matter because the way we labeled the roots of unity $1^{n/N}$ was periodic in $n$. However, if we let $N$ recede to infinity and still label the roots from $n=0$ to $n=N-1,$ then we no longer make it around to the lower half of the unit circle in finitely many steps. In order to include all the roots of unity for very large $N$, we must instead count from $n=-N/2$ to $n=+N/2-1$ (assuming even $N$) in order for the labels $1^{n/N}$ to cover both the upper and lower halves of the unit circle. (This is the first hint of a very general problem with Fourier analysis called aliasing.) Thus, in the limit $N\rightarrow\infty$, we must label our frequency components $n$ as $..., -2, -1, 0, 1, 2, ...$ instead of beginning at $n=0.$
With that understanding and using equation (1), equations (3) and (5) become, respectively, $$\begin{align*} f(t) &= \sum_{n=-\infty}^\infty F_n 1^{nt/T} \tag{8} \\ F_n &= \frac{1}{T} \int_0^T dt\, f(t) 1^{-nt/T} \tag{9} \end{align*}$$ The integral is calculus notation, but in this context, it is merely an alternative way to write the sum over $m$. The $dt$ is part of this shorthand; it stands for the infinitessimal spacing between the time samples $dt=T/N.$ These equations comprise the Fourier series expansion of the periodic function $f(t).$
There is an alternative way that the number of sample points might increase without bound. In practice, the sample spacing $\tau=T/N$ often remains fixed, so that both $T$ increases proportionally to $N.$ In this case, we replace (1) and (2) by $$\begin{align*} t &= m\tau \tag{1a} \\ \nu\tau &= n/N \tag{2a} \\ \end{align*}$$ Here $1/\tau$ is the largest possible frequency, corresponding to one cycle per sample spacing.
With infinitely many discretely sampled times, we hit a new snag in trying to take the limit of equation (3). Instead of discrete frequency components $1^{n/N}$, we now have a continuum of frequency components $1^{\nu\tau}.$ Therefore, we must replace the concept of a finite amplitude at a single frequency $F_n$ by an infinitessimal amplitude $F(\nu)d\nu$ proportional to the bandwidth $d\nu=1/T$ around frequency $\nu.$ Thus, equations (3) and (5) become $$\begin{align*} f_m &= \int_0^{1/\tau} d\nu\,F(\nu) 1^{m\nu\tau} \tag{8a} \\ F(\nu)&=\frac{1}{\tau}\sum_{m=-\infty}^\infty f_m 1^{-m\nu\tau} \tag{9a} \end{align*}$$ Equations (8a) and (9a) are exactly the same as (9) and (8), respectively, with frequency and time swapped provided we replace $f_m$ by $f_m/\tau.$ Another feature of this alternate form which is at first surprising is that $F(\nu)$ is a periodic function of frequency with period $1/\tau.$ (This is another example of the important phenomenon called aliasing.)
Finally, with the continuum interpretation of frequency component amplitude per unit frequency $F(\nu),$ we can take the limit as the sample spacing $\tau\rightarrow 0.$ In this limit, (8a) and (9a) take the appealingly symmetric forms $$\begin{align*} f(t) &= \int_{-\infty}^\infty d\nu F(\nu) 1^{\nu t} \tag{10} \\ F(\nu) &= \int_{-\infty}^\infty dt\, f(t) 1^{-\nu t} \tag{11} \end{align*}$$ These equations comprise the Fourier integral transform. (Once again, we have been forced to symmetrize the limits of the integral in (8a) to start from $-1/(2\tau)$ instead of $0$ to ensure we reach the roots of unity in both halves of the unit circle.)
Besides the transform equations, the fundamental relation that allowed us to invert the DFT, equation (4), takes on very interesting forms in the Fourier series and integral limits. This relation is at the heart of all Fourier analysis. For the series case of continuous periodic functions of time, we can read off the fixed period $T$ while $N\rightarrow\infty$ limit from equation (4): $$\delta_{n0} = \frac{1}{T} \int_0^T dt\ 1^{nt/T}. \tag{12}$$
In the fixed sample spacing $\tau$ while $N\rightarrow\infty$ limit, on the other hand, equation (4) is harder to interpret. We can find how to translate (4) to this case by setting $f_m=1^{m\nu_0\tau}$ in equation (9a), where $\nu_0$ is an arbitrary frequency. From equation (8a), we then have $$1^{m\nu_0\tau} = \int_0^{1/\tau} d\nu\,F(\nu) 1^{m\nu\tau}.$$ But we can make any periodic function of $\nu$ as a sum of coefficients times $1^{m\nu\tau}$, hence this particular function $F(\nu)$ has the surprising property that $$g(\nu_0) = \int_0^{1/\tau} d\nu\,F(\nu) g(\nu)$$ for any function $g(\nu)$ which is periodic with period $1/\tau.$
The only way this can happen is if $F(\nu)$ is zero for all $t$ except $\nu=\nu_0$, where it must be infinite in such a way that its integral is $1.$ This puzzling property defines the Dirac delta function $\delta(\nu-\nu_0),$ so $F(\nu)=\delta(\nu-\nu_0).$ This is still not quite correct, because $F(\nu)$ must be periodic with period $1/\tau.$ We can fix this by summing over all periods to construct a periodic function which has one delta function spike in each period. (Technically we should have done this with the Kronecker delta in (4) as well.) The final result from (9a) is: $$\sum_{n=-\infty}^\infty\delta(\nu-\nu_0+n/\tau) = \frac{1}{\tau} \sum_{m=-\infty}^\infty 1^{-m(\nu-\nu_0)\tau}. \tag{13}$$
In the case of the Fourier integral, we can take the limit of equation (13) as $\tau\rightarrow 0.$ In this limit, $m\tau$ becomes the continuous variable $t,$ and $1/\tau$ becomes the differential $dt.$ Furthermore, we need only consider the $n=0$ term on the left hand side of (13) because there is only a single infinite "period": $$\delta(\nu-\nu_0) = \int_{-\infty}^\infty dt\,1^{-(\nu-\nu_0)t}. \tag{14}$$
One of the primary applications of the FFT is to quickly compute convolution products. Convolution arises in physical processes governed by an impulse response. An example is the chirp sound you hear when you clap your hands at the entrance to a long hollow pipe like a culvert. The clap is the impulse and the chirp is the response. The response is a function of time, which we can denote $g(t)$ or $g_m$ for sampled times. We can regard an arbitrary noise source as a sequence of claps (with a lot of explanation we omit here); this is source is a second function of time $f(t)$ or $f_m.$ In a linear system (like the pipe echo), the complete output $h(t)$ or $h_m$ is simply the sum of the individual impulse responses each starting at different time. That is, $$h_m = \sum_{k=0}^{N-1} f_k\, g_{m-k}. \tag{15}$$ This defines the discrete convolution of the source function $f$ and the impulse response function $g.$ By substituting $k$ for $m-k$ you can see that the convolution operation is symmetric, that is we would get the same result if we swapped $f$ and $g$, source and impulse response. By substituting the impulse $\delta_{k0}$ for $f_k,$ you see that the response $h_m$ is the impulse response $g_m$ as expected.
The DFTs of these three functions, $H_n$, $F_n$, and $G_n$ are related in an especially simple way. Plug (15) into equation (5) to find $$\begin{align*} H_n &= \frac{1}{N}\sum_{m=0}^{N-1}\sum_{k=0}^{N-1}f_k\,g_{m-k}1^{mn/N} \\ &= \frac{1}{N}\sum_{k=0}^{N-1}f_k 1^{kn/N}\sum_{m=0}^{N-1}g_{m-k}1^{(m-k)n/N} \\ H_n &= N F_n G_n \tag{16} \end{align*}$$ Now computing equation (15) directly requires $N^2$ operations, while using equation (16) would require three FFTs (to find $F_n,$ $G_n,$ and then $h_m$) and $2N$ multiplies, which is $3N\log_2N+2N$ operations, potentially far fewer than the direct calculation.
As we take the Fourier series limit of infinite $N$ we need to modify equation (15). The problem is that when we increase $N$, we are decreasing the duration associated with each sample point $m$, which decreases the effect of the source function $f_m,$ which we would expect to decrease the size of the response $g_m.$ What we want is to replace the source function $f$ by a source function per unit time $f(t).$ Thus, we replace the source $f_m$ by $f(t)dt=f(t)T/N$ so that (15) becomes $$h(t) = \int_0^T dt'\,f(t')g(t-t'), \tag{17}$$ where $t'/T=k/N.$ In this case, substituting $\delta(t')$ for $f(t'),$ we see that $g(t)$ is the response to a delta function impulse.
Applying equation (9) to (17), we find the frequency components $$H_n = T F_n G_n. \tag{18}$$ That is, once again the frequency components of the convolution are, apart from a constant factor, the product of the frequency components of the source and impulse response functions.
For the Fourier series case, we can also convolve the frequency components. Although the physical situations in which this occurs are rare, mathematically it does produce an interesting formula. If we define $H_n$ as an infinite sum convolution of $F_n$ and $G_n,$ we can go through the same steps to relate $f(t)$ and $g(t)$ to the corresponding $h(t)$. Namely, using equation (8), $$\begin{align*} H_n &= \sum_{p=-\infty}^\infty F_p G_{n-p} \tag{19} \\ h(t) &= f(t)g(t) \tag{20} \end{align*}$$ This formula needs no normalization constant.
Finally, the Fourier integral case produces the most symmetric convolution transform formula. Once again, our impulse function $g(t)$ will be the response to a $\delta$-function source, so that we just extend the integral in equation (17) to infinity in both directions. Applying equation (11) we find for the Fourier integral case $$\begin{align*} h(t) &= \int_{-\infty}^\infty dt'\,f(t')g(t-t') \tag{21} \\ H(\nu) &= F(\nu)G(\nu) \tag{22} \end{align*}$$
A variant of convolution is autocorrelation. Autocorrelation is an important tool for analyzing noise signals. When a signal is periodic, we can decompose it into frequencies using the Fourier series formalism. However, a noisy signal is by definition aperiodic, although it often persists for an indefinitely long time (compared to the rate it is changing). Hence we turn to the Fourier integral formulation, equations (10) and (11).
We have already remarked that power is the square of amplitude, so consider the square of the modulus of equation (11). From (22), if we convolve a signal $f(t)$ with the complex conjugate of its time-reversal $g(t)=\overline{f(-t)}$, which has Fourier transform $G(\nu)= \overline{F(\nu)}$, then $$H(\nu) = ||F(\nu)||^2 \tag{23}$$ The corresponding function $h(t)$ whose frequency components are $H(\nu)$ is: $$h(t) = \int_{-\infty}^\infty dt'\,f(t')\overline{f(t'-t)} = \int_{-\infty}^\infty dt'\,f(t+t')\overline{f(t')} \tag{24}$$ This definition works as long as the signal $f(t)$ has finite duration, or at least approaches zero both in the distant past and in the distant future. However, for a noise signal $f(t)$ of infinite duration, the integral in equation (24) diverges. This is an important case in practice, so we need to develop an alternative to (24) that works for a noise signal of infinite duration.
Physically, if $||f(t)||^2$ has units of power, then the $h(t)$ defined by (24) has units of energy, while from equation (11), the $H(\nu)$ of (23) has units of energy per unit frequency. But if the signal persists forever, the total energy for all time, per unit frequency or for all frequencies, will also be infinite, so we expect that both (24) and $F(\nu)$ in (23) will diverge. The easiest way to address this problem is to artificially chop the signal off outside some very long time interval $T$, and redefine $h(t)$ to be the mean over that long time interval: $$h(t) = \lim_{T\rightarrow\infty} \frac{1}{T} \int_{-T/2}^{T/2} dt'\,f(t+t')\overline{f(t')} \tag{25}$$ This $h(t)$ is the autocorrelation of the signal $f(t)$; it has units of power, and will be finite even when $f(t)$ lasts indefinitely. In fact, this definition only applies to signals $f(t)$ with infinite duration past and future.
The Fourier transform $H(\nu)$ of the autocorrelation $h(t)$ is called the power spectrum; it has units of power per unit frequency. $H(\nu)$ represents the power per unit frequency in the signal with frequency $\nu$: $$H(\nu) = \int_{-\infty}^\infty dt\,h(t) 1^{-\nu t} \tag{26}$$ From (25), $h(-t)=\overline{h(t)}$ (at least for $t\ll T$) which implies $H(-\nu)=\overline{H(\nu)}.$ For the purposes of the power spectrum, negative and positive frequencies are the same, so the total power per unit frequency at frequency $|\nu|$ is the real number $H(\nu)+H(-\nu).$ (This distinction matters only when the signal $f(t)$ may be a complex number.)
Translating these defnitions back to the DFT case we arrive at the finite versions of these definitions of autocorrelation $h_m$ and power spectrum $H_n$: $$\begin{align*} h_m &= \frac{1}{N}\sum_{k=0}^{N-1} f_{m+k}\overline{f_k} \tag{27} \\ H_n &= \frac{1}{N}\sum_{m=0}^{N-1} h_m 1^{-mn/N} \tag{28} \end{align*}$$ These are how autocorrelation is defined in practice. Autocorrelation adds the $1/N$ normalization in (27) relative to the convolution formula (15), so $h_m$ is an average rather than a sum. (Although we sometimes use the $1/N$ normalization for discrete convolution, when we want to simulate the integral form of convolution. The real question is how we choose to normalize the impulse response.) In (28), the units of $H_n$ are power per mode, rather than power per unit frequency. Since the frequency difference between successive modes is $1/T,$ you can multiply the right hand side of (28) by $T$ to get the power per unit frequency $H(\nu)$ of the integral formula (26).
We can regard a function $f_m$ as the coordinates of a vector in and $N$-dimensional space. If we adopt this viewpoint, then the DFT, equations (3) and (5), amounts to a coordinate transformation. The vector $1^{mn/N}$ for each $n$ is one of $N$ new basis vectors in this space, and $F_n$ are the coordinates of the same vector in this new basis. The inverse coordinate transform is $1^{-mn/N}/N,$ which are the coordinates of the original time basis vectors (which were $\delta_{mn}$) in the frequency basis.
Occasionally you see the normalization $1^{\pm mn/N}/\sqrt{N}$ for these coordinate transforms. This makes the transformation not only orthogonal, but also orthonormal, so that the Fourier transform becomes a true Euclidean rotation operation in the $N$-dimensional function space. Physically, this amounts to normalizing each frequency mode to have unit energy per period, rather than unit amplitude as we have done. While a case can be made for this normalization, it complicates at least half the formulas without providing any significant benefits. As a practical matter, the normalization of Fourier transforms will always be an issue you need to work out for each individual situation.
Notice that in the continuum Fourier integral transform limit, equations (10) and (11), the continuous time and frequency coordinates $f(t)$ and $F(\nu)$ are related by a symmetric, orthonormal integral transform. It may be that this symmetric form is equivalent to the $1/\sqrt{N}$ discrete normalization, since it arises from a blending of the $N\rightarrow\infty$ at constant period $T$ versus at constant spacing $\tau$ limits. (Occasionally you will also see $1/\sqrt{2\pi}$ normalization in Fourier integral formulas, which is normalization to unit energy per radian. This is just misguided both mathematically and physically, and unrelated to the $1/\sqrt{N}$ normalization question.)
Now in this $N$-dimensional Euclidean function space, there is a distance function, or more generally a dot product, which remains invariant under the time-frequency coordinate rotation. This is a special case of our convolution formula. As for autocorrelation, we begin by noting from (5) that the DFT of the conjugate of a time reversed function $\overline{g_{-m}}$ is the conjugate of its DFT $\overline{G_n},$ so the convolution formulas (15) and (16) become $$\begin{align*} h_m &= \sum_{k=0}^{N-1} f_k\overline{g_{k-m}} \\ H_n &= N F_n \overline{G_n} \\ h_m &= N \sum_{n=0}^{N-1} F_n \overline{G_n} 1^{mn/N}, \end{align*}$$ where the final line is equation (5). Now for the special case $m=0,$ this convolution formula is $$\sum_{n=0}^{N-1} F_n \overline{G_n} = \frac{1}{N} \sum_{m=0}^{N-1} f_m\overline{g_m} \tag{29}$$
Aside from the $1/N$ normalization factor, this generalized dot product is an invariant of the time-frequency coordinate rotation. For the case that $g_m=f_m,$ this defines a distance metric in the $N$-dimensional function space: $$\sum_{n=0}^{N-1} ||F_n||^2 = \frac{1}{N} \sum_{m=0}^{N-1} ||f_m||^2 \tag{30}$$ We recognize the right hand side as the time-average signal power over the whole period. Since the average power of each mode $1^{mn/N}$ is $1$, the left hand side is the sum of the average powers of each mode. This shows that each individual frequency mode contributes to the signal power independently of all the other modes, which is a consequence of the orthogonality of the modes or basis functions $1^{mn/N}.$
This page quickly covers most of the formulas we use to decompose a signal into components of different frequencies. The one exception is the relationship between Fourier series and the operations of differentiation and integration that belong to calculus. Seeing how to apply these basic physical ideas to any particular problem is sometimes easy, but just as often very tricky and subtle. No matter how long you work with Fourier transforms, you will almost never get the practical numerical details right on your first attempt. Nevertheless, these tools are critical in so many fields of science and engineering that you need to begin studying the ideas early and practice applying them as frequently as possible.