Fourier series and spectrum of periodic signals

Fourier series and signal spectra

For a deterministic time-domain signal, we only need to know its functional expression to determine a signal at any time, but what we need in various scenarios is often not such an analytical equation, because these complex equations are difficult to obtain quickly and accurately, and in addition, it is difficult to analyze them quickly, and it is also difficult to extract the information contained in them. Therefore, a more efficient tool for signal analysis is needed.

The triangular form of the Fourier series

Fourier had proposed the idea that a linear combination of trigonometric functions could be employed to represent a continuous periodic signal in the time domain. The following conclusion was later obtained after mathematicians studied the related problems: When a periodic signal satisfies the $$Dirichlet$$ condition, it can **uniquely** be represented by a linear combination of trigonometric functions, e.g., the period is $$T$$ and the frequency is $$ \displaystyle\omega = {2\pi\over T}$$ which can be expanded into the following equation: $$ f(t) = a_0 + \sum_{n = 1}^{\infty}\big[a_n \cos{(n\omega t)} + b_n \sin{(n\omega t)}\big] $$ where $$ \begin{aligned} a_0 & = {1\over T} \int_{t_0}^{t_0+T} f(t) \text{d}t\\\\ a_n & = {2\over T} \int_{t_0}^{t_0+T} f(t) \cos{(n\omega t)}\text{d}t\\\ b_n & = {2\over T} \int_{t_0}^{t_0+T} f(t) \sin{(n\omega t)}\text{d}t\\\\ \end{aligned} $$ $Dirichlet$ The conditions are: 1. there are a finite number of interruptions in a cycle 2. the signal is absolutely cumulative within a cycle 3. the number of very large and very small values is finite

For common signals, the$Dirichlet$ The conditions are all met and can be used without validation

If you use the trigonometric formula to combine the same-frequency sine and cosine you can get the following two forms: $$ \begin{aligned} f(t) & = c_0 + \sum_{n = 0}^{\infty}c_n\cos{(n\omega t + \varphi_n)} \\\ f(t) & = d_0 + \sum_{n = 0}^{\infty}d_n\sin{(n\omega t + \theta_n)} \\\\\\ \end{aligned} $$

The relationship between quantities and values in different representations of the Fourier series:

\[\large \begin{cases} & a_0 = b_0 = c_0\\ \\ & c_n = d_n = \sqrt{a_n^2 + b_n^2}\\ \\ & a_n = c_n\cos{\varphi_n} = d_n\sin{\theta_n}\\ \\ & b_n = -c_n\sin(\varphi_n) = d_n\cos{\theta_n}\\ \\ & \tan{\theta_n} = \displaystyle{a_n \over b_n}\\ \\ & \tan{\varphi_n} = \displaystyle-{b_n \over a_n}\\ \end{cases} \]

Using the mathematical tools described above, we are easily able to transform a continuous periodic signal into a trigonometric function, and this expression contains three unknown quantities ---- amplitude, phase, and frequency, and there is a close relationship between the three. Three unknown quantities, at least two relationships are needed to describe, it is easy to find that the most convenient to construct the mathematical relationship is the amplitude and frequency and phase and frequency relationship, because the frequency is generally a monotonic function, while the other two may not be. So the relationship between amplitude and frequency and the relationship between phase and frequency are obtained respectively, and plot them out respectively to get a set of curves ---- amplitude-frequency curve and phase-frequency curve, called amplitude spectrum and phase spectrum, respectively. Drawing process we found that in fact the signal contains a frequency only at some specific frequency value, which means that we draw the image is a discrete image, the amplitude spectrum and the phase spectrum at this frequency are a finite value, drawn out only a line, so called spectral lines, the top of each line connected, we can see the general trend of the spectral lines.

Complex exponential form of Fourier series

The above expression is a trigonometric expression that needs to be solved separately when we need to solve for a specific periodic signal$a_0$、$a_n$cap (a poem)$b_n$, the process is cumbersome.
Is it possible to find a unified expression that makes it possible to find all three at the same time?
The answer is to utilize Euler's formula

\[e^{j\theta} = \cos(\theta) + j \sin(\theta) \]

Euler's formula can represent trigonometric functions as exponential functions over the complex domain.
We can therefore convert the above expression into the form of a complex exponential as follows:

\[\begin{aligned} f(t) = &a_{0}+ \sum_{n = 1}^{\infty} \left( a_{n} \frac{e^{jn\omega t} + e^{-jn\omega t}}{2} + b_{n} \frac{e^{jn\omega t} - e^{-jn\omega t}}{2j}\right)\\ =&a_0 + \sum_{n = 1}^{\infty} ({a_n - jb_n \over 2}e^{jn\omega t} + {a_n + j b_n \over 2}e^{-jn\omega t})\\ \end{aligned} \]

Next, let's go back and look$a_n$ together with$b_n$ The original definition (integral expression), if$n$is an integer, not defined as a natural number as in the previous section, it is easy to obtain that$a_n$ is an even function.$b_n$ is an odd function.
So we might as well define the function

\[F(n\omega) = {a_n - jb_n \over 2} \]

Combined with the parity analysis above there is

\[\begin{aligned} F(-n\omega) & = {a_{-n} - jb_{-n}\over 2} \\ & = {a_{n} + jb_{n}\over 2} \end{aligned} \]

We found$F(n\omega)$ together with$F(-n\omega)$ is conjugate to exactly two terms of the same frequency of the Fourier series expressed in complex numbers. If we define$F(0) = a_0$ , we can then write the complex form of the Fourier series as follows:

\[f(t) = \sum_{n = -\infty}^{\infty} (F(n\omega)e^{jn\omega t}) \]

Among them:

\[F_n(n\omega) = {1\over T} \int_{t_0}^{t_0 + T} f(t) e^{-jn\omega t}\text{d}t \\ \]

This will make it easier for us to solve for the Fourier series, unlike the trigonometric form where several equations need to be solved, butThe tradeoff is the need to integrate functions of a complex variable, which can be cumbersome。

The relationship between the Fourier exponential form and the relevant parameters in the Fourier series:

\[\large \begin{cases} & F_0 = c_0 = d_0 = a_0\\ \\ & F_n = |F_n|e^{j\varphi_n} = \displaystyle{a_n - jb_n \over 2} \\ \\ & F_{-n} = |F_{-n}|e^{-j\varphi_n} = \displaystyle{a_n + jb_n \over 2} \\ \\ & |F_n| = |F_{-n}| = \displaystyle{1\over 2}c_n = \displaystyle{1\over2}d_n = {1\over2}\displaystyle\sqrt{a_n^2 + b_n^2}\\ \\ & |F_n| + |F_{-n}| = c_n \\ \\ & a_n = F_n + F_{-n} \\ \\ & b_n = j( F_n + F_{-n} ) \\ \\ & c_n^2 = d_n^2 = a_n^2 + b_n^2 = 4F_n F_{-n} \end{cases} \]

due to$F(n\omega)$ together with$F(-n\omega)$ is conjugate, so the spectrogram in complex exponential form has a left-right symmetric even function for the magnitude spectrum and an odd function for the phase spectrum, and is generally located in quadrants two and four.

It is worth noting that in the spectrogram of the complex exponential in the amplitude spectrum compared to the corresponding position in the Fourier series spectrogram, the former is highly the latter'shalf, that is, the positive and complex frequency terms add up to a spectrogram in real form.

Note that the appearance of complex frequencies is primarily a mathematical result and has no practical significance.

Relationship between Fourier series and symmetry of functions

The symmetries of periodic functions fall into two main categories:

Symmetry with respect to whole cycles, such as odd and even functions.
Symmetric to half period, odd harmonic function.

even function (math.)

Definition:

\[f(t) = f(-t)\\ \]

On the image: symmetrical about the coordinate axes

The following conclusion exists for even functions:

\[\large \begin{cases} a_{n}= \frac{4}{T}\displaystyle\int_{0}^{\frac{T}{2}}f(t)\cos{ (n\omega t )}\text{d}t \\ \\ b_{n} = 0 \\ \\ c_{n} = d_{n} = a_{n} = 2 F_{n} \\ \\ F_{n} = F_{-n} = \displaystyle\frac{a_{n}}{2} \\ \\ \varphi_{n} = 0 \\ \\ \theta_{n} = \displaystyle\frac{\pi}{2} \\ \\ \end{cases} \]

CONCLUSION: The Fourier series expansion of an even function contains only the cosine term, not the sine term. And the complex exponential form is a real function.

odd function (math.)

Definition:

\[f(t) = -f(-t) \]

Graphically: symmetric about the origin

Conclusions related to odd functions :

\[\large \begin{cases} a_{0} = 0,a_{n} = 0 \\ \\ b_{n} = \frac{4}{T} \displaystyle \int_{0}^{\frac {T}{2}} f(t)\sin(n\omega t) \text{d}t\\ \\ c_{n} = d_{n} = b_{n} = 2 j F_{n}\\ \\ F_{n} = -F_{-n} = - \frac{1}{2} j b_{n}\\ \\ \varphi_{n} = -\frac{\pi}{2}\\ \\ \theta_{n} = 0\\ \\ \end{cases} \]

Conclusion: the odd function of$F_{n}$ is an imaginary function. There is no cosine term in the Fourier series of an odd function, only a sine term. If the odd function is coupled with a DC component, then in addition to the$a_{0}$ let sb. do sth.$0$ There is no change in the other conclusions.

(iii) Odd harmonic functions

Definition:

\[f(t) = -f\left( t \pm \frac{T}{2} \right) \]

On the image: half a cycle of translation and then flipped along the x-axis to coincide with the original function

Conclusions related to odd harmonic functions:

\[\begin{cases} a_{0} = 0 \\ \\ a_{n} = b_{n} = 0 \qquad (n \text{even number}) \\ \\ a_{n} = \displaystyle \frac{4}{T} \displaystyle \int_{0}^{\frac{T}{2}} f(t) \sin(n \omega t ) \text{d} t \qquad (n \text{odd-numbered})\\ \\ a_{n} = \displaystyle \frac{4}{T} \displaystyle \int_{0}^{\frac{T}{2}} f(t) \cos(n \omega t ) \text{d} t \qquad (n \text{odd-numbered})\\ \\ \end{cases} \]

Conclusion:
Odd harmonic functions do not have even-frequency harmonics; only the sine and cosine terms of odd harmonics exist.

Perhaps you're thinking about why the semiperiodic symmetry in the above categorization speaks only of odd harmonic functions, why not even harmonic functions?
This is a good question, because writing an even harmonic function after the definition of an odd harmonic function, we can find that an even harmonic function is a periodic function whose period is half that of the original function, and that the Fourier series is unique when the original function is determined, and so we put it in the first category in the discussion, which is whole-period symmetry.

Parseval's theorem (in calculus)

Parseval's Theorem is a theorem that applies to a wide range of transformations in signal analysis, with the general conclusion that the energy of an energy signal (where energy is a finite value) in the time domain is equal to the energy in the frequency domain, and that the power of a power signal (where power is a finite value) in the time domain is equal to the power in the frequency domain.

For continuous periodic signals here, often the energy is not finite and the power is a finite value, i.e., it is a power signal, which is equal to the sum of the squares of the RMS values of the components of the Fourier series expansion.

Fourier transform of a typical periodic signal

(i) Periodic rectangular pulse signals

Let the periodic rectangular pulse signal$f(t)$ The pulse width of$t$ periodicity$T$ The pulse amplitude is$E$ , then his Fourier series expansion takes the following form:

\[\large \begin{cases} a_{0} = \frac{{E\tau}}{T}\\ \\ a_{n} = \frac{{2 E \tau}}{T} \text{Sa}\left( \frac{{n \pi \tau}}{T} \right) = \frac{{E \tau \omega}}{\pi} \text{Sa}\left( \frac{{n\omega\tau}}{2} \right)\\ \\ b_{n} = 0\\ \\ F_{n} = \frac{{E \tau}}{T} \text{Sa} \left( \frac{{n\omega \tau}}{2} \right)\\ \\ c_{n} = a_{n}\\ \\ c_{0} = a_{0} \\ \end{cases} \]

Conclusion:

The spectrum of a periodic rectangular pulse is discrete, and the larger the repetition period period, the closer the spectral lines are.
The magnitude of the DC component, and each frequency component, is proportional to the pulse amplitude and width, and inversely proportional to the repetition period
The periodic signal contains an infinite number of spectral lines, in which the energy is concentrated within the first crossing of the zeros, the$\omega < \displaystyle{\frac{{2\pi}}{\tau}}$ interior
We call such a region a frequency band with a bandwidth of$B_{\omega} = \displaystyle\frac{{2\pi}}{\tau}$ or$B_{f } = \displaystyle\frac{1}{\tau}$ The bandwidth is only related to the pulse width and is inversely proportional.

Symmetric square wave signals are also a special case of rectangular signals:

It is an alternating positive and negative signal whose DC component$a_{0}$ zero
His pulse width is exactly equal to half of the period, i.e.$\tau = \displaystyle\frac{T}{2}$

The Fourier series form of a symmetric square wave is:

\[\begin{aligned} f(t) & = \frac{{2E}}{\pi} \sum_{n = 1}^{\infty} \frac{1}{n} \sin\left( \frac{{n\pi}}{2} \right) \cos({n \omega t})\\ & = \frac{{2E}}{\pi} \left[ \cos({\omega t}) + \frac{1}{3} \cos(3\omega t +\pi) + \frac{1}{5} \cos({5\omega t}) + \cdots \right] \end{aligned} \]

The harmonic amplitude of a symmetric square wave is measured in$\displaystyle\frac{1}{n}$ Convergence.

(ii) Periodic sawtooth pulse signals

The peak and trough values are$\displaystyle \pm\frac{E}{2}$ The period is$T$ , the signal is an odd function, so only sinusoidal components exist in the Fourier series. The harmonic amplitudes are expressed in terms of$\displaystyle\frac{1}{n}$ The law of convergence of the

\[f(t) = \frac{E}{\pi} \sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n} \sin({n \omega t}). \]

(iii) Periodic triangular pulse signals

peak$E$ The valley value is$0$ The period is$T$ , is an even function with only a cosine component.

\[\begin{aligned} f(t) & = \frac{E}{2} + \frac{4E}{\pi^{2}} \left[ \cos(\omega t) + \frac{1}{3^{2}}\cos({3\omega t}) + \frac{1}{5^{2}} \cos({5 \omega t}) +\cdots \right]\\ \\ & = \frac{E}{2} + \frac{4E}{\pi^{2}}\sum_{n=1}^{\infty} \frac{1}{n^{2}} \sin^{2}\left( {\frac{n\pi}{2}} \right) \cos({n \omega t}) \\ \end{aligned} \]

(iv) Periodic half-wave cosine signals

Even function, only DC, fundamental and even harmonic frequency components are present, and the amplitude of the harmonics is given as$\displaystyle \frac{1}{n^{2}}$ Regular convergence.

\[\begin{aligned} f(t) & = \frac{E}{2} + \frac{E}{2} \left[ \cos({\omega t}) + \frac{4}{3\pi} \cos({2 \omega t}) - \frac{4}{15\pi} \cos({4 \omega t}) +\cdots \right] \\ \\ & = \frac{E}{\pi} - \frac{2E}{\pi} \sum_{n=1}^{\infty} \frac{1}{(n^{2}-1)} \cos\left( \frac{{n\pi}}{2} \right) \cos(n \omega t).\\ \end{aligned} \]

(v) Periodic full-wave cosine signals

The periodic full-wave cosine contains only the DC component and the even-ordered harmonic components, and the amplitude of the harmonics is given in terms of the$\displaystyle{\frac{1}{n^{2}}}$ Regular convergence .

\[\begin{aligned} f(t) & = \frac{2e}{\pi} + \frac{4E}{\pi} \left[ \frac{1}{3}\cos({2 \omega t}) - \frac{1}{15}\cos({4 \omega t}) + \frac{1}{35} \cos({6 \omega t}) \right]\\ \\ & = \frac{2E}{\pi} + \frac{4E}{\pi} \sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{(4n^{2} -1 )} \cos(2n \omega t)\\ \end{aligned} \]

For the above periodic function, the periodic rectangular pulse signal is one of the most important, needs to be scrutinized.

Fourier series examples and finite term approximations are errors introduced

In practical engineering, we can't do the Fourier series expansion to infinite terms, we can only expand to finite terms. How much error exists between the result calculated by the finite term and the real value is our concern.

The error is the sum of the infinite terms that follow:

\[\epsilon_{N}(t) = \sum_{n = N}^{\infty}[ a_{n} \cos(n\omega t) + b_{n} \sin(n \omega t)] \]

The square mean error is:

\[\begin{aligned} E_{N} = \overline{ \epsilon_{n}^2(t) } & = \frac{1}{T} \int_{t_{0}}^{t_{0} + T} \epsilon_{N}(t)^2 \text{d}t \\ & = \overline{f^2(t)} - \bigg[a_{0}^2 + \frac{1}{2} \sum_{n=1}^{N}(a_{n}^2 + b_{n}^2) \bigg]\\ \end{aligned} \]

In the following, we choose a periodic rectangular pulse signal with period 2 and amplitude 1.

ingredient	imagery
Expand to Kibo
Expand to 3rd harmonic
Expand to 5th harmonic
Expanded to 7th harmonic
Expanded to 9th harmonic
Expanded to 11th harmonic

By looking at the set of images above, we can find the following pattern:

The more times the Fourier series is taken, the closer the final waveform is to the original signal
High-frequency signal mainly affects the fast-changing part of the signal; if the signal is pulsed, the high-frequency signal mainly affects the pulse's jumping edge
Low-frequency signals mainly affect the slowly changing part of the signal; if the signal is a pulsed signal, the low-frequency signal mainly affects the top of the pulse.
When the amplitude or phase of any spectral component of the signal changes, the output waveform will generally be distorted, such as in image processing, we find that the amplitude of the spectrum of the two pictures remains unchanged, the phase spectrograms are interchanged and then superimposed, the image itself basically did not change, but the details are slightly different.

Another interesting but distressing thing we noticed. The periodic rectangular pulse signal has a small peak-up at the hopping edge, and this peak-up does not decrease significantly as more Fourier series terms are taken, no matter how many terms are taken.
Regarding such a problem, signal analysis refers to it as the$Gibbs$ impunity. The specifics are:
When performing finite term Fourier series approximations, as more Fourier series components or components are employed, the Fourier series approaches (complete) jumps in the approximate$9%$shows the first overshoot in the oscillatory behavior near the jump point of the oscillation, and the oscillation does not disappear but gets closer and closer to the point, bringing the oscillation integral close to zero (i.e., the oscillation energy is zero)。