“This is the sixth day of my participation in the First Challenge 2022. For details: First Challenge 2022”

A, introducing

1.1 Basic idea of signal decomposition

One of the basic ideas of signal analysis is to represent complex signals with basic signals so that complex signals can be analyzed through the properties of simple signals. The basic signals are required to have: (1) a fairly broad class of signals composed of these basic signals; (2) the response of the linear time-invariant system (LTIS) to the basic signals should be very simple, so that it can have a convenient expression for the response of any input signal

For example, the basic signal can be an impulse function


x ( t ) = ∫ x ( tau ) Delta t. ( t – tau ) d tau x\left( t \right) =\int{x\left( \tau \right)}\delta \left( t-\tau \right) d\tau

The above equation is the impulse decomposition of continuous time signal, using the sampling property of the impulse function

After the signal is impulse decomposed, the impulse response can simply be used to represent the output:


y ( t ) = ∫ x ( tau ) h ( t – tau ) d tau y\left( t \right) =\int{x\left( \tau \right)}h\left( t-\tau \right) d\tau

1.2 Characteristic function of the system

If the response of the system to a signal is only a constant multiplied by the signal, then the signal is said to be an eigenfunction of the system. This constant can be regarded as an amplitude factor and defined as an eigenvalue

1.3 Decomposition of complex exponentials

Based on the idea of 1.1, we investigate the basic signals with similar properties — complex exponentials with excitation x(t)=est S ∈Cx\left(t \right) = E ^{st}\,\, s\in Cx(t)=ests∈C, then the output is expressed as follows:


y ( t ) = ∫ h ( tau ) x ( t – tau ) d tau = e s t ∫ h ( tau ) e – s tau d tau = e s t H ( s ) y\left( t \right) =\int{h\left( \tau \right)}x\left( t-\tau \right) d\tau =e^{st}\int{h\left( \tau \right) e^{-s\tau}}d\tau =e^{st}H\left( s \right)

According to the definition of 1.2, the complex exponential signal is the characteristic function of LTIS. For any given SSS, the constant H(s)H(s)H(s) is the eigenvalue. For general SSS, H(s)H(s)H(s) is a function of SSS, called the system function. When SSS is a pure imaginary number, H(Jw)H(jw)H(jw) H(jw) is called the system frequency response. The Following Fourier analysis to be introduced is based on s= JWS = JWS =jw. When SSS is a general complex number, we consider the Laplace transform, which we will not repeat in this article.

Does the complex exponential satisfy the basic signal requirements of Section 1.1?

X (t)=∑ Kckesktx \left(t \right) = sum_k{c_k}e^{s_kt}x(t)=∑kckeskt, Then the response y(t)=∑kckH(sk)eskty\left(t \right) = sum_k{c_kH\left(s_k \right)}e^{s_kt}y(t)=∑kckH(sk)eskt. The response can be expressed as a linear combination of the same complex exponentials with coefficients related to the input and frequency response

It can be seen that the complex exponential signal perfectly meets the requirements of 1.1, on which the Fourier analysis is established.

Fourier series of periodic signals

2.1 Harmonic complex exponential set

Let x0(t)=ejω 0TX_0 \left(t \right) =e^{j\omega _0t}x0(t)=ejω0t, define the fundamental frequency as ω0\omega _0ω0, the fundamental period as T0T_0T0, let the harmonic signal set:


Bits of k ( t ) = e j k Omega. 0 t . k = 0 . Plus or minus 1 . Plus or minus 2 … \psi _k\left( t \right) =e^{jk\omega _0t}, k=0,\pm 1,\pm 2\cdots

Where k=0k=0k=0 is dc component, k=±Nk=\ PM Nk=±N is NNN sub-harmonic component. Note that in the concentration of harmonic complex exponentials, each signal can be T0T_0T0 periodic, because:


Bits of k ( t ) = e j ( k Omega. 0 ) t ⇒ T k = 2 PI. ∣ k ∣ Omega. 0 = T 0 ∣ k ∣ \psi _k\left( t \right) =e^{j\left( k\omega _0 \right) t}\\\Rightarrow T_k=\frac{2\pi}{|k|\omega _0}=\frac{T_0}{|k|}

Namely every T0T_0T0, equivalent to passed ∣ k ∣ | | k ∣ k ∣ a corresponding harmonic cycle. Therefore, the linear combination of harmonic complex exponentials is periodic T0T_0T0:


x ( t ) = ∑ k = – up + up a k e j k Omega. 0 t     ( 1 ) x\left( t \right) =\sum_{k=-\infty}^{+\infty}{a_k}e^{jk\omega _0t}\,\, \left( 1 \right)

2.2 Fourier series

2.2.1 Representation

This problem is studied under the condition that the signal x(t)x(t)x(t) can be decomposed by complex exponentials. By examining Equation (1), the vast majority of signals in reality are real signals, so x(t)x(t)x(t) is considered to be a real number, which satisfies:


x ( t ) ‾ = ∑ k = – up + up a k e j k Omega. 0 t ‾ \overline{x\left( t \right) }=\overline{\sum_{k=-\infty}^{+\infty}{a_k}e^{jk\omega _0t}}

By x (t) = x (t) ‾ x \ left (t \ right) = \ overline \ {x left (t \ right)} x (t) = x (t) derived:


x ( t ) = ∑ k = – up + up a – k ‾ e j k Omega. 0 t x\left( t \right) =\sum_{k=-\infty}^{+\infty}{\overline{a_{-k}}e^{jk\omega _0t}}

Further:


x ( t ) = a 0 + ∑ k = 1 + up [ a k e j ( k Omega. 0 ) t + a k e j ( k Omega. 0 ) t ‾ ] = a 0 + 2 ∑ k = 1 + up Re { a k e j ( k Omega. 0 ) t } x\left( t \right) =a_0+\sum_{k=1}^{+\infty}{\left[ a_ke^{j\left( k\omega _0 \right) t}+\overline{a_ke^{j\left( k\omega _0 \right) t}} \right]}\\=a_0+2\sum_{k=1}^{+\infty}{\text{Re}\left\{ a_ke^{j\left( k\omega _0 \right) t} \right\}}
  1. If aka_kak is given in polar form, that is, ak=Akejw0a_k=A_ke^{jw_0}ak=Akejw0, then

x ( t ) = a 0 + 2 ∑ k = 1 + up A k cos ⁡ [ ( k Omega. 0 ) t + Theta. k ]       ( 2 ) x\left( t \right) =a_0+2\sum_{k=1}^{+\infty}{A_k\cos \left[ \left( k\omega _0 \right) t+\theta _k \right]}\,\,\,\left( 2 \right)
  1. If aka_kak is given in cartesian coordinates, that is, ak=Bk+jCka_k=B_k+jC_kak=Bk+jCk, then

x ( t ) = a 0 + 2 ∑ k = 1 + up [ B k cos ⁡ ( k Omega. 0 t ) – C k sin ⁡ ( k Omega. 0 t ) ]       ( 3 ) x\left( t \right) =a_0+2\sum_{k=1}^{+\infty}{\left[ B_k\cos \left( k\omega _0t \right) -C_k\sin \left( k\omega _0t \right) \right]}\,\,\,\left( 3 \right)

For periodic functions, eq. (1) is the complex exponential form of Fourier series; Equation (2) is the triangular form of Fourier series (polar coordinates); Equation (3) is the trigonometric form of the Fourier series (in Cartesian coordinates). In general, if a signal can be expanded into a Fourier series, its representation must be one of (1), (2) and (3)

2.2.2 Convergence conditions

Not all periodic signals can be expanded by series, that is, not all signals can be decomposed by complex exponentials. In general, signals that meet the Dirchlet condition must be subject to Fourier analysis. Signals that do not meet the Dirchlet condition do not have Fourier series form, but may have Fourier transforms.

Dirchlet condition (1) The signal is absolutely integrable (2) in any finite interval, the signal has only a finite number of maxima (3) in any finite interval, the signal has only a finite number of discontinuities, and each of these discontinuities has only a finite number of values

2.2.3 Fourier coefficients

If the signal satisfies the Dirchlet condition, the complex exponential can be decomposed into:


x ( t ) = ∑ k = – up + up a k e j k Omega. 0 t x\left( t \right) =\sum_{k=-\infty}^{+\infty}{a_k}e^{jk\omega _0t}

Now the problem lies in the determination of the Fourier coefficient AKa_kak, which can be obtained in the following way:


e – j ( n Omega. 0 ) t x ( t ) = ∑ k = – up + up a k e j ( k – n ) Omega. 0 t     e^{-j\left( n\omega _0 \right) t}x\left( t \right) =\sum_{k=-\infty}^{+\infty}{a_k}e^{j\left( k-n \right) \omega _0t}\,\,

Integrate both sides in the fundamental period:


∫ T e – j ( n Omega. 0 ) t x ( t ) d t = ∑ k = – up + up ∫ T a k e j ( k – n ) Omega. 0 t d t     ⇒ ∫ T e – j ( n Omega. 0 ) t x ( t ) d t = ∑ k = – up + up a k ∫ T [ cos ⁡ ( k – n ) w 0 t + j sin ⁡ ( k – n ) w 0 t ] d t     \int_T{e^{-j\left( n\omega _0 \right) t}x\left( t \right)}dt=\sum_{k=-\infty}^{+\infty}{\int_T{a_ke^{j\left( k-n \right) \omega _0t}dt}}\,\,\\\Rightarrow \int_T{e^{-j\left( n\omega _0 \right) t}x\left( t \right)}dt=\sum_{k=-\infty}^{+\infty}{a_k\int_T{\left[ \cos \left( k-n \right) w_0t+j\sin \left( k-n \right) w_0t \right] dt}}\,\,

Section 2.1 said that harmonic complex index set common cycle is a fundamental cycle, and trigonometric function integral of one cycle is 0, where T = ∣ k – n ∣ TkT = | k – n | T_kT = ∣ k – n ∣ Tk, and so on the left side of the equation in k indicates nk \ ne nk  = 0 n, k = nk nk = = n for TTT, namely:


∫ T e – j ( n Omega. 0 ) t x ( t ) d t = a n T \int_T{e^{-j\left( n\omega _0 \right) t}x\left( t \right)}dt=a_nT

Since k=nk=nk=n, so rewrite as:


a k = 1 T ∫ T e – j ( k Omega. 0 ) t x ( t ) d t     ( 4 ) a_k=\frac{1}{T}\int_T{e^{-j\left( k\omega _0 \right) t}x\left( t \right)}dt\,\, \left( 4 \right)

This is the formula for solving the Fourier coefficient.

Fourier transform

3.1 Periodic rectangular pulse signal

According to Equation (4), its Fourier coefficient is solved, and:


a k = 2 E sin ⁡ ( k Omega. 0 T 1 ) k Omega. 0 T a_k=\frac{2E\sin \left( k\omega _0T_1 \right)}{k\omega _0T}

Look at this from two perspectives:

  1. Think of it as a function of KKK, i.e

a ( k ) = 2 E sin ⁡ ( k Omega. 0 T 1 ) k Omega. 0 T a(k)=\frac{2E\sin \left( k\omega _0T_1 \right)}{k\omega _0T}

At this time of the Fourier coefficient of equidistance is arranged on the KKK shaft, so when the TTT tends to infinity, ∣ ak ∣ | a_k | ∣ ak ∣ tends to zero, namely the aperiodic signal amplitude of Fourier coefficient tends to zero, because of this, just don’t see any information in the amplitude spectrum, so for the non periodic signal, not only pay attention to aka_kak

  1. Think of it as a sampling of the envelope

At this point, it is regarded as:


a k T = 2 E sin ⁡ ( Omega. T 1 ) Omega. ∣ w = k w 0 a_kT=\frac{2E\sin \left( \omega T_1 \right)}{\omega}\mid_{w=kw_0}^{}

Consider the function f(w)=2Esin⁡(ωT1)ωf\left(w \right) =\frac{2E\sin \left(\omega T_1 \right)}{\omega}f(w)=ω2Esin(ωT1), AkTa_kTakT means to sample the position w=kw0w=kw_0w=kw0 on f(w)f(w). Obviously, the above sampling interval is w0=2π/Tw_0=2\ PI /Tw0=2π/T, so with increasing TTT, the phenomenon of sampling densification appears in Figure 2(a)->(c)

The key understanding place comes!!

Note here TTT tends to infinity, ∣ ak ∣ | a_k | ∣ ak ∣ still tends to zero, but it is visible ∣ ak ∣ T | a_k ∣ ak ∣ T | T is limited value (in f (w) (w) f f (w)), Therefore ∣ ak ∣ T | a_k | T ∣ ak ∣ is the significance of T in ∣ ak ∣ | a_k | ∣ ak ∣ tends to 0, through the role of the weighted T, in a limited range shows ∣ ak ∣ | a_k | the relative size of the relationship between ∣ ak ∣, in short, ∣ ak ∣ T | a_k ∣ ak ∣ | T T the invisible aperiodic signal amplification of Fourier coefficient to the visible to the naked eye, this is the introduction of the Fourier transform.

3.2 Fourier transform pairs

We know from section 3.1 that the starting point of Fourier transform is amplitude weighting and sampling of the Fourier coefficients, so X(W)∣ W =kw0=akTX\left(W \right) \mid_{w=kw_0}^{}=a_kTX(W)∣ W =kw0=akT

Thus,


X ( w ) = ∫ – up + up x ( t ) e – j w t d t     ( 5 ) X\left( w \right) =\int\limits_{-\infty}^{+\infty}{x\left( t \right) e^{-jwt}}dt\,\, \left( 5 \right)

Generation x (t) = ∑ k = – up + up akejk omega zero tx \ left (t \ right) = \ sum_ {k = – \ infty} ^ {+ \ infty} {a_k} e ^ {jk \ omega _0t} x (t) = ∑ k = – up + up akejk omega zero t in a quick:


x ( t ) = 1 2 PI. ∫ – up + up X ( w ) e j w t d w     ( 6 ) x\left( t \right) =\frac{1}{2\pi}\int\limits_{-\infty}^{+\infty}{X\left( w \right) e^{jwt}}dw\,\, \left( 6 \right)

Equations (5) and (6) are collectively called a pair of Fourier transform pairs, and equations (5) are called the Integral of the Fourier transform

Four, the connection between Fourier series and Fourier transform

4.1 Signal three parameters

Here, the three parameters of the signal are defined as amplitude, initial phase and frequency (or angular frequency). In Fourier analysis, as long as the three parameters of all the complex exponential signals that comprise the signal are determined, they can be fully characterized. Whether it’s a Fourier series or a Fourier transform, it’s really finding a three-parameter expression to represent a signal.

In Fourier series expansion, the Fourier coefficient represents the amplitude and phase of the complex exponential signal at the frequency w=kw0w=kw_0w=kw0; In the Fourier transform, the Fourier integral X(w)X(w)X(w) represents the three-parameter information of a full-frequency complex exponential signal — the spectrum can be considered formulaic

Specifically, it is listed in the following table:

amplitude phase frequency

a k a_k 		The absolute value The absolute value

X ( w ) X(w) 		Weighted relative value The absolute value The absolute value

In fact, you shouldn’t separate the Fourier transform from the Fourier series in terms of whether the signal is periodic or not. In other words, both the periodic signal and the aperiodic signal have corresponding Fourier transforms and Fourier coefficients, except that the Fourier transform of the periodic signal is a linear combination of impulse functions, and the Fourier coefficients of the aperiodic signal tend to zero, but have a relative magnitude.

4.2 Geometric Intuition

Geometrically, the Fourier transform is a continuous function, because its object is full frequency; The Fourier series is discrete because its object is the partial frequency of the sample.

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