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Overview of Fourier analysis
2022-07-05 22:37:00 【Xiao Qiu HUST】
Fourier transform is mainly divided into two parts: continuous and discrete . Analysis of continuous time signals , From the Fourier series of periodic signals (FS) Expand to the unified Fourier transform (FT), It is a complete system . Fourier analysis of discrete-time signals is very similar to that of continuous time signals , But it's really different , There is no unified expression , The main difference is “ Sum up ” and “ integral ” On .FS,FT,DFS,DTFT,DFT It forms the whole system of Fourier analysis .
No matter what kind of transformation , All satisfied with “ cycle - discrete ”,“ Aperiodic - continuity ” Correspondence of . This relationship is very useful for helping memory .
Analysis method | abbreviation | The transformation process Time domain signal → Frequency domain signal |
---|---|---|
Fourier series | FS | Continuous period → Aperiodic discrete |
The Fourier transform | FT | ( introduce δ ( ω ) \delta(\omega) δ(ω)) Continuous period → Aperiodic discrete Continuous aperiodic → Aperiodic continuous |
Discrete Fourier series | DFS | Discrete period → Periodic discretization |
Discrete time Fourier transform | DTFT | Discrete aperiodic → Periodic continuity |
Discrete Fourier transform | DFT | Discrete aperiodic → Discrete aperiodic ( The essence ) Discrete period → Periodic discretization |
Continuous time Fourier analysis
Fourier series
Continuous time analysis starts with Fourier series .
f ( t ) = f ( t + T 0 ) f ( t ) = ∑ n = − ∞ + ∞ a n e − j n ω 0 f(t) = f(t + {T_0})\;\;\;\;\;f(t) = \sum\limits_{n = - \infty }^{ + \infty } { {a_n}{e^{ - jn{\omega _0}}}} f(t)=f(t+T0)f(t)=n=−∞∑+∞ane−jnω0
Any period is T 0 T_0 T0( The frequency is f 0 f_0 f0, Angular frequency is ω 0 \omega_0 ω0) The periodic signal of , Can use angular frequency ω 0 \omega_0 ω0 Integer multiple complex exponential signal e − j n ω 0 e^{ - jn{\omega _0}} e−jnω0 Linear representation .
Such as cosine signal A c o s ( ω 0 t ) Acos(\omega_0t) Acos(ω0t) Fourier series expansion can be expressed as :
A c o s ( ω 0 t ) = A 2 ( e − j ω 0 t + e j ω 0 t ) Acos(\omega_0t)={A \over 2}\left( { {e^{ - j{\omega_0}t}} + {e^{j{\omega_0}t}}} \right) Acos(ω0t)=2A(e−jω0t+ejω0t)
The amplitude spectrum obtained by Fourier transform of real signal is always even symmetric , The phase spectrum is always odd symmetric . A cosine signal can be decomposed into two conjugate complex exponential signals , As shown in the figure below .
Over time t t t An increase in , They rotate constantly on the complex plane , But their vector sum always falls on the real axis , External performance is a real signal .
That's why “ Negative frequency ” The reason for its existence , there “ negative ” It means that the angular frequency of the complex exponential signal is negative , And “ Positive frequency ” The signal of is conjugate . Complex exponential signals that are conjugate with each other form a real signal .
The Fourier transform
Fourier series is limited to the analysis of periodic signals , For aperiodic signals , The period of the signal can be regarded as infinity , Based on this idea , It can be extended from the analytical synthesis formula of Fourier series to Fourier transform pair . The following is the analytical synthesis formula of Fourier series :
a n = 1 T 0 ∫ T 0 f ( t ) e − j n ω 0 t d t f ( t ) = ∑ n = − ∞ + ∞ a n e j n ω 0 {a_n} = {1 \over { {T_0}}}\int\limits_{ {T_0}} {f(t){e^{ - jn{\omega _0}t}}dt} \;\;\;\;f(t) = \sum\limits_{n = - \infty }^{ + \infty } { {a_n}{e^{jn{\omega _0}}}} an=T01T0∫f(t)e−jnω0tdtf(t)=n=−∞∑+∞anejnω0
a n a_n an Represents the Fourier series with the fundamental wave n n n Coefficient of complex exponential signal of subharmonic relationship . Write them together to get :
f ( t ) = ∑ n = − ∞ + ∞ ( 1 T 0 ∫ T 0 f ( t ) e − j n ω 0 t d t ) e j n ω 0 f(t) = \sum\limits_{n = - \infty }^{ + \infty } {\left( { {1 \over { {T_0}}}\int\limits_{ {T_0}} {f(t){e^{ - jn{\omega _0}t}}dt} } \right){e^{jn{\omega _0}}}} \; f(t)=n=−∞∑+∞⎝⎛T01T0∫f(t)e−jnω0tdt⎠⎞ejnω0
= ∫ − ∞ + ∞ ( ∫ − ∞ + ∞ f ( t ) e − j n ω 0 t d t ) e j n ω 0 d f = \int_{ - \infty }^{ + \infty } {\left( {\int_{ - \infty }^{ + \infty } {f(t){e^{ - jn{\omega _0}t}}dt} } \right){e^{jn{\omega _0}}}df} =∫−∞+∞(∫−∞+∞f(t)e−jnω0tdt)ejnω0df
When T 0 T_0 T0 Towards infinity , f 0 f_0 f0 Tends to infinity , The original sum needs to be changed to integral , The differential variable is from the original 1 / T 0 1/T_0 1/T0 Transformed d f df df.
= 1 2 π ∫ − ∞ + ∞ ( ∫ − ∞ + ∞ f ( t ) e − j n ω 0 t d t ) e j n ω 0 d ω = {1 \over {2\pi }}\int_{ - \infty }^{ + \infty } {\left( {\int_{ - \infty }^{ + \infty } {f(t){e^{ - jn{\omega _0}t}}dt} } \right){e^{jn{\omega _0}}}d\omega } =2π1∫−∞+∞(∫−∞+∞f(t)e−jnω0tdt)ejnω0dω
d f df df To d ω d\omega dω It needs to be multiplied by 2 π 2\pi 2π, So there is one in the synthesis of the last Fourier transform 1 / 2 π 1/2\pi 1/2π The coefficient of .
X ( j ω ) = ∫ − ∞ + ∞ f ( t ) e − j ω t d t f ( t ) = 1 2 π ∫ − ∞ + ∞ X ( j ω ) e j ω t d ω X(j\omega ) = \int_{ - \infty }^{ + \infty } {f(t){e^{ - j\omega t}}dt} \;\;\;\;\;\;f(t) = {1 \over {2\pi }}\int_{ - \infty }^{ + \infty } {X(j\omega ){e^{j\omega t}}d\omega } X(jω)=∫−∞+∞f(t)e−jωtdtf(t)=2π1∫−∞+∞X(jω)ejωtdω
Unity of periodic and aperiodic signals
After introducing the impulse signal δ ( t ) \delta(t) δ(t) after , The Fourier transform of periodic signal and aperiodic signal is unified . The problem of Fourier transform of periodic signals is that the analytical formula of Fourier transform is an integral in infinite range , Periodic signals cannot converge when doing such integrals .
Periodic signals can be decomposed into the sum of multiple complex exponential signals . So as long as we can express the Fourier transform of complex exponential signal, we can get the Fourier transform of periodic signal . Combined with the frequency shift property of Fourier transform :
e j ω 0 t x ( t ) ⇔ X ( j ( ω − ω 0 ) ) {e^{j{\omega _0}t}}x(t) \Leftrightarrow X\left( {j\left( {\omega - {\omega _0}} \right)} \right) ejω0tx(t)⇔X(j(ω−ω0))
We just need to know "1" The Fourier transform of the complex exponential signal can be obtained by the Fourier transform of . The specific solution process has been written , article A relatively complete derivation process is given in , I won't go into details here , The result is the following :
1 ⇔ 2 π δ ( ω ) 1 \Leftrightarrow 2\pi \delta \left( \omega \right) 1⇔2πδ(ω)
This result is also quite natural , hold 2 π δ ( ω ) 2\pi \delta \left( \omega \right) 2πδ(ω) It is obvious that it can be established by substituting it into the Fourier transform synthesis .
F ( A cos ( ω 0 t ) ) = A 2 ⋅ 2 π δ ( ω − ω 0 ) + A 2 ⋅ 2 π δ ( ω + ω 0 ) {\mathcal F}\left( {A\cos ({\omega _0}t)} \right) = {A \over 2} \cdot 2\pi \delta \left( {\omega - {\omega _0}} \right) + {A \over 2} \cdot 2\pi \delta \left( {\omega + {\omega _0}} \right) F(Acos(ω0t))=2A⋅2πδ(ω−ω0)+2A⋅2πδ(ω+ω0)
Therefore, the result of Fourier transform of periodic signal will include 2 π δ ( ω − ω ′ ) 2\pi \delta \left( {\omega - {\omega'}} \right) 2πδ(ω−ω′) Impulse signal of , And the coefficients in front of each impulse signal are exactly the coefficients of the corresponding Fourier series .
Discrete time Fourier transform
Discrete time Fourier transform and continuous time Fourier transform have many similarities , But they are two completely different analysis systems , The biggest difference is from “ integral ” Change into “ Sum up ”.
Discrete Fourier series
The essence of Fourier transform is orthogonal decomposition of signal in frequency domain . In the case of continuous time signals , For a frequency of ω 0 \omega_0 ω0 For periodic signals , Is to project it onto ω 0 \omega_0 ω0、 2 ω 0 2\omega_0 2ω0、 3 ω 0 3\omega_0 3ω0 Wait for these frequencies that have a harmonic relationship with the fundamental frequency .
In the case of continuity , Harmonics are infinite . And in discrete cases , The period is N N N A sequence of numbers , The fundamental frequency can be expressed as 2 π / N 2\pi/N 2π/N, Due to the periodicity of complex exponential signals :
e − j 2 π / N = e − j 2 π ( N + 1 ) / N {e^{ - j2\pi /N}} = {e^{ - j2\pi (N + 1)/N}} e−j2π/N=e−j2π(N+1)/N
So in the case of discrete , One cycle is N Sequence , It can only be decomposed into N Different complex exponential sequences .
X ~ [ k ] = ∑ n = 0 N − 1 x ~ [ n ] W N k n x ~ [ n ] = 1 N ∑ n = 0 N − 1 X ~ [ k ] W N − k n \widetilde X[k] = \sum\limits_{n = 0}^{N - 1} {\widetilde x[n]W_N^{kn}} \;\;\;\;\widetilde x[n] = {1 \over N}\sum\limits_{n = 0}^{N - 1} {\widetilde X[k]W_N^{ - kn}} X[k]=n=0∑N−1x[n]WNknx[n]=N1n=0∑N−1X[k]WN−kn
The above is the analytical formula and synthetic formula of discrete Fourier series , W N k n W_N^{kn} WNkn It's right e − j 2 π k n N e^{-j2\pi\frac{kn}{N}} e−j2πNkn A short note of , Also called rotation factor .
Discrete Fourier series , Both the digital sequence in time domain and the complex exponential sequence in frequency domain are periodic , x ~ [ n ] \widetilde x[n] x[n] and X ~ [ n ] \widetilde X[n] X[n] The wavy lines on the indicate that they are periodic signals .
Discrete Fourier transform
First, the analytical formula and synthetic formula of discrete Fourier transform are given directly :
X [ k ] = ∑ n = 0 N − 1 x [ n ] W N k n , 0 ≤ k ≤ N − 1 X[k] = \sum\limits_{n = 0}^{N - 1} {x[n]W_N^{kn}} ,\;\;0 \le k \le N - 1 X[k]=n=0∑N−1x[n]WNkn,0≤k≤N−1
x [ n ] = 1 N ∑ n = 0 N − 1 X [ k ] W N − k n , 0 ≤ n ≤ N − 1 x[n] = {1 \over N}\sum\limits_{n = 0}^{N - 1} {X[k]W_N^{ - kn}} ,\;\;0 \le n \le N - 1 x[n]=N1n=0∑N−1X[k]WN−kn,0≤n≤N−1
Discrete Fourier transform is also used to deal with the analysis of aperiodic sequences , Here we simply regard aperiodic sequence as periodic sequence , It is equivalent to intercepting a sequence of period length from discrete Fourier series to express . This is it. DFT Inherent periodicity .
Discrete time Fourier transform
The textbook says DFT I also mentioned before DTFT, This is the orthodox method of analyzing the spectrum of finite length sequences . It is also through discrete Fourier series , Make N Going to infinity . Here is DTFT Analytical formula and synthetic formula of :
x [ n ] = 1 2 π ∫ 2 π X ( e j ω ) e j ω n d ω X ( e j ω ) = ∑ n = − ∞ + ∞ x [ n ] e − j ω n x[n] = {1 \over {2\pi }}\int\limits_{2\pi } {X\left( { {e^{j\omega }}} \right){e^{j\omega n}}d\omega } \;\;\;\;X\left( { {e^{j\omega }}} \right) = \sum\nolimits_{n = - \infty }^{ + \infty } {x[n]{e^{ - j\omega n}}} x[n]=2π12π∫X(ejω)ejωndωX(ejω)=∑n=−∞+∞x[n]e−jωn
DTFT The spectrum obtained is a periodic continuous spectrum , And it is based on 2 π 2\pi 2π For cycles , It looks like this :
If we take this finite sequence according to its length N N N Carry out periodic continuation , It is equivalent to combining it with a cycle as N N N Impulse string of ( A sequence of unit impulse functions ) Convolution . Time domain convolution is equivalent to frequency domain multiplication . The period of time domain is N N N In the frequency domain, the period of the impulse string is 2 π / N 2\pi/N 2π/N Impulse string of . therefore Finite sequence DFT Equivalent to DTFT Sampling on the spectrum of .
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