Article directory
1. Write code
2. Write the copy
3. Write the screenplay
4. Write poetry
5. Write a report
This is a proper translation, the data is completely wrong.
6. Check the formula
The time-domain properties of the Fourier transform are as follows:
Symmetry: F ( f ( t ) ) = F ( f ( − t ) ) ∗ \mathcal{F}(f(t)) = \mathcal{F}(f(-t))^*F(f(t))=F(f(−t))∗
Circular convolution: F ( f ( t ) ∗ g ( t ) ) = F ( f ( t ) ) ⋅ F ( g ( t ) ) \mathcal{F}(f(t)\ast g(t)) = \mathcal{F}(f(t)) \cdot \mathcal{F}(g(t))F(f(t)∗g(t))=F(f(t))⋅F(g(t))
Time translation: F ( f ( t − t 0 ) ) = e − j 2 π f 0 t 0 F ( f ( t ) ) \mathcal{F}(f(t-t_0)) = e^{-j2\ pi f_0t_0} \mathcal{F}(f(t))F(f(t−t0))=e−j2πf0t0F(f(t))
Time expansion: F ( α f ( t ) ) = 1 ∣ α ∣ F ( f ( t α ) ) \mathcal{F}(\alpha f(t)) = \frac{1}{|\alpha|} \ mathcal{F}(f(\frac{t}{\alpha}))F(αf(t))=∣α∣1F(f(at))
Time reversal: F ( f ( − t ) ) = F ( f ( t ) ) ∗ \mathcal{F}(f(-t)) = \mathcal{F}(f(t))^*F(f(−t))=F(f(t))∗
The above is the time domain property of Fourier transform, where, F \mathcal{F}F means Fourier transform,f ( t ) f(t)f ( t ) andg ( t ) g(t)g ( t ) is the time domain signal,∗ \ast∗ means circular convolution,∗ ^*∗ means complex conjugate,f 0 f_0f0is the frequency of the signal, t 0 t_0t0is the amount of time translation, α \alphaα is the amount of time expansion.
Let the original two Gaussian distributions bef ( x ) f(x)f ( x ) andg ( x ) g(x)g ( x ) , the average value of each distribution isμ 1 \mu_1m1和μ 2 \mu_2m2, and the standard deviations are σ 1 \sigma_1p1and σ 2 \sigma_2p2.
Then the result of adding two Gaussian distributions is:
f ( x ) + g ( x ) = 1 2 π σ 1 e − ( x − μ 1 ) 2 2 σ 1 2 + 1 2 π σ 2 e − ( x − μ 2 ) 2 2 σ 2 2 f(x) + g(x) = \frac{1}{\sqrt{2\pi}\sigma_1}e^{-\frac{(x-\mu_1)^2 }{2\sigma_1^2}} + \frac{1}{\sqrt{2\pi}\sigma_2}e^{-\frac{(x-\mu_2)^2}{2\sigma_2^2}}f(x)+g(x)=2 p.mp11e−2 p12( x − m1)2+2 p.mp21e−2 p22( x − m2)2Through simple mathematical calculations, the average value μ \mu of this result is obtainedμ and standard deviationσ \sigmaσ for:
μ = μ 1 σ 2 2 + μ 2 σ 1 2 σ 1 2 + σ 2 2 \mu = \frac{\mu_1\sigma_2^2 + \mu_2\sigma_1^2}{\sigma_1^2 + \ sigma_2^2}m=p12+p22m1p22+m2p12 σ = σ 1 2 σ 2 2 σ 1 2 + σ 2 2 \sigma = \sqrt{\frac{\sigma_1^2\sigma_2^2}{\sigma_1^2 + \sigma_2^2}} p=p12+p22p12p22
with f ( t ) f(t)f ( t ) represents the gate function, useg ( t ) g(t)g ( t ) represents the Singh function, and the convolution result is: ( f ∗ g ) ( t ) = ∫ − ∞ ∞ f ( τ ) g ( t − τ ) d τ (f*g)(t)=\ int_{-\infty}^\infty f(\tau)g(t-\tau)d\tau(f∗g)(t)=∫−∞∞f ( τ ) g ( t−τ ) d τ due tof ( t ) f(t)f ( t ) is a gate function, which has the following form: f ( t ) = { 1 , t ∈ [ − 1 2 , 1 2 ] 0 , t ∉ [ − 1 2 , 1 2 ] f(t)=\ begin{cases} 1, & t\in[-\frac{1}{2},\frac{1}{2}] \ 0, & t\notin[-\frac{1}{2},\frac {1}{2}] \end{cases}f(t)={
1,t∈[−21,21] 0,t∈/[−21,21] g ( t ) g(t) g ( t ) is a sinc function which has the following form: g ( t ) = sinc ( t ) = sin ( π t ) π tg(t)=\text{sinc}(t)=\frac{\ sin(\pi t)}{\pi t}g(t)=sinc(t)=t _sin ( t ) _Substitute into the result of convolution: ( f ∗ g ) ( t ) = ∫ − 1 2 1 2 sin ( π ( t − τ ) ) π ( t − τ ) d τ (f*g)(t)=\ int_{-\frac{1}{2}}^{\frac{1}{2}}\frac{\sin(\pi (t-\tau))}{\pi (t-\tau)}d \tau(f∗g)(t)=∫−2121π ( t−t )sin ( π ( t−t ) )d τ forsin \sinThe periodicity of the sin function can be obtained: ( f ∗ g ) ( t ) = 1 π ∫ − 1 2 1 2 sin ( π t ) t − τ d τ (f*g)(t)=\frac{1 }{\pi}\int_{-\frac{1}{2}}^{\frac{1}{2}}\frac{\sin(\pi t)}{t-\tau}d\tau(f∗g)(t)=Pi1∫−2121t−tsin ( t ) _The above d τ is the result of convolution, and the specific value can be calculated by other methods, such as variable transformation or calculation of mathematical formulas.
7. Write couplets
8. Write articles
One problem with writing articles is that no matter how many words are generated in English, you can give them directly, but Chinese is very limited.
9. Make a table
10. Make a plan
