Then the Fourier Transform will continue to be described later (although the Fourier Transform is very messy), discuss the relationship between the Laplace Transform and the Fourier Transform
It is known that the Fourier transform is based on the Fourier integral. In addition to satisfying the Dirichlet condition, a function must also be absolutely integrable in the interval ( -∞ , + ∞ ), that is, the value of the integral cannot be equal to infinity Big.
Absolute integrability is a very strong condition, and some very simple functions (such as linear functions, sine and cosine functions, etc.) do not satisfy this condition, so the Fourier transform has the following two defects
One: After the introduction of the delta function, the scope of the Fourier transform has been widened a lot, so that the "slowly increasing" function can also perform the Fourier transform, but it is still powerless for the exponential and growth functions.
Two: The Fourier transform must be defined on the entire real axis, but in actual engineering, there is no concept of time t <0, usually starting from t = 0, only the part corresponding to t> 0 function.
Assuming that there is a function f (t) that satisfies the conditions of Fourier transform, then there is Fourier transform
$$
\mathscr{L}\left[ f\left( t \right) \right] =\int_{-\infty}^{+\infty}{f\left( t \right) e^{-jwt}dt}\,\, \text{式}1
$$
In order to solve the above two defects, the Fourier transform can be processed as follows:
To solve the first problem, we can multiply the function f (t) by an attenuation factor (a very small fraction) e -βt , and obtain f (t) e -βt .
To solve the second problem, we can multiply the function f (t) by a unit step function u (t). When t <0, u (t) = 0, and when t> 0, u (t) = 1.
In summary, f (t) u (t) e -βt can be obtained , and then the Fourier transform of f (t) u (t) e -βt can be obtained:
$$
\ mathscr {L} \ left [f \ left (t \ right) \ right] = \ int _ {-\ infty} ^ {+ \ infty} {f \ left (t \ right) u \ left (t \ right) e ^ {-\ beta t} e ^ {-jwt} dt}
\\
\ text {Because the unit step function is multiplied} u \ left (t \ right) \ text {, you can divide it into} Two \ text {partial calculation}
\\
\, \, \ int_0 ^ {+ \ infty} {f \ left (t \ right) * 1 * e ^ {-\ beta t} e ^ {-jwt} dt} = \ int_0 ^ {+ \ infty} {f \ left (t \ right) e ^ {-\ left (\ beta + jw \ right) t} dt}
\\
\, \, \ int _ {-\ infty} ^ 0 {f \ left (t \ right) * 0 * e ^ {-\ beta t} e ^ {-jwt} dt} = 0
\\
\ text {due to (} 0 \ text {,}-\ infty \ text {）} District \ text {integral between points} = 0 \ text {, so the Fourier transform of} f \ left (t \ right) \ text {can be simplified to \ text {as follows}
\ \
\ mathscr {L} \ left [f \ left (t \ right) \ right] = \ int_0 ^ {+ \ infty} {f \ left (t \ right) e ^ {-\ left (\ beta + jw \ right) t} dt}
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\ text {令}-\ left (\ beta + jw \ right) = s \, \, \ text {you can get the Laplace transform formula of} f \ left (t \ right) \ text {}
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F \ left (s \ right) = \ int_0 ^ {+ \ infty} {f \ left (t \ right) e ^ {-st} dt}
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\ text {称} F \ left (s \ right) is Laplace transform of f \ left (t \ right) \ text {, remember} as F \ left (s \ right) = \ mathscr {L} \ left [f \ left (t \ right) \ right] \ text {. }
$$
The above is the relationship between Laplace transform formula and Laplace transform and Fourier transform.
Reference video: https://www.bilibili.com/video/BV16x411M7HR