| Transfer Functions |
The meaning of s
- In Fourier transform the complementary variable usually has a clear physical meaning.
E.g. if working in time t? or f
Diffraction in optics, where FTs are used, has a similar relationship between spatial distributions and spatial frequency
- although laplace transform look very similar (and many result can be easily obtained by following methods for deriving FTs ), the complementary variable s does not have the same physical significance.
It is a mathematical method of solving problems suing transforms
- since we spent a significant time on the ft, I will not spend so much time on the details of derving LTs
Integrals are usually straightforward
I will discuss only transformer will need here
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