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alibaig1991 January 26, 2018 09:17
Why we use finite number of Terms in Fourier Decomposition of Error in Von Neumann St
 
In Fourier decomposition of error in Von Neumann stability analysis, we assume that the domain is of finite length, and therefore, we use discrete Fourier representation. My question is why we use finite number of harmonics instead of infinite number of harmonics?

FMDenaro January 26, 2018 10:42
Quote:
Originally Posted by alibaig1991 (Post 679511)
In Fourier decomposition of error in Von Neumann stability analysis, we assume that the domain is of finite length, and therefore, we use discrete Fourier representation. My question is why we use finite number of harmonics instead of infinite number of harmonics?

The answer is twofold.
1) the finite lenght L, once subdivided by a finite number of N step of size h, introduces the Nyquist cut-off frequency Kc=L/2h. No Fourier components are represented beyond Kc.

2) The von Neumann analysis is linear, therefore we introduce one generic components having wavenumber ranging from the lowest to the highest represented.

alibaig1991 January 26, 2018 14:46
Quote:
Originally Posted by FMDenaro (Post 679522)
The answer is twofold.
1) the finite lenght L, once subdivided by a finite number of N step of size h, introduces the Nyquist cut-off frequency Kc=L/2h. No Fourier components are represented beyond Kc.

2) The von Neumann analysis is linear, therefore we introduce one generic components having wavenumber ranging from the lowest to the highest represented.
Thank you Sir.


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