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Why we use finite number of Terms in Fourier Decomposition of Error in Von Neumann St

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Old   January 26, 2018, 09:17
Default Why we use finite number of Terms in Fourier Decomposition of Error in Von Neumann St
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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?
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Old   January 26, 2018, 10:42
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Originally Posted by alibaig1991 View Post
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.
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Old   January 26, 2018, 14:46
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Originally Posted by FMDenaro View Post
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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