Why should we use only dyadic length signal in discrete wavelet transform?
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Why should the signal length be dyadic while taking discrete wavelet transform? What happens if the signal length is not dyadic. I am working on compressing real time sensor signals containing 2000 samples in 20ms duration.Can some one help me with valuable answers. Many thanks in advance! my mail id is charanchakravarthy@gmail.com
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Qudus Thanni
2024년 5월 28일
You need to have your data in the nearest power of two for it to have a dyadic length
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Prasanna
2024년 9월 16일 5:30
Hi,
When performing a Discrete Wavelet Transform (DWT), having a dyadic length (a length that is a power of two) for the signal simplifies the computation and ensures efficient decomposition and reconstruction. This is due to the following factors:
- Effective decomposition: The DWT algorithm typically works by recursively decomposing the signal into approximation and detail coefficients. If the signal length is a power of 2, it allows for efficient computation and splitting at each level of decomposition.
- Orthogonality and symmetry: dyadic lengths ensures that the wavelet coefficients accurately represent the signal without redundancy.
- Boundary effects: Non-dyadic lengths can introduce boundary effects, where the edges of the signal do not align well with the wavelet functions. This can lead to inaccuracies in the transformed coefficients and, consequently, in the reconstructed signal.
If the signal length is not dyadic, you can still perform the DWT, but you might need to pad the signal with zeros to the nearest power of two. This padding can introduce some artifacts, but it’s a common practice to handle non-dyadic lengths.
For more information regarding DWT, you can refer the following documentations:
- DWT - https://www.mathworks.com/help/wavelet/ref/dwt.html
- Border effects - https://www.mathworks.com/help/wavelet/ug/dealing-with-border-distortion.html
Hope this helps!
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