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Bo_Q_

Beginner

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07-21-2014
09:03 AM

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I have a seemingly naive question: why the length of the output of cosine transform might be longer than n? For example, for staggered cosine transform, the output length is 3n/2. But just looking at its equation, the output should have the same length as the input, which is n. Thanks!

Bo

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Chao_Y_Intel

Employee

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07-23-2014
08:11 PM

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Bo,

I think you are talking the dpar/spar parameter for these functions. Actually, they are not used for the computed result. They are just used for store the intermediate sin/cos data that are used in the TT computation.

For the computation function:

d_forward_trig_transform(double f[], DFTI_DESCRIPTOR_HANDLE *handle, MKL_INT ipar[], double dpar[], MKL_INT *stat);

Both the input and output is F[], which is the same size.

Thanks

Chao

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Chao_Y_Intel

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07-21-2014
07:11 PM

125 Views

Bo,

Are you talking to some specific MKL questions, or this just a general question regarding the DCT transform?

Thanks,

Chao

Bo_Q_

Beginner

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07-21-2014
07:37 PM

125 Views

Thanks for your reply. I am only interested in using DCT but the question is applicable to sine transforms as well. Here are the defination of the transforms:

For instance, for the staggered cosine transform, its output is F(k) with k = 0...n-1, that is n elements in total. But if you look at the documentation here:

You will find the output has 3n/2 elements. Why?

Bo

Chao_Y_Intel

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07-23-2014
08:11 PM

126 Views

Bo,

I think you are talking the dpar/spar parameter for these functions. Actually, they are not used for the computed result. They are just used for store the intermediate sin/cos data that are used in the TT computation.

For the computation function:

d_forward_trig_transform(double f[], DFTI_DESCRIPTOR_HANDLE *handle, MKL_INT ipar[], double dpar[], MKL_INT *stat);

Both the input and output is F[], which is the same size.

Thanks

Chao

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