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DFT output maximum size

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Syswip

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Hi folks,

As I know the output of the DFT is N times more than input.
I'm doing 16 point DFT. My input samples are in range -32768..32767.
It means that the maximum value of the output must be 32767 * 16 = 524272.

But I found the input sample sequence which DFT result is more than the
maximum value mentioned above.

Is it possible?.

Thank you very much.
Tiksan.
 

This is the equation for output port width: bxk = bxn+ log2 (maximum point size) +1.

For you, bxk = 16+log2(16)+1 = 21 bits. (note that [-32768, 32767] is the range of representation for 16 bit 2's complement).

Finally, range of representation for 21 bits (again in 2's complement) = [-2^(n-1), 2^(n-1)-1] = [-1048576, 1048575]
 

Thank you guitarguy12387,

Could you please give me the source of this equation?
bxk = bxn+ log2 (maximum point size) +1

Why +1? I want to understand.

Thank you very much,
Syswip.
 

Thank you very much guitarguy12387.

I'll try to understand the theory.

Bests,
Tiksan.
 

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