signal power estimation

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Dear kaz

It's ok. But what about working with 16 bits words ?

If you add up 2^20 samples of 4 you will get 4*2^20 that needs more than 21 bits.. or not ?

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4*2^20 requires 3 bits + 20 bits = 23 bits.

in your case you have to multiply say 8 bits * 8bits => 16 bits +16bits => 17 bits + 20bits => 37 bits

You can imagine that by cascading pairs of additions:

sample1(17bits) + sample2(17 bits) needs 18 bits (res1)

sample3(17bits) + sample4(17 bits) needs 18 bits (res2)

res1(18bits) + res2(18 bits) needs 19 bits

...

thus you imagine 20 cascaded stages of addition needed for 2^20 samples