hamming window overlap #2

Continuing an old post regarding Hamming window overlaps:
http://www.dsprelated.com/showthread/comp.dsp/105321-1.php

The first answer gives:

"a 50% overlap would give a sum of 1.08".

If I wish to reconstruct the input signal perfectly (so the amplitudes are
exactly the same), then can I merely subtract the extra ~".08" from all
the samples?


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On 7/1/15 9:39 AM, mavavilj wrote:
> Continuing an old post regarding Hamming window overlaps:
> http://www.dsprelated.com/showthread/comp.dsp/105321-1.php
>
> The first answer gives:
>
> "a 50% overlap would give a sum of 1.08".
>
> If I wish to reconstruct the input signal perfectly (so the amplitudes are
> exactly the same), then can I merely subtract the extra ~".08" from all
> the samples?

scale the Hamming window a little so that instead of

           { 0.54*cos(pi*t) + 0.46      |t|<1
    w(t) = {
           { 0                          |t|>1

(the traditional definition for a normalized-length Hamming)

scale it so that it is

           { 0.50*cos(pi*t) + 0.426     |t|<1
    w(t) = {
           { 0                          |t|>1


then it will be complementary with windows spaced and overlapped every 
unit of t and share all other properties of the Hamming (regarding the 
relative sizes of the sidelobes).

still won't get rid of the discontinuity inherent to the Hamming.  is 
that what you want?  perhaps a Hann window would be better for you.


-- 

r b-j                  rbj@audioimagination.com

"Imagination is more important than knowledge."