[linux-audio-dev] jamin and FFT filtering

[Steve Harris]

On Wed, Aug 11, 2004 at 12:24:36 +0200, Alfons Adriaensen wrote:
> On Wed, Aug 11, 2004 at 09:32:11AM +0100, Steve Harris wrote:
> 
> > I just checked jamin, and there is some ripple around the impulse (as you
> > would expect), but it peaks at around -56dB, and I couldn't find traces of
> > a pre echo visibly, or audibly when amplified. There is what looks like
> > post echo though (at around -66dB, 4.5ms after), which is odd, I would
> > have expected the response to be symmetrical?
> 
> Yes, that's odd.
>  
> > The wave which makes up the ripple looks like a fs/2 sinewave.
> 
> It's not clear to me what exactly you did to obtain these results.

Its a full JAMin running with a flat freq response. I was testing it to see
how bad the pre-echo was.
 
> I assume it's of the form 
> 
> 1.  input --> FFT --> complex x(f)   
> 
> 2.  y(f) = x(f) * r(f),  with r(f) a real symmterical function
> 
> 3.  y(f) --> IFFT --->  real output
> 
> Note that for symmetry, r(f) must be of the form 
> 
>  1 + a * cos (b * f), not 1 + a * sin (b * f)  

In this example r(f) = 1, its just a flat response.
 
> Using a sine 'ripple' would only show up in the imaginary component
> the output, which is probably not even computed.

Yes, its a half-complex FFT.
 
> If the ripple is at "fs/2", it would be removed anyway by any windowing
> that is apllied before overlapping parts are added together.
> Could you try a cosine ripple at "fs/4" or "fs/8" ?

Applied where?
 
> BTW, when do you apply windowing : before step 1, after step 3,
> or both ? What is the FFT size ?

Before 1, the FFT is 2048 bins, 8x overlap.

- Steve