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## Re: Why does noise floor disappear depending on n bits?

 From: Doug Stewart Subject: Re: Why does noise floor disappear depending on n bits? Date: Sat, 27 Nov 2010 07:33:09 -0500

```On Sat, Nov 27, 2010 at 12:08 AM,  <address@hidden> wrote:
> In analyzing a system with pure tone signal that is 2.36 Vpk at 31.250 Hz,
> synchronous sampling at 1000 S/s data acquisition, digitized to 24 bits
> within +-2.5V.  Also, there is approx 5 nV/rtHz added to the signal.
>
> When a 8192-point FFT is done, some strange things happen when plotting
> the resulting FFT in dB scale.
>
> The result looks about right, but digitize using 23 bits, again about
> right notice the noise floor is slightly increasing. 22 bits, still ok, 21
> ok, and 20 ok.
>
> BUT! when I digitize with 19 bits, the noise floor of the FFT drops to
> zilch, like -300 dB?!  Then 18 bits ok again, but then at 17 bits, again
> with the no noise floor.
>
> What am I doing wrong here?  Anyone have an explanation?
>
> section of code looks like
>
> f=31.25;A=2.36;t=[1:8192]/1000;
> sig=A*cos(2*pi()*t);
> vnoise=5e-9;
> vn=sqrt(500)*vnoise*randn(1,8192);
> sn=sig+nv;
> n=24;Vmax=2.5;Vmin=-2.5;
> dV=(Vmax-Vmin)/2^n;
> s=dV*round((sig1n-Vmin)/dV)+Vmin;
> b=sqrt(2)*fft(s)/8192;
> blog=20*log10(abs(b(1:4096)));
> blog=blog-max(blog);
>
> Any idea why certain n-bits causes the expected noise floor to simply
> disappear?
>
>
> Robert
>
>

What value does    sig1n   have?

Doug

```