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

[macy]

should have read:
  s=dV*round((sn-Vmin)/dV)+Vmin;

> 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
>