filtering random processes

Started by January 16, 2010
```Let's assume on an input of the passband filter there is a complex white
noise.

As on known spectral density on an input it is possible to calculate noise
level on an output of the passband filter?
%=================================================================
clear all
fd=1000;

sko=95;

N=2^16;
t=0:1/fd:(N-1)*1/fd;

smes_s=sko*(randn(1,N)+j*randn(1,N));

sp0=((fft(smes_s)))/N;

sp1=sp0.*conj(sp0);
sp_noise=mean(sp1(2:end)); %measurement of spectral density

%==============================================================
Fpr=10;
Fpod=12.5;
[L1,f,a] = remezord([Fpr Fpod],[1 0],[0.1 0.000001],fd);
h1 = remez(L1, f , a);

fop=105/fd;
opora=exp(sqrt(-1)*2*pi*fop*(1:length(h1)));
Hn=h1.*opora;

D=round(fd/(2*Fpod))-1;
D=1;
n_f=upfirdn(smes_s,Hn,1,D);
n_f=n_f(L1:end);

prakt_ur=mean(abs(n_f))
%=================================================================

FR0=((fft(Hn,N)))/N;
FR=FR0.*conj(FR0);
filt_1=sp_noise*FR;
Disp=sqrt(sum(filt_1))/sqrt(2) %It is impossible a correct level=
prakt_ur????????

Help to find a mistake please

```
```alex65111 wrote:
> Let's assume on an input of the passband filter there is a complex white
> noise.

How do you apply a complex signal to an ordinary bandpass filter?

> As on known spectral density on an input it is possible to calculate noise
> level on an output of the passband filter?

There must be a typo in "As on known spectral density on an input ..." I
don't want to guess what you intended to write.

...

Jerry
--
Engineering is the art of making what you want from things you can get.
&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;
```
```On Jan 16, 9:46&#2013266080;pm, Jerry Avins <j...@ieee.org> wrote:
> alex65111 wrote:
> > Let's assume on an input of the passband filter there is a complex white
> > noise.
>
> How do you apply a complex signal to an ordinary bandpass filter?

from a POV of simulation, it's no big deal.  it turns out that if the
impulse response is purely real ( Im{ h(t) } = 0 ) then, using
superposition, you run the real part though and get a real response.
running the imaginary part of the input (with the j attached) will
result in a purely imaginary response.  add the two response up and
you get a meaningful and complex output.

r b-j

```
```robert bristow-johnson wrote:
> On Jan 16, 9:46 pm, Jerry Avins <j...@ieee.org> wrote:
>> alex65111 wrote:
>>> Let's assume on an input of the passband filter there is a complex white
>>> noise.
>> How do you apply a complex signal to an ordinary bandpass filter?
>
> from a POV of simulation, it's no big deal.  it turns out that if the
> impulse response is purely real ( Im{ h(t) } = 0 ) then, using
> superposition, you run the real part though and get a real response.
> running the imaginary part of the input (with the j attached) will
> result in a purely imaginary response.  add the two response up and
> you get a meaningful and complex output.

The separation in time amounts to using two separate filters. Used that
way, they become a complex filter. As I see it, that approach begs the
question.

Jerry
--
Engineering is the art of making what you want from things you can get.
&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;
```
```On 17 Jan, 00:22, "alex65111" <alex65...@list.ru> wrote:
> Let's assume on an input of the passband filter there is a complex white
> noise.
>
> As on known spectral density on an input it is possible to calculate noise
> level on an output of the passband filter?

Yes it is.

There is a standard result from statistical data analysis that
expresses the autocovariance Syy(w) of the output signal y in
terms of the autocovariance Sxx(w) of the input signal x and
the filter response H(w) as something like

Syy(w) = c Sxx(w)|H(w)|^2

where c is some scaling coefficient possibly differnet from 1.

Find that equation in some textbook, figure out the factor c and
use it to find the power of y.

Rune
```
```On 1/17/2010 3:21 AM, Rune Allnor wrote:
> On 17 Jan, 00:22, "alex65111"<alex65...@list.ru>  wrote:
>> Let's assume on an input of the passband filter there is a complex white
>> noise.
>>
>> As on known spectral density on an input it is possible to calculate noise
>> level on an output of the passband filter?
>
> Yes it is.
>
> There is a standard result from statistical data analysis that
> expresses the autocovariance Syy(w) of the output signal y in
> terms of the autocovariance Sxx(w) of the input signal x and
> the filter response H(w) as something like
>
> Syy(w) = c Sxx(w)|H(w)|^2
>
> where c is some scaling coefficient possibly differnet from 1.
>
> Find that equation in some textbook, figure out the factor c and
> use it to find the power of y.
>
> Rune

It might be easier than that.   If the bandwidth of the noise at the
input of the filter is known then the output of the noise power is
reduced by the ratio of the input and output noise bandwidth.   This
affects SNR directly, i.e., a 2:1 reduction in noise bandwidth yields a
3dB increase in SNR.   The shape of the filter may affect the result a
bit for obvious reasons.

--
Eric Jacobsen
Minister of Algorithms
Abineau Communications
http://www.abineau.com
```
```As soon as [FR0 = ((fft (Hn, N)))/N] has led to a kind
[FR0 = ((fft (Hn, N)))] in the further counted value of noise level
after the passband filter on an output of the peak detector (Disp) has
coincided with experimentally measured (prakt_ur).

All thanks for the help
```