From: alex65111 on
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

From: Jerry Avins on
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.
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From: robert bristow-johnson on
On Jan 16, 9:46 pm, Jerry Avins <j...(a)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

From: Jerry Avins on
robert bristow-johnson wrote:
> On Jan 16, 9:46 pm, Jerry Avins <j...(a)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.
�����������������������������������������������������������������������
From: alex65111 on
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