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

# filtering random processes

Started by ●January 16, 2010

Reply by ●January 16, 20102010-01-16

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. �����������������������������������������������������������������������

Reply by ●January 16, 20102010-01-16

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. r b-j

Reply by ●January 17, 20102010-01-17

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. �����������������������������������������������������������������������

Reply by ●January 17, 20102010-01-17

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

Reply by ●January 17, 20102010-01-17

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. > > RuneIt 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

Reply by ●January 17, 20102010-01-17