filtering random processes
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From: alex65111 on 16 Jan 2010 18:22 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 16 Jan 2010 21:46 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. �����������������������������������������������������������������������
From: robert bristow-johnson on 16 Jan 2010 22:34 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 16 Jan 2010 23:26 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 17 Jan 2010 13:57
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 |