magnitude response of IdQ-QdI
in [DSP]
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From: Jerry Avins on 25 May 2010 10:41 On 5/25/2010 10:19 AM, Clay wrote: > On May 25, 10:11 am, Clay<c...(a)claysturner.com> wrote: >> On May 24, 3:44 pm, "jacobfenton"<jacob.fenton(a)n_o_s_p_a_m.gmail.com> >> wrote: >> >>> I am trying to find the mathmatical magnitude response of the following FM >>> demodulation equation: >> >>> I[n-1]*(Q[n]-Q[n-2])-Q[n-1]*(I[n]-I[n-2]) >>> ----------------------------------------- >>> I[n-1]^2+Q[n-1]^2 >> >>> How do I represent I and Q in terms of some x[n] to find the z transform of >>> the equation? >>> I know I[n]=x[n]*cos(phi) and Q[n]=x[n]*-sin(phi). >>> But phi is also a function of 'n'. Not sure what to do here. >> >>> Thanks. >> >>> -Jacob Fenton >> >> First let's assume your analytic signal is truly analytic, then feed a >> sinusoid into the system and see what you get: >> >> Thus I(n) is A*cos(2*pi*n*f/fs) and Q(n)=A*sin(2*pi*n*f/fs) At a particular phase. In general, I(n) is A*cos(2*pi*n*f/fs + phi) and Q(n)=A*sin(2*pi*n*f/fs + phi) Since the phase can be arbitrarily chosen, it might as well be set to zero, as here. >> plug it in and reduce (you only need a few trigonometric identities), >> and you will get >> >> sin(2*pi*f/fs) for your result >> >> fs is the sample rate, f is the frequency and A is the arbitrary >> amplitude. >> >> IHTH, >> Clay > > I left out a factor of two, the result is 2*sin(2*pi*f/fs) Shouldn't that be A*2*sin(2*pi*f/fs)? Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
From: jacobfenton on 25 May 2010 11:23 >On May 24, 3:44=A0pm, "jacobfenton" <jacob.fenton(a)n_o_s_p_a_m.gmail.com> >wrote: >> I am trying to find the mathmatical magnitude response of the following F= >M >> demodulation equation: >> >> I[n-1]*(Q[n]-Q[n-2])-Q[n-1]*(I[n]-I[n-2]) >> ----------------------------------------- >> =A0 =A0 =A0 =A0 =A0 =A0 I[n-1]^2+Q[n-1]^2 >> >> How do I represent I and Q in terms of some x[n] to find the z transform = >of >> the equation? >> I know I[n]=3Dx[n]*cos(phi) and Q[n]=3Dx[n]*-sin(phi). >> But phi is also a function of 'n'. Not sure what to do here. >> >> Thanks. >> >> -Jacob Fenton > >First let's assume your analytic signal is truly analytic, then feed a >sinusoid into the system and see what you get: > > >Thus I(n) is A*cos(2*pi*n*f/fs) and Q(n)=3DA*sin(2*pi*n*f/fs) > >plug it in and reduce (you only need a few trigonometric identities), >and you will get > >sin(2*pi*f/fs) for your result > >fs is the sample rate, f is the frequency and A is the arbitrary >amplitude. > >IHTH, >Clay > > Thanks for your repsponse. -JF
From: Clay on 25 May 2010 11:41 On May 25, 10:41 am, Jerry Avins <j...(a)ieee.org> wrote: > On 5/25/2010 10:19 AM, Clay wrote: > > > > > > > On May 25, 10:11 am, Clay<c...(a)claysturner.com> wrote: > >> On May 24, 3:44 pm, "jacobfenton"<jacob.fenton(a)n_o_s_p_a_m.gmail.com> > >> wrote: > > >>> I am trying to find the mathmatical magnitude response of the following FM > >>> demodulation equation: > > >>> I[n-1]*(Q[n]-Q[n-2])-Q[n-1]*(I[n]-I[n-2]) > >>> ----------------------------------------- > >>> I[n-1]^2+Q[n-1]^2 > > >>> How do I represent I and Q in terms of some x[n] to find the z transform of > >>> the equation? > >>> I know I[n]=x[n]*cos(phi) and Q[n]=x[n]*-sin(phi). > >>> But phi is also a function of 'n'. Not sure what to do here. > > >>> Thanks. > > >>> -Jacob Fenton > > >> First let's assume your analytic signal is truly analytic, then feed a > >> sinusoid into the system and see what you get: > > >> Thus I(n) is A*cos(2*pi*n*f/fs) and Q(n)=A*sin(2*pi*n*f/fs) > > At a particular phase. In general, I(n) is A*cos(2*pi*n*f/fs + phi) and > Q(n)=A*sin(2*pi*n*f/fs + phi) > > Since the phase can be arbitrarily chosen, it might as well be set to > zero, as here. > > >> plug it in and reduce (you only need a few trigonometric identities), > >> and you will get > > >> sin(2*pi*f/fs) for your result > > >> fs is the sample rate, f is the frequency and A is the arbitrary > >> amplitude. > > >> IHTH, > >> Clay > > > I left out a factor of two, the result is 2*sin(2*pi*f/fs) > > Shouldn't that be A*2*sin(2*pi*f/fs)? > Jerry, The "A" part cancels out as that is the whole point of the denominator term. Thus you get a normalized (amplitude independent) frequency measure. If you have a strongly AGCed receiver, then you can dispense with the denominator as it becomes nearly constant. Cool! Clay
From: Jerry Avins on 25 May 2010 16:16 On 5/25/2010 11:41 AM, Clay wrote: > On May 25, 10:41 am, Jerry Avins<j...(a)ieee.org> wrote: >> On 5/25/2010 10:19 AM, Clay wrote: >> >> >> >> >> >>> On May 25, 10:11 am, Clay<c...(a)claysturner.com> wrote: >>>> On May 24, 3:44 pm, "jacobfenton"<jacob.fenton(a)n_o_s_p_a_m.gmail.com> >>>> wrote: >> >>>>> I am trying to find the mathmatical magnitude response of the following FM >>>>> demodulation equation: >> >>>>> I[n-1]*(Q[n]-Q[n-2])-Q[n-1]*(I[n]-I[n-2]) >>>>> ----------------------------------------- >>>>> I[n-1]^2+Q[n-1]^2 >> >>>>> How do I represent I and Q in terms of some x[n] to find the z transform of >>>>> the equation? >>>>> I know I[n]=x[n]*cos(phi) and Q[n]=x[n]*-sin(phi). >>>>> But phi is also a function of 'n'. Not sure what to do here. >> >>>>> Thanks. >> >>>>> -Jacob Fenton >> >>>> First let's assume your analytic signal is truly analytic, then feed a >>>> sinusoid into the system and see what you get: >> >>>> Thus I(n) is A*cos(2*pi*n*f/fs) and Q(n)=A*sin(2*pi*n*f/fs) >> >> At a particular phase. In general, I(n) is A*cos(2*pi*n*f/fs + phi) and >> Q(n)=A*sin(2*pi*n*f/fs + phi) >> >> Since the phase can be arbitrarily chosen, it might as well be set to >> zero, as here. >> >>>> plug it in and reduce (you only need a few trigonometric identities), >>>> and you will get >> >>>> sin(2*pi*f/fs) for your result >> >>>> fs is the sample rate, f is the frequency and A is the arbitrary >>>> amplitude. >> >>>> IHTH, >>>> Clay >> >>> I left out a factor of two, the result is 2*sin(2*pi*f/fs) >> >> Shouldn't that be A*2*sin(2*pi*f/fs)? >> > > Jerry, > > The "A" part cancels out as that is the whole point of the denominator > term. Thus you get a normalized (amplitude independent) frequency > measure. If you have a strongly AGCed receiver, then you can dispense > with the denominator as it becomes nearly constant. Cool! You're solving the demodulator equation. I was solving something else. as my next post probably made evident. My bad! Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
From: Jerry Avins on 25 May 2010 16:18
On 5/25/2010 4:16 PM, Jerry Avins wrote: ... > You're solving the demodulator equation. I was solving something else. > as my next post probably made evident. My bad! The "next" post didn't show up. Never mind! Jerry -- Engineering is the art of making what you want from things you can get. ����������������������������������������������������������������������� |