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magnitude response of IdQ-QdI

Started by jacobfenton May 24, 2010
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



jacobfenton <jacob.fenton@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.
There is not enough information here. If you know the statistics of phi(n), you can then compute the correlations between pairs of signals such as Q[n] and Q[n-2], I[n] and Q[n], and you can then come up with an analytic form for the magnitude of the above ratio. (Or if you happen to know all these signals are uncorrelated then the answer is simple, but they are almost certainly not.) Steve

jacobfenton 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
~ 2WT School math.
> Not sure what to do here.
Break your stupid head against the wall.
On May 24, 3:44&#2013266080;pm, "jacobfenton" <jacob.fenton@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]) > ----------------------------------------- > &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; 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) 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
On May 25, 10:11&#2013266080;am, Clay <c...@claysturner.com> wrote:
> On May 24, 3:44&#2013266080;pm, "jacobfenton" <jacob.fenton@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]) > > ----------------------------------------- > > &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; 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) > > 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) Clay
On 5/25/2010 10:19 AM, Clay wrote:
> On May 25, 10:11 am, Clay<c...@claysturner.com> wrote: >> On May 24, 3:44 pm, "jacobfenton"<jacob.fenton@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. &#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 May 24, 3:44=A0pm, "jacobfenton" <jacob.fenton@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
On May 25, 10:41&#2013266080;am, Jerry Avins <j...@ieee.org> wrote:
> On 5/25/2010 10:19 AM, Clay wrote: > > > > > > > On May 25, 10:11 am, Clay<c...@claysturner.com> &#2013266080;wrote: > >> On May 24, 3:44 pm, "jacobfenton"<jacob.fenton@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]) > >>> ----------------------------------------- > >>> &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080;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
On 5/25/2010 11:41 AM, Clay wrote:
> On May 25, 10:41 am, Jerry Avins<j...@ieee.org> wrote: >> On 5/25/2010 10:19 AM, Clay wrote: >> >> >> >> >> >>> On May 25, 10:11 am, Clay<c...@claysturner.com> wrote: >>>> On May 24, 3:44 pm, "jacobfenton"<jacob.fenton@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. &#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 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. &#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;