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Date: Thu, 10 Dec 2015 20:49:38 +0000
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From: Andy Talbot <andy.g4jnt@gmail.com>
To: rsgb_lf_group@blacksheep.org
Subject: Re: LF: LF EbNaut test from JN80 on 137370
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Yep, same here.
It feels correct: I = COS,  Q = SIN

'jnt


On 10 December 2015 at 20:34, Paul Nicholson <vlf0403@abelian.org> wrote:

>
> In vlfrx-tools I use
>
>  I = s(t) * cos( 2pi * Fo * t);
>
>  Q = s(t) * sin( 2pi * Fo * t);
>
> where s(t) is the incoming signal and Fo is the
> local oscillator frequency.   Is that right, or
> should there be a minus sign in there somewhere?
>
> If s(t) is VO1NA at 137370.00000 and I set Fo to
> 137369.9, then when I put I and Q into EbNaut, it
> needs an offset of +0.1 Hz to decode as you would
> expect with the LO being 0.1Hz on the low side.
>
> For more general mixing and frequency shift,
> to select the upper sideband of the mixer output
> I use
>
>    I - j * Q
>
> is that the right way round?   When I take I/Q data
> from an RTL2832 I find I have to invert Q to get the
> correct sideband, ie I + j * Q.   That's the only
> independent source of I/Q signal that I have available
> to try.
>
> --
> Paul Nicholson
> --
>
>

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<div dir=3D"ltr">Yep, same here.<div>It feels correct: I =3D COS, =C2=A0Q =
=3D SIN</div><div><br></div><div>&#39;jnt</div><div><br></div><div class=3D=
"gmail_extra"><br><div class=3D"gmail_quote">On 10 December 2015 at 20:34, =
Paul Nicholson <span dir=3D"ltr">&lt;<a href=3D"mailto:vlf0403@abelian.org"=
 target=3D"_blank">vlf0403@abelian.org</a>&gt;</span> wrote:<br><blockquote=
 class=3D"gmail_quote" style=3D"margin:0 0 0 .8ex;border-left:1px #ccc soli=
d;padding-left:1ex"><br>
In vlfrx-tools I use<br>
<br>
=C2=A0I =3D s(t) * cos( 2pi * Fo * t);<br>
<br>
=C2=A0Q =3D s(t) * sin( 2pi * Fo * t);<br>
<br>
where s(t) is the incoming signal and Fo is the<br>
local oscillator frequency.=C2=A0 =C2=A0Is that right, or<br>
should there be a minus sign in there somewhere?<br>
<br>
If s(t) is VO1NA at 137370.00000 and I set Fo to<br>
137369.9, then when I put I and Q into EbNaut, it<br>
needs an offset of +0.1 Hz to decode as you would<br>
expect with the LO being 0.1Hz on the low side.<br>
<br>
For more general mixing and frequency shift,<br>
to select the upper sideband of the mixer output<br>
I use<br>
<br>
=C2=A0 =C2=A0I - j * Q<br>
<br>
is that the right way round?=C2=A0 =C2=A0When I take I/Q data<br>
from an RTL2832 I find I have to invert Q to get the<br>
correct sideband, ie I + j * Q.=C2=A0 =C2=A0That&#39;s the only<br>
independent source of I/Q signal that I have available<br>
to try.<br>
<br>
--<br>
Paul Nicholson<br>
--<br>
<br>
</blockquote></div><br></div></div>

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