Hi Jason,
now to comments to your comments:
On 12.01.2016 18:43, Jason Matusiak wrote:
> Thanks for the quick response Marcus!!!!
>
> Since my Latex isn't very good (as in pretty much non-existent). Let me
> see if I can rewrite what you recommended in my dumbed down language and
> see if I am close.:
>
> *I have two dongles, dongle 1 will be my modified dongle, dongle 2 will
> be my un-modified dongle.
>
> *Put a a known reference tone into each of the dongles where Ftune =
> Fref - Foffset
> ** Foffset should be roughly a third of the sample rate
Exactly what I had in mind.
> **An example at a sample rate of 1.024Msps would be a reference tone at
> 98MHz, and then tune the dongles to 97.659MHz
>
> *I'll now see a baseband signal for both dongles whose offsets won't be
> exactly the same.
Yep.
>
> *Multiple the resulting signals found above against each other (offset,1
> * offset,2)
> *Pass that through a LPF with a cutoff of Fsample/4, or 256khZ in this
> case
> **This will give the difference between the frequencies at frequency at
> Foffset,1 +/- Foffset,2
>
> *perform a QT freq sync or a quad demod into a QT time sink to compare.
>
> Is that close?
For close being identical , yes :)
> I think I am missing something in there, and I have a
> feeling that it has to do with the multiplication step as that makes the
> least amount of sense to me. Any way to enlighten me on what I am
> missing above? Thanks!
That's modulation. So, the math behind that is:
Let's consider both tones to be cosines. Thanks to Euler, we know we can
express a cosine as ($j$ is the imaginary unit, $j=\sqrt{-1}$)
$\cos x = \frac{1}{2}\left(e^{j x}+e^{-j x}\right)$; therefore, $\cos
\left( f_1 x \right) = \frac{1}{2}\left(e^{j f_1 x}+e^{-j f_1 x}\right)$.
Now, $\cos \left( f_1 x \right) \cos \left( f_2 x \right) = \frac14
\left(e^{j f_1 x}+e^{-j f_1 x}\right)\left(e^{j f_2 x}+e^{-j f_2 x}\right)$.
Let's expand the multiplication of the two (), and use the fact that
$a^b a^c = a^{b+c}$:
\documentclass{article} \usepackage[utf8x]{inputenc}
\usepackage{amsmath} \usepackage{amsfonts} \usepackage{amssymb}
\usepackage[usenames,dvipsnames]{xcolor} \usepackage{trfsigns}
\DeclareMathOperator*{\argmin}{arg\,min} \usepackage{tikz}
\usepackage{circuitikz} \usepackage[binary-units=true]{siunitx}
\sisetup{exponent-product = \cdot} \DeclareSIUnit{\dBm}{dBm}
\newcommand{\imp}{\SI{50}{\ohm}} \newcommand{\wrongimp}{\SI{75}{\ohm}}
\pagestyle{empty} \begin{document} \begin{align*} e^{j f_1 x}e^{j f_2
x}&+e^{j f_1 x}e^{-j f_2 x}&+ e^{-j f_1 x}e^{j f_2 x}&+ e^{-j f_1
x}e^{-j f_1 x}\\ =e^{j f_1 x+j f_2 x}&+e^{j f_1 x-j f_2 x}&+ e^{-j f_1
x+j f_2 x}&+ e^{-j f_1 x+-j f_2 x}\\ = e^{j ( f_1 +f_2)x}&+e^{j( f_1 -
f_2)x}&+ e^{j( -f_1+ f_2)x}&+ e^{j(- f_1 - f_2)x}\\ = {\color{blue}e^{j
( f_1 +f_2)x}}&+{\color{OliveGreen}e^{j( f_1 -
f_2)x}}&+{\color{OliveGreen} e^{-j( f_1- f_2)x}}&+ {\color{blue} e^{-j(
f_1 + f_2)x}}\\ \end{align*} \end{document}
Now, let's sort this, and lo!
\documentclass{article} \usepackage[utf8x]{inputenc}
\usepackage{amsmath} \usepackage{amsfonts} \usepackage{amssymb}
\usepackage[usenames,dvipsnames]{xcolor} \usepackage{trfsigns}
\DeclareMathOperator*{\argmin}{arg\,min} \usepackage{tikz}
\usepackage{circuitikz} \usepackage[binary-units=true]{siunitx}
\sisetup{exponent-product = \cdot} \DeclareSIUnit{\dBm}{dBm}
\newcommand{\imp}{\SI{50}{\ohm}} \newcommand{\wrongimp}{\SI{75}{\ohm}}
\pagestyle{empty} \begin{document} \begin{align*} = {\color{blue}e^{j (
f_1 +f_2)x}}+ {\color{blue} e^{-j( f_1 + f_2)x}}
&+{\color{OliveGreen}e^{j( f_1 - f_2)x}}+{\color{OliveGreen} e^{-j( f_1-
f_2)x}}\\ = \left( {\color{blue}e^{j ( f_1 +f_2)x} + e^{-j( f_1 +
f_2)x}} \right)&+\left({\color{OliveGreen}e^{j( f_1 - f_2)x} +e^{-j(
f_1- f_2)x}}\right)\\ = 2 \cos\left((f_1+f_2)x\right) &+
2\cos\left((f_1-f_2)x\right)\text. \end{align*} \end{document}
Which means that
\documentclass{article} \usepackage[utf8x]{inputenc}
\usepackage{amsmath} \usepackage{amsfonts} \usepackage{amssymb}
\usepackage[usenames,dvipsnames]{xcolor} \usepackage{trfsigns}
\DeclareMathOperator*{\argmin}{arg\,min} \usepackage{tikz}
\usepackage{circuitikz} \usepackage[binary-units=true]{siunitx}
\sisetup{exponent-product = \cdot} \DeclareSIUnit{\dBm}{dBm}
\newcommand{\imp}{\SI{50}{\ohm}} \newcommand{\wrongimp}{\SI{75}{\ohm}}
\pagestyle{empty} \begin{document} \begin{align*} \cos \left( f_1 x
\right) \cos \left( f_2 x \right) &= \frac14 \left[ 2
\cos\left((f_1+f_2)x\right) + 2\cos\left((f_1-f_2)x\right)\right]\\
&=\frac12 \cos\left((f_1+f_2)x\right) +
\frac12\cos\left((f_1-f_2)x\right) \end{align*} \end{document}
Now you see where the low pass filter comes into play:
it filters out the $\cos\left((f_1+f_2)x\right)$ component, leaving you
with $\frac12 \cos\left((f_1-f_2)x\right)$, which is an oscillation with
the difference frequency.
Best regards,
Marcus
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