30 Dec
2016
30 Dec
'16
7:20 p.m.
Here are some real beginner questions that I would like
to have a better understanding of concerning SDR in general and
the rtl dongles:
I understand that the I and Q signals are both 8-bit
numbers so each one can have 2^8 possible levels. Is the in-phase
value related to the amplitude of the signal such that frequency
variations within the pass band don't change it much?
Is the Q or Quadrature value something that varies with
whether or not the frequency is higher or lower than the center
frequency set in to the device?
I could imagine that if the Q value responds to changes
in frequency that one could get the effect of a discriminator
circuit.
I bought a couple of the rtl dongles and tried out the
rtl-fm program to receive local FM signals and it worked quite
well.
Finally, what determines the pass-band? It did seem to
get smaller if I tried a 12,000 HZ sample rate. I also was
surprised at how accurate the frequency is.
Other than the fact that I am at the low end of the
learning curve, I see all kinds of possibilities.
Martin McCormick WB5AGZ
31 Dec
31 Dec
midnight
Hi Martin,
let me quickly address this in-text
On 12/30/2016 07:20 PM, Martin McCormick wrote:
Here are some real beginner questions that I would
like
to have a better understanding of concerning SDR in general and
the rtl dongles:
I understand that the I and Q signals are both 8-bit
numbers so each one can have 2^8 possible levels. Is the in-phase
value related to the amplitude of the signal such that frequency
variations within the pass band don't change it much?
Short answer: Yep.
Long answer: analog filters are never perfectly flat; in fact, the
flatter they are, the more expensive. But: if you use a bandwidth of
multiple MHz, and your signal of interest shifts within that by a
couple 100 kHz, you can basically assume flatness.
Is the Q or Quadrature value something that varies with
whether or not the frequency is higher or lower than the center
frequency set in to the device?
You should not consider I and Q to be independent
things. I and Q are
*two* physical analog signals, but IQ **together** is the baseband
signal that represents the passband (==RF) signal that you want to observe.
If there is a gain difference, we call that /IQ imbalance/, and it's
detrimental to any phase-sensitive modulation; therefore, receivers are
designed to minimize that effect. I and Q analog signal chains are
always designed to be as identical as possible! In fact, the only
difference is that the I signal is the RF signal mixed with a *cosine*
of the RF center frequency (and then low-pass filtered), and the Q
signal is the RF signal mixed with a *sine* of the same frequency – the
two oscillators used for mixing are 90° out-of-phase, which is why the
first one is called *I*nphase, and the second *Q*uadrature (if you draw
a constellation diagram, the Q axis is orthogonal to the I axis).
I could imagine that if the Q value responds to changes
in frequency that one could get the effect of a discriminator
circuit.
No, sorry. As explained, I and Q are exactly the same, but for a 90°
oscillator phase shift. That's how you convert a *real-valued passband*
to a *complex baseband* signal (just to give you two terms to look out for).
I bought a couple of the rtl dongles and tried out
the
rtl-fm program to receive local FM signals and it worked quite
well.
Finally, what determines the pass-band? It did seem to
get smaller if I tried a 12,000 HZ sample rate. I also was
surprised at how accurate the frequency is.
So: The passband that you can observe
is from -f_sample/2 to +f_sample/2
around the frequency you tune to. Filtering is adjusted to give you a
perfect-as-feasible part of the spectrum that fits into that.
Other than the fact that I am at the low end of the
learning curve, I see all kinds of possibilities.
:) Don't worry, you seem to
be doing fine so far.
I don't really know from which background you're coming from, but if
you're rather curious and want to learn about *why* we use SDR, and how
that actually works, I'd recommend something like [1] (which also
explains in detail what I and Q are). Beware: It's pretty math-y !
That's why I like SDR: It's really just doing math, but that math
actually does something to a signal that converts an intangible RF wave
to something very /practical/ (e.g. audible, if audio) and
/understandable, rather than just observable/ (as math concepts).
If you're not *that* curious (which I really couldn't blame you for), a
simple explanation for I and Q is that if you simply mix something with
a tone of its center frequency, than half of the signal (the upper
sideband) ends up on low, positive frequencies, whereas the other half
ends up on the negative frequencies, theoretically: mixing with a tone
shifts by the tone's frequency, and if you mix with the center
frequency, what was originally right and left of that center frequency
in spectrum ends up around 0Hz. Since with "normal" real signals, you
can't tell positive from negative frequencies (I can look at a cosine of
frequency f or -f for as long as I want, cos(f t) == cos (-f t)), you
need a way to tell positive from negative frequencies. The combination
of the two mixing products of sine and cosine of the same frequency does
that.
Best regards,
Marcus
[1] Free PDF:
http://www.afidc.ethz.ch/A_Foundation_in_Digital_Communication/Getting_The_…
5 Jan
5 Jan
6:07 p.m.
Thanks and sorry for my delay. I will respond to several points.
=?UTF-8?Q?Marcus_M=c3=bcller?= <marcus.mueller(a)ettus.com> writes:
Ah! That is more or less what I expected. I knew there
had to be sines and cosines in there since we are dealing with a
cyclic pattern and essentially adding and subtracting vectors
which produce positive, 0 and negative values depending upon the
instantaneous values in the two readings.
I am glad to read these things because it is one thing to
be totally mystified by something as opposed to at least slightly
understanding what is going on.
I did try values lower than 12 KHZ but rtl_fm was
reporting with floating-point issues and not running so I don't
know if that was because I was trying to do something akin to
dividing by 0 or trying to get the square root of a negative
number.
Again, another light bulb moment. Briefly, my math
background is best described as basic. I needed to take college
algebra, trig and 6 hours of what was called Technical Calculus when
I was attempting to become a vocational teacher during the
1980's. As a computer experimenter and amateur radio operator who
also happens to be blind, the math started out a little rough in
high school and early college but I then began to get the hang of
it and ended up doing reasonably well by the end of the calculus
course. A word to the wise is to learn those trig identities
well. They will make your life much easier.
To make a long story short, life took some other
interesting turns and I worked for 25 years for Oklahoma State
University's IT department in Network Operations, riding herd on
unix systems and building all sorts of scripts and a few C
programs for us to do our jobs better and more quickly. I retired
in 2015, but it was mostly lots of fun, sort of the ultimate game.
Back to the real topic at hand. You and another person in
a completely unrelated discussion group just described the
phasing method for transmitting and receiving single sideband
signals which has always mystified me but is starting to come in
to focus. The common method for sending and receiving SSB signals
is to have a very expensive filter as you mentioned and then set
the injection center frequency for the signal right on either the
upper edge of the filter so that only lower sideband gets through
or on the lower edge so that only USB signals get through.
This works great but if you switch sidebands, the signal
is now off frequency unless one adjusts a mixer frequency to
compensate for the width of the rejection filter.
Since we have digital signals in SDR, the phasing method
should work really well as the old way is done with discrete
components. I would imagine that a phasing modulator/demodulator
probably only works right at the one frequency it was designed for
and starts to be less effective at other frequencies.
Anyway, many thanks for the good explanations and an
apology for the rather long message.
Martin WB5AGZ
Hi Martin,
let me quickly address this in-text
On 12/30/2016 07:20 PM, Martin McCormick wrote:
Is the in-phase
Makes perfects sense to me.
value related to the amplitude of the signal such
that frequency
variations within the pass band don't change it much?
Short answer: Yep.
Long answer: analog filters are never perfectly flat; in fact, the
flatter they are, the more expensive. But: if you use a bandwidth of
multiple MHz, and your signal of interest shifts within that by a
couple 100 kHz, you can basically assume flatness. Is the Q
or Quadrature value something that varies with
whether or not the frequency is higher or lower than the center
frequency set in to the device?
You should not consider I and Q to be independent
things. I and Q are
*two* physical analog signals, but IQ **together** is the baseband
signal that represents the passband (==RF) signal that you want to
observe.
If there is a gain difference, we call that /IQ imbalance/, and it's
detrimental to any phase-sensitive modulation; therefore, receivers are
designed to minimize that effect. I and Q analog signal chains are
always designed to be as identical as possible! In fact, the only
difference is that the I signal is the RF signal mixed with a *cosine*
of the RF center frequency (and then low-pass filtered), and the Q
signal is the RF signal mixed with a *sine* of the same frequency ? the
two oscillators used for mixing are 90? out-of-phase, which is why the
first one is called *I*nphase, and the second *Q*uadrature (if you draw
a constellation diagram, the Q axis is orthogonal to the I axis). I could
imagine that if the Q value responds to changes
in frequency that one could get the effect of a discriminator
circuit.
No, sorry. As explained, I and Q are exactly the same, but for a 90?
oscillator phase shift. That's how you convert a *real-valued passband*
to a *complex baseband* signal (just to give you two terms to look out
for). I bought
a couple of the rtl dongles and tried out the
rtl-fm program to receive local FM signals and it worked quite
well.
Finally, what determines the pass-band? It did seem to
get smaller if I tried a 12,000 HZ sample rate. I also was
surprised at how accurate the frequency is.
So: The passband that you can observe
is from -f_sample/2 to +f_sample/2
around the frequency you tune to. Filtering is adjusted to give you a
perfect-as-feasible part of the spectrum that fits into that. Other
than the fact that I am at the low end of the
learning curve, I see all kinds of possibilities.
:) Don't worry, you seem to
be doing fine so far.
I don't really know from which background you're coming from, but if
you're rather curious and want to learn about *why* we use SDR, and how
that actually works, I'd recommend something like [1] (which also
explains in detail what I and Q are). Beware: It's pretty math-y !
That's why I like SDR: It's really just doing math, but that math
actually does something to a signal that converts an intangible RF wave
to something very /practical/ (e.g. audible, if audio) and
/understandable, rather than just observable/ (as math concepts).
If you're not *that* curious (which I really couldn't blame you for), a
simple explanation for I and Q is that if you simply mix something with
a tone of its center frequency, than half of the signal (the upper
sideband) ends up on low, positive frequencies, whereas the other half
ends up on the negative frequencies, theoretically: mixing with a tone
shifts by the tone's frequency, and if you mix with the center
frequency, what was originally right and left of that center frequency
in spectrum ends up around 0Hz. Since with "normal" real signals, you
can't tell positive from negative frequencies (I can look at a cosine of
frequency f or -f for as long as I want, cos(f t) == cos (-f t)), you
need a way to tell positive from negative frequencies. The combination
of the two mixing products of sine and cosine of the same frequency does
that.
2992
days inactive
2998
days old
2 comments
2 participants
participants (2)
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Marcus Müller -
Martin McCormick