# Just Getting Started with the SDR Dongles.

Marcus Müller marcus.mueller at ettus.com
Fri Dec 30 23:00:01 UTC 2016

```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?
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  (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

 Free PDF:
http://www.afidc.ethz.ch/A_Foundation_in_Digital_Communication/Getting_The_Book.html
```