Hi Luca,
just to get a feeling
Changing the power of the input signal does not change anything.
In your received signal, there seem to be high-power peaks. How large is
the digital signal amplitude during these?
Since the mirrored peak seems to be just a bit weaker, my suspicion
still is too high input power.
In any case I can’t understand why the replica is there (?).
/If, and only if/ my suspicion is correct the problem is the inherent
nature of semiconductor amplifiers, and the awesomeness that is math:
Your amplifier has the job of taking the input signal amplitude x and
map it to the output y, effectively making it a function $y$
mathematically
$y_\text{ideal} = a x$,
$a$ being the amplification.
In reality, $y$ is better represented as an exponential function.
However, using the power of Taylor, and a bit of analysis, without loss
of generality, we can represent that as a power series:
$y_\text{real} = a_0 x^0 + a_1 x^1 + a_2 x^2 + a_3 x^3 + a_4 x^4\dots$
Amplifier producers try to keep $a_n, , n>1$ as small as technically
possible.
Obviously, we want all $a_n$ to be as close to 0 as possible, but for
$n= 1$, where it represents the linear amplification we want;
$y_\text{real}$ would then collapse into the function we want. Hence, we
want to truncate the series after the 2nd element.
$a_0$ usually isn’t as bad for RF signals, it’s “just” DC offset,
something relatively easy to cancel (e.g. by capacitive coupling or
other means of high-pass filtering).
Remember that I mentioned that we need to apply Taylor’s Theorem to the
exponential, which develops the original function into a series /around
a specific point/, ie. if we truncate the series for $n > 1$, because we
trust the manufacturer and want to use the amplifier just as it was
/actually /(instead of /approximately/) linear, that’s only a valid
thing to do if we don’t drive the amplifier with $x$ that are larger
than what the manufacturer promised to amplify linearly; obviously, for
sufficiently large $x$, $x^2$ and higher powers can get substantially
bigger than $a_1 x^1$, and then you see interesting behaviour. (you’ll
notice that when looking at amplifier datasheets, their functions
typically look a bit like a quadratic with negative coefficient for
$x^2$ – that’s actually the case for semiconductors, usually).
Now, that will distort your ampitude, but you see frequency magic.
That’s only natural, however: If you look at this, $y$ contains a $x^2$
term. Assume furthermore that $x=\cos(ft)$. Trigonometrics has it that
$\cos^2(ft) = \frac 12 \left( 1+ \cos(2ft)\right)$
so that’s where a component with twice the frequency comes into your
$y$, additional to the “regular” amplified original signal, so we can
say:
$y(t) = \alpha \cos(ft) + \beta\cos(2ft) + \epsilon$,
subsuming all the third order and above terms as well as DC in
$\epsilon$.
Now, the signal with the “fake” 2f component gets fed into a mixer. A
mixer is just an amplifier where the manufacturer did his best to keep
all $a_n$ small but $a_2$, so to enable people to use the squaring to
produce its output $z$; in our case, we get funky frequencies. Remember
that $f$ is our input frequency, and let $f_{LO}$ be our local
oscillator, ie. the center frequency that becomes 0Hz in baseband:
$z(t) = \left(y(t) + \cos(f_{LO} t) \right)^2 =\left(\alpha \cos(ft) +
\beta\cos(2ft) + \cos(f_{LO} t) \right)^2$
That formula is going to be fun to expand, calculate the powers of the
trigonometric functions etc.
However, the gist is that you’ll simply have to assume that now all
frequencies mix with their multiple, so the result doesn’t only contain
$f$, and $f-f_{LO}$ (that being what we want to have in baseband!), but
also $2f - f_{LO}$ and $2f_{LO}-2f$, which is possibly the mirror you’re
seing in complex baseband.
Best regards,
Marcus