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