This question apparently is not at all about negative frequencies which is the point I addressed earlier in a separate answer but about the signal at the image
frequency that occurs as a result of the
It appears from the comments by the OP on an earlier answer of mine that his question is
about an entirely different problem. The OP apparently
creates a BPSK signal $x(t)\cos(2\pi f_0 t)$
at a low frequency $f_0$ Hz and then mixes it with a high-frequency signal at frequency
$f_1$ Hz where $f_1 \gg f_0$. This creates two BPSK signals at carrier frequencies
$f_1 \pm f_0$ respectively and the question is whether this duplication
can be avoided somehow in the mixing
process, or whether one must filter the output to leave only one BPSK signal.
The trigonometric identity
$2\cos(A)\cos(B) = \cos(A+B) + \cos(A-B)$ shows that mixing (multiplying)
signal $x(t)\cos(2\pi f_0 t)$ with $2\cos(2\pi f_1 t)$ gives
$$2\cos(2\pi f_1 t)x(t)\cos(2\pi f_0 t)
= x(t)\cos(2\pi (f_1+f_0)t) + x(t)\cos(2\pi (f_1-f_0)t),$$
that is, two BPSK signals at carrier frequencies $f_1+f_0$ Hz and
$f_1-f_0$ Hz. If one desires to have only the BPSK signal
at $f_1+f_0 = f_c$ Hz, say, then the BPSK signal at the
image frequency $f_1-f_0$
must be filtered out by passing the mixer output through
a bandstop filter that removes the image signal.
Can the filtering to remove the image frequency
be avoided by clever design of the mixer?
Mathematically Yes, but the costs might be prohibitive
and the physical circuitry requiring very careful design
and constant retuning. The practical details are best
discussed over on the sister site electronics.SE.
Suppose we create two BPSK signals $x(t)\cos(2\pi f_0 t)$
and $x(t)\sin(2\pi f_0 t)$ on phase-orthogonal carriers
at frequency $f_0$ and mix them (separately) with phase-orthogonal
carriers at frequency $f_1$. Then, their difference
$$x(t)\cos(2\pi f_0 t)\cos(2\pi f_1 t) - x(t)\sin(2\pi f_0 t)\sin(2\pi f_1 t)
= x(t)\cos(2\pi (f_1+f_0)t)$$
which is a single BPSK signal at the desired frequency $f_1+f_0 = f_c$!!
However, notice that we need two baseband modulators and two RF mixers
instead of one of each plus a filter. Also, we need precise matching
of hardware so that the two signals are generated with equal amplitudes,
are amplified exactly equally, are modulated onto phase-orthogonal carrier
signals of precisely equal amplitudes and precise phase difference that must
be maintained through oscillator drift, ambient temperature changes, etc.
Why is all this so important? Well, both mixer outputs
$x(t)\cos(2\pi f_0 t)\cos(2\pi f_1 t)$ and $x(t)\sin(2\pi f_0 t)\sin(2\pi f_1 t)$
contain two BPSK signals, one at carrier frequency $f_1+f_0 = f_c$
and the other at carrier frequency $f_1-f_0$. When we take the difference,
the BPSK signals at carrier frequency $f_1+f_0 = f_c$ add constructively
and appear at the output of the subtractor, while the BPSK signals at
at carrier frequency $f_1-f_0$ cancel out. This is easy to do mathematically
or MATLABitacally or in software radio, but much more complicated to
achieve with analog circuits where slight differences in gains, phase shifts,
etc in the two allegedly identical circuits make it difficult to
achieve the exact cancellation of the undesired BPSK signal.
In short, getting a single BPSK signal via this method is by no means
as simple in practice as the straightforward mathematical result suggests.
As noted previously, questions about feasibility are best discussed over