Fourier transform of a sum of signals
The Fourier transform is linear, which means that if you sum two signals, then also their spectrum will be the sum.

$$ \mathscr{F} [x(t) + y(t)] = \mathscr{F}[x(t)] + \mathscr{F}[y(t)] $$

If you consider the power, you can have two cases:


The signals have different frequency
Then their power just sums up, because:
$$ P_{tot} = P_1 cos (\omega _1 t + \phi _1) + P_2 cos (\omega _2 t + \phi _2) $$
The signals have the same frequency
Then if they are in phase you have:
$$ P_{tot} = P_1 cos (\omega _0 t + \phi _0) + P_2 cos (\omega _0 t + \phi _0) = 
(P_1 + P_2) \cdot cos(\omega _0 t + \phi_0) $$
If they're out of phase, the resulting power will be lower and precisely:
$$ P_{tot} = P_1 cos (\omega _0 t + \phi _1) + P_2 cos (\omega _0 t + \phi _2) $$
and setting arbitrarily \$\phi_1 = 0\$ we obtain:
$$ P_{tot} = P_1 cos (\omega _0 t) + P_2 cos (\omega _0 t + \Delta \phi) $$
which for \$\Delta \phi = \pi\$ gives the subtraction of the signals.