discrete signals - Quadrature modulation

I have a question:

What is the difference between

$ I \cos(t) - Q \sin(t) $

and

$ I \cos(t) + Q \sin(t) $

except that they are phase shifted 90 degrees ?

On the internet I can find both modulators with adding quadrature data and substracting quadrature data.

thank you

Answer

If we take the complex baseband signal as $S(t) = I(t) + jQ(t)$ (which of course means that we have two separate wires on which the $I(t)$ and $Q(t)$ voltages appear), then we have a nice signal representation for the transmitted signal

$$s(t) = I(t) \cos(2\pi f_c t) - Q(t) \sin(2\pi f_ct) = \operatorname{Re}\left(S(t)e^{j2\pi f_c t}\right).$$ If we continue to fool ourselves by thinking of $S(t)$ having been modulated onto the complex carrier signal (complex sinusoid) $e^{j2\pi f_c t}$, then demodulation is "easy". Simply multiply $S(t)e^{j2\pi f_c t}$ by $e^{-j2\pi f_c t}$ to demodulate the signal and recover $S(t)$. Of course, we don't really have $S(t)e^{j2\pi f_c t}$ in all its complex glory: what we have as the received signal is $\operatorname{Re}\left(S(t)e^{j2\pi f_c t}\right)$. But let us not let that phase us. Let's just multiply $\operatorname{Re}\left(S(t)e^{j2\pi f_c t}\right)$ by $$e^{-j2\pi f_c t} = \cos(2\pi f_c t) - j\sin(2\pi f_c t)$$ which requires two separate multipliers and two separate wires to output the real and imaginary parts. On the real part wire, we get (after ignoring noise, multiplicative factors etc as in MBaz's answer) \begin{align*} s(t)\cos(2\pi f_ct) &= I(t)\cos^2(2\pi f_ct)-Q(t)\sin(2\pi f_ct)\cos(2\pi f_ct)\\ &=I(t)\quad +I(t)\cos(2\pi 2f_ct)-Q(t)\sin(2\pi 2f_ct).\tag{1} \end{align*} while on the imaginary part wire we get \begin{align*} s(t)\left[-\sin(2\pi f_ct)\right] &= -I(t)\cos(2\pi f_ct)\sin(2\pi f_ct)+Q(t)\sin^2(2\pi f_ct)\\ &= Q(t) \quad -I(t)\sin(2\pi 2f_ct)- Q(t)\cos(2\pi 2f_ct).\tag{2} \end{align*} In both $(1)$ and $(2)$, "low-pass" filtering is used to remove the double-frequency terms and leave us with just $I(t)$ and $Q(t)$ on two separate wires, i.e. the complex baseband signal.

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So at the transmitter, why multiply $I(t)$ by $\cos(2\pi f_c t)$ and $Q(t)$ by $-\sin(2\pi f_ct)$ and then add to get the transmitted signal $$s(t) = I(t) \cos(2\pi f_c t) - Q(t) \sin(2\pi f_ct)?$$ Why not multiply $Q(t)$ by $+\sin(2\pi f_ct)$ and then add to get $$s(t) = I(t) \cos(2\pi f_c t) + Q(t) \sin(2\pi f_ct)?\tag{3}$$ Sure we can, and now we have the nice representation $\operatorname{Re}\left(S(t)e^{-j2\pi f_c t}\right)$ which works as well except that the company VP will ask you during your presentation why you are using negative frequencies in your carrier signal. At the receiver, we demodulate by multiplying $s(t)$ (as given in $(3)$) by $e^{+j2\pi f_c t}$ so as to get (in the Q branch) \begin{align*} s(t)\sin(2\pi f_ct) &= I(t)\cos(2\pi f_ct)\sin(2\pi f_ct)+Q(t)\sin^2(2\pi f_ct)\\ &= Q(t) \quad +I(t)\sin(2\pi 2f_ct)- Q(t)\cos(2\pi 2f_ct).\tag{4} \end{align*} Note that no complementation/polarity reversal of $Q(t)$ is necessary if we follow a consistent approach and think of everything as complex signals.