Electronic – In Analog to Analog conversion , why we multiply instead of adding

Your perfectly single-sideband suppressed-carrier modulated sinusoid certainly has a phase which can be measured. However, what you cannot tell is what the contributions of that measured phase from the audio input and the RF oscillator were.

There is another form of single-sideband modulation, in which not only one sideband but also the carrier component is transmitted. This provides a reference which can be used to synchronize the receive LO to the transmit one - normally done to insure exact tuning, but it would also give you the ability to recover the original audio phase.

It is also quite possible, especially with modern DSP gear, to transmit two separate audio channels, one on each side band. This is commonly called independent sideband modulation (ISB).

Many spread spectrum implementations are DSP based and capable of receiving multiple channels at once - GPS being a good example.

I am not sure what the phrase to correct frequency offset in the title of this question means. Does it mean that the carrier frequency is supposed to be \$10\$ MHz but actually is \$10.001\$ MHz, that is, off by \$1\$ kHz, and what is wanted is a method to fix this problem? If so, the method described below will not work.

Frequency translation by substantial amounts, e.g. changing a \$10\$ MHz to, say, \$455\$ kHz, is generally accomplished by heterodyning or mixing the signal with another carrier signal at a different frequency and bandpass filtering the mixer output. Suppose that the QAM signal at carrier frequency \$f_c\$ Hz is $$x(t) = I(t)\cos(2\pi f_c t) - Q(t)\sin(2\pi f_c t)$$ where \$I(t)\$ and \$Q(t)\$ are the in-phase and quadrature baseband data signals. The spectrum of the QAM signal occupies a relatively narrow band of frequencies, say, \$\left[f_c-\frac{B}{2}, f_c+ \frac{B}{2}\right]\$ centered at \$f_c\$ Hz. Multiplying this signal by \$2\cos(2\pi\hat{f}_ct)\$ and applying the trigonometric identities

$$\begin{align*}2\cos(C)\cos(D) &= \cos(C+D) + \cos(C-D)\\ 2\sin(C)\cos(D) &= \sin(C+D) + \sin(C-D) \end{align*}$$

gives us

$$\begin{align*} 2x(t)\cos(2\pi \hat{f}_ct) &= \quad \left(I(t)\cos(2\pi (f_c +\hat{f}_c) t) - Q(t)\sin(2\pi (f_c+\hat{f}_c)t)\right)\\ &\quad +\ \left(I(t)\cos(2\pi (f_c-\hat{f}_c)t) - Q(t)\sin(2\pi(f_c- \hat{f}_c)t)\right) \end{align*}$$

which is the sum of two QAM signals with identical data streams but different carrier frequencies shifted up and down by \$\hat{f}_c\$ Hz from the input carrier frequency \$f_c\$. The frequency spectra of these two QAM signals occupy bands of width \$B\$ Hz centered at \$f_c+\hat{f}_c\$ and \$f_c-\hat{f}_c\$ respectively, and if

$$f_c-\hat{f}_c + \frac{B}{2} < f_c+\hat{f}_c - \frac{B}{2} \Rightarrow \hat{f_c} > \frac{B}{2},$$ then bandpass filtering can be used to eliminate one of the two QAM signals while retaining the other. Needless to say, if the frequency shift is much larger than the QAM signal bandwidth, that is, if \$\hat{f}_c \gg B/2\$, then the task of designing and implementing the bandpass filter is easier. Note also that this method cannot be used to correct small frequency offsets because the two QAM signals produced at the mixer output will have overlapping spectra and cannot be separated by filtering.