Electronic – What to use as an external 100 Hz modulation source for an RF Signal Generator

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.

The PAM signal \$s(t)\$ is a weighted sum of functions \$h(t)\$, where the weights are the samples of the signal \$m(t)\$:

$$s(t)=\sum_km(kT_s)h(t-kT_s)$$

This can be modeled as a multiplication of \$m(t)\$ by a comb of Dirac impulses, convolved with \$h(t)\$:

$$s(t)=\left(m(t)\sum_k\delta(t-kT_s)\right)*h(t)\tag{1}$$

From (1) it follows that the spectrum \$S(f)\$ is given by

$$S(f)=\left(M(f)*f_s\sum_k\delta(f-kf_s)\right)\cdot H(f)= f_s\sum_kM(f-kf_s)H(f)\tag{2}$$

where I've made use of the fact that convolution in one domain corresponds to multiplication in the other domain, and that a Dirac comb in one domain corresponds to a Dirac comb in the other domain (you can find this in most Fourier transform tables). \$M(f)\$ and \$H(f)\$ are of course the spectra of \$m(t)\$ and \$h(t)\$, respectively. So the spectrum \$S(f)\$ is the sum of shifted spectra \$M(f-kf_s)\$, multiplied by the spectrum \$H(f)\$. In order to sketch \$S(f)\$ you need to know \$M(f)\$ and \$H(f)\$:

$$M(f)=\frac{A_m}{2}[\delta(f-f_m)-\delta(f+f_m)]\\ H(f)=T\frac{\sin(\pi Tf)}{\pi fT}e^{-j\pi Tf}$$

For sketching \$S(f)\$ you simply ignore the phase term \$e^{-j\pi Tf}\$ of \$H(f)\$, so you just need to know that the magnitude \$|H(f)|\$ is the magnitude of a sinc function with \$H(0)=T\$ and with zeros at \$f_k=k/T\$, \$k=\pm 1,\pm 2,\ldots\$ (note that \$T\neq T_s\$!).

For (b) just remove all shifted spectra (that's what the ideal low-pass reconstruction filter does), so from (2) you're left with \$f_sM(f)H(f)\$ in the frequency range \$[0,f_s/2]\$.

For question 2 you just need to show that if \$s_1(t)\$ and \$s_2(t)\$ are the PAM signals corresponding to signals \$m_1(t)\$ and \$m_2(t)\$, respectively, then \$as_1(t)+bs_2(t)\$ is the PAM signal corresponding to the signal \$am_1(t)+bm_2(t)\$ for arbitrary constants \$a\$ and \$b\$. This is also obvious because the generation of the PAM signal only involves multiplication and convolution, so it is a linear process.