I don't understand I in your formula.
Normally (traditionally) AM is: -
y(t) = [A + M cos(ωm t + φ)] . sin(ωc t)
where
- y(t) is the final modulated signal
- M is the amplitude of the modulating cosine (or sine to answer your
question)
- A is the amplitude of the carrier sine (or cosine to reinforce the
answer!!)
- φ is the phase displacement of the modulating sinewave but is
irrelevant all but mathematically
- ωm and ωc are the frequencies of modulation and carrier.
Maybe I just don't recognize your formula but the answer is, like Jim Dearden implies swap them up or use the same because carrier and modulator are not going to be the same frequency when dealing with AM.
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.
Best Answer
It doesn't matter what equation you use to describe AM, the actual physical modulation itself has an unsuppressed carrier, and a certain amount of power. It's a very simple modulation, one of the earliest, and as such is inefficient, but easy to use.
If you want a suppressed carrier, that's a different type of modulation, which needs a more complicated demodulator to read it.
You could make \$k_a\$ vary in the range 0 to 1, if you also modified \$A_m\$, and interpretted \$k_a\$ as something other than the modulation that you put on the carrier with a conventional modulator, and recovered from the carrier with a conventional detector.
So you see it's easier to interpret \$k_a\$ conventionally, because then it maps onto what's typically happening as we use the system.