Find. (a) the full name of AM; (b) the full name of FM; (c) a sketch of the AM signal produced by modulating a 500 Hz carrier with $v(t)$, explained; (d) a sketch of the FM signal produced by modulating a 1 kHz carrier with $v(t)$, explained.
Approach. (a)/(b) are direct definitions. For (c)/(d), write the standard modulated-signal equation for each scheme, substitute the given $v(t)$ and carrier, and plot the result against the message over one full message period $T_m = 1/f_m = 20\text{ ms}$; a carrier amplitude $A_c$ (AM) and a frequency-sensitivity constant $k_f$ (FM) are not given in the source and are assumed to illustrate the shape — flagged below.
Check: the question gives no carrier amplitude for AM and no frequency-deviation constant for FM, both of which are needed to draw an actual waveform. Assumed $A_c=15\text{ V}>V_m$ (so the AM envelope never crosses zero, i.e. no over-modulation) and $k_f=20\text{ Hz/V}$ (giving a modest ±200 Hz deviation about the 1 kHz FM carrier) purely to make the sketches concrete; the QUALITATIVE shape of each sketch (what varies, what stays constant) does not depend on the exact assumed values, and is the substance of the answer.
(a) AM. AM stands for Amplitude Modulation — the message signal varies the AMPLITUDE of a constant-frequency carrier.
(b) FM. FM stands for Frequency Modulation — the message signal varies the (instantaneous) FREQUENCY of a constant-amplitude carrier.
(c) AM signal. Standard (double-sideband, transmitted-carrier) AM is
$$s_{\text{AM}}(t) = \big[A_c + v(t)\big]\cos(2\pi f_c t), \qquad f_c = 500\text{ Hz}, \ A_c = 15\text{ V (assumed)}$$
Substituting $v(t)=10\sin(100\pi t)$ (taking $\phi=0$ for the sketch, without loss of generality) gives a carrier at 500 Hz (10 cycles per message period, since $f_c/f_m=10$) whose PEAK-TO-PEAK amplitude swells and shrinks in step with $v(t)$: the upper and lower envelopes trace $\pm[A_c+v(t)]$ exactly, so the carrier oscillates fast inside a slowly-varying "balloon" shaped like the message.
(d) FM signal. FM's instantaneous phase advances at a rate proportional to the message:
$$s_{\text{FM}}(t) = A_c\cos\!\left(2\pi f_c t + 2\pi k_f\!\int_0^t v(\tau)\,d\tau\right), \qquad f_c = 1000\text{ Hz}, \ k_f = 20\text{ Hz/V (assumed)}$$
so the instantaneous frequency is $f_i(t) = f_c + k_f v(t)$: it rises above 1 kHz whenever $v(t)>0$ (carrier cycles bunch closer together, visibly denser in the sketch) and falls below 1 kHz whenever $v(t)<0$ (cycles spread out, visibly sparser) — while the AMPLITUDE stays fixed at $A_c$ throughout, the opposite of the AM case.
Fig. Q4(c) — AM: the 500 Hz carrier's envelope (dashed) tracks $A_c+v(t)$; the carrier itself (solid) stays inside that envelope at all times.
Fig. Q4(d) — FM: constant-amplitude 1 kHz carrier with cycles compressed where $v(t)$ is near its positive peak and stretched where $v(t)$ is near its negative peak.
Part
Result
(a)
AM = Amplitude Modulation
(b)
FM = Frequency Modulation
(c)
Envelope $\pm[A_c+v(t)]$ about a 500 Hz carrier (amplitude varies, frequency fixed)
(d)
Instantaneous frequency $f_c+k_fv(t)$ about 1 kHz (frequency varies, amplitude fixed)
