Final answer:
The FM signal is represented by ΦPM(t) = Acos[2π(f_c + Δf cos(π10³t))t], with a peak frequency deviation of 4kHz. The frequency deviation constant (k) is 2000 Hz/V. The bandwidth of the FM signal is approximately 10kHz.
Step-by-step explanation:
The question involves writing the time-domain expression for a frequency modulated (FM) signal and determining the frequency deviation constant as well as the bandwidth of the FM signal. Considering the given modulating signal m(t) = 2cos(2π10³t) and the carrier frequency of 1 MHz, the resulting FM signal ΦPM(t) would have the form:
ΦPM(t) = Acos[2π(f_c + Δf cos(π10³t))t]
Where A is the amplitude of the carrier wave, f_c is the carrier frequency, and Δf is the peak frequency deviation. In this case, f_c is 1MHz and Δf is 4kHz. To find the frequency deviation constant k, we use the relation Δf = k · A_m, where A_m is the amplitude of the modulating signal. This would give us k = 4kHz / 2 = 2000 Hz/V.
The bandwidth of the FM signal can be estimated using Carson's rule, which states that the bandwidth of the FM signal B_{FM} is approximately B_{FM} = 2(Δf + f_m), where f_m is the maximum frequency of the modulating signal. In this case, we have f_m = 1kHz, therefore B_{FM} is approximately 2(4kHz + 1kHz) = 10kHz. Thus, the FM signal has a bandwidth of approximately 10kHz.