Final answer:
The amplitude spectra of xc₁ (t) and xc₂ (t) are identical because the amplitude spectrum of an FM signal only depends on frequency deviation and the modulating frequency, not the modulating signal's shape. The phase spectrum can be computed using a Fourier transform.
Step-by-step explanation:
We previously computed the spectrum of the FM signal defined by xc₁ (t)= Ac cos(2πfct + βsin(2πfmt)). Now, let us assume the modulated signal is given by xc₂ (t) = Ac cos(2πfct + βcos(2πfmt)). The question is to show that the amplitude spectra of xc₁ (t) and xc₂ (t) are identical.
The amplitude spectrum of an FM signal is independent of the modulating signal's shape; it only depends on the frequency deviation and the modulating frequency. We know from Fourier analysis that shifting a signal in time, which is analogous to changing the phase of the modulating signal from sine to cosine, or vice versa, does not affect the amplitude spectrum of the modulated signal. As a result, the amplitude spectra of xc₁ (t) and xc₂ (t) are identical because both signals have the same frequency deviation and modulating frequency.
To compute the phase spectrum of xc₂ (t), one would need to perform a Fourier transform which would reveal the phase information as a function of frequency. Comparing this to the phase spectrum of xc₁ (t), we would observe that the two spectra would differ due to the different modulating functions (sine versus cosine), which will introduce different phase shifts for each of the frequency components of the modulating signal.
Your question is incomplete, but most probably your full question can be seen in the attachment.