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Consider the frequency-modulated (FM) signal \[f(t)=A_c\cos\big(2\pi f_c t+3\sin(2\pi f_1 t)+4\sin(6\pi f_1 t)\big).\] where \(A_c\) and \(f_c\) are, respectively, the amplitude and frequency (in Hz) of the carrier waveform. The frequency \(f_1\) is in Hz, and assume that \(f_c>100f_1\). The peak frequency deviation of the FM signal, in Hz, is _______.

1. 15f₁
2. 12f₁
3. 4f₁
4. 2f₁

Correct answer: 15f₁

Solution

For an FM signal, instantaneous frequency is \(f_i(t)=f_c+\frac{1}{2\pi}\frac{d}{dt}[\text{phase deviation}]\). Differentiating the phase deviation gives \(6\pi f_1\cos(2\pi f_1 t)+24\pi f_1\cos(6\pi f_1 t)\), so the frequency deviation is \(3f_1\cos(2\pi f_1 t)+12f_1\cos(6\pi f_1 t)\). The peak possible deviation is the sum of the maxima: \(3f_1+12f_1=15f_1\).

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