Question

$\text{The message signal m(t), which is shown in the figure below, is applied to a frequency modulator}\text{If the frequency sensitivity of the modulator is 50 Hz/V, then the maximum phase deviation}\text{of the resultant FM signal will be}$

• A. $\frac{8\pi }{5}radius$
• B. $2\pi radius$
• C. $2\pi radius$
• D. $\frac{4\pi }{5}radius$

Answer 1

The correct option is D $\frac{4\pi }{5}radius$

$\text{The maximum phase deivation of an FM signal can be given by,}△{\varphi }_{max}=2\pi k_f{\left|{\int }_{\infty }^{t}m\right(t\left)dt\right|}_{max}\text{The given m(t) signal is varying only on positive side}\text{So,}{\left|{\int }_{\infty }^{i}m\right(t\left)dt\right|}_{max}=\text{area under m(t)}=[(\frac{1}{2}\times 2\times 2)+(\frac{1}{2}\times 6\times 2)]\text{mV-sec=8 mV-sec}\text{Hence,}\Delta {\varphi }_{max}=2\pi \times 50\times 8\times {10}^{-3}radians=\frac{16\pi \times 50}{1000}=\frac{4\pi }{5}radians$