A signal x(t) given to FM and PM results in equal frequenccy deviation. The relation between $k_{p} a n d k_{f} i f x (t) = r (t) - r (t - 1) - 2 u (t - 2) + u (t - 3)$ is

k_{f} = k_{p}

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can't be established

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k_{f} = k_{p} / 2 π

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k_{f} = k_{p} / π

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Solution

The correct option is B can't be established
$[Δ f]_{F M} = [k_{f} x (t)]_{m a x}$

$[Δ f]_{P M} = {[\frac{k_{p}}{2 π} \frac{d x (t)}{d t}]}_{m a x}$

$∴ [Δ f]_{P M} = k_{f} (1)$

$[Δ f]_{P M} = \frac{k_{p}}{2 π} (\infty)$

As we know that amplitude of impluse is $\infty$ .

$∴ A n s = R e l a t i o n b e t w e e n k_{f} a n d k_{p} c a n^{'} t b e e s t a b l i s h e d .$

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