The message signal m(t), which is shown in the figure below, is applied to a frequency modulator (with a frequency sensitivity of $k_{f}$ ) and a phase modulator (with a phase sensitivity of $k_{p}$ ) separately.

If the maximum frequency deviation of the modulation signal in both the cases (FM and PM) is same, then ratio $\frac{k_{p}}{k_{f}}$ will be equal to

4 π r a d / H z

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8 π r a d / H z

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2 π r a d / H z

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10 π r a d / H z

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Solution

The correct option is A $4 π r a d / H z$
${(Δ f_{max})}_{F M} = A_{m} k_{f}$

${(Δ f_{max})}_{P M} = \frac{k_{p}}{2 π} {∣ ∣ ∣ \frac{d m (t)}{d t} ∣ ∣ ∣}_{max}$

So, ${(Δ f_{max})}_{F M} = \frac{k_{p}}{2 π} (\frac{A_{m}}{2}) = \frac{k_{p} A_{m}}{4 π}$

If, ${(Δ f_{max})}_{P M} = {(Δ f_{max})}_{P M}$

Then, $\frac{k_{p} A_{m}}{4 π} = k_{f} A_{m}$

$\frac{k_{p}}{k_{f}} = 4 π r a d / H z$

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The correct option is A 4πrad/Hz (Δfmax)FM=Amkf (Δfmax)PM=kp2π∣∣∣dm(t)dt∣∣∣max So, (Δfmax)FM=kp2π(Am2)=kpAm4π If, (Δfmax)PM=(Δfmax)PM Then, kpAm4π=kfAm kpkf=4πrad/Hz