A carrier signal is frequency modulated by a sinusoidal message signal of 15 kHz, resulting in a maximum frequency deviation of 50 kHz. If the amplitude of the message signal is increased by a factor of 3 and it's frequency is decreased to 5 kHz, then the bandwidth of the modified FM signal (according to Carson's rule) will be equal to ______ kHz.
310

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Solution

The correct option is A 310
$Δ f_{1} = A_{m 1} k_{f} = 50 k H z$

$Δ f_{1} = A_{m 1} k_{f} = (3 A_{m 1}) k_{f} = 150 k H z$

$f_{m 1} = 15 k H z a n d f_{m 2} = 5 k H z$

$β_{2} = \frac{Δ f_{2}}{f_{m 2}} = \frac{150}{5} = 30$

According to Carson's rule,
${(B W)}_{F M (n e w)} = (1 + β_{2}) 2 f_{m 2} = (31) (2 \times 5) k H z = 310 k H z$

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The correct option is A 310Δf1=Am1kf=50 kHz Δf1=Am1kf=(3Am1)kf=150 kHz fm1=15 kHzand fm2=5 kHz β2=Δf2fm2=1505=30 According to Carson's rule, (BW)FM(new)=(1+β2)2fm2=(31)(2×5) kHz=310 kHz