For an FM carrier with a carrier frequency of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The frequency deviation $\Delta f$ is related to the channel spacing $(CS)$ by the formula $\Delta f = \frac{CS}{2}$.Given the carrier frequency s

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(B) The frequency deviation $\Delta f$ is related to the channel spacing $(CS)$ by the formula $\Delta f = \frac{CS}{2}$.Given the carrier frequency spacing (channel spacing) $CS = 200 	ext{ kHz}$ and the modulating frequency $f_m = 10 	ext{ kHz}$.First,calculate the frequency deviation: $\Delta f = \frac{200 	ext{ kHz}}{2} = 100 	ext{ kHz}$.The modulation index $m_f$ is defined as the ratio of frequency deviation to the modulating frequency: $m_f = \frac{\Delta f}{f_m}$.Substituting the values: $m_f = \frac{100 	ext{ kHz}}{10 	ext{ kHz}} = 10$.

For an $FM$ carrier with a carrier frequency of $200 \text{ kHz}$ and a modulating frequency of $10 \text{ kHz}$,the modulation index is...

Similar Questions

An antenna behaves as a resonant circuit only when its length is

While tuning in a certain broadcast station with a receiver,we are actually

If the maximum and minimum amplitudes of a modulated wave are $25 \ V$ and $5 \ V$ respectively,then the modulation index is

If the length of an antenna is $150 \,cm$,the transmission frequency is . . . . . . $MHz$. (Speed of light in vacuum $= 3 \times 10^8 \,ms^{-1}$)

The modulation index used in amplitude modulation of a message signal of peak voltage $10 \,V$ is $0.5$. By how much should the peak voltage of the carrier signal be changed, so as to obtain the modulation index of $0.8$, keeping the peak voltage of the message signal the same as before?

Vedclass Test Series

Exam Paper Generator

Online Exam Module