A signal of 0.1 , kW is transmitted in a cabl | vedclass

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

Question

(D) The total attenuation in the cable is given by: $	ext{Total Attenuation} = (	ext{Attenuation per km}) 	imes (	ext{Length}) = -5 \, dB/km 	ime

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) The total attenuation in the cable is given by: $	ext{Total Attenuation} = (	ext{Attenuation per km}) 	imes (	ext{Length}) = -5 \, dB/km 	imes 20 \, km = -100 \, dB$.The formula for gain in $dB$ is: $	ext{Gain} = 10 \log_{10}(\frac{P_0}{P_i})$.Here,$	ext{Gain} = -100 \, dB$,$P_i = 0.1 \, kW = 100 \, W = 10^2 \, W$.Substituting the values: $-100 = 10 \log_{10}(\frac{P_0}{10^2})$.$-10 = \log_{10}(\frac{P_0}{10^2})$.$10^{-10} = \frac{P_0}{10^2}$.$P_0 = 10^{-10} 	imes 10^2 = 10^{-8} \, W$.Comparing $P_0 = 10^{-8} \, W$ with $10^{-x} \, W$,we get $x = 8$.

$A$ signal of $0.1 \, kW$ is transmitted in a cable. The attenuation of the cable is $-5 \, dB$ per $km$ and the cable length is $20 \, km$. The power received at the receiver is $10^{-x} \, W$. The value of $x$ is ....... .
$[\text{Gain in } dB = 10 \log_{10}(\frac{P_0}{P_i})]$

Similar Questions

An amplitude modulated wave is represented by $10[1+0.6 \sin(40 \times 10^3 t)] \sin(4 \times 10^6 t)$ volt,where $t$ is in seconds. The ratio of the upper sideband frequency to the lower sideband frequency is:

$A$ sinusoidal wave $y(t) = 40 \sin(10 \times 10^6 \pi t)$ is amplitude modulated by another sinusoidal wave $x(t) = 20 \sin(1000 \pi t)$. The amplitude of the minimum frequency component of the modulated signal is ...........

When the angle of incidence is $74^{\circ}$,the critical frequency of the $F$ layer is $50 \text{ MHz}$. What is the Maximum Usable Frequency $(MUF)$ in $\text{ MHz}$ (in $.5$)?

$A$ square wave of the modulating signal is shown in the figure. The carrier wave is given by $C(t) = 5 \sin(8\pi t) \text{ V}$. The modulation index is:

In amplitude modulation,if a message signal of $5 \ kHz$ is modulated by a carrier wave of frequency $900 \ kHz$,then the frequencies of the side bands are

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$A$ signal of $0.1 \, kW$ is transmitted in a cable. The attenuation of the cable is $-5 \, dB$ per $km$ and the cable length is $20 \, km$. The power received at the receiver is $10^{-x} \, W$. The value of $x$ is ....... .$[\text{Gain in } dB = 10 \log_{10}(\frac{P_0}{P_i})]$