For radio transmission in free space signal reduced proporti

For radio transmission in free space, signal reduced proportion to the power is in square of the distance from the source, whereas in wire transmission, the attenuation is a fixed number of dB per kilometer. The following table used to show reduction relative to some reference for free space radio and uniform wire. Fill in the missing numbers to complete the table.

Solution

Given
For the radio transmission, we have loss (L) is proportional to square of distance, so
L = alpha*d^2
==>L = alpha + 20logd.
From the given table,
L=-6,d=1,-6 =alpha+ 20log 1 ==> -6=alpha*20log1 =>alpha= -6 ;
so the equatio is
L=-6+20logd;
For wire transmission:
Given in the table,the loss for 1 kilometer is –3 dB.

L Radio Wire
1 -6 -3
2 -6+20*log2=-6+20*.301029=-6+6.0205=0.02058~0.0206 (-3*2)=-6
4 -6+20log4=-6+20*.602059=6.0411 (-3*4)=-12
8 -6+20log8=-6+20*0.903089=12.061 (-3*8)=-24
16 -6+20log16=-6+20*1.204119=18.082 (-3*16)=-48
 For radio transmission in free space, signal reduced proportion to the power is in square of the distance from the source, whereas in wire transmission, the at

Get Help Now

Submit a Take Down Notice

Tutor
Tutor: Dr Jack
Most rated tutor on our site