In a 50% modulated signal, the carrier is suppressed for transmis
A. 88.9%
B. 11%
C. 72%
D. 18%
Please scroll down to see the correct answer and solution guide.
Right Answer is: A
SOLUTION
Concept:
The total transmitted power for an AM system is given by:
\({P_t} = {P_c}\left( {1 + \frac{{{μ^2}}}{2}} \right)\)
Pc = Carrier Power
μ = Modulation Index
Analysis:
The above expression can be expanded to get:
\({P_t} = {P_c} + P_c\frac{{{μ^2}}}{2}\)
When the carrier is suppressed, the total power transmitted becomes:
\({P_t} = P_c\frac{{{μ^2}}}{2}\)
The amount of power saving is of PC
The saving in transmitted power is, therefore:
\(P_S = \frac{{{P_C}}}{{{P_C}\left( {1 + \frac{{{{\rm{μ }}^2}}}{2}} \right)}} \times 100 \)
\(P_S=\frac{2}{2+μ ^2}\times100\)
Calculation:
With μ = 0.5, the percentage of power-saving will be:
\(P_S=\frac{2}{2+0.5 ^2}\times100\)
\(P_S=\frac{2}{2.25}\times100\)
PS = 88.9 %