`100%` मॉडुलन के लिए पार्श्व बैण्ड `(P_(SB))` द्वारा वहन की जाने वाली सामर्थ्य

Options

A. `P_(SB)=P`
B. `P_(SB)=3p`
C. `P_(SB)=(P)/(3)`
D. `P_(SB)=0`

Answer

C

Explanation

`P_(SB)=(1)/(R)((m_(a)E_(c))/(2sqrt(2)))^(2)+(1)/(lambda)((m_(a)E_(a))/(2sqrt(2)))^(2) " " (becauseP_(c)=(E_(c)^(2))/(2R))`
`P_(SB)=(m_(a)^(2)E_(c)^(2))/(4R)=(m_(a)^(2)P_(c))/(2)=(P_(c))/(2) " " (becausem_(a)=1)`
और `P=P_(c)[1+(m_(e)^(2))/(2)]`
यहाँ `m_(a)=1rArrP=P_(c)[1+(1)/(2)]=(3)/(2)P_(c)`
अतः `(P_(SB))/(P)=(P_(c)//2)/((3)/(2)P_(c))=(1)/(3)P`