Question

An intelligence signal is amplified by a 65% efficient amplifier before being combined with a 250W carrier to generate an AM signal. If it is desired to operate at 50% modulation, what must be the dc input power to the final intelligence signal amplifier

Answer 1

"192.3 watt" is the right answer.

Step-by-step explanation:

Given:

Efficient amplifier,

= 65%

or,

= 0.65

Power,

`P_c=250 \ watt`

As we know,

⇒
`P_t=P_c(1+(\mu^2)/(2) )`

By putting the values, we get

`=P_c(1+(1)/(2) )`

`=1.5 \ P_c`

Now,

⇒
`P_i=(P_t-P_c)`

`=1.5 \ P_c-P_c`

`=(P_c)/(2)`

DC input (0.65) will be equal to "
`((P_c)/(2) )`".

hence,

The DC input power will be:

=
`(250)/(2)* (1)/(0.65)`

=
`(125)/(0.65)`

=
`192.3 \ watt`