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

Answer:

"192.3 watt" is the right answer.

Explanation:

Given:

Efficient amplifier,

= 65%

or,

= 0.65

Power,

$P_c=250 \ watt$

As we know,

⇒ $P_t=P_c(1+\frac{\mu^2}{2} )$

By putting the values, we get

        $=P_c(1+\frac{1}{2} )$

        $=1.5 \ P_c$

Now,

⇒ $P_i=(P_t-P_c)$

        $=1.5 \ P_c-P_c$

        $=\frac{P_c}{2}$

DC input (0.65) will be equal to "$(\frac{P_c}{2} )$".

hence,

The DC input power will be:

= $\frac{250}{2}\times \frac{1}{0.65}$

= $\frac{125}{0.65}$

= $192.3 \ watt$