Question

et to 12.7 dB, the amplifier has Determine Ps and P, in the circuit below. The attenuator is set to 12.7 dB a gain of 13.8 dB, and the power meter reading is +2 dBm.

Answer 1

Answer:-

Given That:-

$P_{s}$ and  $P_{1}$ in the circuit below.

ТАР,

The attenuator is set to 12.7 dB, the amplifier has a gain 13.8 dB, and the power meter reading is +2 dB m.

Pi=?

P =?

P3 =?

given data

Attenuation =12.7 dB

Amplifier gain = 13.8 dB

power deliver to Transistor = 2d Bm.

=-28dB

So  

P3 = –28dB.

Attenuation =

= 10 log P1 – 10 log P2

  

= P1(dB) – P2(dB)

12.7 = P1(dB) - P2(dB) - (1)

gain

= 10 log P3 - 10log P2

13.8 = P3(dB) - P2(dB) - (2)

From(2)

13.8 = -28 – P2 dB)

P (dB) = -28 – 13.8 = -41.8dB.

From(1)

12.7 = Pi(dB)-(-41.8)

= P1(dB) + 41.8

PidB) = 12.7 – 41.8 = -29.1dB

P= -29.1dB

P = -41.8dB

P3 = –28dB

P1 = 1.23 x 10-3 watts

P = 66 x 10-watts

P3 = 1.5 x 10-3 watts