Introduction to Electric Circuits

228 Two-port networks

1300f~

o I I o

r~

880t2

Figure 9.21

0 0

75f~ ~]

Figure 9.22

75f~

,

Solution

The network is now as shown in Fig. 9.23.

Figure 9.23

75t21~ RG

1300t2

I2

R1 75t2

RL

Before insertion we see from Fig. 9.22 that

V2 (=Vb say)= [75/(75 + 75)]V 1 --0.5V 1

After insertion we see from Fig. 9.23 that V2 (= Va say) = I2RL. Now

I2 = [R1/(R2 -[- R L -[- R1)111

But

11 = V1/[RG + {RI(R2 + RI~)/(R~ + R2 + RI~)}]

= VI(R 1 + R 2 + RL)//[RG(R1 + R 2 + RL) + RI(R2 + RL)]

SO

I2 = [R~//(R2 + RL + R1)][VI(R1 + R2 + RL)//{RG(R1 + R2 + RL) + RI(R2 + RL)}]

= RIV~//[RG(R~ + R2 + RL) + RI(R2 + RL)]

and

Va- I2RL = RLR~V~//[RG(R~ + R2 + RL) + RI(R2 + RL)]

Putting in the values we have

V a -- 75 • 880V,/[75(880 + 1300 + 75) + 880(1300 + 75)1

= 0.0479V1

From Equation (9.68), the insertion loss is 20 log (Ib/la). This is equivalent to