written 2.5 years ago by |
Solution:
KVL to 100pV1(s)−10I1(s)−10[I1(s)−I3(s)]=0V1(s)=20I1(s)−10I3(s)...(1)KVL to 100p(2)V2(s)−10[I2(s)+I3(s)]=0V2(s)=10I2(s)+10I3(s).....(2)KVL to 100p(3)−10I3(s)−10[I3(s)+I2(s)]−10[I3(s)−I1(s)]=0−30I3(s)+10I1(s)−10I2(s)=030I3(s)=10I1(s)−10I2(s)I3(s)=13I1(s)−13I2(s).....(3)
put (3) in (1) V1(s)=20I1(s)−10[13I1(s)−13I2(s)]=503I1(S)+103I2(S)....(4) put (3) in (2) V2(s)=10I2(s)+10[13I1(s)−13I2(s)]=103I1(s)+203I2(s).....(5) Rearranging eq-n (5) 103I1(S)=V2(S)−203I2(S)∴I1(s)=310V2(s)−2I2(s)...(6)
put V1(S)=503[310V2(S)−2I2(S)]+103I2(S)=5V2(S)−1003I2(S)+103I2(S)V1(S)=5V2(S)−30I2(S).....(7)
Hence for first network, transmission parameters are
[A1B1C1D1]=[5+300.32]
for resulting network, when similar identical network connected in cascade, then transmission parameter for that network are
[A2B2C2D2]=[A1B1C1D1]=[5303/102]
Now, by property of cascade connection, transmission parameters of overall resulting network is given by
[ABCD]=[A1B1C1D1]×[A2B2C2D2]=[5300.32][5300.32]
[ABCD]=[342102.113]