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Find the transmission parameter of the resulting Ckt. when both are in a cascade connection.

Find the transmission parameter:

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Solution:

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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)]=030I3(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]

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