Question

Q4: Sketch and analyze the following transmission system as noted. Two 4 port networks connected in tandem . Left to right: Network A then Network B . Network A consists of an series impedance Znetwork-17.3 +j 10.0 Ω Network B consists of a parallel impedance Zaetwok2 -35.4-j35.4 S2 Part A: Sketch the transmission system in a proper block diagram clearly labeling all components, voltages, and currents Part B: Solve for the ABCD constants in Network A Part C: Solve for the ABCD Part D: Solve for the equivalent ABCD constants of Network A+Network B Part E: Solve for the equivalent ABCD constants of Network B +Network A constants in Network B

Answer 1

Part (A) The two 4 port networks are connected in tandem. The sketch of transmission system in block diagram with labelling of all the components, voltages and currents is shown in figure below: 6 + NETWORK NETWORK В LA B] C D [AB 10 Figure 1 Here the voltages are V1, V2,V,', and V,' and currents are 11,12,1', and I,'. The relation between voltages are as follows: V2 =V,' ....... (1) The relation between currents are as follows: ....... (2)

Part (B) Consider network A. It consists of series impedance as shown: owm 6 + + 12 1 R. L 16 o In order to find ABCD parameters, the ABCD parameter equations are as follows: Vi = AV, - BI ....... (3) In = CV, - DI,

Refer to the figure shown above, apply KVL in the loop as follows: -V, +(R+ joL)I, + V2 = 0 V =V, +(R+joL)1, Vi =V, +(R+ jol)(-12) (:17 =-12) Vi =V2 -(R+ jol)12 Substitute the given values: ...... (4) In =-12 Compare equation (3) and (4) as follows: A=1 B=17.3+j10 C=0 D=1

Therefore, equation becomes, [V] [i (17.3+j10) V21 [1] [O 1 1 -12] ..... (5) *** Thus, the ABCD parameter matrix is given by, 1 (17.3 +j10) 0 1 Part (C) Consider network B. It consists of parallel impedance as shown: Α 3 + I - L

In order to find ABCD parameters, the ABCD parameter equations are as follows: Vi' = A'V2'- B'I,' ....... (6) I'=C'V2'-D'I,' Refer to the figure shown above, the voltage Vi' and V,' are equal. Vi'=V2' Vi'=V2'-(0)1, ...... (7) Apply KCL at node A as follows: I'+1,'= TI I'= V-11 I/'= (35.4 - 335.4) -12 I,'=(0.014 + j0.014)V2'-13' ....... (8)

Comparing equation (7) and (8) with equation (6) as follows: A'=1 Bl=0 D'=1 Therefore, equation becomes, 1'] (0.014+ j0.014) 11-12] Thus, the ABCD parameter matrix is given by, (0.014 + j0.014) 1:

Part (D) Refer to the figure (1), the network A+ network B shows cascade connection. In order to find equivalent ABCD parameter consider the following equations in matrix form of network A from equation (5) as follows: 03-6* [V] [i (17 7.3 +j10) V2 [11] LO 11-12 Substitute the values of V, and I, from equation (1) as follows: 10,7 [1 (17.3 + j10) |

Substitute the value from equation (9) as follows: [V] [1 (17.3 + j10) 1 0 V,'] 1, 0 1 (0.014 + j0.014) 11-12') [1x1+(17.3 + j10)(0.014 + 30.014) 1x0+(17.3+j10)x1] V,'] 1 0x1+1x(0.014+ 30.014) 0x0+1x1 -IZ! [1+0.102 + j0.382 (17.3+j10)]V,'] (0.014+ 30.014) 1 JL-1,'] [(1.102 + j0.382) (17.3+j10)][ V,'] |(0.014 + j0.014) 1 L-1,'] Therefore, equivalent ABCD parameter matrix is, (0.014 + j0.014)

Part (E) Refer to the following figure, the network B + network A shows cascade connection. NETWORK 6 + to + A B LA B LA B] ! 10 Here the voltages are V1, V2,V,', and V,' and currents are 11, 12, 1,', and I2'. The relation between voltages are as follows: V,' =V, ....... (10) The relation between currents are as follows: IL'=-1, ....... (11)

In order to find equivalent ABCD parameter consider the following equations in matrix form of network B from equation (9) as follows: [V] | 1 0TV,'] (1,'] [(0.014+ j0.014) 11-12') Substitute the values of V,' and I,' from equation (10) and (11) as follows: V 1 0 || [1] [(0.014 + 30.014) 11,

Substitute the value from equation (5) as follows: [V] 1 0 1 (17.3 + j10)TV, [1] [(0.014 + j0.014) 1] 0 1 -12] [ 1x1+0x0 1x(17.3 + j10)+0x1 TV21 |(0.014 + j0.014)x1+1x0 (0.014 + j0.014)(17.3+j10)+1x1 (-12) [ 1 (17.3 + j10) TV, |(0.014 + j0.014) 0.102 + j0.382 +1 -1, [ 1 (17.3+j10) v, 1 |(0.014 + j0.014) 1.102 + j0.382 || -1, Therefore, equivalent ABCD parameter matrix is, 1 (17.3+j10) (0.014 + j0.014) 1.102 + j0.382