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Note for Network Theory - NT by Bhubanjit Sahu

  • Network Theory - NT
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  • Biju Patnaik University of Technology Rourkela Odisha - BPUT
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Note for Network Theory - NT by Bhubanjit Sahu

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NATIONAL INSTITUTE OF SCIENCE & TECHNOLOGY, PALUR HILLS, BERHAMPUR, ODISHA -761 008 Solutions 1i. From the given equations I1=2V1+V2, I2= 2V1+3V2. Compare with the Y- parameters governing equations I1=Y11V1+ Y12V2, I2=Y21V1+ Y22V2 Y12=1, ∆Y= 4 Z12 = − Y12 ∆Y = −1 4 Answer: D 1ii. Given network is KVL equations for the two loops V1=Z1I1+ Z1 I2, V2=Z1I1+ (Z1+ Z2)I2  Z11 Z  21 Z12   Z1 = Z 22   Z1  Z1 + Z 2  Z1 Answer: D 1iii. Answer: D 1iv. Answer: D 1v. According to the Foster First form realization If there is a pole at S=jω=0, the first element C0 is present If there is a pole at S=jω= ∞ , the last element Ln is present Consider this example Z ( s ) = (s 2 + 1)(s 2 + 9) s ( s 2 + 4) If, we substitute s=0 in Z(s),then Z(s)= ∞ . That is capacitor in s- domain will be 1 . So, it shows the property of a capacitor Z (s) = sC ∞ Similarly, If, we substitute s= ∞ ,in Z(s),then Z(s)= . So, ∞ DEPARTMENT OF ELECTRICAL ENGINEERING 2016-2017

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NATIONAL INSTITUTE OF SCIENCE & TECHNOLOGY, PALUR HILLS, BERHAMPUR, ODISHA -761 008 s 2 (1 + 1 s Z (s) = 2 ) s 2 (1 + ss 2 (1 + 9 s 4 1 s (1 + ) s = 2 (1 + )(1 + 4 9 ) s 2 , Now substitute s= ∞ ) s s2 That is inductor in s- domain will be Z ( s) = sL . So, it shows the property of a inductor There is a pole at S=jω=0, the first element C0 is present 2 ) 2 There is a pole at S=jω= ∞ , the last element Ln is present. So, box 1 contains only series LC circuit. Answer: D 1vi. The given circuit diagram is From the circuit diagram V1=0, I2= 0  Z11 Z  21 Z12   0 ∞ undefined  Y11 Y12   = =     Z 22   Finite ∞  Y21 Y22  undefined Finite  0  ∞   h11 h12  0 0 A B  0 = = C D   finite 0  h    21 h22  0 0    So, Answer :C 1vii. Answer :C 1viii. The given network is Apply KVL to the loop DEPARTMENT OF ELECTRICAL ENGINEERING 2016-2017

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NATIONAL INSTITUTE OF SCIENCE & TECHNOLOGY, PALUR HILLS, BERHAMPUR, ODISHA -761 008 1 1 + (100 µ ) S (100 µ ) S 2 V1 ( S ) = 10k + (100 µ ) S S+2 V1 ( S ) = (100 µ ) S 1 S +1 = V2 ( S ) = 10k + (100 µ ) S (100 µ ) S V2 ( S ) S + 1 = V1 ( S )1 S + 2 V1 ( S ) = 10k + Answer: D 1ix. Given network is Apply KVL equations V1=12I1+6I2 and V2=20I1+10I2  Z11 Z12  12 6  Z =   21 Z 22  20 10 ∆Z = 0 − Z 21 Y21 = =∞ ∆Z So, Answer is does not exist. Answer: D 1x. The given network V1=AV2+BI2 I1=CV2+DI2 V2=Z2(I1+I2)………(i) DEPARTMENT OF ELECTRICAL ENGINEERING 2016-2017

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NATIONAL INSTITUTE OF SCIENCE & TECHNOLOGY, PALUR HILLS, BERHAMPUR, ODISHA -761 008 I C = 1 I2 = 0 V2 putting , I 2 = 0, in equation(i ) V2 = Z 2 I1 V ⇒ Z2 = 2 I2 = 0 I1 Z2 = 1 1 = C 0..025∠45 o Z 2 = 40∠ − 45 o Answer: B 2. Write all the answers 2 X 10 = 20 MARKS a) The parameter type and the matrix representation of the relevant two port parameters that describe the circuit shown in figure 2.a are Solutions: From the given network, I1=0, V2=0. So, from the two port network only the suitable parameters are g-parameters. Z- parameters V1   Z11 Z12   I1    V  =  Z  2   21 Z 22   I 2  T- parameters V1   A B   V2   I  =    1  C D   − I 2  H- parameters V1   h11  I  = h  2   21 h12   I1  h22  V2  Y- parameters  I1  Y11 Y12  V1     I  = Y  2   21 Y22  V2  Inverse T- parameters V2   A′ B ′   V1   I  =    2  C ′ D ′ − I1  g- parameters or inverse g- parameters  I1   g11 g12  V1    V  =  g  2   21 g 22   I 2  Figure 2.a Note: No need to find allparameters, just observe the circuit diagram I1=0 and V2=0 so, from the table of two – port network parameters left hand side values observe it. Which is suitable directly write those parameters. So, here g- parameters are suitable. Find out those parameters.  I1   g11 V  =  g  2   21 g12  V1  g 22   I 2        g = I1  g = I1 0 0 = =    12 I  11 V 1 V =0 2 V =0     2 1    V2 V2    g g 0 0 = = = =    22 I  21 V 1 2     I 2 =0 V1 = 0 g so, [g ] =  11  g 21 g12  0 0 = g 22  0 0 Only exits g-parameters. b) Check whether 4S + 1 is a Brune’s function or not. Z(S) = S +2 Solution: Brune’s function means Positive real function. i. If the function is brune’s function Z(s) is real for real s, i.e. Z(σ) is real DEPARTMENT OF ELECTRICAL ENGINEERING 2016-2017

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