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
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
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
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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