S.E. Sem. III [EXTC] Circuit and Tranmiion Line Time : Hr.] Prelim Quetion Paper Solution [Mark : 80 Q.(a) Tet whether P() = 5 4 45 60 44 48 i Hurwitz polynomial. (A) P() = 5 4 45 60 44 48 5 45 44 4 60 48 40 40 0 48 48 0 0 A complete row become zero we will form an auxiliary equation. A() = 48 48 A() = 96 0 Replacing zero th row with A() 5 45 44 4 60 48 40 40 0 48 48 96 0 0 48 A all the element in the t column are poitive given f n i a Hurwitz polynomial. Q.(b) The contant of a tranmiion line are R = 6 /km, L =. mh/km, G = 0.5 0 6 /km, C = 0.005 0 5 F/km. Determine the characteritic impedance, propagation contant and attenuation contant at KHz. (A) Given : R = 6 / km, L =. mh/km, G = 0.5 0 6 /km C = 0.005 0 5 F/km, F = KHz To find : Zo,, z = R jl = 6 j 0. 0 = 6 j.8 z = 5.06 66.5 [5] [5]
: S.E. CTL y = G jc = 0.5 0 6 j 0 0.005 0 5 = 0.5 0 6 j.4 0 4 =.4 0 4 89.95 z Characteritic impedance = zo = y = 5.0666.5 4.40 89.95 = 9.00.7 Propagation contant = z.y = 4 (5.0666.5) (.4 0 89.95) = 0.04 j 0.067 / km = j = 0.04 Np/km, = 0.067 rad/km Q.(c) Determine the hort circuit admittance parameter of the network hown : (A) Apply RCL at V V V V V = V V V V = V V = 0.5 V.5 V Apply KCL at V VV I = V I = = 0.5V.5V I = V (0.5) V (.75) V V I I I V V V I V V () [5]
Prelim Quetion Paper Solution Comparing with I = y V y V y = 0.5 y =.75 Apply KCL at V V V I = I = V 0.5V.5V I = 0.5 V.5 V y = 0.5, y =.5 Q.(d) State and prove final value theorem of Laplace tranform. [5] (A) If L {f(t)} = F() then Limf(t) Lim F() t 0 Proof : We know that L {f(t)} = F() f(0) F() = L {f(t)} f(0) F() = = = t Lim f (t) e f(0) 0 0 t Lim [f (t)e ] f(0) 0 0 f(t) f(0) 0 = f(t) 0 f(0) = Lim f(t) f(0) f(0) = t Lim f(t) t Q.(a) The network hown in figure, a teady tate i reached with the witch open. At t = 0, the witch i cloed. Determine V c (0 ), i (0 ), i (0 di ), di (0 ) and (0 ). (A) Step : At t = 0, witch i open i (0 ) = 0 00V 0 i (t) 0 H i (t) 0 F [0]
: S.E. CTL finding i (0 ) 00 0 i 0 i = 0 00 = 0 i i = 00 0 = 0 A 00 V i (0 ) = 0 A V C (0 ) = 0 i (0 0 ) = 0 V C (0 ) = 00 V Step : At t = 0 i (0 ) = 0 A Apply KVL to the outer loop 00 0i 00 0 00 V 0 i = 00 00 i (0 ) =.67 A Step : At t > 0 Apply KVL to loop () di 00 0 i = 0 00 V di = 00 0 i (0 ) 0x0 = 00 di =. A/ Apply KVL to outer loop 00 0 i i0 6 0 Differentiate w.r.t. t di 0 0 06 i = 0 di (0 ) 0 =0 6 i (0 ) 0 0 i i 0 0 V C i i 0 0 0 A 0 00 V 0 H F 4
Prelim Quetion Paper Solution 6 di = 0 (.67) 0 di = 8500 A/ V V V Q.(b) Find the network function,, I I V (A) Let V = I b = V S V a = V I b S = SS V a = 4S I a = V a S = ( 4S )S I a = S 8S I = I a I b = S 8S S I = 8S 4S V = V a I i S = 4S (8S 4S)S = 4S 6S 4 8S V = 6S 4 S V 6S S 4 I = 8S 4S V I = V & 8S 4S I H H I = 0 V F F V 4 V 6S S for the network hown : [5] I S V a H I = 0 V S S V I a I b 5
: S.E. CTL Q.(c) In the circuit hown in figure, find V x. [5] (A) V x V V 0A V V 4 V V x V 4 = 0 V V x = V V () Relation between V & V V V = V V 0V = () Supernode equation VV V V 0 Vx = 0 V V V V 0 (V V ) = 0 (from ) 4V V V = 0 () KCL to V VV V = 0 V 0V V = 0 (4) Solving (), () & (4) V = V V = 0 V V = 4 V V x = V V V x = 4 V x = 6 V V x V 0A V x 6
Prelim Quetion Paper Solution Q.(a) Find the voltage acro 5 reitor in the network hown below. If K = 0.8 i coefficient of coupling. (A) For a magnetically coupled circuit, M = K 50 X m = K X X = 0.8 50 = 5.66 L L The equivalent circuit in term of dependent ource can be drawn a j5 j5.66 j0 j5.66 I 50 0 V Applying KVL to Meh, 50 0 j5i (I I ) j4 (I I ) j5.66i = 0 50 0 = ( j) I ( j.66) I () ( j) I ( j.66) I = 50 0 Applying KVL to Meh, j4 (I I ) (I I ) j0i j5.66i 5I = 0 j4 I j4i I I j0i j5.66i 5I = 0 j4 I j4i I I j0i j5.66i 5I = 0 ( j.66) I (8 j6) I = 0 () By Cramer rule, I = j 500 j.66 0 j j.66 j.66 8j6 00 j5 I j4 I = 8.6 4.79 A V = 5I = 5 (8.6 4.79) = 4. 4.79 V j j0 5 5 [8] 7
: S.E. CTL Q.(b) Check the poitive real function : (i) F() = (A) (i) F 6 5 9 4 6 5 94 Step : Let F (ii) F() = N 6 5 D 94 6 7 5 F 7 The function f() ha pole at = 7 & = & zero at = 5 and = All the pole & zero are in the left half of plane. Step : A there i no pole on jw axi, the reidue tet i not carried out. Step : Even part of N() = m = 5 Odd part of N() = n = 6 Even part of D() = m = 4 Odd part of D() = n = 9 A(w ) = m m n n = ( 5) ( 4) (6) (9) = jw = 4 5 70 = jw A(w ) = w 4 5w 70 A (w ) i poitive for all w 0 A all the three condition are atified, the function i a poitive real function. 6 7 (ii) F N 6 7 Step : Let F D 6 7 F From above both the pole are lying at = Now we will check for N() N() = 6 7 The Routh array of N() i given by 7 6 9 0 6 0 A all element in t column are ve, N() i Hurwitz polynomial. t condition i atified. [8] 8
Prelim Quetion Paper Solution Step : A there i no pole on jw axi, reidue tet need not to be carried out. Step : Even part of N() = m = 6 Odd part of N() = n = 7 Even part of D() = m = Odd part of D() = n = A(w ) = m m n n = (6 ) ( ) ( 7) () =jw = 4 4 5 = jw A(w ) = 4w 4 5w We can ay that A (w ) i poitive for all w 0 A all three condition are atified, function f() i a poitive real function. Q.(c) Lit the type of damping in erie R-L-C circuit and mention the condition for each damping. (A) Nature of Root Sytem Repone. Negative unequal Real Over damped pt pt ke ke Critically damped. Negative Real equal pt pt ke kte. Complex conjugate 4. Conjugate imaginary Underdamped t e k cotk int Ocillatory kcotkin t [4] where = R L and 0 = LC 9
: S.E. CTL Q.4(a) For the network hown, determine the current i(t) when the witch i cloed at t = 0 with zero initial condition. (A) Step At t = 0, witch i open i(0 ) = 0 A V c (0 ) = 0V Step At t > 0, witch i cloed. Applying KVL e 6 5 5i() i() i() = 0 5e 6 = i() 5 5e = 5 6 ( 5 6) i() = i() = i() = Let i () = 5 5r(t ) i(t) 5e 5e ( 5 6) 5e ( )( ) 5e i() ( )( ) 5 i() = A B C = A( ) ( ) B ( ) C ( ) A = 6, B =, C = i(c) = 5e i () = 5e 6 ( ) ( ) 6 H F 6 [8] 0
Prelim Quetion Paper Solution 5e 5e 5e i() = 6 ( ) ( ) Taking invere laplace tranform i(t) = 5 5 5 6 (t) (t) u(t ) e u(t ) e u(t ) Q.4(b) In the given network witch i cloed at t = 0. Solve for V, dv, (A) dv at t = 0. Step : At t = 0, witch i open. i L (0 ) = 0 A V c (0 ) = 0V & V(0 ) = 0V Step : At t = 0, witch i cloed. 5A Step : At t > 0 Applying KCL to V t = 0 i L (0 ) = 0A V (0 ) = 0V, V(0 ) = 0V V 6 dv V 0. 0 0 0.5 = 5 () At t = 0 V(0 ) dv 0 6 00. 0 (0 ) = 5 6 dv 0.0 (0 ) = 5 0 5A 0 0.5H 0 5A 0 0.5H V(t) V(t) 5A 0.F 0.F [8]
: S.E. CTL dv (0 ) = 5 0. 0 6 dv (0 ) = 50 0 6 v/ Diff. equation () w.r.t. t dv V(0 ) 0. 0 6 dv = 0 0 0.5 6 dv 0. 0 = dv dv = dv 0 6 50 0 0 0.0 = 50 0 V/ 6 I Q.4(c) Obtain pole-zero plot for I. I [4] (A) By C.D.R. 4000 0 I = S I 4000 0 00 0S S 40000S I I = S 4000 0S 00S 0S S I I = 0S 0S4000 I 4000 0S 0(S 400) I = 0 (S S 400) I I = S S400 S400 Z = 400 P = 0.5 7.0j P = 0.5 7.0j I I 0 50 F 0 4000 S 00 0 H I 00 0 S
400 00 00 00 Prelim Quetion Paper Solution Q.5(a) Syntheize the driving point function uing Foter-I and Foter-II [0] ( ) ( 9) form : Z() = ( 4) (A) Foter I form : A degree of numerator i greater than degree of denominator, diviion i firt carried out. 4 4( )( 9) 4 40 6 Z() = ( 4) 4 4 4 4 4 40 6 4 4 6 4 6 Z() = Quotient Remainder 4 6 = 4 Divier 4 By partial fraction expanion k k 0 Z() = 4 4 4()(9) k z() 9 0 0 4 ( 4)z() 4( 4 )( 4 9) 5 k ( 4) 4 9 5 Z() 4 4 t term 4 inductor of 4H nd term 9 capacitor of F 9 rd term parallel LC network j j0 j0 j0 0 00 00 00 400 j0 j0 j0 = 4 ( 4) 4 6
: S.E. CTL For parallel LC network c Z () LC LC Comparing the term 5 C F and L H 5 4 Correponding network i a follow. Foter II form : It i obtain by partial fraction expanion of admittance function ( 4) Y() 4( )( 9) By partial fraction expanion k k Y() 9 ( ) ( 4) k y() 8( 9) 64 L FI Realization 9 ( 94) 5 k y() 8( 9 ) 64 9 5 y() 9 Above two term preent admittance of erie LC network for erie LC L network Y () LC 5 F F LC 88 By comparion FII H H Realization 5 L H C 4H F F 9 5 H 4 F 5 4
Prelim Quetion Paper Solution 5 L H C F 5 88 Correponding network i a hown. Q.5(b) Obtain hybrid parameter of the inter-connected network. [0] (A) Separate the two network 0 8 t network V x = 0 I V x 4 Apply KVL to loop of I V I V 0I 4(I I ) V x = 0 I V V = 0 I 4(I I ) (0I ) V x V = 0I 4I 4I 0I V = 44I 4I Comparing with V = z I z I z = 44 z = 4 Apply KVL to loop of I V 8I 4(I I ) V x = 0 V = 8I 4(I I ) (0I ) V = 4I I Comparing with V = z I z I z = 4 z = z = z z z z = 44 4 4 z = 9 Now finding h-parameter from Z-parameter z h = = 9 z =.66 z h = = 4 z = 0. z 4 h = = =.8 z 0 4 0 V x 5 4 V x 8 5
: S.E. CTL h = h h z h h V = = 0.08.66 0. =.8 0.08 0 I I I 5 Apply KVL to I 5(I I ) I (I I ) = 0 0I 5I I = 0 5I I I = 0 I = 0.5 I 0. I Apply KVL to I V 0I 5(I I ) = 0 V = 0I 5I 5(0.5I 0.I ) V =.5I I Z =.5 & Z = Apply KVL to I V 4I (I I ) = 0 V = 4I I (0.5I 0. I ) V = I 5.6 I Z = Z = 5.6 z = Z Z Z Z =.5 5.6 = 69 Converting Z to h parameter '' '' h h. 0.78 = '' '' h h 0.78 0.78 h parameter of the interconnected network i given by, ' '' ' '' h h h h h h = ' '' ' '' h h h h h h h h 44.98 0.5 = h h.0 0.6 I V 4 I () 6
Prelim Quetion Paper Solution Q.6(a) The characteritic impedance of a high frequency line i 00. If it terminated by a load impedance of 00 j00. Uing mith chart, find out : (i) VSWR, (ii) Reflection coefficient, (iii) Impedance at (/0) th of wave length away from load, (iv) VSWR minimum and VSWR maximum away from the load. (A) Z L = 00 j00 00 j00 Z = = j L 00 i) VSWR =.6 ii) Reflection coefficient = 0.4464 th iii) Impedance at of away from load =.5 j0.4 0 iv) VSWR minimum = 0.5 0.088 = 0.8 VSWR maximum = 0.5 0.6 = 0.088 (Smith chart i attached behind.) Q.6(b) Find the Thevenin equivalent acro the terminal XY for the circuit hown in figure. (A) A A V x = 5 Relation between node V & V V V = 0 ix V V V = 0 5 V V = V V V = 0 Supernode equation i x 0i x 5 V V i x 5 0i x 0 0 () V V = () 5 0 Solving () and () V = 0 V 0V 0V X Y V th X Y [0] [0] 7
: S.E. CTL V = 0 V Relation between V & V th V V th = 0 0 0 = V th V th = 0 V Finding I SC 0 i x 0V A I = A i x = I I = I () Apply KVL to I 5(I ) 0 ( I ) 0 (I I SC ) = 0 5I 0I SC = 5 () Apply KVL to I SC 0(I SC I ) 0 = 0 0I 0I SC = 0 () Solving () to () I = A I SC = A Find R th Vth R th = = 0 I = 0 SC Thevenin' equivalent circuit i a follow 0 V 0 i x I 5 I 0 I SC x y I SC 8