WIXLIB

GATE SOLVED PAPER - EESimilarly,2014-34 sin 2wh 2 sin 2wh w2wsin w/2hrect (t ) * sin C c w m 2pw/2rect c t m * 4 sin C b 2w l p4andTherefore, we have the fourier transformsF "x (2t), F "rect (t ),sin w/2h w/22 sin (2w)wThus, we obtain the fourier transform of signal y (t ) asandF {x (t/2)} F {rect (t/4)} Sol. 10For the signal f t h 3 sin 8pt 6 sin 12pt sin 14pt , the minimum samplingfrequency (in Hz) satisfying the Nyquist criterion is .t.cQ. 10omF {y (t )} F {x (2t) * x (t/2)} F {x (2t)} .F {x (t/2)}sin w/2h 2 sin (2w) ww/2 42 sin b w l sin (2w)2wCorrect answer is 14 Hz.Given the signal,buf t h 3 sin 8pt 6 sin 12pt sin 14ptAccording to Nyquist criterion the sampling frequency isywf s 2f maxwhere fmax is the maximum frequency of the message signal. For the given signal,we have maximum frequency asmw max 14por2pf max 14porf max 7Thus, the sampling frequency isfs 2 # 7orfs 14i.e. minimum sampling frequency is 14 Hz.Q. 11In a synchronous machine, hunting is predominantly damped by(A) mechanical losses in the rotor(B) iron losses in the rotor(C) copper losses in the stator(D) copper losses in the rotorSol. 11Correct option is (D).After sudden change of the load in three phase synchronous machine, the rotorhas to search or hunt for its new equilibrium position. It causes the hunting,which can be damped by rotor copper losses.Q. 12A single phase induction motor is provided with capacitor and centrifugal switchin series with auxiliary winding. The switch is expected to operate at a speed of0.7 Ns, but due to malfunctioning the switch fails to operate. The torque-speedcharacteristic of the motor is represented by

2014-3Correct answer is (C).If the switch operates correctly, then the torque-speed characteristic ismywSol. 12but.comGATE SOLVED PAPER - EEBut, the switch does not operate due to malfunctioning. So, there will be nodiscontinuity in the characteristic curve. Thus, we have the modified characteristiccurve of the motor asQ. 13The no-load speed of a 230 V separately excited dc motor is 1400 rpm. Thearmature resistance drop and the brush drop are neglected. The field current iskept constant at rated value. The torque of the motor in Nm for an armaturecurrent of 8 A is .

GATE SOLVED PAPER - EESol. 132014-3Correct answer is 12.55 N-m.A dc motor circuit is shown below.GivenNo load speed,N 1400 rpmIa 8 AV 230 VSo, we obtain the torque of the motor ast V Ia 2pV Iaw60 # N 2p230 # 860 # 1400 9.55 # 230 # 81400t.comArmature current,Excitation voltage, 12.55 N-mIn a long transmission line with r , l , g and c are the resistance, inductance,shunt conductance and capacitance per unit length, respectively, the condition fordistortionless transmission is(A) rc lg(B) r l/cCorrect option is (A).For a transmission line to be distortionless, the required condition isR GLCmSol. 14(D) g c/lyw(C) rg lcbuQ. 14G shunt conductance (in siemens/meter)C capacitance (in Farad/meter)R Resistance (in ohm/meter)L inductance (in henries per meter)So, for the given parameters of long transmission line, the required condition fordistortionless transmission becomesr gclwhereorQ. 15rc lgFor a fully transposed transmission line(A) positive, negative and zero sequence impedances are equal(B) positive and negative sequence impedances are equal(C) zero and positive sequence impedances are equal(D) negative and zero sequence impedances are equal

GATE SOLVED PAPER - EE2014-3Correct option is (B).A fully transposed transmission line is completely symmetrically and thereforethe phase impedance offered by it is independent of phase sequence of a balancedset of current. In other word impedance offered by it to positive and negativesequence currents are equal.Q. 16A 183 bus power system has 150 PQ buses and 32 PV buses. In the general case, toobtain the load flow solution using Newton-Raphson method in polar coordinates,the minimum number of simultaneous equations to be solved is .Sol. 16Correct answer is 332.Given,number of PQ buses 150number of PV buses 32For PQ buses, 2 equations are formed (one for P and other for Q)For PV buses only 1 equation is formed (for Q only)Thus, the minimum number of simultaneous equations, required to obtain theload, isTotal number of equations Number of equation for PQ buses omSol. 15Number ofequation for PV buses.t.c 2 # (number of PQ buses) 1 # (number of PV buses)The signal flow graph of a system is shown below. U s h is the input and C s h isthe output.mywQ. 17bu 2 # (150) 1 # (32) 332Assuming, h1 b1 and h 0 b 0 - b1 a1 , the input-output transfer function,C s hof the system is given byG s h U s h(B) G s h 2 a1 s a 0(A) G s h 2 b 0 s b1s b1 s b 0s a 0 s a1(C) G s h Sol. 17b1 s b 0s 2 a1 s a 0(D) G s h Correct option is (C).We have the signal flow graph of the system asa 0 s a1s 2 b 0 s b1

GATE SOLVED PAPER - EE2014-3omNow, we have to obtain the transfer function,C s hG s h U s hFrom the SFG, we have two forward paths given asp1 h 0 1 1 1 1 1 h20s ssp 2 h1 1 1 h1ssThe graph determinant is given bybut.cT 1 - (sum of all loop gains) (sum of all gain products of 2 non-touching loops)- (sum of all gain products of 3 non-touching loops) .So,T 1 - a - a1 - a2 0 kss 1 a1 a20ssAlso, we have the determinants for the forward paths p1 and p2 asywT1 1mT 2 1 - a - a1 k 1 a1ssThus, using Mason’s gain formula, we obtain the transfer function as/ p k D k p1 D 1 p 2 D 2T.F. DDh 0 1 h1 1 a12 hsask s1 a1 a20ssh 0 h1 s a1h 2s a1 s a 0Since, we haveh1 b1andh 0 b 0 - b 1 a1Substituting it in equation (i), we getT.F. orG s h b 0 b1 ss a1 s a 02C s h 2 b 0 b1 sV s hs a1 s a 0

GATE SOLVED PAPER - EEQ. 182014-3A single-input single output feedback system has forward transfer function G s hand feedback transfer function H s h. It is given that G s h H s h 1. Which ofthe following is true about the stability of the system ?(A) The system is always stable(B) The system is stable if all zeros of G s h H s h are in left half of the s -plane(C) The system is stable if all poles of G s h H s h are in left half of the s -plane(D) It is not possible to say whether or not the system is stable from theinformation givenSol. 18Correct option is (A).For the given system, we haveG s h H s h 1So, we may easily conclude that the Nyquist-plot intersect the negative real axisbetween 0 and - 1 point, i.e. the Nyquist plot does not enclose the point - 1, 0hor in other words number of encirclements is zero.omN 0From Nyquist criterion, we have the relation.(i)but.c.(ii)N P-Zwhere P is the number of open loop poles in right-half of s -plane and Z is thenumber of closed loop poles (roots of characteristic equation) in right-half of s-plane. Since, for a stable closed loop system, we must have.(iii)Z 0Thus, from equations (i), (ii), and (iii), we getAn LPF wattmeter of power factor 0.2 is having three voltage settings 300 V, 150V and 75 V, and two current settings 5 A and 10 A. The full scale reading is 150.If the wattmeter is used with 150 V voltage setting and 10 A current setting, themultiplying factor of the wattmeter is .mQ. 19ywP 0i.e. no poles of open loop system lies in right-half of s -plane, or in other words,all the poles lies in left-half of s -plane.Hence, the given system is stable if all poles of G s h H s h are in left half of the s-plane.Sol. 19Correct answer is 2.In low power factor (LPF) wattmeter, the multiplying factor is given bypower settingM.F. power for maximum difflection voltage range usedh # current range usedh # (power factor) Full scale readingFrom the given data, we haveFull scale reading 150Voltage range used 150 VCurrent range used 10 APower factor 0.2Substituting these values in above expression, we getM.F. 150 # 10 # 0.2 2150

GATE SOLVED PAPER - EEThe two signals S1 and S2, shown in figure, are applied to Y and X deflectionplates of an oscilloscope.omQ. 202014-3mywbut.cThe waveform displayed on the screen isSol. 20Correct option is (A).We consider the two signals S1 and S2 applied to Y and X deflection plates forthe period T as

GATE SOLVED PAPER - EE2014-3A state diagram of a logic gate which exhibits a delay in the output is shown inthe figure, where X is the don’t care condition, and Q is the output representingthe state.mQ. 21ywbut.comobserving the waveforms, we condude thatFor 0 t T/2 , 0 S1 1 and S2 1At t T , S1 S2 02For T t T , 0 S1 - 1 and S2 - 12Since, S1 is applied to Y -deflection plate and S2 is applied to X -deflection plate.So, we haveFor 0 t T/2 ; 0 y 1, x 1At t T/2 ; x y 0For T/2 t T ; - 1 y 0 , x - 1Thus, the waveform displayed on the CRT screen isThe logic gate represented by the state diagram is(A) XOR(B) OR(C) AND(D) NANDSol. 21Correct option is (D).We have the state diagram of a logic gate,

GATE SOLVED PAPER - EE2014-3For the state diagram, we form the truth table (irrespective of the current state)asABX001011101110Thus, from the truth table, it is clear that the logic gate is NAND gate.An operational amplifier circuit is shown in the figure.t.comQ. 22mSol. 22ywbuThe output of the circuit for a given input vi is(A) -b R2 l vi(B) -b1 R2 l viR1R1(C) b1 R2 l vi(D) Vsat or - VsatR1Correct option is (D).We have the op-amp circuit asFor an operation amplifier, we know that1. If vi Vsat ; then output voltage will be Vsat2. If vi - Vsat ; then the output voltage will be - Vsat3. If - Vsat vi Vsat ; then the output voltage varies according to the appliedinput.For the given op-amp circuit, we assume that- Vsat vi Vsat

GATE SOLVED PAPER - EE2014-3So, we obtain the output for the corresponding input vi . Applying KCL at noninverting terminal of first op-amp, we havevi - 0 vi - x 0R1R2orvi b 1 1 l xR2R1 R 2or.(i)x b R1 R 2 l v iR1where x is the voltage at non-inverting terminal of second op-amp. Again, applyingKCL at non-inverting terminal of second op-amp, we getx - 0 x - v0 0RRv 0 2xor[substituting equation (i)]v 0 2vi R1 R2hR1 2vi b1 R2 lR1There is no single option so our assumption is wrong and vi Vsat and vi - Vsat. So, the output voltage is Vsat and - Vsat .omorIn 8085A microprocessor, the operation performed by the instruction LHLD2100 H is(A) Hh ! 21 H , Lh ! 00 H(B) Hh ! M 2100hH , Lh ! M 2101 Hh(C) Hh ! M 2101 Hh, Lh ! M 2100 Hh(D) Hh ! 00 H , Lh ! 21 HSol. 23Correct option is (C).For the 8085A microprocessor, given instruction isLHLD 2100 HThis operation loads registers register L and H with the content in memory atlocations 2100 H and 2101 H respectively, i.e.mywbut.cQ. 23 Hh ! M 2101hH and Lh ! M 2100hHQ. 24A non-ideal voltage source Vs has an internal impedance of Zs . If a purely resistiveload is to be chosen that maximizes the power transferred to the load, its valuemust be(A) 0(B) real part of Zs(D) complex conjugate of Zs(C) magnitude of ZsSol. 24Correct option is (C).Consider the circuit shown below with non-ideal voltage source Vs , an internalimpedance Zs Rs jXs , and purely resistive load RL .

GATE SOLVED PAPER - EE2014-3Current through the circuit is given byVsRs RL jXsVsI Rs RL h Xs h2I orTherefore, the power transferred to load isP I 2 RL V s2 RL Rs RL h2 X s2hFor maximum power, we havedP 0dRLor[ Rs RL h2 X s2 ] 1 - 2RL Rs RL h 02 Rs RL h2 X s2hR s2 R L2 2Rs RL X s2 - 2Rs RL - 2R L2 0omorR s2 X s2 R L2orThe torque speed characteristics of motor TM h and load TL h for two cases areshown in the figures (a) and (b). The load torque is equal to motor torque atpoints P , Q , R and S .mywbuQ. 25t.corRL R s2 X s2i.e. to maximise the power transferred to pure resistive load (RL ), its value mustbe equal to the magnitude of Zs of internal circuit.The stable operating points are(A) P and R(C) Q and RSol. 25(B) P and S(D) Q and SCorrect option is (B).Since, increase in torque accelerates the motor, so the slope of torque speedcharacteristic should be negative to restore the speed to original value, which inturn makes the system stable.Thus, for stable operating point, the slope of torque-speed characteristic shouldbe negative.Hence, the points P and S are stable operating points.

GATE SOLVED PAPER - EE2014-3Q. 26 - Q. 55 Carry two marks each.Q. 26Sol. 262Integration of the complex function f z h 2 z , in the counterclockwisez -1direction, around z - 1 1, is(A) - pi(B) 0(C) pi(D) 2piCorrect option is (C).Given the complex function,z2z -1and the contour, along which integration to be performed, isC : z - 1 1f z h 2t.comHere, z - 1 1 is a circle with radius 1 and centre at point (1,0). So, we sketchthe contour C asbuNow, we have the complex functionf z h z2z -12mywSo, the poles and zeros of the function arepoles 1, - 1zeros 0 , 0Therefore, only pole (z 1) lies within the curve C . So, the residue at z 1 poleis2Residue lim zz " 1 z 1h2 lim z 1 11 1 2z"1 z 1Thus, the integral of complex function along the contour is obtained as# f z hdzC 2pi (sum of residues) 2pi b 1 l2 piQ. 27The mean thickness and variance of silicon steel laminations are 0.2 mm and0.02 respectively. The varnish insulation is applied on both the sides of thelaminations. The mean thickness of one side insulation and its variance are 0.1mm and 0.01 respectively. If the transformer core is made using 100 such varnishcoated laminations, the mean thickness and variance of the core respectively are(A) 30 mm and 0.22(B) 30 mm and 2.44(C) 40 mm and 2.44(D) 40 mm and 0.24

GATE SOLVED PAPER - EESol. 27Q. 28Sol. 282014-3Correct option is (D).The function f x h ex - 1 is to be solved using Newton-Raphson method. If theinitial value of x 0 is taken as 1.0, then the absolute error observed at 2 nd iterationis .Correct answer is 0.06005.We have the function,f x h ex - 1So, its first derivative is obtained asf l x h exBy Newton-Raphson method, we havef xi hxi 1 xi f l xi hSince, we have the initial value,omx0 1So, we obtain the 1st iteration asf x 0hf l x 0hx1 x0 - e ex1 1 - e -1 1 e-1 .3678ex1 x 0 -0t.c0f x1hf l x1hx1 x1 - e ex x1 - 1 e-x 0.3678 - 1 e-0.3678buTherefore, the 2nd iteration isQ. 29Sol. 29mywx 2 x1 -111 .06005The Norton’s equivalent source in amperes as seen into terminals X and Y is.Correct answer is 2.To obtain the Norton equivalent along X -Y terminal, we redraw the given circuitas

GATE SOLVED PAPER - EE2014-3t.comIn the network, 5 W resistor is short-circuit. So, the equivalent circuit isNow, writing the KVL equation in two loops, we have5 - 5I1 - 5 I1 - Isc h 0- 5 Isc - I1h - 2.5Isc 2.5 0Solving equation (i),bu.(i).(ii)10I1 - 5Isc 52I1 Isc 1yw.(iii)Solving equation (ii),moror7.5Isc - 5I1 2.53Isc - 2I1 13Isc - Isc - 1 1[using equation (iii)]2Isc 2Thus,Isc 1 Ai.e. the Norton equivalent source as seen into terminals X and Y is 1 A.orQ. 30The power delivered by the current source, in the figure, is .Sol. 30Correct answer is 3 W.To obtain the required unknown, we redraw the given circuit as

GATE SOLVED PAPER - EE2014-3Applying KCL at node Vx , we haveVx - 1 Vx 2112Vx - 1 2orVx 32Therefore, the power delivered by current source is obtained asomor Voltage across current source # current throughcurrent sourcet.cmywbuQ. 31 3#2 3W2A perfectly conducting metal plate is placed in x -y plane in a right handedcoordinate system. A charge of 32pe 0 2 coulombs is placed at coordinate 0, 0, 2h. e 0 is the permittivity of free space. Assume i , j , k to be unit vectorsalong x , y and z axes respectively. At the coordinate 2 , 2 , 0h, the electricfield vector E (Newtons/Coulomb) will beSol. 31(A) 2 2 k(B) - 2k(C) 2k(D) - 2 2 kCorrect option is (D).We have the given system of a charge and a perfectly conducting metal plate asshown below.

GATE SOLVED PAPER - EE2014-3but.comDue to presence of conducting plane, we can assume an image charge for thesystem as shown belowywSo, we can observe from the figure that electric field vector along x -y planeis cancelled, i.e. zero. Therefore, the electric field vector acts along negative z-direction and given as".(i)E d E1 E2 n - kth22mwhere E1 and E2 are the field intensity due to the charge Q and - Q at point( 2 , 2 , 0). So, we obtainQ32pe 0 2E1 1 2 14pe 0 r4pe 0 ( ( 2 ) 2 ( 2 ) 2 (2) 2) 2 2Similarly, we get, E2 2Substituting these values in equation (i), we obtain"2 2 - 2ktE eQ. 322 - kto h2A series RLC circuit is observed at two frequencies. At w 1 1 krad/s , we notethat source voltage V1 100 0c V results in a current I1 0.03 31c A . Atw 2 2 krad/s , the source voltage V2 100 0c V results in a current I2 2 0c A. The closest values for R, L, C out of the following options are(A) R 50 W ; L 25 mH ; C 10 mF ; (B) R 50 W ; L 10 mH ; C 25 mF ;(C) R 50 W ; L 50 mH ; C 5 mF ;(D) R 50 W ; L 5 mH ; C 50 mF ;

GATE SOLVED PAPER - EECorrect option is (B).Given the current and voltage at different frequencies asAt w 1 1 krad /s,V1 100 0I1 0.03 31cAt w 2 2 krad /s,V2 100 0I 2 2 0cFrom the given data, it is clear that voltage and current are in phase at w 2 2 .So, it satisfies the resonance condition. Therefore, we haveR V 100 50 W2IAlso, for the resonance circuit, we havewL 1wC1or103 # 2 # L 2 # C # 103or.(i)4 # 106 L 1CAgain, at w 1 krad/s, we havec

pq prqs rs pq pr qs 0 22 22 22 2 - h h R T S S S S V X W W W W If rs pq 22 pr qs 22 2 - h h - hps rq 2 0 or ps-rq 0 then rank of matrix B is 1, otherwise the rank is 2. Thus, from the above results, we conclude that If ps rq- 0, then rank of matrix A and B is 1. If ps rq- ! 0, then rank of A and B is 2. i.e. the rank of .