EC-GATE-2014 PAPER-01 Q. No. 1 5 Carry One Mark Each

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1 EC-GATE-4 PAPE- Q. No. 5 Carry One Mark Each. Choose the most appropriate phrase from the options given below to complete the following sentence. The aircraft take off as soon as its flight plan was filed. (A) is allowed to (B) will be allowed to (C) was allowed to (D) has been allowed to Answer: (C). ead the statements: All women are entrepreneurs. Some women are doctors Which of the following conclusions can be logically inferred from the above statements? (A) All women are doctors (B) All doctors are entrepreneurs (C) All entrepreneurs are women (D) Some entrepreneurs are doctors Answer: (D). Choose the most appropriate word from the options given below to complete the following sentence. Many ancient cultures attributed disease to supernatural causes. However, modern science has largely helped such notions. (A) impel (B) dispel (C) propel (D) repel Answer: (B) 4. The statistics of runs scored in a series by four batsmen are provided in the following table, Who is the most consistent batsman of these four? Batsman Average Standard deviation K. 5. L M N (A) K (B) L (C) M (D) N Answer: (A) If the standard deviation is less, there will be less deviation or batsman is more consistent 5. What is the next number in the series? Answer: 75 India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India

2 EC-GATE-4 PAPE difference Q. No. 6 Carry One Mark Each 6. Find the odd one from the following group: W,E,K,O I,Q,W,A F,N,T,X N,V,B,D (A) W,E,K,O (B) I,Q,W,A (B) F,N,T,X (D) N,V,B,D Answer: (D) W E K O Q W A F N T X N V B D Difference of position: D 7. For submitting tax returns, all resident males with annual income below s lakh should fill up Form P and all resident females with income below s 8 lakh should fill up Form All people with incomes above s lakh should fill up Form, except non residents with income above s 5 lakhs, who should fill up Form S. All others should fill Form T. An example of a person who should fill Form T is Answer: (B) (A) a resident male with annual income s 9 lakh (B) a resident female with annual income s 9 lakh (C) a non-resident male with annual income s 6 lakh (D) a non-resident female with annual income s 6 lakh esident female in between 8 to lakhs haven t been mentioned. 8. A train that is 8 metres long, travelling at a uniform speed, crosses a platform in 6 seconds and passes a man standing on the platform in seconds. What is the length of the platform in metres? Answer: 56 For a train to cross a person, it takes seconds for its 8m. So, for second 6 seconds. Total distance travelled should be 84. Including 8 train length so length of plates India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India

3 EC-GATE-4 PAPE The exports and imports (in crores of s.) of a country from to 7 are given in the following bar chart. If the trade deficit is defined as excess of imports over exports, in which year is the trade deficit /5th of the exports? Exports Im ports (A) 5 (B) 4 (C) 7 (D) 6 Answer: (D) imports exp orts 4, exp orts , , 5 7,. You are given three coins: one has heads on both faces, the second has tails on both faces, and the third has a head on one face and a tail on the other. You choose a coin at random and toss it, and it comes up heads. The probability that the other face is tails is Answer: (B) (A) /4 (B) / (C) / (D) / India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India

4 EC-GATE-4 PAPE- Q. No. 5 Carry One Mark Each. For matrices of same dimension M, N and scalar c, which one of these properties DOES NOT ALWAYS hold? (A) (M T ) T M (B) (cm T ) T c(m) T (C) (M + N) T M T + N T (D) MN NM Answer: (D) Matrix multiplication is not commutative in general.. In a housing society, half of the families have a single child per family, while the remaining half have two children per family. The probability that a child picked at random, has a sibling is Answer:.667 Let E one children family E two children family and A picking a child then by Baye s theorem, required probability is.x E P.667 A x. +.x (Here x is number of families). C is a closed path in the z-plane given by z. The value of the integral dz is Answer: (C) (A) 4π ( + j) (B) 4π( j) (C) 4π ( + j) (D) 4π( j) Z j is a singularity lies inside C : Z By Cauchy s integral formula, Z Z + 4j dz π j. Z Z + 4j C Z + j Z j [ ] [ ] πj 4 + j + 4j 4π + j z z + 4j z + j C 4. A real (4 4) matrix A satisfies the equation A I, where I is the (4 4) identity matrix. The positive eigen value of A is. Answer: A I A A if λ is on eigen value of A then is also its eigen value. Since, we λ require positive eigen value. λ is the only possibility as no other positive number is self inversed India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 4

5 EC-GATE-4 PAPE Let X, X, and X be independent and identically distributed random variables with the uniform distribution on [, ]. The probability P{X is the largest} is Answer: For maximum power transfer between two cascaded sections of an electrical network, the relationship between the output impedance Z of the first section to the input impedance Z of the second section is Answer: (C) (A) Z Zl (B) Z Zl (C) Z Z (D) Z Z Two cascaded sections Section Z ZL Z Section Z Output impedance of first section Z Input impedance of second section For maximum power transfer, upto st section is Z Z * L Z Z Z * L 7. Consider the configuration shown in the figure which is a portion of a larger electrical network i 5 i i 4 i i i6 For Ω and currents i A, i 4 -A, i 5-4A, which one of the following is TUE? (A) i 6 5 A (B) i Answer: (A) 4A (C) Data is sufficient to conclude that the supposed currents are impossible (D) Data is insufficient to identify the current i,i, and i 6 India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 5

6 EC-GATE-4 PAPE- Given i A i A i 4 5 4A KCL at node A, i + i4 i i A. KCL at node B, i + i5 i i 4 A KCL at node C, i + i i 6 6 i 5A i 4 A Ω i i Ω i 5 B Ω i C i 6 8. When the optical power incident on a photodiode is µ W and the responsivity is.8a / W, Answer: 8 the photocurrent generated ( in µ A) is. esponsivity( ) p I I 8µ A Ip P 9. In the figure, assume that the forward voltage drops of the PN diode D and Schottky diode D are.7 V and. V, respectively. If ON denotes conducting state of the diode and OFF denotes non-conducting state of the diode, then in the circuit, kω Ω Ω D D (A) both D and D are ON (C) both D and D are OFF Answer: (D) Assume both the diode ON. Then circuit will be as per figure ().7 I 9.mA k.7. ID ma Now, I I I D D.7 ma Not possible D is OFF and hense D ON (B) D is ON and D is OFF (D) D is OFF and D is ON V V K K Figure I Ω D D Ω I I D D.7V.V India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 6

7 EC-GATE-4 PAPE- If fixed positive charges are present in the gate oxide of an n-channel enhancement type MOSFET, it will lead to (A) a decrease in the threshold voltage (B) channel length modulation (C) an increase in substrate leakage current (D) an increase in accumulation capacitance Answer: (A). A good current buffer has (A) low input impedance and low output impedance (B) low input impedance and high output impedance (C) high input impedance and low output impedance (D) high input impedance and high output impedance Answer: (B) Ideal current Buffer has Zi Z. In the ac equivalent circuit shown in the figure, if i in is the input current and F is very large, the type of feedback is D D υ out M M small signal input i in F (A) voltage-voltage feedback (B) voltage-current feedback (C) current-voltage feedback (D) current-current feedback Answer: (B) Output sample is voltage and is added at the input or current It is voltage shunt negative feedback i.e, voltage-current negative feedback. In the low-pass filter shown in the figure, for a cut-off frequency of 5kHz, the value of in kω is. Vi kω C + nf Vo India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 7

8 Answer:.8 f 5KHz Cut off frequency ( LPF) 5KHz π C.8 kω 9 π 5 EC-GATE-4 PAPE In the following circuit employing pass transistor logic, all NMOS transistors are identical with a threshold voltage of V. Ignoring the body-effect, the output voltages at P, Q and are, 5V 5V 5V 5V P Q (A) 4 V, V, V (C) 4 V, 4 V, 4 V Answer: (C) Assume al NMOS are in saturation V V V For m ( 5 Vp ) ( 5 Vp ) ( p ) > ( p ) D DS GS T 5 V 4 V Sat ID k V GS VT ( p ) I K 4 V... For m, D D D D ( Q ) ( Q ) I K 5 V I K 4 V... I I ( 4 Vp ) ( 4 VQ ) V V & V + V 8 p Q p Q V V 4V p Q (B) 5 V, 5 V, 5 V (D) 5 V, 4 V, V For m, D D D I K 5 V I I ( 4 VQ ) ( 4 V ) V V 4V Q 5V 5V 5V V V V 4V p Q 5V M M M P Q India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 8

9 EC-GATE-4 PAPE The Boolean expression ( X + Y)( X + Y) + ( X + Y) + X simplifies to Answer: (A) (A) X (B) Y (C) XY (D) X+Y Given Boolean Expression is ( X + Y)( X + Y) + XY + X As per the transposition theorem ( A + BC) ( A + B)( A + C) so, X + Y X + Y X + YY X + ( + )( + ) X Y X Y XY X X XY.X X + X + Y.X X + XX. + Y.X X + + Y.X Apply absorption theorem X + Y X. X 6. Five JK flip-flops are cascaded to form the circuit shown in Figure. Clock pulses at a frequency of MHz are applied as shown. The frequency (in khz) of the waveform at Q is. J4 Q4 >clk K4 J >clk K Q J Q > clk K J Q > clk K J > clk K clock Answer: 6.5 Given circuit is a ipple (Asynchrnous) counter. In ipple counter, o/p frequency of each flip-flop is half of the input frequency if their all the states are used otherwise o/p frequency input frequency of the counter is modulus of the counter So, the frequency at Q input frequency 6 6 H z 6.5kHz 6 7. A discrete-time signal [ ] ( π ) (A) periodic with period π. x n sin n,n being an integer,is (B) periodic with period (C) periodic with period π /. (D) not periodic Answer: (D) π. India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 9

10 Assume x[ n ] to be periodic, (with period N) x[ n] x[ n + N] ( ) sin π n sin π n + N EC-GATE-4 PAPE- Every frigonometric function repeate after π interval. sin π n + π k sin π h + π N k π k π N N π Since k is any integer, there is no possible value of k for which N can be an integer, thus non-periodic. y t band- 8. Consider two real valued signals, x(t) band-limited to [ 5 Hz, 5 Hz] and limited to [ khz, khz ]. For z ( t) x ( t ). y( t ), Answer: x( t ) is band limited to [ 5Hz, 5Hz] y( t) is band limited to [ Hz, Hz] z( t) x ( t ).y( t) the Nyquist sampling frequency (in khz) is Multiplication in time domain results convolution in frequency domain. The range of convolution in frequency domain is [ 5Hz, 5 Hz] So maximum frequency present in z(t) is 5Hz Nyquist rate is Hz or khz 9. A continuous, linear time-invariant filter has an impulse response h(t) described by Answer: 45 { for t h t otherwise When a constant input of value 5 is applied to this filter, the steady state output is. x ( t) h ( t) y ( t) y( t) x ( t )* h( t) x ( t ) 5 h ( t ) t t τ y t.5.d 45 steady state output India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India

11 EC-GATE-4 PAPE- The forward path transfer function of a unity negative feedback system is given by Answer:.5 G s K ( s + )( s ) The value of K which will place both the poles of the closed-loop system at the same location, is. Given G( s) H s K ( s+ )( s ) + G s H s Characteristic equation: K + ( s+ )( s ) 9 The poles are s, ± 4K 4 If 9 K, then both poles of the closed loop system at the same location. 4 So, 9 K.5 4. Consider the feedback system shown in the figure. The Nyquist plot of G(s) is also shown. Which one of the following conclusions is correct? + k G ( s) ImG ( jω) + eg ( jω) Answer: ( D) (A) G(s) is an all-pass filter (B) G(s) is a strictly proper transfer function (C) G(s) is a stable and minimum-phase transfer function (D) The closed-loop system is unstable for sufficiently large and positive k For larger values of K, it will encircle the critical point (-+j), which makes closed-loop system unstable. India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India

12 EC-GATE-4 PAPE- In a code-division multiple access (CDMA) system with N 8 chips, the maximum number of users who can be assigned mutually orthogonal signature sequences is Answer: 7.99 to 8. chip rate Spreading factor(sf) symbol rate This if a single symbol is represented by a code of 8 chips Chip rate 8 symbol rate 8 symbol rate S.F (Spreading Factor) 8 symbol rate Spread factor (or) process gain and determine to a certain extent the upper limit of the total number of uses supported simultaneously by a station.. The capacity of a Binary Symmetric Channel (BSC) with cross-over probability.5 is Answer: Capacity of channel is -H(p) H(p) is entropy function With cross over probability of.5 H( p) log + log.5.5 Capacity S S S S. If the port- of the two-port is short circuited, the S parameter for the resultant one-port network is 4. A two-port network has sattering parameters given by [ S] Answer:(B) s s s + s s + s ( A) s s s + s s s ( C) b a Two port Network s s s s s + s ( B) s s s + s s s ( D) a b b s a + s a b s a + s a b s s a b s s a ; b s a a By verification Answer B satisfies. India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India

13 EC-GATE-4 PAPE The force on a point charge +q kept at a distance d from the surface of an infinite grounded metal plate in a medium of permittivity is (A) (C) Answer:(C) q 6π d ( d) towards the plate QQ F 4π 9 9 F 4 π 6 π d Since the charges are opposite polarity the force between them is attractive. q (B) 6π d (D) q 4π d Q.No Carry Two Marks Each away from the plate towards the plate +q d d q metal plate 6. The Taylor series expansion of sin x + cos x is x A x x... Answer: (A) + + x C x x x x sin x + cos x x !! x + x x +... x B x + x +... x D x x For a Function g(t), it is given that t + ( τ) τ is y t g d,then y t dt jωt ω g t e dt ωe for any real value ω. If (A) (B)-j (C) - j (D) j Answer: (B) Given jwt w ω ( ( ω) ) g t.e dt.e let G j g t dt India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India

14 EC-GATE-4 PAPE- t y t g z.dz y t g t * u t u t in unit step function Y jω G j ω.u jω jωt Y j y t.e dt ( ω ) w Y( j) y( t) dt ω.e + πδ( ω) j ω ω j j 8. The volume under the surface z(x, y) x + y and above the triangle in the x-y plane defined y x and x is. Answer: 864 by { } x ( + ) Volume Z x, y dydx x y dydx x y x y x xy +.dx x dx 864 x 9. Consider the matrix: J 6 Which is obtained by reversing the order of the columns of the identity matrix I 6. Let P I6 + α J 6, where α is a non-negative real number. The value of α for which det(p) is. Answer: Consider, ( i) Let P I + αj ii Let P I J 4 + α 4 α + α α P α α α α α India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 4

15 EC-GATE-4 PAPE- α P α α α α ( ) ( ) ( ) α α α α α S im ilarly, if P I + α J then w e get P α P α, α is non negative α 6 6 α. A Y-network has resistances of Ω each in two of its arms, while the third arm has a resistance of Ω in the equivalent network, inω among the three resistances is. Answer: 9.9Ω the lowest value Ω X Ω Z Ω Ω Star Connection Ω Ω Y Delta Connection X 9.9Ω y Ω z Ω ( )( ) + ( ) + ( ) X Ω ( )( ) + ( ) + ( ) y Ω ( )( ) + ( ) + ( ) z Ω i.e, lowest value among three resistances is 9.9Ω. A V rms source supplies power to two loads connected in parallel. The first load draws kw at.8 leading power factor and the second one draws kva at.8 lagging power factor. The complex power delivered by the source is (A) (8 + j.5) kva (C) ( + j.5) kva (B) (8 - j.5) kva (D) ( - j.5) kva India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 5

16 EC-GATE-4 PAPE- Answer: (B) + V L o a d I L o a d II Load : P kw cosφ.8 SI P jq j7.5 KVA Q P tan φ 7.5 KVA Load : S KVA Q cosφ.8 sin φ S P cosφ S P.8 P 8kw Q 6KVA SI P + jq 8 + j6 Complex power delivered by the source is SI + SII 8 j.5 KVA. A periodic variable x is shown in the figure as a function of time. The root-mean-square (rms) value of x is. x T / T / t Answer:.48 India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 6

17 EC-GATE-4 PAPE- T rms ( ) x x t dt T x t t t T T T t T T T.t.dt + ( ).dt T T T T 4 t. T T 4 T x rms..48 T 8 6 X (, ) T T t. In the circuit shown in the figure, the value of capacitor C(in mf) needed to have critically damped response i(t) is. 4Ω 4 H C i( t) + V O Answer: mf By KVL, di( t) v( t) i( t) + L. i( t) dt dt + C Differentiate with respect to time,.di ( t) di( ti) i( t) +. + dt L dt LC d i( t) di( t) i( t) +. + dt L dt LC D, 4 ± L L LC D, ± L L LC For critically damped response, 4L C L LC Given, L4H; 4Ω 4 4 C mf ( 4) F India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 7

18 EC-GATE-4 PAPE A BJT is biased in forward active mode, Assume VBE.7V,kT / q 5mV and reverse saturation current A. The transconductance of the BJT (in ma/v) is. Answer: IS KT VBE.7V, 5mV, Is q Transconductance, g V BE /VT IC IS e m IC V I ma g T.7/5mV e 44.65mA C m VT 5mV 5.785A / V 5. The doping concentrations on the p-side and n-side of a silicon diode are 6 cm and 7 cm, respectively. A forward bias of. V is applied to the diode. At T K, the intrinsic carrier concentration of silicon ni.5 cm and kt 6mV. concentration at the edge of the depletion region on the p-side is (A) Answer:(A) 9. cm (B) 6 cm (C) ni V bi /VT Electron concentration, n e N A (.5 ) e 6 9. / cm./6mv q 7 cm (D) The electron 6.5 cm 6. A depletion type N-channel MOSFET is biased in its linear region for use as a voltage controlled resistor. Assume threshold voltage 8 VTH.5V,VGS.V, VDS 5V,W / L, COX F / cm and µ n 8cm / V s. in Ω is. Answer:5 The value of the resistance of the voltage controlled resistor W L 8 Given VT.5V; VGS V; VDS 5V; ; Cθ f / cm x µ n 8cm / v s W I µ C ( V V ) V V L D n x GS T DS DS I W r µ C V V V V D ds n x GS T DS DS VDS VDS L India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 8

19 EC-GATE-4 PAPE- r ds W W µ C ( V V ) µ C V L L n x GS T n x DS W µ C V V V L n x GS T Ds 5Ω In the voltage regulator circuit shown in the figure, the op-amp is ideal. The BJT has V.7V and β, and the zener voltage is 4.7V. For a regulated output of 9 V, the BE value of inω is. VI V V 9 V kω + kω Vz 4.7 V Answer:9 Given VBE.7V, β, VZ 4.7V, V 9V V 9 + k V Vz + k 9Ω ( ) V V i V z + 9V K V 8. In the circuit shown, the op-amp has finite input impedance, infinite voltage gain and zero input offset voltage. The output voltage V is (A) I ( + ) out (B) I (C) I l l + V out (D) I ( + ) India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 9

20 EC-GATE-4 PAPE- Answer: (C) Given, Z i A L V i V / / I I... + KCL at inverting node V V V + Zi V V + V + I + V I V V I + V 9. For the amplifier shown in the figure, the BJT parameters are VBE.7 V, β, and thermal voltage 5mV. v / v of the amplifier is. VT The voltage gain i VCC + V v i µ F kω C 5kΩ µ F v o kω S Ω E k Ω C E mf Answer: VBE.7V, β, VT 5mV DC Analysis: k VB V k + k V.7.V I E E..77 ma + k India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India

21 I I B C EC-GATE-4 PAPE- A.6 ma 5mV re.98ω.77 ma A C V i e s V V β 5k V β r + + β.98 + A The output F in the digital logic circuit shown in the figure is X Y XO AND F Z Answer: (A) ( A) F XYZ + XYZ ( B) F XYZ + XYZ ( C) F XYZ + XYZ ( D) F XYZ + XYZ X Y XO K XNO F Z XNO Assume dummy variable K as a output of XO gate K X Y XY + XY ( ) F K. K Z ( KZ K.Z) + K. KZ + K.K.Z ( ) + K.Z K. K and K.K K Put the value of K in above expression F XY + XY Z XYZ + XYZ India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India

22 EC-GATE-4 PAPE Consider the Boolean function, F( w, x, y,z) wy + xy + wxyz + wxy + xz + xyz. which one of the following is the complete set of essential prime implicants? Answer: (D) (A) w,y, xz, x z (B) w,y,xz (C) y,x yz (D) y,xz,xz Given Boolean Function is F( w, x, y, z) wy+ xy+ wxyz + wxy+ xz + xyz By using K-map wx yz x z y xz So, the essential prime implicants (EPI ) are y, xz, xz 4. The digital logic shown in the figure satisfies the given state diagram when Q is connected to input A of the XO gate. S CLK D > Q Q A S D Q > Q S S S S S S S Answer: (D) Suppose the XO gate is replaced by an XNO gate. Which one of the following options preserves the state diagram? (A) Input A is connected to Q (B) Input A is connected to Q (C) Input A is connected to Q and S is complemented (D) Input A is connected to Q India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India

23 EC-GATE-4 PAPE- The input of D flip-flop is ( ) D Q s + Q s A Q The alternate expression for EX-NO gate is A B A B A B So, if the Ex-O gate is substituted by Ex-NO gate then input A should be connected to Q ( ) D Q S + Q S Q S + Q.S A Q Q S + Q.S i 9 transform of x[n] 4. Lex x[ n] u ( n) u( n ) (A) is z > (B) is z 9 Answer: (C) n n < (C) Given x[ n] u[ n] u[ n ] h n 9 for u[ n] oc in z > 9 9 (ight sided sequence, Thus overall oc in < z < 9 n oc in exterior of circle of radius ) Consider a discrete time periodic signal x[n]. The egion of Convergence (OC) of the z- is > z > (D) does not exist. 9 πn sin s. Let a k be the complex Fourier series coefficients of x[n]. The coefficients { a k } are non-zero when k Bm ±, where m is any integer. The value of B is. Answer: πn Given x n sin ; N 5 [ ] Fourier series co-efficients are also periodic with period N x[ n] e e j j π π j n i n a ; a a a + a9 j j j a a + a a + or a a + a a + k m + or k.m B India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India

24 EC-GATE-4 PAPE A system is described by the following differential equation, where u(t) is the input to the system and y(t) is the output of the system.. + y t 5y t u t When y() and u(t) is a unit step function, y(t) is (A) Answer: (A). +.8e 5t (B)..e 5t (C).8 +.e 5t (D).8.8e 5t Given y( t) + 5y( t) u ( t) and y( ) ; u ( t) is a unit step function. Apply Laplace transform to the given differential equation. S y( s) y( ) + 5y( s) s dy y( s)[ s+ 5] + y( ) L s y( s) y( ) L u ( t) s dt s + y s s y s ( s + 5) ( + ) s A B + s s+ 5 s s+ 5 A ; B y( s) + 5s 5 s 5 ( + ) Apply inverse Laplace transform, 4 5t y( t) + e 5 5 y t.+.8e 5t 46. Consider the state space model of a system, as given below. x x x. x x 4 u; y [ ] x +. x x x The system is (A) controllable and observable (B) uncontrollable and observable (C) uncontrollable and unobservable (D) controllable and unobservable Answer: (B) India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 4

25 EC-GATE-4 PAPE- From the given state model, A B 4 c Controllable: Qc c B AB A B Observable : Q if Q controllable c 4 8 Qc Qc uncontrollable C CA CA If Q observable Q Q 4 Observable The system is uncontrollable and observable [ ] 47. The phase margin in degrees of G ( s) asymptotic Bode plot is. Answer: 48 G( s) ( s+.)( s+ )( s+ ) G( s) s s. + [ + s] +.. G( s) + s + s +.s [ ][ ][ ] By Approximation, G( s) [ s + ] ( s +.)( s + ) + ( s + ) calculated using the India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 5

26 Phase Margin θ 8+ GH EC-GATE-4 PAPE- ωωgc.99 8 tan Phase Margin 95.7 Asymptotic approximation, Phase margin φ ω gc ω ω ω ω ω ω gc.9949r / sc 48. For the following feedback system G ( s) response is required to be less than seconds. ( s + ) + ( s + ). The % settling time of the step r + C( s ) G ( s) y Which one of the following compensators C(s) achieves this? A s + 5 Answer: (C). B 5 + s ( C) ( s + 4) s + 8 D 4 s + By observing the options, if we place other options, characteristic equation will have rd order one, where we cannot describe the settling time. If C( s) ( s + 4) is considered The characteristic equation, is s + s + + s + 8 s 5s + + Standard character equation s + ξω s +ω Given, % settling time, 4 n n ω ; ξω.5 n ξw < ξ > n n n w 49. Let x be a real-valued random variable with E[X] and E[X ] denoting the mean values of X and X, respectively. The relation which always holds true is Answer: (B) ( [ ]) A E X > E X ( B) E X ( E[ X] ) ( C) E X ( E[ X] ) ( D) E X > ( E[ X] ) { } V x E x E x i.e., var iance cannot be negative { E( x) } E x India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 6

27 EC-GATE-4 PAPE Consider a random process X( t) sin ( π t + ϕ ), where the random phase ϕ is uniformly distributed in the interval [,π ]. The auto-correlation E X ( t ) X( t ) Answer: (D) ( A) cos( π ( t + t )) ( B) sin ( π( t t )) ( C) sin ( π ( t + t )) ( D) cos( π( t t )) Given X(t) sin ( π t + φ ) φ in uniformly distributed in the interval [,π ] π [ ] E x(t )x(t ) sin(π t + θ) sin π t + θ f ( θ)dθ π sin ( π t + θ) sin ( π t + θ)..dθ π π π sin( π (t + t ) + θ)dθ + cos( π(t t )dθ π π First integral will result into zero as we are integrating from to π. Second integral result into cos{ π(t t )} [ ] E X(t )X(t ) cos π(t t 5. Let Q( γ ) be the BE of a BPSK system over an AWGN channel with two-sided noise power spectral density N /. The parameter γ is a function of bit energy and noise power spectral density. A system with tow independent and identical AWGN channels with noise power spectral density N/ is shown in the figure. The BPSK demodulator receives the sum of outputs of both the channels. φ π f ( θ φ ) π θ / BPSK Modulator AWGN Channel AWGN Channel + BPSK Demodulator / Answer:.44 If the BE of this system is Q( b γ ), then the value of b is. E E Bit error rate for BPSK Q. Q NO N O E Y N O India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 7

28 EC-GATE-4 PAPE- Function of bit energy and noise P Counterllation diagram of BPSK Channel is SD N A WGN which implies noise sample as independent Let x + n + n x + n O φ ( t) a a φ ( t) where x x n n + n E Now Bit error rate Q N O x + + x + n noiseinchannel x + n + x + n + n E is energy in x N E N is PSD of h O 4E [as amplitudes are getting doubled] O N [independent and identical channel] O 4E E Bit error rate Q Q b or.44 N N O O noiseinchannel 5. A fair coin is tossed repeatedly until a Head appears for the first time. Let L be the number of tosses to get this first Head. The entropy H(L) in bits is. Answer: In this problem random variable is L L can be,,... P{ L } P{ L } 4 P{ L } 8 H{ L} log + lgo + log [ Arithmatic gemometric series summation]. + India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 8

29 5. In spherical coordinates, let ˆ ˆ a.a θ φ Answer: 55.5 EC-GATE-4 PAPE- E sin θ cos ( ωt β r) aˆ θv / m and r denote until vectors along the θ, φ directions..65 H sin θcos( ωt β r) aˆ φa / m r represent the electric and magnetic field components of the EM wave of large distances r from a dipole antenna, in free space. The average power (W) crossing the hemispherical shell located at r km, θ π / is E θ sin e r θ Jβr.65 Jβr HQ sin θe r * Pavg E H s Q.ds θ (.65) r Pavt ( 6.5) sin dθdφ s sin θ r sin θdθdφ s π π.5 sin d d.5. θ Q ( ) θ θ φ π P 55.5 w 54. For a parallel plate transmission line, let v be the speed of propagation and Z be the characteristic impedance. Neglecting fringe effects, a reduction of the spacing between the plates by a factor of two results in (A) halving of v and no change in Z (B) no change in v and halving of Z (C) no change in both v and Z (D) halving of both v and Z Answer: (B) 76 d Zo log r r d distance between the two plates so, z o changes, if the spacing between the plates changes. V independent of spacing between the plates LC India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 9

30 EC-GATE-4 PAPE- λ 55. The input impedance of a section of a lossless transmission line of characteristic 8 impedance 5Ω is found to be real when the other end is terminated by a load L Z + jx Ω. if X is, Answer: 4 Given, l λ s Z 5Ω Z o in ( l λ ) Z Z + JZ L o 8 o Zo + KZL Ω the value of ZL + J5 ZL + J5 5 JZ L Zin JZ 5+ JZ 5 JZ mg L L in L L L in 5Z + 5Z + J 5 Z Zin ZL Given, Z eal So, I Z 5 Z Z 5 + X 5 L L L 5 X 5 4Ω inω is India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India