Section-I: General Ability. 1. By giving him the last of the cake, you will ensure lasting in our house today.

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2 EC-GATE 08 Section-I: General Ability. By giving him the last of the cake, you will ensure lasting in our house today. The words that best fill the blanks in the above sentence are (A) peas, piece (B) piece, peace (C) peace, piece (D) peace, peas Key: (B). Even though there is a vast scope for its, tourism has remained a/an area. The words that best fill the blanks in the above sentence are (A) improvement, neglected (B) rejection, approved (C) fame, glum (D) interest, disinterested Key: (A) 3. What is the value of...? Key: (D) (A) (B) 7 4 (C) 3 (D) If the number 75? 43 is divisible by 3 (? Denotes the missing digit in the thousands place), then the smallest whole number in the place of? is. Key: (B) (A) 0 (B) (C) 5 (D) 6 5. A.5 m tall person is standing at a distance of 3 m from a lamp post. The light from the lamp at the top of the post casts her shadow. The length of the shadow is twice the height. What is the height of the lamp post in meters? Key: (B) (A).5 (B) 3 (C) 4.5 (D) 6 6. A coastal region with unparalleled beauty is home to many species of animals. It is dotted with coral reefs and unspoilt white sandy beaches. It has remained inaccessible to tourists due to poor connectivity and lack of accommodation. A company has spotted the opportunity and is planning to develop a luury resort with helicopter service to the nearest major city airport. Environmentalists

3 Key: (B) EC-GATE 08 are upset that this would lead to the region becoming crowded and polluted like any other major beach resorts. Which one of the following statements can be logically inferred from the information given in the above paragraph? (A) The culture and tradition of the local people will be influenced by the tourists. (B) The region will become crowded and polluted due to tourism. (C) The coral reefs are on the decline and could soon vanish. (D) Helicopter connectivity would lead to an increase in tourists coming to the region. 7. A cab was involved in a hit and run accident at night. You are given the following data about the cabs in the city and the accident. Key: (A) (i) 85% of cabs in the city are green and the remaining cabs are blue. (ii) A witness indentified the cab involved in the accident as blue. (iii) It is known that a witness can correctly identify the cab colour only 80% of the time. Which of the following is closest to the probability that the accident was caused by a blue cab? (A) % (B) 5% (C) 4% (D) 80% 8. Leila aspires to buy a car Rs. 0,00,000 after 5 years. What is the minimum amount in Rupees that she should deposit now in a bank which offers 0% annual rate of interest, if the interest was compounded annually? Key: (B) (A) 5,00,000 (B) 6,,000 (C) 6,66,667 (D) 7,50, Two alloys A and B contain gold and copper in the ratios of :3 and 3:7 by mass, respectively. Equal masses of alloys A and B are melted to make an alloy C. The ratio of gold to copper in alloy C is. Key: (B) (A) 5:0 (B) 7:3 (C) 6: (D) 9:3 0. The Cricket Board has long recognized John s potential as a leader of the team. However, his onfield temper has always been a matter of concern for them since his junior days. While this 3

4 Key: (C) EC-GATE 08 aggression has filled stadia with die-hard fans, it has taken a toll on his own batting. Until recently, it appeared that he found it difficult to convert his aggression into big scores. Over the past three seasons though, that picture of John has been replaced by a cerebral, calculative and successful batsman-captain. After many years, it appears that the team has finally found a complete captain. Which of the following statements can be logically inferred from the above paragraph? (i) Even as a junior cricketer, John was considered a good captain. (ii) Finding a complete captain is challenge. (iii) Fans and the Cricket Board have differing views on what they want in a captain. (iv) Over the past three seasons John has accumulated big scores. (A) (i), (ii) and (iii) only (C) (ii) and (iv) only (B) (iii) and (iv) only (D) (i), (ii), (iii) and (iv) Section-II: Electronics & Communication Engineering. A binary source generates symbols X {,} which are transmitted over a noisy channel. The probability of transmitting X = is 0.5. Input to the threshold detector is R = X + N. The probability density function f N (n) of the noise N is shown below. f N (n) 0.5 n If the detection threshold is zero, then the probability of error (correct to two decimal places) is. Key: (0.5) Ep: Given P P 0.5, threshold is 0V When = is transmitted, error is going to take place if received voltage is less than 0V. When = - is transmitted, error is going is take place if received voltage is greater than 0V. P P R 0.P P R 0.P 0 error Given that R= X+M, where M is a Random noise 4

5 EC-GATE 08 Perror P N 0 P P R 0 P P M.P P M P Using pdf figure, fm P M P M n Shaded portion represent the PM and PM Pe In the circuit shown below, the op-amp is ideal and Zener voltage of the diode is.5 volts. At the input, unit step voltage is applied, i.e., IN (t) u(t) volts. Also, at t = 0, the voltage across each of the capacitors is zero. F V 0V t 0 vn t k F v OUT (t) Key: (C) The time t, in milliseconds, at which the output voltage crosses 0V is OUT (A).5 (B) 5 (C) 7.5 (D) 0 Ep: Given V t ut z IN V.5V i C.V m t C C ; VC t Voltage across at second capacitor is.5,, also, t = 0 voltage across each capacitor is zero. By K.V.L, VC.5 V0 0, given V0 0V V 7.5 C t msec m vn t k F F v OUT (t) 5

6 EC-GATE Consider the following amplitude modulated signal: Key: (*) Ep: s(t) cos(000 t) 4cos(400 t) cos(800 t) The ratio (accurate to three decimal places) of the power of the message signal to the power of the carrier signal is. Given, s t cos 000 t 4cos 400 t cos 800 t side band carrier side band 4cos 400t cos 400t cos 400 t 4 cos 400t cos 400t Now at this point, some data is missing, if we compare this with standard AM signal equation c a s t A k m t cos 400 t if k, m t cos 400t, PM a if ka, mt cos 400t, PM 8 if ka, mt cos400t, PM 3 4 Like this infinite possibilities eist Hence many answers are possible. 4. A p-n step junction diode with a contact potential of 0.65 V has a depletion width of m at equilibrium. The forward voltage (in volts, correct to two decimal places) at which this width reduces to 0.6 m is. Key: (0.46) Ep: We know that in pn-junction W Vbi Vf Given, W m,v 0 f W 0.6m, V? f W W

7 EC-GATE The points P, Q, and R shown on the Smith chart (normalized impedance chart) in the following figure represent: (A) P: Open Circuit, Q: Short Circuit, R: Matched Load (B) P: Open Circuit, Q: Matched Load, R: Short Circuit (C) P: Short Circuit, Q: Matched Load, R: Open Circuit P Q R (D) P: Short Circuit, Q: Open Circuit, R: Matched Load Key: (C) Ep: Constant resistance circle are given by r y r r r center, 0, radius, r is normalized resistance constant reactance circle are given by r r y, in normalized reactance y r 0 P Q r r 5 R r P is short circuit R in open circuit, Q is matched load. At point P, normalized reactance is zero At point Q, normalized reactance is zero At point R, normalized reactance is zero 6. A lossy transmission line has resistance per unit length R = 0.05 / m. The line is distortionless and characteristic impedance of 50. The attenuation constant (in Np/m, correct to three decimal places) of the line is. Key: (0.00) Ep: Given R 0.05 m. Condition for distortion less transmission line 7

8 EC-GATE 08 R L RC LG G C R R G G L L zo C C Alternation constant RG Three are two photolithography systems: one with light source of wavelength 56nm (System ) and another with light source of wavelength 35 nm(system ). Both photolithography system are otherwise identical. If the minimum feature sizes that can be realized System and System are Lmin and L min respectively, the ratio L min / L min (correct to two decimal places) is. Key: (0.48) Ep: CD K Lmin NA CD is the minimum feature size NA=Numerical aperture Lmin 56 k k and NA NA 0.48 L 35 min 8. Let (t) be a periodic function with period T = 0. The Fourier series coefficients for this series are denoted by a k, that is Key: (C) (t) a ke k jk t T The same function (t) can also be considered as periodic function with period T 40. Let b k be the Fourier series coefficients when period is taken as T. If a k k 6, then b k k is equal to (A) 56 (B) 64 (C) 6 (D) 4 Ep: Only for case study let us consider the periodic waveform (t) as a square wave. 8

9 EC-GATE 08 A (t) T 0 t t a e k jk t T k We know a k is the eponential Fourier series coefficient and is given by T jk0 t ak te dt T 0 Since the waveform repeat itself, we can say if we evaluate a k using the period, as any integer multiple of To like To, 3To the coefficient will not change, we can notice this in the above integration. So we can say, if T=0, the E.F.S coefficient is a, k then if T=40, the E.F.S coefficient is b k a Since, ak bk ak bk 6 k k T' 40 k For further more clarity, if we find the D.C. component (k=0) of (t), then if we consider T 0, T 0, T 40, etc, the value of a o remains same. 9. Let M be a real 4 4 matri. Consider the following statements: Key: Ep: S: M has 4 linearly independent eigenvectors S: M has 4 distinct eigenvalues. S3: M is non-singular (invertible) Which one among the following is TRUE? (A) S implies S (C) S implies S (C) (B) S implies S3 (D) S3 implies S We know that if a matri A has n distinct eigen values then A has n linearly independent eigenvectors. S implies S. 9

10 EC-GATE A discrete-time all-pass system has two of its poles at and 30. Which one of the following statements about the system is TRUE? Key: (B) (A) It has two more poles at and 40. (B) It is stable only when the impulse response is two-sided. (C) It has constant phase response over all frequencies. (D) It has constant phase response over the entire z-plane. Ep: For the given all pass system the poles are located at z and z 30 P 30 P 0.50 If we are interested for stability of the given system, then the associated ROC must include the unit circle as shown in the figure. For the above highlighted ROC, with respect to the pole z 0.50, the ROC is outside the circle. So its impulse response will be right sided. Similarly for the pole z 30, the ROC is inside the circle of radius, so its impulse response is left sided, will contain un term. So, overally the impulse response should be both sided for the given stability. So Option B is Correct. By observing the pole pattern of an All Pass System, we can comment the location of zeros. But we never can comment regarding rest of the poles, if present. Hence option A is wrong. We know for an All Pass System, the magnitude response is constant for all frequency or in entire z-plane but the phase response changes, Hence option C and option D are wrong.. A traffic signal cycles from GREEN to YELLOW. YELLOW to RED and RED to GREEN. In each cycle, GREEN is turned on for 70 seconds. YELLOW is turned on for 5 seconds and the RED is turned on for 75 seconds. This traffic light has to be implemented using a finite state machine (FSM). The only input to this FSM is a clock of 5 second period. The minimum number of flipflops required to implement this FSM is. 0

11 Key: (5) Ep: Green 70 seconds Yellow 5 seconds Red 75 seconds So to complete one cycle i.e., EC-GATE 08 Green yellow Red Green yellow red. Then required time is =50 sec. Since clock period is 5 sec; the number of clock needed for one cycle of 50 sec is 30, so we can regard the FSM as MOD 30 counter. n 30 min n 5 so, minimum 5 flip flops are required.. Two identical nmos transistor M and M are connected as shown below. The circuit is used as amplifier with the input connected between G and S terminals and the output taken between D and S terminals. V bias and V D are so adjusted that both transistors are in saturation. The id transconductance of this combination is defined as gm while the output resistance is v v DS r 0, id where i D is the current flowing into the drain of M. Let g m, g m be transconductances and r 0,r 0 be the output resistance of transistor M and M, respectively. Which of the following statements about estimates for g m and r 0 is correct? (A) gm gmg m.r0 and r0 r0 r 0. V D GS D Key: (C) Ep: (B) gm gm gm and r0 r0 r 0. (C) gm gm and r0 r 0.g m.r 0. (D) gm gm and r0 r0 g i D m ro rms v I DS The given circuit is MOS cascode io gm v i D AC equivalent circuit of MOS cascade is G S V bias M M

12 EC-GATE 08 V gs m gs d g v ro i D V gs v i m i ds g v ro Writing a node equation for the node, we have Vgs Vgs gmvgs gmv i... ro ro Observe that the voltage d, s vgs Equation () can be simplified as gm vgs gmv i... ro ro Since g m, r r o o g v g v...(3) m gs m i Node equation at node d is vgs i0 gmvgs gm vgs ro ro i g v m gs From equation (3) and (4) i0 ID gm gm v v Similarly i R 0 v I gs v I gs D i V gs g v ro m gs v s i r o

13 Voltage at the source node, which is gs o EC-GATE 08 vgs can epressed in terms of i is v i r and... v i g v r i r... m gs o o Substituting () in () v i r r g r r o o m o o R r r r r g o o o o o m In above epression the last term will dominate, thus Ro g m.ro r o k k 3. Consider matri A k k k and vector. The number of distinct real values of k for which the equation A = 0 has infinitely many solutions is. Key: () Ep: Given that AX=0 has infinitely many solutions. Then A 0 K K 0 K K K 3 K K K K K K K 0 3 K K 0 K K 0 K 0,0, Number of distinct values of K =. [i.e., 0 and ]. 4. Considered a binary channel code in which each codeword has a fied length of 5 bits. The Hamming distance between any pair of distance codewords in this code is at least. The maimum number of codewords such a code can contain is. Key: () Ep: Hamming distance d implies that two code words differ in d bits. Total no. of code formed using 5 bits 5 3. Total no. of code with unit hamming distance 5c 0 Total no. of code words with at least two hamming Distance =3-0=. 3

14 EC-GATE Taylor series epansion of f() t e dt around = 0 has the form 0 f () a0 a a... The coefficient a (correct to two decimal places) is equal to. Key: (0) Ep: t f e dt 0 f ' e '' f e.e f '' 0 0 By Taylor s series, of f() around =0, we have f f 0 f ' 0 f '' 0... ; where the coefficient of! But given the coefficient of f'' 0 a! 0 a '' 0 f 0 0! a 0. in f a f'' 0! 6. A good transimpedance amplifier has (A) low input impedance and high output impedance (B) high input impedance and high output impedance (C) high input impedance and low output impedance (D) low input impedance and low output impedance Key: (D) Ep: Trans impedance amplifier contain, input of Norton's circuit and output of Thevenin circuit R o I Is R s Ri RmI R L A good transimpedance amplifier has low input impedance 4

15 7. Consider 3 0 EC-GATE 08 p(s) s a s a s a with all real coefficients. It is known that its derivative p'(s) has no real roots. The number of real roots of p(s) is (A) 0 (B) (C) (D) 3 Key: (B) Ep: Given p'(s) has no real roots. Discriminant of p'(s) < 0. p(s) will have eactly one real root. 8. Let X,X, X3 and X 4 be independent normal random variables with zero mean and unit variance. Key: (0.5) The probability that X 4 is the smallest among the four is. Ep: Since X,X,X3 and X 4 are independent normal random variables with zero mean and unit variance, then P X is the smallest P X is the smallest P X is the smallest 3 PX4 is the smallest The ABCD matri for a two-port network is defined by: V A B V I C D I I I V V 5 Key: (4.8) The parameter B for the given two-port network (in ohms, correct to two decimal places) is. Ep: The ABCD matri equation is V AV BI I CV DI If we make output port short i.e. V 0, then V B I Making the output short, the network becomes 5

16 EC-GATE 08 V I X I 5 I 5 I I Writing KVL on the input loop V I I 0 V I I... Writing KCL at node X I I I 0 5 I 0.4I I I.4I... Putting equation () in equation () V.4I I.8I I V B 4.8 I 0. In a p-n junction diode at equilibrium, which one of the following statements is NOT TRURE? Key: (D) (A) The hole and electron diffusion current components are in the same direction. (B) The hole and electron drift current components are in the same direction. (C) ON an average, holes and electrons drift in opposite direction. (D) ON an average, electrons drift and diffuse in the same direction. Ep: Since the concentration of holes on the p side is much greater than that in the n side, a very large hole diffusion current tends to flow across the junction from the p to the n material. Hence an electric field must build up across the junction in such a direction that a hole drift current will tend to flow across the junction from n to p side in order to counterbalance the diffusion current. 6

17 EC-GATE 08. The logic function f(x,y) realized by the given circuit is V DD X X Y Y f(x,y) (A) NOR (B) AND (C) NAND (D) XOR Key: (D) Ep: If we focus the N-MOS section f X,Y X X Y Y XY XY f X.Y X Y XY X Y XY. The Nyquist stability criterion and the Routh criterion both are powerful analysis tools for determining the stability of feedback controllers. Identify which of the following statements is FALSE: (A) Both the criteria provide information relative to the stable gain range of the system. (B) The general shape of the Nyquist plot is readily obtained from the Bode magnitude plot for all minimum-phase systems. 7

18 Key: (B) EC-GATE 08 (C) The Routh criterion is not applicable in the condition of transport lag, which can be readily handled by the Nyquist criterion. (D) The closed-loop frequency response for a unity feedback system cannot be obtained from the Nyquist plot. Ep: Analysis of option A By observing the encirclement about the point -+j0 in Nyquist plot we find the range of k for which system will be stable similarly in Routh criterion by observing the sign change we find the range of k for which system will be stable Hence option A represents a correct statement. Analysis of option B Shape of Nyquist plot can t be commented readily form bode plot, first we have to find the transfer function and by using this we have to draw the Nyquist plot, so readily we can t comment. Hence option B represents a false statement. Analysis of option C e s For transportation log Nyquist plot is suitable as its magnitude is for all frequency and phase is, so no approimation required but in Routh criterion we need to do approimation first, so its not suitable. Hence option C represents a correct statement. Analysis of option D With the help of Nyquist plot we cannot comment closed loop frequency response for this we need M, N circles, Nichols chart etc. Hence option D represents a correct statement. 3. A function F(A, B, C) defined by three Boolean variables A, B and C when epressed as sum of products is given by F A.B.C A.B.C A.B.C Where, A, B, and C and the complements of the respective variables. The product of sums (POS) form of the function F is (A) F (A B C). A B C. A B C (B) FA B C. A B C. A B C (C) F (A B C).(A B C).(A B C). A B C.(A B C) (D) f A B C. A B C.(A B C).(A B C).(A B C) Key: (C) 8

19 Ep: F A.B.C ABC A B C A B C m 0,,4 M,3,5,6,7 EC-GATE 08 A B CA B CA B CA B CA B C 4. Let the input be u and the output be y of a system, and the other parameters are real constants. Identify which among the following systems is not a linear system: (A) (B) Key: (C) 3 d y d y dy du d u a 3 a a3y b3u b b (with initial rest conditions) dt dt dt dt dt t (t ) y(t) a u( )d 0 (C) y = au + b, b 0 (D) y = au Ep: If we observe all the 4 options carefully, we can notice option C (i.e., y au b, b 0 ) is non linear system due to the addition of non-zero constant b, since it does not satisfy homogeneity and additivity as follows aku b kau b au b au b a u u b Option A is linear differential equation with constant co-efficient, so it is linear. Integrator system is generally linear, so option B represents a linear system. y au, is a perfect linear system. 5. Let f(,y) a by, where a and b are constants. If y f f at = and y =, then the relation y between a and b is b b (A) a (B) a 4 (C) a b (D) a = 4b Key: (D) Ep: Let f, y a by y a by y y 9

20 EC-GATE 08 a by f, y y f a by f a b & y y y f f Given y,y a by a b y y,y a a b b 4 a a b b 4 a a 4 3b 3a b a 4b. 6. In the circuit shown below, the (W/L) value for M is twice that for M. The two nmos transistors are otherwise identical. The threshold voltage V T for both transistors is.0 V. Note that V GS for M must be >.0 V. Current through the nmos transistors can be modeled as W IDS Co VGS VT V DS VDS L for VDS VGS VT W I DS Co VGS V T / L for VDS VGS VT 3.3V.0V M V M The voltage (in volts, accurate to two decimal places) at V is. 0

21 EC-GATE 08 Key: (0.466) Ep: Let K C W L n o n W W Given that L L So, K K n n GS T For M, V V V For M, V V V V V V s GS T V 3.3 V V V DS GS T So, M will be in linear region and M will be in saturation region. I D I D K V V V V K V V n GS T DS DS n GS T K n V V Kn V V V V V V 4V 3V 6V 0; V V V V 4 3 V V V V GS T V 0.46V 3 7. A junction is made between p - Si with doping density 7 3 NA 0 cm. NA cm and p Si with doping density Given: Boltzmann constant k = J.K, electronic charge q = C. Assign 00% acceptor ionization. At room temperature (T = 300 K), the magnitude of the built-in potential (in volts, correct to two decimal places) across this junction will be. V V Key: (0.9) Ep: Setting J P = 0 VT dp E P d P P

22 EC-GATE 08 dp dv dv VT E d d P V V V VT ln P Built in potential, kt N.38 3 A Vbi n n 00 V 0.9V q NA In the circuit shown below, a positive edge-triggered D Flip-Flop is used for sampling input data D in using clock CK. The XOR gate output 3.3 volts for logic HIGH and 0 volts for logic LOW levels. The data bit and clock periods are equal and the value of T / TCK 0.5, where the parameters T and T CK are shown in the figure. Assume that the Flip-Flop and the XOR gate are ideal. D in D Q DFlip-Flop CLK X CK D in T CK CK T T T If the probability of input data bit (D in ) transition in each clock period is 0.3, the average value (in volts, accurate to two decimal places) of the voltage at node X, is. Key: (0.845) Ep: If P : Probability of Output V ON state P 0 : Probability of Output V 0OFF state V X : Voltages corresponding to logic i.e. 3.3V V X0 : Voltage corresponding to logic i.e. 0 V T ON : The period when Output T OFF : The period when Output V V 0 Then we can say the average value of V P V TON P0 V TOFF avg 0 V is The above relation is somewhat similar to the average probability of error calculation in digital modulation Schemes. Let us draw the wave form of V and find the unknown parameter of the above equation.

23 EC-GATE 08 Form the given logic circuit we can say that the Output of the e or D in and previous state of the D flip flop, because the elements are given as ideal. T CLK D in T T T T 3.3V T T CLK CLK V P V TON P V TOFF avg T T o T T TCLK TCLK V. When there V is a transition of D in 9. The circuit shown in the figure is used to provide regulated voltage (5V) across the k resistor. Assume that the Zener diode has a constant reverse breakdown voltage for a current range, starting from a minimum required Zener current, IZ min ma to its maimum allowable current. The input voltage V I may vary by 5% from its nominal value of 6V. The resistance of the diode in the breakdown region is negligible. R V I 5V k 3

24 EC-GATE 08 The value of R and the minimum required power dissipation rating of the diode, respectively, are (A) 86 and 0 mw (B) 00 and 40 mw (C) 00 and0mw (D) 86 and 40mW Key: (C) 5 Ep: Given Izmin ma; IL 5mA V 6 5% of 6V I I I I 5 7mA z L R V I R 5V I z I k P V I 50 P min min z zmin 0mW For the circuit given in the figure, the magnitude of the V loop current (in amperes, correct to three decimal places) 0.5 second after closing the switch is. Key: (0.36) H Ep: V t 0 H assume switching is happening at t 0 We can redraw the above circuit as t 0 V il t H t/ il t il il 0 il e ; t 0 switch is in open state, so the loop is not closed hence, L At t 0, i 0 0A 4

25 EC-GATE 08 At t switch is closed and inductor is short Circuited V i L A Time constant L R tn L, R tn Rn So we have i 0 0 i L L t 0 e ; t 0 t/ il t il il 0 il e ; t 0 e t ; t il t 0.5 e e amp 3. Consider a white Gaussian noise process N(t) with two-sided power spectral density S N (f) = 0.5 Key: (B) Ep: W/Hz as input to a filter with impulse response 0.5 Y(t). The power in Y(t) in watts is e t / (where t is in seconds) resulting in output (A) 0. (B) 0. (C) 0.33 (D) 0.44 Given SM t 0.5 w / Hz. 5

26 EC-GATE 08 Nt h t 0.5.e t yt y PSD S M f M S t H jt.s t Sy t 0.5 H t Power in y t area under s t s y t dt y 0.5 H t dt 0.5 H t dt Applying parsevals theorem, H t dt h t dt t Power in y t e dt t e dt 3. The position of a particle y(t) is described by the differential equation: t e dt a. 0. Gaussian function Area PSD S y f d y dy 5y. dt dt 4 The initial conditions are y(0) = and the particle at t is. dy dt t 0 0. The position (accurate to two decimal places) of Key: (-0.) Ep: Given, d y dy 5y ; dt dt 4 6

27 EC-GATE 08 where y(t) represents the position of a particle d y dy 5 y 0 dt dt 4 The auiliary equation of the D.E is m m i m icomple Roots are comple The required solution of equation () is t y t e C cost +C sint dy Given y0 & 0 dt t0 From ;if y 0 ; then C dy dy if 0at t 0; then e C sin t C cos t C cos t C sin t e dt dt 0 C C C C C The position of the particle at t is t t t y e cos t sin t t e e A random variable X takes value 0.5 and 0.5 with probabilities and 3, respectively. The noisy 4 4 observation of X is Y = X + Z, where Z has uniform probability density over the interval (, ). X and Z are independent. If the MAP rule based detector outputs ˆX as 0.5, Y ˆX 0.5, Y, Then the value of (accurate to two decimal places) is. Key: (-0.75) P 0.5, T , y m 4 Ep: Given, Noise is with uniform density function as shown 7

28 EC-GATE 08 fz g pdf of y when 0.5is transmitted z y f y/ p pdf of y when 0.5is transmitted y f y 0.5 y p g.5 y In a general rule, threshold shift towards that side where probability is less Using MAP rule, 0.5 T P y P 0.5 T 0.5 P y P 0.5 T Equality can be used to find threshold 3 5 Xt.5 T T T T T 34. The state equation and the output equation of a control system are given below: 4.5 u, y The transfer function representation of the system is 8

29 EC-GATE 08 3s 5 3s.875 (A) (B) s 4s 6 s 4s 6 4s.5 6s 5 (C) (D) s 4s 6 s 4s 6 Key: (A) Ep: 4.5 A 4 0 B 0 C T F C si A B s 4.5 si A 4 s s.5 si A ss s 4 s.5 s 4s 6 s 4s 6 4 s 4 s 4s 6 s 4s 6 s I A B s I A 0 s s 4s 6 8 s 4s 6 C s I A B s s 4s s 5 s 4s 6 s 4s 6 s 4s 6 3s 5 s 4s 6 So the closed loop transfer function is Ts 3s 5 s 4s 6 9

30 EC-GATE The figure below shows the Bode magnitude and phase plots of a stable transfer function n0 G(s) 3 s d s d s d 0 G j 36 db 0 db 0 db k Gs G j Key: (0.) Consider the negative unity feedback configuration with gain k in the feedforward path. The closed loop is stable for k < k 0. The maimum value of k 0 is. Ep: Basically here we are supposed to find the value of K K0 for which system will be marginally stable. We know for marginal stable, pc gc, if we notice both the magnitude and phase plot carefully we can say, the frequency of which the phase is 80 in the phase plot the corresponding gain is 0dB in the magnitude plot. So for marginal stable, at this frequency, the gain plot should touch 0 db line, so this to happens, if we select K 0. 0dB, then each point in the gain plot will be subtracted by + 0dB, including the point where gain is + 0dB, it becomes 0 db & the system becomes marginal stable. Since K=0., it is positive phase plot remains unaffected so no change in so that we have pc gc. So if K 0., pc gc which is the condition for marginal stable. if K 0., system stable K 0., unstable K 0., marginal stable pc but gc changed 30

31 EC-GATE The contour C given blow is on the comple plane z jy, where j. y C Key: () Ep: dz The value of the integral j is. z Let f z C z z z The singular points of f(z) are z = -, both lies inside C. By Cauchy s integral formula, dz dz j C z j C z z dz dz j C z C z j j j (Since the closed curve C is clock wise & j j j j j 'C ' is anti clock wise) 37. A band limited low pass signal (t) of bandwidth 5 khz is sampled at a sampling rate f s. The signal (t) is reconstructed using the reconstruction filter H(f) whose magnitude response is shown below: H(f ) C C K f (khz) Key: (3) The minimum sampling rate f s ( in khz) for perfect reconstruction of (t) is. Ep: It is given that follows: t is a low pass signal, and is band limited to 5 khz, let its spectrum Xf is as 3

32 EC-GATE 08 Xf 5 5 f (khz) If we sample frequency. t ideally, the sampled signal is If we think to sample it on critical sampling rate s m Xs f X f nfs, where f s is the sampling n f f 0kHz, Then the spectrum of the sampled signal will be X s (f ) fs 3kHz f (khz) If we pass now Xs f through the non ideal filter (shown in dotted line), at the output of the filter along with the original signal, we get some undesired frequency of 5 to 8 khz. However, if we select fs 3kHz, then the filter will capture only the desired part, as follows, (this idea we can get by observing the spectrum when we have selected f s = 0 khz). s f fs 3kHz f (khz) 3

33 EC-GATE Consider the network shown below with R, R and R3 3. The network is connected to a constant voltage source of V. R R R R R R 3 R 3 V R R R Key: (8) The magnitude of the current (in amperes, accurate to two decimal places) through the source is. Ep: By observing the network we can say, it holds a symmetry and hence we can say Node A and Node B are equipotential. Similarly Node C and Node D are also equipotential. When nodes are equipotential, the element connected across these nodes (R in this case) can be replaced by either short circuit, or open circuit or any other value of resistor without affecting the network. If we replace the R by short circuit, then the network can be redrawn as A R R R R R R 3 R 3 C R 3 R R R I B D V R R R R 3 R 3 R R I R R eq V R R R R R R R R R eq 3 3 R R R3 R R R3 R 3 R, R, R3 values are given 3/4 3 4/ 8 8 I 8A R / 8 eq 33

34 EC-GATE A dc current of 6 A flows through the circuit shown. The diode in the circuit is forward biased and it has an ideality factor of one. At the quiescent point, the diode has a junction capacitance of 0.5 nf. Its neutral region resistances can be neglected. Assume that the room temperature thermal equivalent voltage is 6 mv. 5sin( t)mv 00 V For Key: (6.4) Ep: 6 0 rad/s, the amplitude of the small-signal component of diode current (in A, correct to one decimal place) is. nv 6m T 3 rd 0 k 6 ID 60 C d F A.C equivalent r d j 3 Xd j0 C r.x j0 0 Z r X r X 0 0 j j 6 3 d d d d 3 3 d d 5m I j j j I 6.4 A C d 40. Let X[K] k,0 k 7 be 8-point DFT of a sequence [n], Where X[k] N jnk/n. [n] e n0 The value (correct to two decimal places) of 3 [n] is. n0 34

35 Key: (3) Ep: It is given that EC-GATE 08 X k k, 0 k 7 for 8 point DFT of n So Xk,, 3, 4, 5, 6, 7, 8 We are asked to find 3 n0 n For N-point DFT we know that X 0 N n0 n N X n N n n0 Centralordinate Propertyof DFT If we put N=8 in the above equations 8 X 0 n n0 8 n X 4 n n0 n0 X 0 X X0 X4 3 n For the circuit given in the figure, the voltage V C (in volts) across the capacitor is 00 k 5sin(5t)V F V C 00 k Key: (C) (A).5 sin(5t 0.5 ) (B).5 sin(5t 0.5 ) (C).5 sin(5t 0.5 ) (D).5 5sin(5t 0.5 ) Ep: Using sine function as reference phasor, if we transform the given time domain network into its equivalent phasor domain, it will be as follows 35

36 EC-GATE 08 R 5 0 V C jc Here R 00k, C F and 5rad / sec Using the voltage division rule for V C we have / jc VC R / jc jrc C j j V t.5 sin 5t 0.5 R An op-amp based circuit is implemented as shown below. 3k V k A 5V 5V o Key: (0.5) In the above circuit, assume the op-amp to be ideal. The voltage (in volts, correct to one decimal place) at node A, connected to the negative input of the op-amp as indicated in the figure is. R f 3k Ep: Given V0 3 R k V 5V and V 5V sat sat As V V V 5V 0 sat 0 36

37 EC-GATE 08 Applying KVL at point A Rf 3k VA VA V 3 V 5 0 A A 6 3VA 6 VA 0.5V 3 V R k V A 5V V o 5V 43. A solar cell of area.0 cm, operating at.0 sun intensity, has a short circuit current of 0 ma, and an open circuit voltage of 0.65 V. Assuming room temperature operation and thermal equivalent voltage of 6 mv, the open circuit voltage (in volts, correct to two decimal places) at 0. sun intensity is. Key: (0.608) Ep: A.0cm ; IL 0mA; Voc 0.65; VT 6mV J I 0m A/cm V J V n ; J J ' J 0. 0mA/cm 4mA/cm L L L oc T L S A Js L oc Voc T oc V VT 6m oc e e VT J V V L e 0.65 ; n J' L V V T e 3.39 V 6m oc Voc VT e 44. The cutoff frequency of TE 0 mode of an air filled rectangular waveguide having inner dimensions a cm b cm (a > b) is twice that of the dominant TE 0 mode. When the waveguide is operated at frequency which is 5% higher than the cutoff frequency of the dominant mode, the guide wavelength is found to be 4 cm. The value of b (in cm, correct to two decimal places) is. Key: (0.6) C Ep: Cut off frequency for TE0 b C Cut off frequency for TE0 a It is given that C C a b a b 37

38 EC-GATE 08 5 C Given waveguide frequency f 4 a (5% higher than the cut off frequency of dominant mode) 5 f fc 4 0 guide fc f 0 50 guide 4cm cm a 0.5cm b 0.75cm 45. The distance (in meters) a wave has to propagate in a medium having a skin depth of 0.m so that the amplitude of the wave attenuates by 0 db, is Key: (C) Ep: (A) 0. (B) 0.3 (C) 0.46 (D).3 Skin depth 0.m f Attenuation constant Amplitude of wave varies as z f 0 0. e z e 00 z n 0 z z 0.46m The logic gates shown in the digital circuit below use strong pull-down nmos transistors for LOW logic levels at the outputs. When the pull-downs are off, high-value resistors set the output logic levels to HIGH (i.e., the pull-ups are weak). Note that some nodes are intentionally shorted to 38

39 EC-GATE 08 implement wired logic. Such shorted nodes will be HIGH only if the outputs of all the gates whose outputs are shorted are HIGH. X 0 X X Y X 3 Key: (8) Ep: The numbers of distinct values of X3XXX 0 (out of the 6 possible values) that give Y =. X 0 X0 X X X 0 X X G XG 0 F 0 fied Y 3 G X X.X X 3 X3 If we evaluate F by referring the above logic circuit F X X G So the node F is always 0. Irrespective any possible combinations of X 0,X,X,X 3 Since already one of the input to the OR gate is logic 0 (fied) and we are epecting Y=, so we must need the other input of the or gate is logic for which X3. So for Y=. We need the fied value of X3, with this X 0,X,X can take any possible combinations, So the number of possible combination of X 0,X,X is X3. Hence there are 8 combinations of X 0,X,X,X 3 for which output Y= 3 8, for each of the combination 47. A ROM array is built with the help of diodes as shown in the circuit below. Here W0 and W are signals that select the word lines and B0 and B are signals that are output of the sense amps based on the stored data corresponding to the bit lines during the read operation. During the read operation, the selected word line goes high and the other word line is in a high impedance state. 39

40 EC-GATE 08 B0 B Sense amps W0 W B B 0 W0 D00 D0 W D0 D Bits stored in the ROM Array VDD As per the implementation shown in the circuit diagram above, what are the bits corresponding to D ij (where i = 0 or and j = 0 or ) stored in the ROM? (A) Key: (A) Ep: 0 0 (B) 0 0 (C) 0 0 (D) 0 0 Form the given statement; it is clear that, if we want to read the stored information in the ROM word line value should be logic. So in all the 4 options, the top row of matri indicate if W0, what are the value of B0 and B similarly the bottom row indicates if W. what are the value of B 0& B. If we make W0, it s corresponding diode will be ON. And V DD will be available on B, 0 since the diode corresponding to W is in off state W now, by default B 0. So when W0, then B0 and B 0. Similarly when W, then B0 0 and B So the matri is B B 0 B B 0 W0 0or W0 0 W 0 W A curve passed through the point ( =, y = 0) and satisfies the differential equation dy y y. The equation that describes the curve is d y (A) y ln (B) y ln (C) y ln (D) ln y 40

41 EC-GATE 08 Key: (A) Ep: Given D.E is dy y y d y dy y d y dy y y d dy dt dy dt Let y t y y d d d d dt From t d dt d e I.F d t ; which is linear D.E. I.F e P & Q The solution of eq is e e t. d e t. e c e y e c 3 Given that eq(3) passes through the point (,0). C e from 3 From 3 ; ye e e y y y e e n. e 49. A uniform plane wave traveling in free space and having the electric field 8 ˆ ˆ z E a a cos t ( z) V / m E 4

42 EC-GATE 08 is incident on a dielectric medium (relative permittivity >, relative permeability = ) as shown in the figure and there is no reflected wave. z Free space Dielectric Medium( ) r Key: () The relative permittivity (correct to two decimal places of the dielectric medium is. 8 Ep: z Given E a a cos 3 0 t z V m The wave is parallel to the, z plane Since, there is no reflection, Angle of incidence in Brewster angle is tan tan i B tan i Free space i i z 50. For a unity feedback control system with the forward path transfer function K G(s) s(s ) The peak resonant magnitude M r of the closed-loop frequency response is. The corresponding value of the gain K (correct to two decimal places) is. Key: (5.0) Ep: The O.L.T.F of the unity ve feed back system is K Gs s s The C.L.T.F is Ts Gs K G s s s K By comparing with standard nd order equation 4

43 EC-GATE 08 s, we have ns n n K and n n n K Let us obtain for the resonant peak Mr let , by solving this we get K A four-variable Boolean function is realized using 4 multipleers as shown in the figure. I 0 I 0 F(U,V,W,X) V CC I I 4 4 MUX MUX I I I 3 S I 3 S0 S S0 U V W X The minimized epression for F(U,V,W,X) is (A) UV (C) UV UV W (B) UV UVWX WX UV W (D) UV UVWX WX 43

44 Key: (C) Ep: 0 I 0 I I I 3 S S Y EC-GATE 08 Y 4 I 4 Y V FU,V,W,X cc 0 MUX 0 0 I 0 I I 3 MUX Output of the first MUX Y UV.0 UV. U V. UV.0 UV U V Output of second MUX F W X Y WXY WX.0 WX.0 W Y X X WY YW UV U V W U V W X Although option C and option D are representing same function, but we can say the minimized epression is option C. 5. Let c(t) Accos( fct) and m(t) cos( fmt). It is given that fc 5 f m. The signal c(t) + m (t) is Key: (D) Ep: applied to the input of a non-linear device, whose output v 0(t) is related to the input v (t) as v (t) av (t) bv (t), where a and b are positive constants. The output of the non-linear i 0 i i device is passed through an ideal band-pass filter with center frequency f c and bandwidth 3f m, to produce an amplitude modulated (AM) wave. If it is desired to have the sideband power of the AM wave to the half of the carrier power, then a/b is (A) 0.5 (B) 0.5 (C) (D) Given c t A cos f t, m t cos f t f c 5f C c m m mt vi t av t bv t i i v0 t ct a and b are constants 44

45 EC-GATE 08 v0 t a Ac cos f ct cos fmt b Ac cos f ct cos f m t ba ba b b m c c c m c c a cos f t a A cos f t cos 4f t cos 4f t b A cos f t cos f t. c m c After passing through BPF of center frequency f c and bandwidth yt a A cos f t carrier Carrier power It is given that b A cos f tcos f t c c c m c aa c, side bands side band power = ba c sideband power carrier power bac b a aac a 4 b 3f m, 53. The input 4 sinc (t) is fed to a Hilbert transformer to obtain y(t), as shown in the figure below: 4sinc(t) Hilbert Transform y(t) Here sinc() = sin( ). The value (accurate to two decimal places) of Key: (8) Ep: By Parseval s theorem we can say y(t) dt is. yt dt Y d The impulse response of the Hilbert transform is ht, t H jsgn( ) If t k sinc(at) j j k then X rect a a -a X k/a a and Fourier transform 45

46 EC-GATE 08 4sinc t 4sinct ht t yt y t 4sin c t Y rect. jsgn t* h t X.H t 4 Y j j * j - j Y - Y 4 - Finally, y t dt Y d area of Y Red (R), Green (G) and Blue (B) Light Emitting Diodes (LEDs) were fabricated using p-n junctions of three different inorganic semiconductors having different band-gaps. The built-in voltage of red, green and blue diodes are V R, VG and V B, respectively. Assume donor and acceptor doping to be Key: (B) the same (N A and N D, respectively) in the p and n sides of all the three diodes. Which one of the following relationships about the built-in voltages in TRUE? (A) VR VG VB (B) VR VG VB (C) VR VG VB (D) VR VG VB Ep: Built in potential of pn diode V N.N A D bi kt n n i bi A D i V kt n N N kt n n... Where n i can be defined as 46

47 Eg kt i C V n N.N e i Eg kt n k ' e... So substitute equation () in () V kt n N N kt n k' E bi A D g V M E ;where E is band gap bi g g EC-GATE 08 All diodes NA and N D are same, so that M same V E ; E bi g g hc We know that E E E because g Red g green g blue Red green blue V V V R G B 55. Let Key: (4.50) r y z and 3 3 z y yz y. Assume that and y are independent variables. At (,y,z) = (,, ), the value (correct to two decimal places) of Ep: Given, r y z... i 3 3 and z y yz y... ii r z..., y are independent from i z z 3z y y 0 formii z 3z y y z y 3z y r y form & ; 3z y r,, r is. 47

48 EC-GATE 08 48