: 2 : GATE Ans: (D) Sol: N. n q (D) V. = 25 mv n (10 3 ) Another Method: = V. n N. n q. ACE Engineering Academy C N

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2 : : GATE An n + n Silicon device is fabricated with uniform and non-degenerate donor doping concentrations of N D = 0 8 cm and N D = 0 cm corresponding to the n + and n regions respectively. At the operational temperature T, assume complete impurity ionization, kt/q = mv, and intrinsic carrier concentration to be n i = 0 0 cm. What is the magnitude of the built-in potential of this device? (A) V (B) V (C) 0.88 V (D) 0.7 V Ans: (D) Sol: N E C EF KT n n N C D N E C EF KT n n N E E KT F F n n N V o.q KT n N C D N n N D D KT N D V o n q ND = mv n (0 ) = 0.7 V Another Method: KT N AND Vo n q ni KT N D n q n i NA KT N D n q ND = 0.7 V D D

3 : : GATE - 07_EC 0. A periodic signal x(t) has a trigonometric Fourier Series expansion x t a a cos n t b sin n 0 n 0 n 0t n If x(t) = x( t) = x(t / 0 ), we can conclude that (A) a n are zero for all n and b n are zero for n even (B) a n are zero for all n and b n are zero for n odd (C) a n are zero for n even and b n are zero for n odd (D) a n are zero for n odd and b n are zero for n even 0. Ans: (A) Sol: Given that x (t) = x( t) i.e odd signal, so the signal contains only b n terms and x (t) = x t i.e Half wave symmetry, so the signal contains only odd harmonics. o So the resultant signal contains only b n terms with odd harmonics. 0. Three fair cubical dice are thrown simultaneously. The probability that all three dice have the same number of dots on the faces showing up is (up to third decimal place). 0. Ans: 0.07 Sol: The probability that all three dice have the same number of dots on the faces showing up is For a narrow base PNP BJT, the excess minority carrier concentrations (n E for emitter, p B base, n C for collector) normalized to equilibrium minority carrier concentrations (n E0 for emitter, p B0 for base, n C0 for collector) in the quasi-neutral emitter, base and collector regions are shown below. Which one of the following biasing modes is the transistor operating in? Normalized Excess Carrier Concentration 0 n E n E0 Emitter (p) p p 0 B B0 Base (N) X and Y axes are not to scale n n C C0 Collector (P)

4 : 4 : GATE 07 (A) Forward active (B) Saturation (C) Inverse active (D) Cutoff 04. Ans: (C) Sol: At C - B junction, due to FB a lot of holes are injected into the base from collector. At E-B junction, due to RB very few holes are injected into the base from Emitter (or) At collector Base junction, ratio of Excess minority carrier concentration to equilibrium minority carrier concentration is in order of 0 (very high).this is possible when the junction is forward bias (injection) At emitter base junction, ratio of Excess minority carrier concentration to equilibrium minority carrier concentration is in order of (negligible). This is possible when the junction is reverse Bias (no injection) 0. Consider the following statements for continuous-time linear time invariant (LTI) systems. I. There is no bounded input bounded output (BIBO) stable system with a pole in the right half of the complex plane. II. There is no causal and BIBO stable system with a pole in the right half of the complex plane. Which one among the following is correct? (A) Both I and II are true (B) Both I and II are not true (C) Only I is true (D) Only II is true 0. Ans (D) Sol: (I) For example consider a pole location at right half of complex plane, if it is anti-causal, ROC is left sided, and Roc includes j axis, so It is a BIBO stable, so statement I is false. (II) If a causal system having a pole on right side of s-plane it is compulsory unstable because ROC is not including j-axis. So statement II is true.

5 : : GATE - 07_EC 06. For the operational amplifier circuit shown, the output saturation voltages are V. The upper and lower threshold voltages for the circuit are, respectively. + V in + V out 0k k + V (A) + V and V (B) +7 V and V (C) +V and 7V (D) +V and V 06. Ans: (B) Sol: V in V out + 0k V k + v Replying KCL at node (V) V V V out 0 0K k V 6 = V out V V out 6 UTP: If V out = + V sat = + V UTP = 7 V LTP: If V out = V sat = V LTP = V

6 : 6 : GATE Consider the following statements about the linear dependence of the real valued functions y =, y = x and y = x. over the field of real numbers. I. y, y and y are linearly independent on x 0 II. y, y and y are linearly dependent on 0 x III. y, y and y are linearly independent on 0 x IV. y, y and y are linearly dependent on x 0 Which one among the following is correct? (A) Both I and II are true (B) Both I and III are true (C) Both II and IV are true (D) Both III and IV are true 07. Ans: (B) Sol: y =, y = x, y = x, Let the wronskian determinant be W 0 0 x 0 x x x 0 0 y, y and y are Linearly Independent for any x (, +) 08. A good trans-conductance amplifier should have (A) high input resistance and low output resistance (B) low input resistance and high output resistance (C) high input and output resistances (D) low input and output resistances 08. Ans: (C)

7 : 7 : GATE - 07_EC 09. Consider the D-Latch shown in the figure, which is transparent when its clock input CK is high and has zero propagation delay. In the figure, the clock signal CLK has a 0% duty cycle and CLK is a one-fifth period delayed version of CLK. The duty cycle at the output of the latch in percentage is. T clk CLK CLK CLK CLK D Q D-Latch CK output T clk / 09. Sol: T clk Clk (input) Clk (input) T clk T T clk T clk T T clk T clk T clk T 0 clk T 0 clk So duty cycle is 0%

8 : 8 : GATE The rank of the matrix M 0 is 6 6 (A) 0 (B) (C) (D) 0. Ans: (C) Sol: 0 0 M det (M) = 0 (first and third rows are proportional) Rank of M < 0 Let M be a submatrix 0 det (M ) = 0 0 Rank of M =. The Miller effect in the context of a Common Emitter amplifier explains (A) an increase in the low-frequency cutoff frequency (B) an increase in the high-frequency cutoff frequency (C) a decrease in the low-frequency cutoff frequency (D) a decrease in the high-frequency cutoff frequency. Ans (D) Sol: Due to miller effect the effective input Capacitance seen through the base terminal will increase which in turn reduces the higher cutoff frequency. Let (X, X ) be independent random variables, X has mean 0 and variance, while X has mean and variance 4. The mutual information I(X ; X ) between X and X in bits is.. Ans: 0 Sol: I(X, X ) = H (X ) H(X /X ) = H (X ) H (X ) [H(X /X ) = H(X ), since X & X are independent] = 0

9 : 9 : GATE - 07_EC

10 : 0 : GATE 07. Consider a single input single output discrete-time system with x[n] as input and y[n] as output, where the two are related as n xn, n xn, for 0 n 0 y n x otherwise, Which one of the following statements is true about the system? (A) It is causal and stable (B) It is causal but not stable (C) It is not causal but stable (D) It is neither causal nor stable. Ans: (A) Sol: The given system is y(n) = n x (n), 0 n 0 x(n) x(n ), other wise Present output depends on present input and past input, so it is a causal system For a bounded input, bounded output yields, so it is a stable system 4. In the circuit shown, the positive angular frequency (in radians per second) at which the magnitude of the phase difference between the voltages V and V equals 4 radians, is. V 00 cost H V Sol: V s s s s s Vi V () s V i s Vi V s s s V V i s s s () s

11 : : GATE - 07_EC V V s s s V j tan V j j The given condition is tan = 4 tan = 4 = tan 4 = rad/sec.. The voltage of an electromagnetic wave propagating in a coaxial cable with uniform characteristic impedance is V(l) = e l + jt Volts, where l is the distance along the length of the cable in metres, = (0. + j40) m is the complex propagation constant, and = 0 9 rad/s is the angular frequency. The absolute value of the attenuation in the cable in db/metre is.. Ans: Sol: Given = (0. + j40)m = + j = 0.Np/m ( Np = db) db = = dB/m 6. In a digital communication system, the overall pulse shape p(t) at the receiver before the sampler has the Fourier transform P(f). If the symbols are transmitted at the rate of 000 symbols per second, for which of the following cases is the inter symbol interference zero? (A) (B) p(f) p(f). 0. f(khz) f(khz)

12 : : GATE 07 (C) p(f) (D) p(f) f(khz) f(khz) 6. Ans: (B) Sol: Given f s =/T s = k symbols/sec Condition for zero ISI is given by n P (f n / T s ) = T s (constant) The above condition is satisfied by only option (b) n P (f nk) f (khz) 0 P(f nk) n Option (A) is correct if pulse duration is from to + Option (C) is correct if the transition is from 0.8 to., 0.8 to. Option (D) is correct if the triangular duration is from to +

13 : : GATE - 07_EC 7. Consider a stable system with transfer function p s G s q s b s a p q s... b... a p q where b,..,b p and a a q are real valued constants. The slope of the Bode log magnitude curve of G(s) converges to 60 db/decade as. A possible pair of values for p and q is (A) p = 0 and q = (B) p = and q = 7 (C) p = and q = (D) p = and q = 7. Ans: (A) Sol: To get a slope of 60dB/decade, required (p q) = 8. The clock frequency of an 808 microprocessor is MHz. If the time required to execute an instruction is.4 s, then the number of T-states needed for executing the instruction is (A) (B) 6 (C) 7 (D) 8 8. Ans: (C) Sol: f clk = MHz T clk f clk 0.s (no. of T-states) (T clk ) =.4 s.4s Number of T-states = 7 0.s 9. Which of the following can be the pole-zero configuration of a phase-lag controller (lag compensator)? (A) s-plane j Pole 0 zero (B) s-plane j Pole 0 zero

14 : 4 : GATE 07 (C) s-plane j Pole 0 zero (D) s-plane j Pole 0 zero 9. Ans: (A) s Sol: TF for lag compensator s For > lag compensator 0 0. Which one of the following statements about differential pulse code modulation (DPCM) is true? (A) the sum of message signal sample with its prediction is quantized (B) The message signal sample is directly quantized, and its prediction is not used (C) The difference of message signal sample and a random signal is quantized (D) The difference of message signal sample with its prediction is quantized 0. Ans: (D) Sol: In DPCM the difference of the message signal sample value and the output of prediction filter block is quantized. Consider a wireless communication link between a transmitter and a receiver located in free space, with finite and strictly positive capacity. If the effective areas of the transmitter and the receiver antennas, and the distance between them are all doubled, and everything else remains unchanged, the maximum capacity of the wireless link (A) increases by a factor of (B) decreases by a factor of (C) remains unchanged (D) decreases by a factor of

15 : : GATE - 07_EC. Ans: (C) Sol: Channel capacity (C) = C depends on S S B log N 4 4 Pt A et A PtG tg r S = p r (received signal power) = 4d 4d Given that (A et ) = ( A et ) (A er ) = (A er ) d = d A A d P r = P r (received power remains unchanged) et er Pr C remains unchanged er. The open loop transfer function G s s p s s s where p is an integer, is connected in unity feedback configuration as shown in the figure. + G(s) Given that the steady state error is zero for unit step input and is 6 for unit ramp input, the value of the parameter p is.. Ans: Sol: To get steady state error zero for unit step input and 6 for unit ramp input, the type of the system is one.

16 : 6 : GATE 07. In the latch circuit shown, the NAND gates have non-zero, but unequal propagation delays. The present input condition is: P = Q = 0, If the input condition is changed simultaneously to P = Q =, the outputs X and Y are P X Q Y (A) X =, Y = (B) either X =, Y = 0 or X = 0, Y = (C) either X =, Y = or X = 0, Y = 0 (D) X = 0, Y = 0. Ans: (B) Sol: Take t pd of gate < t pd of gate If we take option (A): Since output of st NAND is, it is input to second NAND, so output of second NAND has to be 0 (but given ), so not satisfied. Take option (B): 0 Since output of st NAND is 0, it is input to second NAND, so output of second NAND has to be (given also ) Hence satisfied (Or)

17 : 7 : GATE - 07_EC 0 Output of st NAND is, it is input to second NAND, so output of second NAND has to be 0 (given also 0 ) Hence satisfied Take option C: Already option (a) is not possible, hence c is also not possible Take option D: X (0) Y(0) Since output of first NAND is taken as 0, it is input to second, so output of second NAND has to be (but given 0 ) So not satisfied 4. Consider the matrix A It is given that A has only one real eigen value. Then the real eigen value of A is (A). (B) 0 (C) (D)

18 : 8 : GATE Ans: (C) Sol: A A I = C C + C + C + C 4 + C = 0 = 0 = A bar of Gallium Arsenide (GaAs) is doped with Silicon such that the Silicon atoms occupy Gallium and Arsenic sites in the GaAs crystal. Which one of the following statements is true? (A) Silicon atoms act as p-type dopants in Arsenic sites and n-type dopants in Gallium sites (B) Silicon atoms act as n-type dopants in Arsenic sites and p-type dopants in Gallium sites (C) Silicon atoms act as p-type dopants in Arsenic as well as Gallium sites (D) Silicon atoms act as n-type dopants in Arsenic as well as Gallium sites. Ans: (A) Sol: Substituting a Gallium site by a si atom produces a free electron so n-type Substituting an Arsenic site by a si atom produces a hole so p-type.

19 : 9 : GATE - 07_EC 6 Let x(t) be a continuous time periodic signal with fundamental period T = seconds, Let a k be the complex Fourier series coefficients of x(t), where k is integer valued. Consider the following statement about x(t): I. The complex Fourier series coefficients of x (t) are a k where k is integer valued II. The complex Fourier series coefficients of x (t) are a k where k is integer valued III. The fundamental angular frequency of x (t) is 6 rad/s For the three statements above, which one of the following is correct? (A) only II and III are true (B) only I and III are true (C) only III is true (D) only I is true 6. Ans: (B) Sol: x(t) a k, o = x(at) a k, a o x(t) a k, o = 6 so I and III are true

20 : 0 : GATE A 4-bit shift register circuit configured for right-shift operation is D in A, A B, BC, CD, is shown. If the present state of the shift register is ABCD = 0, the number of clock cycles required to reach the state ABCD = is D in Clock A B C D 7. Ans: 0 Sol: Clk A B C D required state 8. Let h[n] be the impulse response of a discrete-time linear time invariant (LTI) filter. The impulse response is given by h = 0 for n < 0 and n >. 0 ;h ;h ;and hn Let H () be the Discrete-Time Fourier transform (DTFT) of h[n], where is the normalized angular frequency in radians. Given that H ( 0 ) = 0 and 0 < 0 <, the value of 0 (in radians) is equal to. 8. Ans:.094,, Sol: h n j j j j j j j H e e e e e e e

21 : : GATE - 07_EC H j j j e e cos e j e cos j H e = 0 when + cos = 0 cos cos rad 9. A finite state machine (FSM) is implemented using the D flip-flops A and B and logic gates, as shown in the figure below. The four possible states of the FSM are Q A Q B = 00, 0, 0, and. D A Q A Q D Q Q B B CK Q X IN CK Q Assume that X IN is held at a logic level throughout the operation of the FSM. When the FSM is initialized to the state Q A Q B = 00 and clocked, after a few clock cycles, it starts cycling through (A) all of the four possible states if X IN = (B) three of the four possible states if X IN = 0 (C) only two of the four possible states if X IN = (D) only two of the possible states if X IN = 0 9. Ans: (D) Sol: If X IN = 0 D B Q A.0 CLK

22 : : GATE 07 Clk D A = Q A Q B D B = Q A Q B Thus number of possible states are two If X IN = D B Q A. Q A Clk D A = Q A Q B D B = Q A Q A Q B Thus number of possible states are three 0. A linear time invariant (LTI) system with the transfer function K s s s s s G is connected in unity feedback configuration as shown in the figure. + G(s) For the closed loop system shown, the root locus for 0 < K< intersects the imaginary axis for K =.. The closed loop system is stable for (A) K >. (B) < K <. (C) 0 < K < (D) no positive value of K 0. Ans: (A) Sol: For CL system stability system gain is greater than.

23 : : GATE - 07_EC j K= K=. +j 0 K=0 + K=0 + K= j K=.. Which one of the following gives the simplified sum of products expression for the Boolean function F = m 0 + m +m + m, where m 0,m,m and m are minterms corresponding to the inputs A, B, and C with A as the MSB and C as the LSB? (A) (C) AB AC. Ans: (B) Sol: F = m(0,,, ) A A BC ABC (B) A C AB ABC AB ABC (D) ABC A C ABC B C B C BC B C A F ABC A C AB A C. Let f (x) = AB ABC x x e for real x. From among the following. Choose the Taylor series approximation of f (x) around x = 0, which includes all powers of x less than or equal to. (A) +x+x + x (B) +x+ x +x (C) +x + x 7 6 x (D) +x+x + 7x

24 : 4 : GATE 07. Ans: (C) Sol: Given f(x) = x x e () x x x We know that e x...!! e x x x x x x x x x x x x x x 7 6 x!!. For the DC analysis of the common-emitter amplifier shown, neglect the base current and assume that the emitter and collector currents are equal. Given that V T = Mv, V BE = 0.7V, and the BJT output resistance r 0 is practically infinite. Under these conditions the midband voltage gain magnitude, A v = v 0 /v i V/V, is V cc = V x 6 7 k 0F R R c k C 0 F + C 8k R L V o v i 47 k R R E k C E 00F. Ans: 8 Sol: V k 4.7V R th i b 0 4.7V 4V k i E i C ma Figure: DC Analysis

25 : : GATE - 07_EC AC analysis A v = g Rc // m R L k // 8k 8 4. Let X(t) be a wide sense stationary random process with the power spectral density S x (f) as shown in figure (a), where f is in Hertz (Hz). The random process X(t) is input to an ideal low pass filter with the frequency response, f Hz H(f) = 0, f Hz As shown in Figure (b). The output of the low pass filter is Y(t). S x (f) exp( f ) x(t) Ideal lowpass y(t) filter h(t) Cutoff = / Hz f 0 (a) (b) Let E be the expectation operator and consider the following statements: I. E (X(t)) = E(Y(t)) II. E (X (t)) = E(Y (t)) III. E(Y (t)) = Select the correct option: (A) only I is true (B) only II and III are true (C) only I and II are true (D) only I and III are true 4. Ans: (A) Sol (I) E[x(t) ] = E[y(t) ] Proof: E [y(t) ] = H (0) E [x(t)] = E[x(t)] [ H (0) = ]

26 : 6 : GATE 07 (II) [x (t)] E [y (t)] Proof: / 0. t s f df s f H(f ) df S f df e E y E y x t S f df () () x x () x / () (III) E [y (t) ] Proof: From (II) E [x (t) ] = but E[y (t)]

27 . In the figure shown, the npn transistor acts as a switch. +V : 7 : GATE - 07_EC 0V V in (t) V k + T V in (t) t(in seconds) 4.8k For the input V in (t)as shown in the figure, the transistor switches between the cut-off and saturation regions of operation, when T is large. Assume collector-to-emitter voltage at saturation V CE(sat) = 0.V and base-to-emitter voltage V BE = 0.7V.The minimum value of the common-base current gain () of the transistor for the switching should be. Ans: min = Sol: I B 0.08mA k 0. IC sat ma 4.8k I k = I C-sat + I B = ma ma =.08 ma For saturation: I C-sat < I E IC IE sat ma.08ma > 0.90

28 : 8 : GATE 07

29 : 9 : GATE - 07_EC 6. An optical fiber is kept along the Z direction. The refractive indices for the electric fields along X and Y direction in the fiber are n x =.000 and n y =.00, respectively (n x n y due to the imperfection in the fiber cross-section). The free space wavelength of a light wave propagating in the fiber is. m. If the light wave is circularly polarized at the input of the fiber, the minimum propagation distance after which it becomes linearly polarized, in centimeters, is. 6. Ans: 0.7 Sol: For to have linear polarization, phase difference has to be 0 or 80. Given the light wave is circularly polarized that is initial phase difference is 90. So z z = z z px py n f x c f c n X n X n y c n y n y z ( px = c/n x and py = c/n y ) z z z = //(n x n y ) = /4(n x n y ) = cm 7. A three dimensional region R of finite volume is described by x + y z ; 0 z, Where x, y, z are real. The volume of R (up to two decimal places) is 7. Ans: Sol: Given: x + y z ; 0 z Put x = rcos, y = rsin, z = z dxdydz = rdrddz

30 Volume V dxdydz, where R: x + y z ; 0 z () The region of integration is 0 r 0 z V R r0 0 z0 r 0 = rdrddz 0 z 0 : 0 : GATE For the circuit shown, assume that the NMOS transistor is in saturation. Its threshold voltage W V tn = V and its trans-conductance parameter ncox ma / V. Neglect channel length L modulation and body bias effects. Under these conditions the drain current I D in ma is V DD = 8V M R R D k M R R s I D k 8. Ans: ma Sol: V G 8 V s = I D.k V

31 : : GATE - 07_EC V GS = I D k I D n C ox W L V V GS I 000I D = (4 I D.k) tn I.k D 0 D I D = 6 + I.0 8I.k D ID 0 ID 6 0 I D 0 4 D = 8 ma/ m A 6 0 I D = 8 ma when V s = 8 V (not possible) I D = ma when V s = V (possible) 6 9. The Nyquist plot of the transfer function S G K s s s does not encircle the point ( +j0) for K = 0 but does encircle the point ( + j0) for K = 00. Then the closed loop system (having unity gain feedback) is (A) stable for K = 0 and stable for K = 00 (B) stable for K = 0 and unstable for K = 00 (C) unstable for K = 0 and stable for K = 00 (D) unstable for K = 0 and unstable for K = Ans: (B) Sol: For given system, P = 0 For stability N = 0 For k =0, No encirclements about (, j0). Hence the system is stable For K = 00, encircles the point (, j0). Hence the system is unstable.

32 : : GATE 07

33 : : GATE - 07_EC 40. In binary frequency shift keying (FSK), the given signal wave forms are u 0 (t) = cos (0000t); 0 t T, and u (t) = cos (000t); 0 t T, where T is the bit-duration interval and t is in seconds. Both u o (t) and u (t) are zero outside the interval 0 t T. With a matched filter (correlator) based receiver, the smallest positive value of T (in milliseconds) required to have u 0 (t) and u (t) uncorrelated is (A) 0. ms (B) 0. ms (C) 0.7 ms (D).0 ms 40. Ans: (B) Sol: Method - I 0000 f = 0k f 000 k f [Condition for un-correlation in coherent FSK] Tb f T b f f T b = 0.ms Method - II T 0 T 0 t f u 0 u 0 [If two signals are un-correlated] cos 4000t cos000t 4000T sin000t sin dt 0 0 sin 000T = 0 when 000T =

34 : 4 : GATE 07 At T 000 T, = = T = 0.ms 4. Which one of the following is the general solution of the first order differential equation dy dx x y where x, y are real? (A) y = + x + tan (x + c), where c is a constant (B) y = + x + tan (x + c), where c is a constant (C) y = x + tan (x + c), where c is a constant (D) y = x + tan (x + c), where c is a constant 4. Ans: (D) dy Sol: x y () dx Put x + y = t dy dx dy dx dt dx dt dx dt dx t dt dx t dt dx t Integrating both sides dt t dx tan t = x + c

35 : : GATE - 07_EC tan (x+y ) = x + c (or) x + y = tan (x + c) (or) y = x + tan (x+c) is the solution. 4. Let I = c (z dx+y dy + x dz) where x, y, z are real, and let C be the straight line segment from point A: (0,, ) to point B: (4,, ). The value of I is 4. Ans: Sol: Given: I C B zdx ydy xdz I d xz ydy A 4,, 0,, d xz ydy 4,, xz y 0,, = ( 8 + ) (4) = 4. Two discrete-time signals x [n] and h [n] are both non-zero only for n = 0,,,and are zero otherwise. It is given that x[0] =, x [] =, x [] =, h[0] =, Let y[n] be the linear convolution of x [n] and h [n]. Given that y [] = and y [] = 4, the value of the expression (0y[] + y[4]) is 4. Ans: Sol: x(n) = {,,} h(n) = {,x,y} y(n) = x ( n) h(n) x x x x y y y y

36 : 6 : GATE 07 y(n) = {, +x, x + y +, x + y, y} y() = = +x x = y() = 4 = x+y+ y = y(n) = {,, 4,, } 0 y () + y(4) = 0 + = 44. The amplifier circuit shown in the figure is implemented using a compensated operational amplifier (op-amp), and has an open-loop voltage gain, A 0 = 0 V/V and an open-loop cutoff frequency, f c = 8 Hz. The voltage gain of the amplifier at khz in V/V, is R = k V i + R = 79k V o 44. Ans: 44.6 Sol: 79k k + V o + V I Method-I: A s Ao s p 0 s f c

37 : 7 : GATE - 07_EC V V o V o 80 Vo 80 Vo VI A s 80 A s Vo VIA s 80 V V o I A s A s s 80 0 c s 0 / c s c V V o s 0 80 c A Method-II: A CL A / A jf / f A f c (+A) = 8 0 A CL c 80 jf /0k 0kHz 80 A CL f k 80 k j 0k

38 : 8 : GATE In the circuit shown the voltage V IN (t) is described by: 0, for t 0 V IN t Volts, for t 0 where t is in seconds. The time (in seconds) at which the current I in the circuit will reach the value Ampere is + V IN (t) H 4. Ans: 0.40 Sol: By applying the laplace transform with initial conditions I H V(s) I(s) S + s s Nodal in S-D V s s V s s 0 V s s I s I s s V s z V s s V s s s 0 0 s s i(t) = t e A for 0 t

39 : 9 : GATE - 07_EC when i(t) = A, then t = e t = loge = 0.40sec 46. As shown a uniformly doped silicon (Si) bar of length L = 0. m with a donor concentration N D = 0 6 cm is illuminated at x = 0 such that electron and hole pairs are generated at the rate of G L = G Lo x, 0 x L, where G Lo = 0 7 cm s, Hole lifetime is 0 4 s, electronic L charge q = C, hole diffusion coefficient D P = 00cm /s and low level injection condition prevails. Assuming a linearly decaying steady state excess hole concentration that goes to 0 at x = L, the magnitude of the diffusion current density at x = L/, in A/cm is. Light Si(N D =0 6 cm ) x= 0 L= 0. m 46. Ans: 6 dp Sol: J Dpq dx P x G Lo x x L po p J = (+) (+) L no x 0 L p no L = 0 cm x L 0 p no =.60 A/cm = 6 A/cm

40 : 40 : GATE As shown, two Silicon (Si) abrupt p-n junction diodes are fabricated with uniform donor doping concentrations of N 4 D 0 cm and N 6 D 0 cm in the n-regions of the diodes, and uniform acceptor doping concentrations of N A = 0 4 cm and N A = 0 6 cm in the p-regions of the diodes, respectively. Assuming that the reverse bias voltage is >> built-in potentials of the diodes, the ratio. C of their reverse bias capacitances for the same applied reverse bias, is C p n cm cm C Diode p n cm cm C Diode 47. Ans: 0 Sol: C A For abrupt junction, vo V q R N A N D Since V o << V R V o +V R = V R C C 6 N A ND VR N A N D N A ND 0 V R 4 N N 0 A D 48. The expression for an electric field in free space is E E 0 (xˆ ŷ jẑ)e -j( t-kxky), where x, y, z represent the spatial coordinates, t represents time, and, k are constants. This electric field (A) does not represent a plane wave. (B) represents a circularly polarized plane wave propagating normal to the z-axis. (C) represents an elliptically polarized plane wave propagating along the x-y plane. (D) represents a linearly polarized plane wave.

41 : 4 : GATE - 07_EC 48. Ans: (C) xˆ ŷ jẑ e jtkxky Sol: E = E 0 e jkr = e j ( kx + ky) kr = k ( x + y ) propagation vector (kr) = k xˆ ŷ (kr) = k â p kr xˆ kr â P ŷ For plane wave â p.ê 0 xˆ ŷ â p.ê.e Eo o Eo j0 kr kr xˆ ŷ jẑ â p.ê 0 given is a plane wave. jtkxky As E E xˆ ŷ jẑe o For the given wave, plane of incidence is xy-plane. E in xy plane is parallel polarized and along z is perpendicular polarized E E xy E E z E T E E r E E and phase difference is 90 ; i.e. given is a r Elliptically polarized plane wave

42 : 4 : GATE Which one of the following options correctly describes the locations of the roots of the equation s 4 + s + = 0 on the complex plane? (A) Four left half plane (LHP) roots (B) One right half plane (RHP) root, one LHP root and two roots on the imaginary axis (C) Two RHP roots and two LHP roots (D) All four roots are on the imaginary axis 49. Ans: (C) Sol: CE S 4 + S + = 0 S 4 S S S S 0 S 4 S S 0 0 ROZ sign changes and ROZ poles in right half of S-plane And symmetrical poles in the LHS-plane. 0. The following FIVE instructions were executed on an 808 microprocessor. MVI A, H MVI B, 78H ADD B CMA ANI H The Accumulator value immediately after the execution of the fifth instruction is (A) 00H (B) 0H (C) H (D) H 0. Ans: (B) Sol: MVI A, H: A H MVI B, 78H: B 78 H

43 : 4 : GATE - 07_EC ADD B: 7 8 AB H A AB H CMA: A = 00 0 A A 4 H ANI H: H A 0 H. Starting with x =, the solution of the equation x + x =, after two iterations of Newton- Raphson s method (up to two decimal places) is. Ans: Sol: Let f(x) = x + x, x 0 = f (x) = x + The first approximation is x x = 0.7 f f x x 0 0 f f

44 : 44 : GATE 07 The second approximation is x x f f x x f 0.7 f = x = A half wavelength dipole is kept in the x-y plane and oriented along 4 o from the x-axis. Determine the direction of null in the radiation pattern for 0. Here the angle (0 ) is measured from the z-axis, and the angle (0) is measured from the x-axis in the x-y plane. (A) = 90 o, = 4 o (B) = 4 o, = 90 o (C) = 90 o, = o (D) = 4 o, = o. Ans: (A) Sol: As the antenna is placed in xy- plane which is horizontal plane i.e y =4 x Antenna in As there is no field along antenna i.e null along antenna, = 4 as 0 given for the given antenna null is at = 90, = 4

45 : 4 : GATE - 07_EC. The dependence of drift velocity of electrons on electric field in a semiconductor is shown below. The semiconductor has a uniform electron concentration of n = 0 6 cm and electronic charge q =.60 9 C. If a bias of V is applied across a m region of this semiconductor, the resulting current density in this region, in ka/cm, is. Drift velocity (cm/s) 0 7 linear constant 0 0 Electric field (V/cm). Ans:.6 Sol: E V 4 0 V / cm 6 d 0 m V d (cm/s) At E = 0 4 v/cm V d = 0 6 cm/s J = E = n q n E = n q V d = =.6 KA /cm 4. A continuous time signal x(t) = 4cos(00t) + 8cos(400t), where t is in seconds, is the input to a linear time invariant (LTI) filter with the impulse response h (t) sin(00t), t 600, t 0 t 0 Let y(t) be the output of this filter. The maximum value of y(t) is

46 : 46 : GATE Ans: 8 Sol: x(t) = 4cos (00t) + 8 cos (400t) h(t) = sin 00t t 600 ; ; t 0 t 0 H() The input signal frequencies are 00, 00 Hz The o/p signal is = 4 cos (00t) = 8 cos (00t) The maximum value y(t) is 8. The figure shows an RLC circuit excited by the sinusoidal voltage 00cos(t) Volts, where t is in seconds. The ratio amplitude of V amplitude of V is. V 4 H 00cost V F 6. Ans:.6 Sol: V s i V s 4 s 6 4 s s V s s 4 () V 6 i s s 9 s

47 : 47 : GATE - 07_EC s 6 s 6 s 4 s V s V i 9 s 6 s s 6 s V s V i () 4 s s 6 V (s) V (s) j 4 j 6 j V j V j 4 j 6 j V j V rad / sec j 4 j =.6

48 : 48 : GATE 07 General Aptitude 6. In the summer, water consumption is known to decrease overall by %. A water Board official states that in the summer household consumption decreases by 0% while other consumption increases by 70%. Which of the following statements is correct? 8 (A) The ratio of household to other consumption is 7 (B) The ratio of household to other consumption is 7 7 (C) The ratio of household to other consumption is 8 (D) There are errors in the officials statement. 6. Ans: (D) Sol: H = house hold consumption P = other consumption House hold consumption decreases by 0% H Other consumption increases by 70% 70 P 00 80H 70P 7 H P H + 70 P = 7 H + 7P 80 H 7 H = 7 P 70P H = 9 P There is a negative ratio so, there are errors in the official s statement % of deaths on city roads may be attributed to drunken driving. The number of degrees needed to represent this as a slice of a pie chart is (A) 0 (B) 44 (C) 60 (D)

49 : 49 : GATE - 07_EC 7. Ans: (B) Sol: Sum of angles in a pie chart = 60 o The relation between angle and percentage is 00 % = 60 o % =.6 o 40% =? 40.6 = 44 o 8. Some tables are shelves. Some shelves are chairs. All chairs are benches. Which of the following conclusions can be deduced from the preceding sentences? (i) At least one bench is a table (ii) At least one shelf is a bench (iii) At least one chair is a table (iv) All benches are chairs (A) Only i (B) Only ii (C) Only ii and iii (D) Only iv 8. Ans: (B) Sol: From given statements the following venn diagrams are possible T = tables, S = shelves, C = chairs and B = benches T S C B T S C B (a) (b) T S C B (c) From all of the above diagrams, conclusion (ii) only deduced from the statements.

50 : 0 : GATE I made arrangements had I informed earlier. (A) could have, been (B) would have, being (C) had, have (D) had been, been 9. Ans: (A) Sol: Conditional tense Type Past perfect (could have) + perfect conditional (had + V ) 60. She has a sharp tongue and it can occasionally turn (A) hurtful (B) left (C) methodical (D) vital 60. Ans: (A) Sol: Hurtful It is a supporting sentence. The word sharp tongue strengthens the latter part of the sentence it can occasionally turn hurtful 6. If you are looking for a history of India, or for an account of the rise and fall of the British Raj, Or for the reason of the cleaving of the subcontinent into two mutually antagonistic parts and the effects this mutilation will have in the respective sections, and ultimately on Asia, you will not find it in these pages; for though I have spent a lifetime in the country, I lived too near the seat of events, and was too intimately associated with the actors, to get the perspective needed for the impartial recording of these matters. Here, the word antagonistic is closest in meaning to (A) impartial (B) argumentative (C) separated (D) hostile 6. Ans: (D) Sol: Antagonistic means showing dislike or opposition. So the word closest in meaning is hostile (not friendly, having or showing unfriendly feelings, unpleasant or harsh) 6. S, T, U, V, W, X, Y, and Z are seated around a circular table. T s neighbours are Y and V. Z is seated third to the left of T and second to the right of S. U s neighbours are S and Y; and T and W are not seated opposite each other. Who is third to the left of V? (A) X (B) W (C) U (D) T

51 : : GATE - 07_EC 6. Ans: (A) Sol: From the given data, eight persons are seated around a circular table as follows Y T V S U Y (or) (or) V T Y Y U S S Z T U Y T S V X Z W X is third to the left of V 6. A contour line joins locations having the same height above the mean sea level. The following is a contour plot of a geographical region. Contour lines are shown at m intervals in this plot Q P The path from P to Q is best described by (A) Up-Down-Up-Down (B) Down-Up-Down-Up (C) Down-Up-Down (D) Up-Down-Up

52 : : GATE Ans: (C) Sol: Contour lines can be observed to cross region with height from P to Q is as follows (P) (Q) Down Up Down The path from P to Q is Down-Up-Down option (C) is satisfies this path 64. Trucks (0 m long) and cars ( m long) go on a single lane bridge. There must be a gap of at least 0 m after each truck and a gap of at least m after each car. Trucks and cars travel at a speed of 6km/h. If cars and trucks go alternately, what is the maximum number of vehicles that can use the bridge in one hour? (A) 440 (B) 00 (C) 70 (D) Ans: (A) Sol: Length of truck + gap required = = 0 m Length of car + gap required = + = 0 m Total distance is need for truck and car for passing alternatively = = 0 m Given, speed = 6 kmph 6 0 m / sec 8 Let x be the number of repetitions of (Truck + car) in one hour 0 x 0 m /s x = 70 numbers of (Trucks + cars) 0 The maximum number of vehicles = = There are Indians and Chinese in a group of 6 people. How many subgroups of this group can we choose so that every subgroup has at least one Indian? (A) 6 (B) (C) 48 (D) Ans: (A) Sol: Sub group has at least one Indian means minimum one Indian and maximum three (or) more

53 : : GATE - 07_EC Sub groups containing only Indians = C +C +C = + + = 7 In the sub group one Indian and remaining are Chinese = C [C +C +C ] = [ + + ] = 7 = In the sub group two Indians and remaining are Chinese = C [C +C +C ] = [ + + ] = 7 = In the sub group three Indians and remaining are Chinese = C [C +C +C ] = [ + + ] = 7 Total number of sub groups = = 6

54 : 4 : GATE 07