EC-Paper Code-B GATE 2011

Transcription

1 . No. 5 Carry One Mark Each. Consider the following statements regarding the complex Poynting vector P for the power radiated by a point source in an infinite homogeneous and lossless medium. Re ( P ) denotes the real part of P. S denotes a spherical surface whose centre is at the point source, and ndenotes ɵ the unit surface normal on S. Which of the following statements is TRUE? (A) Re ( P ) remains constant at any radial distance from the source (B) Re ( P ) increases with increasing radial distance from the source (C) Re ( P ).nds remains constant at any radial distance from the source s (D) Re ( P ).nds decreases with increasing radial distance form the source s Re P.nds ˆ gives average power and it decreases with increasing radial S Exp: - distance from the source. A transmission line of characteristic impedance 50Ω is terminated by a 50Ω load. When excited by a sinusoidal voltage source at 0GHz, the phase difference π between two points spaced mm apart on the line is found to be radians. The 4 phase velocity of the wave along the line is (A) Answer: - (C) m /s (B) Exp: - Z0 = 50Ω ; ZL = 50Ω 8. 0 m / s (C) For 4 π radians the distance is mm m /s (D) 3 0 m /s 8 The phase velocity ω π 0 β π vp = = = 6 0 =.6 0 m / s 3 3. An analog signal is band-limited to 4kHz, sampled at the Nyquist rate and the samples are quantized into 4 levels. The quantized levels are assumed to be independent and equally probable. If we transmit two quantized samples per second, the information rate is bits / second. (A) (B) (C) 3 (D) 4 Exp: - Since two samples are transmitted and each sample has bits of information, then the information rate is 4 bits/sec. EC-Paper Code-B GATE 0

2 4. The root locus plot for a system is given below. The open loop transfer function corresponding to this plot is given by (A) s ( s ) ( ) ( ) G s H s = k s s 3 (B) ( s ) ( ) ( ) G s H s = k s s s 3 G s H s = k s s s s 3 (C) (D) ( ) ( ) ( ) ( s ) ( ) ( ) G s H s = k s s s 3 Answer: - (B) Exp: - 'x' indicates pole 'O' indicates zero The point on the root locus when the number of poles and zeroes on the real axis to the right side of that point must be odd n 5. A system is defined by its impulse response h ( n) u ( n ) Answer: - (B) (A) stable and causal (C) stable but not causal =. The system is (B) causal but not stable (D) unstable and non-causal Exp: - h ( n) = n u ( n ) h ( n ) is existing for n> ; so that h ( n) = 0;n < 0 causal n n h ( n) = u ( n ) = = System is unstable n= n= n= 6. If the unit step response of a network is ( e αt ), then its unit impulse response is (A) α e αt (B) Exp: - S ( t) step response Impulse response αt αt α e (C) ( ) e d d αt h t = ( S t ) = ( e ) = α e dt dt o 3 α (D) ( α ) αt jω 0 e αt σ 7. The output Y in the circuit below is always when P R Y

3 (A) two or more of the inputs P,,R are 0 Answer: - (B) (B) two or more of the inputs P,,R are (C) any odd number of the inputs P,,R is 0 (D) any odd number of the inputs P,,R is Exp: - The output Y expression in the Ckt Y P PR R = So that two or more inputs are, Y is always. 8. In the circuit shown below, capacitors C and C are very large and are shorts at vo the input frequency. v i is a small signal input. The gain magnitude at 0M vi rad/s is 5V 0µ H kω kω C.7V v i ~ kω C v o kω (A) maximum (B) minimum (C) unity (D) zero Exp: - In the parallel RLC Ckt L = 0µ H and C = nf ω g = = = = 6 9 LC rad/ s 0Mrad /s So that for a tuned amplifier, gain is maximum at resonant frequency 9. Drift current in the semiconductors depends upon Answer: - (C) (A) only the electric field (B) only the carrier concentration gradient (C) both the electric field and the carrier concentration (D) both the electric field and the carrier concentration gradient Exp: - Drift current,j n P = σ E J = nµ pµ qe So that it depends on carrier concentration and electric field.

4 0. A Zener diode, when used in voltage stabilization circuits, is biased in (A) reverse bias region below the breakdown voltage (B) reverse breakdown region (C) forward bias region (D) forward bias constant current mode Answer: - (B) Exp: - V For Zener diode Voltage remains constant in break down region and current carrying capacity in high.. The circuit shown below is driven by a sinusoidal input v V cos ( t /RC) steady state output v o is i =. The p R C v i ~ R C v o (A) ( V p /3) cos ( t /RC) (B) ( V p /3) sin ( t /RC) (C) ( V p /) cos ( t /RC) (D) ( V p /) sin ( t /RC) Exp: - v0 z = v z z z i R = R jcw where z = R j ω and z c = R j ω c t R Given w = vi = vp cos RC RC z = j z = R = R R ( j) j ω c jr R v0 j = = = v R i 3 R ( j) j

5 . Consider a closed surface S surrounding volume V. If r is the position vector of a point inside S, with n ɵ the unit normal on S, the value of the integral 5r.ndS ɵ is (A) 3V (B) 5V (C) 0V (D) 5V Exp: - Apply the divergence theorem 5r.n.dx = 5.rdV S v = 5 ( 3) dv = 5 V (.r = 3 and r is the position vector ) v s 3. The modes in a rectangular waveguide are denoted by TE TM mn mn where m and n are the eigen numbers along the larger and smaller dimensions of the waveguide respectively. Which one of the following statements is TRUE? (A) The TM 0 mode of the wave does not exist (B) The TE 0 mode of the wave does not exist (C) The TM 0 and the TE 0 modes both exist and have the same cut-off frequencies (D) The TM 0 and TM 0 modes both exist and have the same cut-off frequencies Exp: - TM 0 mode doesn t exist in rectangular waveguide. dy 4. The solution of the differential equation ky, y ( 0) (A) Answer: - (C) Exp: - Given y ( 0) x = ce ky (B) x = C and dy ky, dx = = = kx c ln y kx c y e e When y ( 0) = C, y cy = ke (C) dy = kdx y 0 = ke y = c e kx ( k = C) c dx = = is y kx = ce (D) y = ce kx 5. The Column-I lists the attributes and the Column-II lists the modulation systems. Match the attribute to the modulation system that best meets it Column-I Column-II P Power efficient transmission of signals Conventional AM Most bandwidth efficient transmission of voice signals FM R Simplest receiver structure 3 VSB S Bandwidth efficient transmission of signals with Significant dc component (A) P-4;-;R-;S-3 (C) P-3;-;R-;S-4 4 SSB-SC (B) P-;-4;R-;S-3 (D) P-;-4;R-3;S-

6 Answer: - (B) Exp: - Power efficient transmission FM Most bandwidth efficient SSB-SC Transmission of voice signal Simplest receives structure conventional AM Bandwidth efficient transmission of VSB Signals with significant DC component d y dy = describes a system with an dt dt input x(t) and an output y(t). The system, which is initially relaxed, is excited by a unit step input. The output y(t) can be represented by the waveform 6. The differential equation 00 0 y x ( t) (A) y ( t) (B) y ( t) t t y ( t) y ( t) (C) (D) t t 00d y Exp: - y = x ( t) dt 0dy dt Apply L.T both sides ( 00s 0s ) Y ( s) = x ( t) u ( t) x ( s) s = = 3 Y s = ( ) s 00s 0s So we have poles with positive real part system is unstable.

7 7. For the transfer function positive frequency has the form G jω = 5 jω, the corresponding Nyquist plot for jω jω j5 (A) 5 σ (B) σ jω jω (C) /5 σ (D) /5 σ Exp: - As we increases real part 5 is fixed only imaginary part increases. 8. The trigonometric Fourier series of an even function does not have the (A) dc term (C) sine terms Answer: - (C) Exp: - f ( t ) is even function, hence b k = 0 Where 'b k ' is the coefficient of sine terms (B) cosine terms (D) odd harmonic terms 9. When the output Y in the circuit below is, it implies that data has Data D D Y Clock (A) changed from 0 to (B) changed from to 0 (C) changed in either direction Exp: - When data is 0, is 0 And is first flip flop Data is changed to is first D is connected to nd flip flop So that = So that the inputs of AND gate is y = '' (D) not changed

8 0. The logic function implemented by the circuit below is (ground implies logic 0) 4 MUX I 0 I I Y F I 3 S S 0 (A) F = AND ( P,) (B) F = OR ( P,) (C) F = XNOR ( P,) (D) F = XOR ( P,) Exp: - From the CKT O is connected to 'I 0 ' & 'I 3 ' And is connected to I & I F P P P = = XOR ( P,). The circuit below implements a filter between the input current i i and the output voltage v o. Assume that the opamp is ideal. The filter implemented is a L i i R v o (A) low pass filter (C) band stop filter Exp: - When W=0; inductor acts as a S.C V0 = 0 (B) band pass filter (D) high pass filter And when ω =, inductor acts as a O.C V0 = ir So it acts as a high pass filter.. A silicon PN junction is forward biased with a constant current at room temperature. When the temperature is increased by 0ºC, the forward bias voltage across the PN junction (A) increases by 60mV (C) increases by 5mV Exp: - For Si forward bias voltage change by -.5mv / 0 C For (B) decreases by 60mV (D) decreases by 5mV 0 0 C increases, change will be.5 0 = 5mV

9 3. In the circuit shown below, the Norton equivalent current in amperes with respect to the terminals P and is j30ω P O 6 0 A 5Ω j50 Ω 5Ω (A) 6.4 j4.8 (B) 6.56 j7.87 (C) 0 j0 (D) 6 j0 Exp: - When terminals P & are S.C Then the CKT becomes j30ω P A 5Ω IN = I SC 5Ω From current Division rules I N = = 5 5 j30 40 j30 ( 6) ( 5) ( ) = = 6.4 j j3 4. In the circuit shown below, the value of R L such that the power transferred to R L is maximum is (A) 5Ω (B) 0 Ω (C) 5 Ω (D) 0 Ω 0Ω 0Ω 0Ω R L 5V V A Answer: - (C) Exp: - For maximum power transmission For the calculation of R TH R L = R * TH

10 0Ω 0Ω 0Ω R TH TH R = = 5Ω 5. The value of the integral 3z 4 dz where c is the circle z ( c z 4z 5) (A) 0 (B) /0 (C) 4/5 (D) 3z 4 Exp: - dz = 0 z 4z 5 C ( z 4z 5 = z = 0) z = ± j will be outside the unit circle So that integration value is zero. = is given by. No. 6 5 Carry Two Marks Each 6. A current sheet J = 0u ɵ y A/m lies on the dielectric interface x=0 between two dielectric media with ε r = 5, µ r = in Region - (x<0) and ε r = 5, µ r = in Region - (x>0). If the magnetic field in Region- at x=0 - is H = 3uɵ 30uɵ A /m the magnetic field in Region- at x=0 is x > 0 Region : ε, µ = x J x = 0 y x < 0 ( Region ) : ε r, µ r = (A) H ɵ ɵ ɵ =.5ux 30uy 0uz A /m (B) H ɵ ɵ ɵ = 3ux 30uy 0uzA /m (C) H =.5uɵ 40uɵ A /m (D) H = 3uɵ 30uɵ 0uɵ A /m x y Exp: - H H = J a H = H 0uˆ = 30u 0u ˆ t t n t t z y z And Bn = Bn µ H = µ H µ H = µ H Normal component in x direction H 3 uˆ = x x r r =.5u ˆ ; H =.5u ˆ ˆ x 30uy 0uz A / m x y z x y

11 7. A transmission line of characteristic impedance 50W is terminated in a load impedance Z L. The VSWR of the line is measured as 5 and the first of the voltage maxima in the line is observed at a distance of 4 λ from the load. The value of ZL is (A) 0Ω (B) 50 Ω (C) ( 9.3 j46.5) Ω (D) ( 9.3 j46.5) Exp: - Voltage maximum in the line is observed exactly at 4 λ Therefore 'z L ' should be real Ω VSWR z 0 = L z L 50 z = = 0Ω ( Voltage minimum at load) 5 8. X(t) is a stationary random process with autocorrelation function R τ = exp π r. This process is passed through the system shown below. The x power spectral density of the output process Y(t) is (A) ( 4π f ) exp ( π f ) (B) ( 4π f ) exp ( π f ) (C) ( 4π f ) exp ( π f ) (D) ( 4π f ) exp ( π f ) Exp: - The total transfer function H(f) = ( jπf ) X F = R ( τ) S ( f) S f H f S f x = ( 4π f ) e πf t F f π π e e x X ( t ) x H ( f ) = jπf Y ( t ) 9. The output of a 3-stage Johnson (twisted ring) counter is fed to a digital-toanalog (D/A) converter as shown in the figure below. Assume all the states of the counter to be unset initially. The waveform which represents the D/A converter output V o is Vref D / A Converter D D D 0 V o Clock 0 Johnson Counter

12 V o V o (A) (B) V o V o (C) (D) Exp: - For the Johnson counter sequence DDD0 V Two D flip-flops are connected as a synchronous counter that goes through the following B A sequence The combination to the inputs D A and D B are (A) DA = B; DB = A (B) DA = A; DB = B (C) DA = ( AB A B ); DB = A (D) Exp: - (present) B A (next) B A D B D A D = ; D = A A B A B B B D A = A B 0 D B B A

13 3. In the circuit shown below, for the MOS transistors, µ C = 00µ A / V and the threshold voltage V V. T = The voltage V x at the source of the upper transistor is n ox 6V 5V W / L = 4 V x W / L = Answer: - (C) (A) V (B) V (C) 3V (D) 3.67V Exp: - The transistor which has W 4 L = DS = x and VGS 5 Vx V 6 V = VGS VT = 5 Vx = 4 Vx VDS > VGS VT So that transistor in saturation region. The transistor which has W L = Drain is connected to gate So that transistor in saturation > > ( V = V ) VDS VGS VT The current flow in both the transistor is same DS GS w VGS V T w VGS V T µ n c0x n c 0x. L = µ L ( 5 V ) x Vx 4 = ( V = V 0) 4 GS 4 Vx 8Vx 6 = Vx Vx 3Vx 30Vx 63 = 0 Vx = 3V x 3. An input x ( t) = exp ( t) u ( t) δ ( t 6) is applied to an LTI system with impulse response h ( t) = u( t). The output is is (A) exp ( t) u ( t) u ( t 6) (B) exp ( t) u ( t) u ( t 6) (C) 0.5 exp ( t) u ( t) u( t 6) (D) 0.5 exp ( t) u ( t) u( t 6)

14 Exp: - x ( s) = e s 6s Y ( s) = H ( s) ( s) and H ( s) = s e = s s s 6s e = s s ( ) ( s ) t y t = 0.5 e u t u t 6 6s 33. For a BJT the common base current gain α = 0.98 and the collector base junction reverse bias saturation current I CO = 0.6µ A. This BJT is connected in the common emitter mode and operated in the active region with a base drive current I B =0 A. The collector current I C for this mode of operation is (A) 0.98mA (B) 0.99mA (C).0mA (D).0mA Exp: - = β ( β) I I I C B CB0 B = µ = µ I CB0 C I 0 A,I 0.6 A α 0.98 = β = = = 49 α 0.98 =.0mA 34. If F ( s) L f ( t) = = ( ) s s 4s 7 then the initial and final values of f(t) are respectively (A) 0, (B),0 (C) 0,/7 (D) /7,0 Answer: - (B) Exp: - ( ) s s Lt f t = Lt = t 0 s s 4s 7 ( ) s s Lt f t = Lt = 0 t s 0 s 4s In the circuit shown below, the current I is equal to 4 0ºV I ~ j4ω 6Ω j4ω 6Ω 6Ω (A) 40ºA (B).00ºA (C).80ºA (D) 3.0ºA Answer: - (B) Exp: - Apply the delta to star conversion The circuit becomes

15 4 0ºV ~ I j4ω j4ω Ω Ω Ω The net Impedance = ( j4 ) ( j4) I = = A 4 6 = = 7 Ω A numerical solution of the equation f ( x) = x x 3 = 0 can be obtained using Newton- Raphson method. If the starting value is x = for the iteration, the value of x that is to be used in the next step is (A) (B) (C).694 (D).306 Answer: - (C) f ( xn ) Exp: - xn = xn f ( x ) f ( 3) n = = and ( ) xn = =.694 x f = = 37. The electric and magnetic fields for a TEM wave of frequency 4 GHz in a homogeneous medium of relative permittivity ε r and relative permeability µ r = are given by j ωt80πy j( ωt80π y ˆ ) E = E e u V /m H = 3e ˆu A /m p z x Assuming the speed of light in free space to be 3 x 0 8 m/s, the intrinsic impedance of free space to be 0π, the relative permittivity ε r of the medium and the electric field amplitude E p are (A) ε r = 3, E p = 0π (B) ε r = 3, E p = 360π (C) ε r = 9, E p = 360π (D) ε r = 9, E p = 0π Exp: - E µ = η = = 0π H E P 3 µ µ r r r = η = 0π Only option D satisfies r

16 38. A message signal m ( t) = cos00π t 4cos π t modulates the carrier c ( t) = cosπ f t where f MHZ c c = to produce an AM signal. For demodulating the generated AM signal using an envelope detector, the time constant RC of the detector circuit should satisfy (A) 0.5 ms < RC < ms (B) µ s << RC < 0.5 ms (C) RC Answer: - (B) << µ s (D) RC >> 0.5 ms Exp: - Time constant should be length than f And time constant should be far greater than 4000a fm = = 000 a << Rc < f 000 C µ s << RC << 0.5ms m f c 39. The block diagram of a system with one input it and two outputs y and y is given below. s y s y A state space model of the above system in terms of the state vector x and the output vector y = y y is (A) x = x u; y = x (B) x = x u; y = x 0 (C) x = x u; y = x 0 0 (D) x = x u; y = x 0 Answer: - (B) Exp: - Draw the signal flow graph T

17 x / S 4 y x / S y From the graph xɺ = x 4 & y = x ; y = x ɺ y x = x u; = x y 40. Two systems H (z) and H (z) are connected in cascade as shown below. The overall output y(n) is the same as the input x(n) with a one unit delay. The transfer function of the second system H (z) is X ( n ) H ( z) = ( 0.4z ) ( 0.6z ) H ( z ) y ( n) ( ) ( ) ( 0.6z ) 0.6z (A) z 0.4z z 0.4z (C) Answer: - (B) Exp: - The overall transfer function = z ; H ( z) H z H z ( ) ( 0.4z ) ( ) z 0.6z (B) 0.4z (D) z 0.6z = z ( unit day T.F z ( 0.6z ) ( ) = = z H ( z) 0.4z = z ) 4. An 8085 assembly language program is given below. Assume that the carry flag is initially unset. The content of the accumulator after the execution of the program is MVI RLC MOV RLC RLC ADD RRC A,07H B,A B (A) 8CH (B) 64H (C) 3H (D) 5H

18 Answer: - (C) Exp: - MVI A, 07 H The content of A RLC The content of A MOV B, A The content of B RLC The content of B RLC The content of B ADD B RRC A B H 4. The first six points of the 8-point DFT of a real valued sequence are5, j3,0,3 j4, 0 and 3 j4.. The last two points of the DFT are respectively Answer: - (C) (A) 0, -j3 (B) 0, j3 (C) j3, 5 (D) j3, 5 Exp: - DFT points are complex conjugates of each other and they one symmetric to the middle point. = * = * = * = * x 0 x 7 x x 6 x x 5 x 3 x 4 * * Last two points will be x ( 0 ) and x = j3 and For the BJT L in the circuit shown below, β =, V = 0.7V. The switch BEon= 0.7V, V CEsat is initially closed. At time t = 0, the switch is opened. The time t at which leaves the active region is 5V 0.5mA 5V t = 0 5µ F 4.3kΩ 0V (A) 0 ms (B) 5 ms (C) 50 ms (D) 00 ms

19 Answer: - (C) Exp: - Apply KVL at the BE junction I E = = = ma 4.3kΩ 4.3kΩ Always I E Cap = ma ; At collector junction = ( β = ;I = I ) I 0.5mA ma ICap = 0.5 = 0.5mA always constant VCE = VC VE VC = VCE VE = 0.7 ( 4.3) = ( V = I R ) V = 5V = V V C cap cap t = ICap Or c Cap E C E E E VCap C t = = I = 50ms 44. In the circuit shown below, the network N is described by the following Y matrix: 0.S 0.0S Y =. the voltage gain 0.0S 0.S V V is 5V V C 4.3kΩ 5V 0.5mA t = 0 V = I R 0V E E E 5mF 5Ω I I 00V V N V 00Ω (A) /90 (B) /90 (C) /99 (D) / Exp: - N = 00V 5I ; V = 00I V I = Y3 V Y4V 0.0V = 0.0V 0.V = V 45. In the circuit shown below, the initial charge on the capacitor is.5 mc, with the voltage polarity as indicated. The switch is closed at time t=0. The current i(t) at a time t after the switch is closed is i( t) 3 i t = 5exp 0 t A (A) 3 (B) = ( ) i t 5exp 0 t A 3 (C) = ( ) i t 0exp 0 t A 3 (D) = ( ) i t 5exp 0 t A 00V 0Ω 50µ F

20 Exp: - =.5mC V C initial 6 i t = = 50V Thus net voltage = = 50V 50 0 F 50 0 t t = exp ( 0 t) A =5 exp ( ) 0 t A 46. The system of equations x y z = 6 Answer: - (B) x 4y 6z = 0 x 4y λ z = µ has NO solution for values of λ and µ given by (A) λ = 6, µ = 0 (B) λ = 6, µ 0 (C) λ 6, µ = 0 (D) λ 6, µ 0 Exp: - Given equations are x y z = 6, x 4y 6z = 0and x 4y λ z = µ If λ = 6 and µ = 0, then x 4y 6z = 0 x 4y 6z = 0 infinite solution If λ = 6 and µ 0, the x 4y 6z = 0 x 4y 6z = µ ( µ 0) no solution If λ 6 and µ = 0 x 4y 6z = 0 x 4y λ z = 0 will have solution λ 6 and µ 0 will also give solution 47. A fair dice is tossed two times. The probability that the second toss results in a value that is higher than the first toss is Answer: - (C) (A) /36 (B) /6 (C) 5/ (D) ½ Exp: - Total number of cause = 36 Total number of favorable causes = = 5 Then probability 5 5 = = 36 (, ) (, ) ( 3, ) ( 4, ) ( 5,) ( 6, )

21 (, ) (, ) ( 3, ) ( 4, ) ( 5,) ( 6, ) (,3 ) (,3 ) ( 3,3 ) ( 4,3 ) ( 5,3) ( 6,3 ) (,4 ) (,4 ) ( 3,4 ) ( 4,4 ) ( 5,4) ( 6,4 ) (,5 ) (,5 ) ( 3,5 ) ( 4,5 ) ( 5,5) ( 6,5 ) (,6 ) (,6 ) ( 3,6 ) ( 4,6 ) ( 5,6) ( 6,6 ) Common Data uestions: 48 & 49 The channel resistance of an N-channel JFET shown in the figure below is 600 Ω when the full channel thickness (t ch ) of 0µ mis available for conduction. The built-in voltage of the gate P N junction (V bi ) is - V. When the gate to source voltage (V GS ) is 0 V, the channel is depleted by µ m on each side due to the builtin voltage and hence the thickness available for conduction is only 8µ m Gate V GS Source t ch P N Drain P 48. The channel resistance when V GS = -3 V is (A) 360Ω (B) 97Ω (C) 000Ω (D) 3000Ω Answer: - (C) Exp: - Width of the depletion largew α Vbi VGS W 3 = w W = w = ( µ m) = µ m So that channel thickness = 0 4 = 6µ m 8µ m 750 6µ m? 8 rd = 750 = 000 Ω The channel resistance when V GS = 0 V is (A) 480Ω (B) 600Ω (C) 750Ω (D) 000Ω

22 Answer: - (C) Exp: -r don α t oh At V GS = 0, t = 0µ m; ( Given r = 600Ω ) ch 0 rd = 600 at 8µ m = 750Ω 8 d Common Data uestions: 50 & 5 The input-output transfer function of a plant 00 H s =. s s 0 ( ) The plant is placed in a unity negative feedback configuration as shown in the figure below. r Σ u H s 00 = s s 0 plant ( ) y 50. The gain margin of the system under closed loop unity negative feedback is (A) 0dB (B) 0dB (C) 6 db (D) 46 db Answer: - (C) Exp: - H ( s) 00 = s s 0 ( ) ω Phase cross over frequency= 90 tan = 80 0 ω tan = ( H ( jw) ) = 00 ( ) j0 j0 0 ω tan = = = 0. 0 GM = 0 log = 0 log0 = 6dB /0 0 ω = 0 rad / sec 5. The signal flow graph that DOES NOT model the plant transfer function H(s) is (A) (C) u u / s / s / s / s 0 0 / s 00 / s 0 00 y 00 y (B) (D) u u / s / s / s 0 / s 00 / s 00 / s 00 y 00 y

23 Exp: -(D) Option (D) does not fix for the given transfer function. Linked Answer uestions:.5 to.55 Carry Two Marks Each Statement for Linked Answer uestions: 5 & 53 In the circuit shown below, assume that the voltage drop across a forward biased diode is 0.7 V. The thermal voltage Vt = kt /q = 5mV. The small signal input v = V cos ω t where V = 00mV. i p p 9900Ω.7V I DC i ac V DC v ac v i 5. The bias current I DC through the diodes is (A) ma (B).8 ma (C).5 ma (D) ma Exp: -I DC = = ma The ac output voltage v ac is (A) 0.5cos ( ω t) mv (B) cos ( ω ) t mv (C) cos ( ω t) mv (D) cos ( ω ) Answer: - (C) η VT 5mV Exp: - AC dynamic resistance, rd = = = 50Ω I ma η = for Si ( forward drop = 0.7V) D t mv The ac dynamic resistance offered by each diode = 50Ω V 4 50Ω V ( ac) ac = i Vac = cos wt mv 3 00 = 00 0 cos wt 0000

24 Statement for Linked Answer uestions: 54 & 55 A four-phase and an eight-phase signal constellation are shown in the figure below. d r I d r I 54. For the constraint that the minimum distance between pairs of signal points be d for both constellations, the radii r, and r of the circles are (A) r = 0.707d, r =.78d (B) r = 0.707d, r =.93d (C) r = 0.707d, r =.545d (D) r = 0.707d, r =.307d Exp:- For st constellation r r = d r d / r = 0.707d = For nd constellation d r cos67.5 = r =.307d r r d / d / d 55. Assuming high SNR and that all signals are equally probable, the additional average transmitted signal energy required by the 8-PSK signal to achieve the same error probability as the 4-PSK signal is Exp: - Energy (A).90 db (B) 8.73 db (C) 6.79 db (D) 5.33 db = r and r Energy ( 0.707d) r = r.307d (.307) ( 0.707) in dd = 0 log = 5.33dB. No Carry One Mark Each 56. There are two candidates P and in an election. During the campaign, 40% of the voters promised to vote for P, and rest for. However, on the day of election 5% of the voters went back on their promise to vote for P and instead voted for. 5% of the voters went back on their promise to vote for and instead voted for P. Suppose, P lost by votes, then what was the total number of voters? (A) 00 (B) 0 (C) 90 (D) 95

25 Exp: - P 40% 60% 6% 6% 5% 5% 49% 5% % = 00% = Choose the most appropriate word from the options given below to complete the following sentence: It was her view that the country's problems had been by foreign technocrats, so that to invite them to come back would be counter-productive. (A) Identified (B) ascertained (C) Texacerbated (D) Analysed Answer: - (C) Exp: -The clues in the question are ---foreign technocrats did something negatively to the problems so it is counter-productive to invite them. All other options are non-negative. The best choice is exacerbated which means aggravated or worsened. 58. Choose the word from the options given below that is most nearly opposite in meaning to the given word: Frequency (A) periodicity (C) gradualness Answer: - (B) (B) rarity (D) persistency Exp: - The best antonym here is rarity which means shortage or scarcity. 59. Choose the most appropriate word from the options given below to complete the following sentence: Under ethical guidelines recently adopted by the Indian Medical Association, human genes are to be manipulated only to correct diseases for which treatments are unsatisfactory. (A) Similar (B) Most (C) Uncommon (D) Available Exp: - The context seeks to take a deviation only when the existing/present/current/ alternative treatments are unsatisfactory. So the word for the blank should be a close synonym of existing/present/current/alternative. Available is the closest of all. 60. The question below consists of a pair of related words followed by four pairs of words. Select the pair that best expresses the relation in the original pair: Gladiator : Arena (A) dancer : stage (C) teacher : classroom (B) commuter: train (D) lawyer : courtroom

26 Exp: - The given relationship is worker: workplace. A gladiator is (i) a person, usually a professional combatant trained to entertain the public by engaging in mortal combat with another person or a wild.(ii) A person engaged in a controversy or debate, especially in public.. No Carry Two Marks Each 6 The fuel consumed by a motorcycle during a journey while traveling at various speeds is indicated in the graph below. 0 Fuel consumption (kilometers per litre) Speed The distances covered during four laps of the journey are listed in the table below Lap Distance (kilometers) Average speed (kilometers per hour) P R S 0 0 From the given data, we can conclude that the fuel consumed per kilometre was least during the lap (A) P (B) (C) R (D) S Exp: - Fuel consumption Actual P 60km /l 5 = l km /l 75 5 = l 90 6 R 75km /l 40 8 = l 75 5 S 30km /l 0 = l 30 3

27 6. Three friends, R, S and T shared toffee from a bowl. R took /3 rd of the toffees, but returned four to the bowl. S took /4 th of what was left but returned three toffees to the bowl. T took half of the remainder but returned two back into the bowl. If the bowl had 7 toffees left, how many toffees-were originally there in the bowl? Answer: - (C) (A) 38 (B) 3 (C) 48 (D) 4 Exp: - Let the total number of toffees is bowl e x R took 3 of toffees and returned 4 to the bowl Number of toffees with R = x 4 3 Remaining of toffees in bowl = x 4 3 Number of toffees with S = x Remaining toffees in bowl = 3 x Number of toffees with 3 T = x Remaining toffees in bowl = 3 x Given, 3 3 x 4 4 = x = 3 x = Given that f(y) = y / y, and q is any non-zero real number, the value of f(q) - f(-q) is (A) 0 (B) - (C) (D) Exp: - Given, f(y) y = f ( q ) ; f ( q) y q q q f ( q) f ( q) = = = q q q q q q = = = q q q 64. The sum of n terms of the series is 4 /8 0 9n n (A) 4 /8 0 9n 0 n (C) 4 /8 0 9n n (B) 4 /8 0 9n 0 (D) n

28 Answer: - (C) Exp: - Let S=4 (..) = ( ) 3 { } n ( 0 ) 4 = = { } = = { } n n n 0 n 0 9n The horse has played a little known but very important role in the field of medicine. Horses were injected with toxins of diseases until their blood built up immunities. Then a serum was made from their blood. Serums to fight with diphtheria and tetanus were developed this way. Answer: - (B) It can be inferred from the passage that horses were (A) given immunity to diseases (C) given medicines to fight toxins (B) generally quite immune to diseases (D) given diphtheria and tetanus serums Exp: - From the passage it cannot be inferred that horses are given immunity as in (A), since the aim is to develop medicine and in turn immunize humans. (B) is correct since it is given that horses develop immunity after some time. Refer until their blood built up immunities. Even (C) is invalid since medicine is not built till immunity is developed in the horses. (D) is incorrect since specific examples are cited to illustrate and this cannot capture the essence.