Full Length Test Electronics and Communication Engineering

Transcription

1 1 [Ans A] PQRS P PQRSS P S QR P S Q QR Q P S R Q P S R (Q P S ) SPQ Full Length Test Electronics and Communication Engineering Answer Keys and Eplanations 2 [Ans A] Green s theorem and stokes theorem convert line integral to surface integral and vice versa Whereas Gauss s ivergence theorem converts from surface to volume and vice versa 3 [Ans *] Range: 5 to 5 y ( ) y( ) y( ) y 4 [Ans A] Maimum value of y over the interval 2 to 5 will be at = 5 Let S n r n / d sin 5 [Ans ] : , info@thegateacademycom Copyright reserved Web:wwwthegateacademycom 1

2 Given ( ) sin y sin sin sin cos sin sin (sin ) sin 6 [Ans C] By KL i 7 [Ans *] Range: 015 to 016 Bandwidth of series RLC circuit R rad sec L rad sec z z 8 [Ans B] n T e n n ( e e ) n (y) n () e e 9 [Ans C] Since, thermal run away is due to minority charge carriers and in FET, the conduction is due to majority carriers, so as temperature increases, mobility decreases and thus no trouble of thermal stability : , info@thegateacademycom Copyright reserved Web:wwwthegateacademycom 2

3 10 [Ans *] Range to L cm cm N cm and N cm L R where qn (N N N ) -cm R 11 [Ans *] Range: 0008 to 001 Open loop gain A da d d d ( ) ( ) : , info@thegateacademycom Copyright reserved Web:wwwthegateacademycom 3

4 12 [Ans *] Range: 32 to 31 k k O Using KCL k k k [ [ ] ] 13 [Ans *] Range: 55 to 58 R R r R Where, ( g R ) and R R and R r R R R k ( ) p k R sec f M z : , info@thegateacademycom Copyright reserved Web:wwwthegateacademycom 4

5 14 [Ans ] X X X X X Minimised form is 15 [Ans ] f X y y 1 st nd rd th f X y nd f X y after clock X 16 [Ans ] MUX M S S M J K = XOR (J, K) J k : , info@thegateacademycom Copyright reserved Web:wwwthegateacademycom 5

6 17 [Ans B] (t) t Since the wave has odd symmetry, hence a 18 [Ans A] h(n), (n) (n )- (z), z - Put z e (e ) [ e ] e [e e ] (e ) e cos (e ) e cos cos 19 [Ans *] Range: 6 to 6 y(t) ( t) (t) ( t) f (t) is band limited to f z Then ( t) is band limited to f Hz Nyquist rate f f a 20 [Ans C] Maimum occurs at n where n is odd ence first maimum occurs at s s s s omparing and : , info@thegateacademycom Copyright reserved Web:wwwthegateacademycom 6

7 21 [Ans ] u S u S S u u [ ] [ ] [ ] [ ] 0 u u 1 One can denote any state by any name; changing and So, that answer is d dt [ ] [ ] ( ) [ ] u u / 22 [Ans *] Range: 4391 to 44 SQNR log ( ) d s log SQNR log d log s 23 [Ans *] Range: 2 to 2 cos t cos t cos t cos t [ cos t] or envelope detection Should be atleast : , info@thegateacademycom Copyright reserved Web:wwwthegateacademycom 7

8 24 [Ans A] e cos y e sin y ( ) at ( )is m harge enclosed ( ) 25 [Ans *] Range: 9 to 9 cos( t ) elocity of EM wave in a los-less medium, - m sec 26 [Ans ], Multiplying both sides by b 0 1 [ (b b ) ( b) ] [ ( b ) ( b b ) b ] 27 [Ans ] Given that a > 0 So, a ( ) nd also g() So a ( ) g() for all R a ( ) g() Has no solution 28 [Ans A] z e ( iy) e ( ) ( y y)(e )(cos y i sin y) (e ),( y ) cos y y sin y- : , info@thegateacademycom Copyright reserved Web:wwwthegateacademycom 8

9 29 [Ans C] or T mode f u Since the wave guide is air filled u c a ence f z G z As f = 4 GHz f, the T mode will propagated u m s / / u ( ) phase velocity group velocity m s 30 [Ans*] Range: 4 to 4 urrent at node e (sin t cos t) d dt e (sin t cos t) e (cos t sin t) d dt *e (cos t sin t)+ L di dt e sin t e cos t k k k k 31 [Ans *] Range: 24 to 24 By KCL at output node So ( ) S M S M : , info@thegateacademycom Copyright reserved Web:wwwthegateacademycom 9

10 32 [Ans *] Range: i i ( ) i ( ) t t ( ) 33 [Ans B] Means : , info@thegateacademycom Copyright reserved Web:wwwthegateacademycom 10

11 34 [Ans B] Effective density of states Function (N ) = Therefore, the electron conduction is given by cm n N ep [ kt ] ep ( cm ) 35 [Ans *] Range: 13 to 13 cm sec cm sec So, diffusion length of holes, L m cm 36 [Ans A] Case I: CB configuration mitter injection efficiency ase transport factor input current n Case II: CE configuration T K T T / ( ) m ( ) ( ) : , info@thegateacademycom Copyright reserved Web:wwwthegateacademycom 11

12 37 [Ans *]Range 450 to 500 k i i and k i ( k ) for and for Regulation Regulation : , info@thegateacademycom Copyright reserved Web:wwwthegateacademycom 12

13 38 [Ans *] Range: 25 to 30 For current through the 3k resistance connected in the collector terminal of pnp transistor is O ( ) m k Therefore, emitter potential of the pnp transistor oltage drop across k Resistor m k current through k resistor m k k Now for Base of npn transistor is grounded ie at 0 Therefore oltage drop across R = ( ) Current through R = 2mA Required value of R m k m k m k Q Q m k m R 39 [Ans C] f R R or R R R R ( )( ) z : , info@thegateacademycom Copyright reserved Web:wwwthegateacademycom 13

14 40 [Ans *]Range: 37 to 4 For proper functioning the clock period should be equal to or greater than all t MO 12 s ns t of N N ns Total t ns s f z M z 41 [Ans *] Range: 1 to 1 A Q X X Y J K Q A = 0 A = 1 A = 0 Q = 0 Q = 1 A = 1 State changes when A = 1 : , info@thegateacademycom Copyright reserved Web:wwwthegateacademycom 14

15 42 [Ans B] LX B, 2100 H LX, 0200 H LX SP, 2700 PUSH B PUSH LX H, 0100 XTHL A L L HLT,E pair remains unchanged L : , info@thegateacademycom Copyright reserved Web:wwwthegateacademycom 15

16 43 [Ans B] (t) (n) { } t h(t) h(n) * + t (n) h(n) y(t) y(n) { } t y(t)dt (from the figure) 44 [Ans *] Range: 79 to 81 To solve this problem we could compute the analytical epression for the inverse Z-T, and then we could evaluate that epression at k=3 An alternative method to recall that (z) f, - f, -z f, -z f, -z f, -z ie, f[k] can be computed by epanding the fraction in power of z This can be done by dividing n(z) by d(z) upto the term z, its coefficient is equal to f[3] z z z z ) z z z z ( z z z z z z z z So, coefficient is 8 z z z z z z z z z z z z z z z z : , info@thegateacademycom Copyright reserved Web:wwwthegateacademycom 16

17 45 [Ans C] y stokes theorem ( ) ds dr i j k y z y z yz i(z ) k( z ),i(z ) k( z )- k (z ) (z )ds (z ) ds 46 [Ans ] ( ( ))u ( ( ))u z ( ( ))u [ ] [ ] [ ] [ ],u- y 47 [Ans *] Range: 30 to 30 Standard form of Phase Lead ompensator G (s) Or T T T T by comparing equation and T s T s sin [ ] = sin [ ] rad : , info@thegateacademycom Copyright reserved Web:wwwthegateacademycom 17

18 48 [Ans ] For stable system s should have roots in left half s plane s s s y (s ),(s y)(s ) - s (s ),s s( y) y- The roots to lie in left hand plane, y And y y y and y 49 [Ans ] For a single tone SSB-SC signal the waveform after carrier reinsertion becomes s (t) s(t) c(l) cos( t t) cos t ( cos t) cos t sin t sin t The output of the demodulation is given by [the envelope will be] (t), cos t-,sin t- cos t [ cos t] [ cos t], inomial epansion- cos t Neglecting dc component the normalized power of detected signal will be P [ ] Given P [ ] : , info@thegateacademycom Copyright reserved Web:wwwthegateacademycom 18

19 50 [Ans A] X (t), (t)- os t S, (t)- f ()d ( )d For m=1, ( S N ) S S ( S N ) / / ( S N ) ( S N ) / ( ) ( ) =31 m 51 [Ans *] Range: 247 to 247 Let the image frequency rejection ratio of the RF amplifier to be added be The rejection ratio ( ) at 1100 khz: f f f k z f f f f Q ( ) ( ) The rejection ratio ( ) at 25 MHz: f k z f f f f Q ( ) ( ) According to requirement ( ) ( ) So, loaded Q required for the RF stage Q (Q ) (Q ) ( ) : , info@thegateacademycom Copyright reserved Web:wwwthegateacademycom 19

20 52 [Ans C] BW of voice signal z The number of quantizing levels Number of bits required for sampling + supervision Number of the bits in each frame For synchronization Required W k z 53 [Ans *] Range: 240 to 240 ( t) cos( t ) verage power density ( )( ) ( ) verage power r ( ) ( ) ( ) watts 54 [Ans ] Airline can be regarded as a loss less line R G R L L rad m ividing by equation R and R ( ) rom equation L R ( ) ( ) n m p m : , info@thegateacademycom Copyright reserved Web:wwwthegateacademycom 20

21 55 [Ans *] Range: 881 to 886 dv d d d ( ) m negry nergy density volume nergy ( ) volume nergy ( ), d- mj J, - 56 [Ans ] They will chime together after the time in minutes equal to L M of L M min hrs min 57 [Ans C] According to the statement, 80% of the total runs were made by spinners So, conclusion I does not follow Nothing about the opening batsmen is mentioned in the statement So, conclusion II also does not follow 58 [Ans ] 1 km = 1000 meter 1 min = 60 second verage speed Total distance Total time Total distance = 12 km = meter Total time = minute = = 1440 seconds verage speed m s 59 [Ans A] 60 [Ans C] CEPQS - E cannot go with S AEPQS - C and P have to be together E cannot go with S ACPRS -It satisfies all the conditions and also there are two boys in the team BPRS - C and P have to be together Hence, C : , info@thegateacademycom Copyright reserved Web:wwwthegateacademycom 21

22 61 [Ans A] Number of males in U P [ of ( of N)] N Total population N Number of males in M P [ of ( of N)] N Number of males in Goa [ of ( of N)] N N N N Total males in these states ( )N N Required ( N ) 62 [Ans C] A cube is cut into 125 smaller cubes Length of cube = l = 5 unit Let upper face be coloured red Then bottom face will be coloured green, two adjacent faces are coloured yellow and blue respectively Two faces are uncoloured Number of cubes uncoloured on all faces = (n ) ( ) Now there are two surfaces which are not coloured There will be 9 cubes at centre on both the uncoloured surfaces each 3 cubes at the common edge of both uncoloured surfaces Total number of uncoloured cubes = = [Ans C] 64 [Ans B] The passage clearly states the unawareness of teachers regarding population education Thus, the teachers should be given a proper orientation on the same : , info@thegateacademycom Copyright reserved Web:wwwthegateacademycom 22

23 65 [Ans C] In statement I nothing is given about c Hence it is not enough to answer the question In statement II nothing is mentioned about a Hence this statement alone cannot answer the question Combining both the statements we get a : b : c = 3 : 15 : 10 a : c = 3 :10 a c a c c Question can be answered using both the statements Hence, C : , info@thegateacademycom Copyright reserved Web:wwwthegateacademycom 23