GATE 2017 [Forenoon (Set - 1)]

Transcription

1 GATE 7 [Forenoon (Set - )].Technical Part. Quetion In a digital communication ytem, the overall pule hape p(t) at the receiver before the ampler ha the Fourier tranform P(f ). If the ymbol are tranmitted at the rate of ymbol per econd, for which of the following cae i the inter ymbol interference zero? (A) P( f) (B) P( f).. f (khz) f (khz) (C) P( f) (D) P( f).. An. (B) Sol. For zero ISI from Nyquit pule haping theorem, the pule haping criterion can be tated a : The um of all the individual pule pectrum alway add upto a contant at each ampling intant m P f m T Contant or T f (khz) P( f).. f (khz) m T P f T f (khz) f (khz) P f T PfP f P f T T f (khz) f (khz) Since, only option B reult in complete flat band. Hence, the correct option i (B). More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

2 GATE 7 [Forenoon (Set - )] Quetion Conider the D-Latch hown in the figure, which i tranparent when it clock input CK i high and ha zero propagation delay. In the figure, the clock ignal CLK ha a 5% duty cycle and CLK i a onefifth period delayed verion of CLK. The duty cycle at the output of the latch in percentage i. CLK T CLK CLK D Q D-Latch Output CLK CK An. Sol. Given : T CLK /5 CLK CLK T CLK CLK D Q D-Latch Output CLK CK T CLK /5 D flip-flop i poitive edge triggered flip-flop. Truth table of D-FF CLK CLK D Q D CLK Output T CLK T CLK /5 TCLK TCK 5T T T For Q, TON 5 TCLK TCK 7T TOFF 5 TON TON % Duty cycle of Q % T T T ON OFF T ( D) Q % % T 7T Hence, the correct anwer i. More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

3 GATE 7 [Forenoon (Set - )] Quetion The open loop tranfer function ( ) G () p ( )( ) where p i an integer, i connected in unity feedback configuration a hown in the figure. G () Given that the teady tate error i zero for unit tep input and i 6 for unit ramp input, the value of the parameter p i. An. Sol. Given : ( ) G () p ( )( ) For tep input : e unit tep input e e k k 6 unit ramp input P P kp k P lim G( ) H( ) ( ) lim P ( )( ) kp lim P 6 P For ramp input : e 6 kv k 6 V kv lim G( ) H( ) 6 ( ) lim P ( )( ) 6 lim P P 6 6 Hence, the correct anwer i. Key Point : For Type ytem, teady tate error i finite when ramp input i applied to the ytem. More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

4 4 GATE 7 [Forenoon (Set - )] Quetion 4 5 The rank of the matrix M i 6 6 (A) (B) (C) (D) An. (C) 5 Sol. Given : M 6 6 Applying elementary tranformation R 5R R and R 5R R 5 M So rank of above matrix i a there are two non-zero row. Hence, the correct option i (C). Quetion 5 Three fair cubical dice are thrown imultaneouly. The probability that all three dice have the ame number of dot on the face howing up i (up to third decimal place). An..7 Sol. Three cubical dice are thrown imultaneouly, o, number of ample pace i ns ( ) 6 6. Favourable outcome i that all dice hould get ame number. So, number of favourable outcome i n(a) = 6 i.e.,,,,,,,, 4, 4, 4, 5,5,5, 6,6,6 No. of favourable outcome So probability of event = No.of ample pace na ( ) 6 PA ( ).7 ns ( ) 6 6 Hence, the correct anwer i.7. Quetion 6 In the latch circuit hown, the NAND gate have non-zero, but unequal propagation delay. The preent input condition i : P = Q =. If the input condition i changed imultaneouly to P = Q =, the output X and Y are P X Q Y More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

5 5 An. Sol. GATE 7 [Forenoon (Set - )] (A) X =, Y = (B) Either X =, Y = or X =, Y = (C) Either X =, Y = or X =, Y = (D) X =, Y = (B) Given : The latch circuit i hown below. P X Q Y If PQ, the latch will have output X Y Let, the input be changed to PQ N Pat Input Preent Input Let, propagation delay of pat output N be le than that of N N Hence, output of Nchange before output of N can change. N X Hence, X and Y Similarly, if propagation delay of N i le We get X and Y Preent Input Hence, the correct option i (B). N Y Preent Output More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

6 6 GATE 7 [Forenoon (Set - )] Quetion 7 Conider the following tatement about the linear dependence of the real valued function y, y x and y x, over the field of real number. I. y, y and y are linearly independent on x II. y, y and y are linearly dependent on x III. y, y and y are linearly independent on x IV. y, y and y are linearly dependent on x Which one among the following i 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 An. (B) Sol. Given : y, y x, y x From Wronkian matrix, f( x) g( x) h( x) If W f '( x) g'( x) h'( x) f ''( x) g''( x) h''( x) then function are linearly dependent. Here f ( x), g( x) x, h( x) x x x W x W So function are linearly independent. Statement (I) & (III) are correct Hence, the correct option i (B). Quetion 8 Which of the following can be the pole-zero configuration of a phae-lag controller (lag compenator)? (A) j (B) j (C) j (D) j Zero - Pole - More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

7 7 GATE 7 [Forenoon (Set - )] An. (A) Sol. Phae lag compenator : For thi compenator, pole i more cloe to origin a compare to zero. And it contain only one pole and one zero. Hence, the correct option i (A). Quetion 9 A bar of Gallium Arenide (GaA) i doped with Silicon uch that the Silicon atom occupy Gallium and Arenic ite in the GaA crytal. Which one of the following tatement i true? (A) Silicon atom act a p-type dopant in Arenic ite and n-type dopant in Gallium ite (B) Silicon atom act a n-type dopant in Arenic ite and p-type dopant in Gallium ite (C) Silicon atom act a p-type dopant in Arenic a well a Gallium ite (D) Silicon atom act a n-type dopant in Arenic a well a Gallium ite An. (A) Sol. Valency of Ga Valency of A 5 Valency of Si 4 In GaA crytal every A atom will form 5 bond with Ga atom and every Ga atom will form bond with A atom. Replacement of by Si atom in GaA : In thi cae Si atom will form bond with Ga atom during formation of bond, four electron of Si will form bond with four neighboring Ga atom, but Si atom will not be able to form bond with fifth Ga atom hence hole will become in exitence in crytal therefore crytal become P-type and Si will behave a acceptor. Replacement of Ga atom by Si atom in GaA : In thi cae Si atom will form bond with A atom during formation of bond, three electron of Si will form bond but fourth electron of Si will not participate. Fourth electron become free to move with in crytal, therefore crytal will tart to behave a an N-type crytal and Si atom will tart to behave a a donor atom. Hence, the correct option i (A). Quetion The Miller effect in the context of a Common Emitter amplifier explain (A) an increae in the low-frequency cutoff frequency (B) an increae in the high-frequency cutoff frequency (C) a decreae in the low-frequency cutoff frequency (D) a decreae in the high-frequency cutoff frequency An. (D) Sol. Small ignal equivalent circuit i hown below, B C C r C gv m RC RL E More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

8 8 GATE 7 [Forenoon (Set - )] Due to the miller effect C i not neglected hence by conidering miller effect above circuit can be redrawn a hown below, B C r C gv m C M RC RL Hence overall input capacitance i given by, C C C Where C M C gmrc RL j M Hence due to miller effect input capacitance increae. Now cut-off frequency i given by, fdb C j Becaue due to miller effect C j increae hence fdb will decreae. Hence, the correct option i (D) Quetion The clock frequency of an 885 microproceor i 5 MHz. If the time required to execute an intruction i.4, then the number of T-tate needed for executing the intruction i (A) (B) 6 (C) 7 (D) 8 An. (C) Sol. Given : Clock frequency of microproceor 885 i 5 MHz. Total time required for intruction i.4 ec. Quetion So, time period of each clock cycle or duration of one T-tate. ec 5MHz So, number of poible T-tate or number of clock cycle Total time required for intruction.4ec 7 Duration of one T-tate.ec Hence, the correct option i (C). Conider the following tatement for continuou-time linear time invariant (LTI) ytem. E I. There i no bounded input bounded output (BIBO) table ytem with a pole in the right half of the complex plane. II. There i no caual and BIBO table ytem with a pole in the right half of the complex plane. More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

9 9 An. Sol. GATE 7 [Forenoon (Set - )] Which one among the following i correct? (A) Both I and II are true (B) Both I and II are not true (C) Only I i true (D) Only II i true (D) Statement II : There i no caual and BIBO table ytem with a pole in the right half of the complex plane. Since, we know that for a caual ytem to be table all pole mut lie on left half of the complex plane. Hence, there i no caual and table ytem with a pole in right half of complex plane. Hence, Statement II i true. Statement I : There i no bounded input bounded output (BIBO) table ytem with a pole in the right half of the complex plane Conider a ytem with one right half pole H() If the ytem i anti-caual ht e u t () t ( ) Hence, ht () dt Hence, an anti-caual ytem i table with pole on right half of complex plane. Hence, Statement I i fale. Hence, the correct option i (D). Quetion For the operational amplifier circuit hown, the output aturation voltage are ± 5 V. The upper and lower threhold voltage for the circuit are, repectively. Vin + k V out 5 k + V (A) + 5 V and An. (B) Sol. Given : V 5 V, V 5 V at 5V (B) + 7 V and V (C) V and 7 V (D) + V and V at More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

10 GATE 7 [Forenoon (Set - )] Vin + k V out V f 5 k + V (i) Calculation of upper threhold voltage ( V UT ) : If V V at 5 V then Vf VUT Upper threhold voltage i given by, Vref V 5 VUT 6 5 VUT 7V (ii) Calculation of lower threhold voltage ( V LT ) : If V V at 5 V then Vf VLT Lower threhold voltage i given by, Vref V 5 VLT 6 5 VLT V Hence, the correct option i (B). Quetion 4 The voltage of an electromagnetic wave propagating in a coaxial cable with uniform characteritic l j t impedance i Vl ( ) e Volt, where l i the ditance along the length of the cable in metre. 9 (. j4) m i the complex propagation contant, and rad/ i the angular frequency. The abolute value of the attenuation in the cable in db/metre i. An..868 Sol. Propagation contant i given by, j. j4m Np rad., 4 m m More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

11 GATE 7 [Forenoon (Set - )] We know that, Np db db So, attenuation in the cable (in db/m) m Hence, the correct anwer i.868. Quetion 5 Which one of the following tatement about differential pule code modulation (DPCM) i true? (A) The um of meage ignal ample with it prediction i quantized (B) The meage ignal ample i directly quantized, and it prediction i not ued (C) The difference of meage ignal ample and a random ignal i quantized (D) The difference of meage ignal ample with it prediction i quantized An. (D) Sol. Tranmitter of DPCM : Figure how the tranmitter of Differential Pule Code Modulation (DPCM) ytem. The ampled ignal i denoted by x( nt ) and the predicted ignal i denoted by xˆ( nt ). The comparator find out the difference between the actual ample value x( nt ) and predicted ample value xˆ( nt ). Thi i known a prediction error and it i denoted by ent ( ). It can be defined a, ent ( ) xnt ( ) xnt ˆ( ).. (i) Sampled ignal + xnt ( ) ent ( ) e q( nt) Quantizer xnt ˆ( ) + + Encoder DPCM ignal Thu, error i the difference between unquantized input ample x( nt ) and prediction of it i.e. xˆ( nt ). The predicted value i produced by uing a prediction filter. The quantizer output ignal e ( nt ) and previou prediction i added and given a input to the prediction filter. Thi ignal i called x ( nt ). Thi make the prediction more and more cloe to the actual ampled ignal. The quantized error ignal eq( nt) i very mall and can be encoded by uing mall number of bit. Thu number of bit per ample are reduced in DPCM. The quantizer output can be written a, e ( nt ) e( nt ) q( nt ).. (ii) q Prediction filter x ( nt ) q Fig. : A Differential pule code modulation tranmitter More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and q q

12 GATE 7 [Forenoon (Set - )] Here, qnt ( ) i the quantization error. A hown in figure, the prediction filter input xq( nt) i obtained by um of xˆ( nt ) and quantizer output e ( nt ) i.e. q x ( nt ) xˆ ( nt ) e ( nt ).. (iii) q q Subtituting the value of e ( nt ) from equation (ii) in the above equation, we get q x ( nt ) xˆ ( nt ) e( nt ) q( nt )... (iv) q Equation (i) i written a, ent ( ) ( ) ˆ xnt xnt ( ) ent ( ) xnt ˆ( ) xnt ( ) (v) Therefore, ubtituting the value of ent ( ) ˆ xnt ( ) from above equation into equation (iv), we get, xq( nt) x( nt) q( nt) (vi) Hence, the correct option i (D). Quetion 6 Conider a wirele communication link between a tranmitter and a receiver located in free pace, with finite and trictly poitive capacity. If the effective area of the tranmitter and the receiver antenna, and the ditance between them are all doubled, and everything ele remain unchanged, the maximum capacity of the wirele link (A) increae by a factor of (B) decreae by a factor of (C) remain unchanged (D) decreae by a factor of An. (C) Sol. Capacity of wirele link i given by, PR C Blog C PA T etaer PR d Where A et = Effective area of tranmitter AeR Effective area of receiver ' AeT AeT ' AeR AeR d' d PA T et' AeR ' PR ' ( d ') PT AeT AeR PTAeT AeR PR ' d d P' P R R Since received power doe not change. So, capacity of wirele link will not change. Hence, the correct option i (C). More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

13 GATE 7 [Forenoon (Set - )] Quetion 7 A periodic ignal x(t) ha a trigonometric Fourier erie expanion x() t a ( a con tb in n t) n n n If xt () x( t) xt, 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 An. (A) Sol. x() t a ( ancontbnin nt).(i) n T x() t x( t) xt xt. (ii) Hence from equation (ii), it i clear that x( t ) i odd half wave ymmetrical ignal. Becaue x( t ) i odd, hence coefficient of coine term i.e. a n will be zero. Becaue x() t ha half wave ymmetry, hence only odd harmonic of b n will exit. Hence, the correct option i (A). Quetion 8 An n + -n Silicon device i fabricated with uniform and non-degenerate donor doping concentration of 8 5 N D cm and N D cm correponding to the n and n region repectively. At the operational temperature T, aume complete impurity ionization, kt / q 5 mv, and intrinic carrier concentration to be n i cm. What i the magnitude of the built-in potential of thi device? (A).748 V (B).46 V (C).88 V (D).7 V An. (D) Sol. 8 5 kt Given : ND / cm, N /cm D, 5mV q and ni /cm Built in voltage i given by, N D Vbi VT ln N D V bi 8 5 Hence, the correct option i (D). 5 ln.7 V More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

14 4 GATE 7 [Forenoon (Set - )] Quetion 9 For a narrow bae PNP BJT, the exce minority carrier concentration ( ne for emitter, pb for bae, nc for collector) normalized to equilibrium minority carrier concentration ( E n for emitter p B for bae, n C for collector) in the quai-neutral emitter, bae and collector region are hown below. Which one of the following biaing mode i the tranitor operating in? (A) Forward active (B) Saturation (C) Invere active (D) Cutoff An. (C) Sol. Figure how the minority carrier ditribution in the pnp tranitor for the invere-active mode. In thi cae, the B-E junction i revere biaed and the B-C junction i forward biaed. Electron from the collector are now injected into the bae. The gradient in the minority carrier electron concentration in the bae i in the oppoite direction compared with the forward-active mode, o the emitter and collector current will change direction. Hence, the correct option i (C). Quetion A good tranconductance amplifier hould have (A) High input reitance and low output reitance (B) Low input reitance and high output reitance (C) High input and output reitance (D) Low input and output reitance An. (C) Sol. Tranconductance of an amplifier i defined a, G Normalized Exce Carrier Concentration m I Output current V Input voltage i A tranconductance amplifier i hown below, p p B B ne ne Emitter (P) Bae (N) Collector (P) X andyaxe are not to cale RS I 5 I L n n C C V V i Ri GV R m i R L More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

15 5 GATE 7 [Forenoon (Set - )] For ideal tranconductance amplifier, V V Hence, R i hould be infinite. and I IL Hence R hould be infinite. Table : i Ideal amplifier S Z i Z Voltage amplifier Tran-reitance amplifier Tran-conductance amplifier Current amplifier Hence, the correct option i (C). Quetion Conider the 5 5 matrix A An. It i given that A ha only one real eigenvalue. Then the real eigenvalue of A i (A).5 (B) (C) 5 (D) 5 (C) Sol. Given : Let X A AIX A X X a b c d More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

16 6 GATE 7 [Forenoon (Set - )] 4 5a a 5 4 b b 4 5 c c 4 5 d d 4 5 e e abc4d 5e a (i) 5abcd 4e b (ii) 4a5bcd e c (iii) a4b5cd e d (iv) ab4c5d e e (v) By adding equation (i), (ii), (iii), (iv) and (v), 5( abcd e) ( abcd e) 5 Hence, the correct option i (C). Quetion Conider a table ytem with tranfer function p p b bp G () q q a aq where b,..., b p and a,..., a q are real valued contant. The lope of the Bode log magnitude curve of G() converge to 6 db/decade a. A poible pair of value for p and q i (A) pand q (B) pand q 7 (C) p andq (D) p andq 5 An. (A) Sol. Given : G () Slope at i 6dB. p p b bp q q a aq Slope at i given by, (Slope) ( q p) [Where q = number of pole, p = number of zero] For option (A), p, q (Slope) ( ) 6 db [correctly matched with given value] For option (B), p, q 7 (Slope) (7 ) db [doe not match with given value] For option (C), p, q (Slope) ( ) db [doe not match with given value] For option (D), p, q 5 (Slope) (5 ) 4 db [doe not match with given value] Hence, the correct option i (A). More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

17 7 GATE 7 [Forenoon (Set - )] Quetion Conider a ingle input ingle output dicrete-time ytem with x[n] a input and y[n] a output, where the two are related a An. yn xn nxn ; for n xn ; otherwie Which one of the following tatement i true about the ytem? (A) It i caual and table (B) It i caual but not table (C) It i not caual but table (D) It i neither caual nor table (A) Sol. Given : nxn ( ) ; n yn ( ) xn ( ) xn ( ); Otherwie i.e. n and n For n : yn ( ) n xn ( ) (i) Let x( n) i bounded i.e. a n, x( n) Where M i finite. M Then a n, x( n ) will approach to a finite value and hence a n, then n x( n) will go to finite value. Hence in thi interval y( n ) i bounded i.e. ytem i table. (ii) Becaue output yn ( ) depend on preent and pat value of input, hence in thi interval ytem i caual. For n and n : (i) Let input x( n ) i bounded. Hence a n, x( n) will go to a finite value and hence yn ( ) [ xn ( ) xn ( )] will go to a finite value. Hence in thi interval ytem i table. (ii) Becaue output y( n) depend on preent and pat value of input, hence ytem i caual. Therefore on conidering both the interval, we can ay that the ytem i caual and table. Hence, the correct option i (A). Quetion 4 In the circuit hown, the poitive angular frequency (in radian per econd) at which the magnitude of the phae difference between the voltage V and V equal radian, i. 4 V H co t V More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

18 8 GATE 7 [Forenoon (Set - )] An. Sol. V I H co t V I jl j V I j V tan 4 V, V tan 4 ( j) V I( j) j. (i) V tan tan 4 V, V tan tan 4 From equation (i) and (ii), VV tan tan tan 4 Quetion 5 An. tan 4 rad/ec Hence, the correct anwer i.. (ii) Let ( X, X ) be independent random variable, X ha mean and variance, while X ha mean and variance 4. The mutual information I ( X; X ) between X and X in bit i. More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

19 9 GATE 7 [Forenoon (Set - )] Sol. Mutual information between two random variable with joint denity function f ( x, y ) i given by, f( x, y) I( XY ; ) f( xy, )log dxdy f ( x ) f ( y ) Given X and Y are independent random variable, f ( xy, ) f( xf ) ( y) f( x) f( y) I( XY ; ) f( xy, )log dxdy f ( x ) f ( y ) I( XY ; ) f( xy, )log() dxdy I( X; Y) Hence, the correct anwer i. Quetion 6 The Nyquit plot of the tranfer function K G () ( )( ) doe not encircle the point ( + j) for K = but doe encircle the point ( + j) for K =. Then the cloed loop ytem (having unity gain feedback) i (A) Stable for K and table for K (B) Stable for K and untable for K (C) Untable for K and table for K (D) Untable for K and untable for K An. (B) Sol. Given : K (i) GH () () ( )( ) (ii) If K i Nyquit plot do not encloe j. (iii)if K i Nyquit plot encloe j. Img Img For K For K (,) Fig. (i) Re Clockwie encirclement N = Fig. (ii) More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and (,) Re

20 GATE 7 [Forenoon (Set - )] Number of encirclement about critical point ( j) in anti-clock wie i given by, N P Z...(i) Where, P = Number of open loop pole in right half of -plane Z = Number of cloed loop pole in right half of -plane For the ytem to be table : Z = From figure (i), N 4 64 Routh Tabulation : In the above Routh table, there i no ign change in firt column. So there i no pole lie in right half of - plane. So, P From equation (i), we get Z Z [ytem i table] From figure (ii), N From equation (i), we get Z Z [ytem i untable] Hence, the correct option i (B). Quetion 7 A finite tate machine (FSM) i implemented uing the D flip-flop A and B, and logic gate, a hown in the figure below. The four poible tate of the FSM are QQ A B =,, and. D Q Q A D Q Q B A X IN B CK Q CK Q CLK Aume that X IN i held at a contant logic level throughout the operation of the FSM. When the FSM i initialized to the tate QQ A B = and clocked, after a few clock cycle, it tart cycling through More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

21 An. Sol. GATE 7 [Forenoon (Set - )] (A) all of the four poible tate if X (B) three of the four poible tate if X IN (C) only two of the four poible tate if X IN (D) only two of the four poible tate if X IN (D) Given : Poible tate of the FSM are QQ A B,,, X IN i at contant logic level throughout the operation. i.e. X IN or contantly from figure, DA QA QB DB QA. XIN (i) Take X IN Q A IN Q B DA QA QB DA QA XIN A Q QB From above truth table we can ee that there are only ditinct tate out of 4 if X IN (ii) Take X IN Q A Q B DA QA QB DA QA XIN A Q QB From above truth table we can ee that there are ditinct tate out of 4 if X Hence, the correct option i (D). Quetion 8 A linear time invariant (LTI) ytem with the tranfer function K ( ) G () ( ) i connected in unity feedback configuration a hown in the figure. G () IN More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

22 GATE 7 [Forenoon (Set - )] For the cloed loop ytem hown, the root locu for < K < interect the imaginary axi for K =.5. The cloed loop ytem i table for (A) K >.5 (B) < K <.5 (C) < K < (D) no poitive value of K An. (A) Sol. Given : K ( ) G () ( ) Characteritic equation i given by, GH ( ) ( ) K ( ) K( ) ( K) ( K) K Routh Tabulation : K K K K For the ytem to be table all the coefficient of firt column of Routh table hould be of ame ign. For tability, K, K and + K > K.5 and K Hence for tability, K.5 Hence, the correct option i (A). Quetion 9 An optical fiber i kept along the ẑ direction. The refractive indice for the electric field along ˆx and ŷ direction in the fiber are nx.5 and ny.5, repectively ( n x n y due to the imperfection in the fiber cro-ection). The free pace wavelength of a light wave propagating in the fiber i.5 m. If the lightwave i circularly polarized at the input of the fiber, the minimum propagation ditance after which it become linearly polarized, in centimetre, i. An..75 Sol. Initially the wave i circularly polarized. So, the initial phae difference between field component i. To become linearly polarized, the wave mut travel a ditance uch that, the phae difference between the field component i hence path difference i. So, Zkx Zky f f Z rx rx C C 4 Z n n x y More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

23 GATE 7 [Forenoon (Set - )] Z.5 n n x y Z.75cm Hence, the correct anwer i.75. Quetion A continuou time ignal x( t) 4co( t) 8co(4 t), where t i in econd, i the input to a linear time invariant (LTI) filter with the impule repone in t ; t ht () t 6 ; t Let y(t) be the output of thi filter. The maximum value of y(t) i An. 8 Sol. Given : x( t) 4co t8co 4 t in t ; t ht () t 6 ; t ht () i a inc function, whoe frequency pectrum i hown below. μm H ( ) Frequency pectrum of x( t) i given below, X ( ) [ ( ) ( )] 4[ ( 4 ) ( 4 )] X ( ) Y( ) H( ) X( ) Y ( ) 4[ ( ) ( )] Take invere Fourier tranform of Y ( ), y( t) 8co t Hence yt () 8 max Hence, the correct anwer i 8. More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

24 4 GATE 7 [Forenoon (Set - )] Quetion A 4-bit hift regiter circuit configured for right-hift operation, i.e. Din A, AB, BC, C D, i hown. If the preent tate of the hift regiter i ABCD =, the number of clock cycle required to reach the tate ABCD = i. An. Sol. For right hift SIPO hift regiter, B A C D n n n B C n n n A per the given circuit A n A n Dn State table for given right hift SIPO hift regiter : Preent State Clock A n B n C n D n A n A n Dn Bn pule Next tate A C B n n n D n C Clock n Hence, the correct anwer i. Quetion For the DC analyi of the Common-Emitter amplifier hown, neglect the bae current and aume that the emitter and collector current are equal. Given that V T 5 mv, V BE =.7 V, and the BJT output reitance r i practically infinite. Under thee condition, the midband voltage gain magnitude A v v v i V, i. V D in Clock A B C D More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and pule

25 5 GATE 7 [Forenoon (Set - )] VCC V 7 k R R C k F F C V i C 47 k R RE k C E R 8k V L μf An. 8 Sol. The midband gain of common emitter configuration i, V Av gmrl' V V Magnitude of voltage gain gmrl' V in Uing DC Analyi (For calculation of g m ) : (i) All capacitor are open circuit. (ii) AC voltage ource i replaced by hort circuit. in V V 7 k k k 8.59 k A 47 k k 4.7 V k Fig. (a) Uing Thevenin theorem, R k th Fig. (b) 47 Vth 4.7 V Applying KVL in the loop hown in figure (b), I E [given I B i negligible] I E ma More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

26 6 GATE 7 [Forenoon (Set - )] Hence, IE IC ma IC ma gm.8a/v VT 5mV Uing AC analyi : (i) All capacitor are hort circuited. (ii) All DC voltage ource i replaced by hort circuit. 7 k k V V in 47 k k 8k Fig. Common emitter without Modified figure i hown below, R E V V in 47 k 7 k 8.59 k R ' k 8 k L.6 k V So, AV gmrl' gmrl' Vin AV Hence, the correct anwer i 8. Quetion The following FIVE intruction were executed on an 885 microproceor. MVI A, H MVIB, 78 H ADD B CMA ANI H The Accumulator value immediately after the execution of the fifth intruction i (A) H (B) H (C) H (D) H More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

27 7 GATE 7 [Forenoon (Set - )] An. Sol. (B) Intruction MVI A, H MVI B, 78 H ADD B CMA Operation Regiter A i loaded with H Regiter B i loaded with 78 H (A+B) A B 78 H + A + H + A All bit of Regiter A complemented individually A = ANI H (Regiter A AND H) A Hence, the correct option i (B). Quetion 4 Let X(t) be a wide ene tationary random proce with the power pectral denity SX ( f ) a hown in figure (a), where f i in Hertz (Hz). The random proce X(t) i input to an ideal lowpa filter with the frequency repone, f Hz H( f), f Hz a hown in figure (b). The output of the lowpa filter i Y(t), Let E be the expectation operator and conider the following tatement : I. EXt [ ( )] EYt [ ( )] II. III. EX t EY t [ ( )] [ ( )] EY [ ( t)] Select the correct option : (A) only I i true (C) only I and II are true SX A= ANDH= ( f) exp( f ) (B) only II and III are true (D) only I and III are true More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and f A A Ideal lowpa X() t filter ht () Y() t cutoff = / Hz

28 8 An. Sol. (A) GATE 7 [Forenoon (Set - )] LTI X() t Y() t Sytem, f H( f), Otherwie The mean value of output random proce i given by, Statement i correct. EY [ ( t)] H() E[ X( t)] EY [ ( t)] E[ X( t)] [From figure, H() ] EY [ ( t)] E[ X( t)] The relationhip between output and input PSD i given by, H( f) f S ( f) H( f) S ( f) YY S ( f) S ( f) YY XX Power of output random proce can be calculated a, P SYY ( f) df XX f / f P e df / [ e ] f / / f / e df ( e ).786 W E Y ( t).786 W Statement i incorrect. Power of input random proce can be calculated a, Pi SXX( f) df f [ e ] Pi e df e df E X () t W EX t EY t [ ( )] [ ( )] Statement i alo incorrect. Hence, the correct option i (A). f f W More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

29 9 GATE 7 [Forenoon (Set - )] Quetion 5 For the circuit hown, aume that the NMOS tranitor i in aturation. It threhold voltage Vth V W and it tranconductance parameter ncox ma/v. Neglect channel length modulation and body L bia effect. Under thee condition, the drain current I D in ma i. V DD = 8 V R = M R D = k I D R = 5 M R S = k An. Sol. V DD = 8 V R = M R D = k I D R = 5 M R S = k Then, 5 VTh 8 5V, 8 R Th 5.875k 5 8V.875 k 5V A I D + V GS k k More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

30 GATE 7 [Forenoon (Set - )] 5V I V GS GS 5 I Given MOSFET i in aturation region, n ID K ( VGS VTN) (5 ) ID ID I (4 I ) If If D D D D D 8 D 6 D I I I I D I 6 D ( I 8)( I ) D D I 8mA, ma D I 8mA V D GS 58V V [Not poible] I D ma V 5V V [Poible] GS I D ma Hence, the correct anwer i. Quetion 6 In binary frequency hift keying (FSK), the given ignal waveform are u ( t ) 5co( t ); t T TN TN and u ( t ) 5co( t ); t T where T i the bit-duration interval and t i in econd. Both u () t and u () t are zero outide the interval t T. With a matched filter (correlator) baed receiver, the mallet poitive value of T (in milliecond) required to have u () t and u () t uncorrelated i (A).5 m (B).5 m (C).75 m (D). m An. (B) Sol. Given : u t 5co t, t T u t 5co t, t T For the ignal to be uncorrelated, YT ( ) Eu( tu ) tt More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

31 GATE 7 [Forenoon (Set - )] For T YT For T YT T YT ( ) 5co t 5co ( tt) dt T 5 YT ( ) co4ttcot t) dt t T T t 5 in 4 in YT ( ) 4 5 in T) in(t in T) in(t YT ( ) 4.5m, m, 5 in (5.5 ).5m, in (5.5 ) For T YT 5 in (6.5 ).75m, in (6.5 ) For T YT Hence, minimum value of T for which YT ( ) i.5 m. Hence, the correct option i (B). Quetion 7 The amplifier circuit hown in the figure i implemented uing a compenated operational amplifier (opamp), and ha an open-loop voltage gain, R k 5 A V/V and an open-loop cut-off frequency, f 8 Hz. R 79 k T C V V in The voltage gain of the amplifier at 5 khz, in V/V, i. An Sol. For practical non-inverting Op-Amp, voltage gain i given by, V AOL Af V A For 5 AOL, i Af OL (i) More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

32 GATE 7 [Forenoon (Set - )] Since, the above circuit i a non-inverting op-amp V R 79 Af 8 V R i V in V 79 k V f k OR V f V AOL A vf 8 A 5 OL 8 5 A (High Gain value) A OL vf R R 79 8 Concept : Gain bandwidth product of cloe loop ytem = Gain bandwidth product of open loop ytem A BW ' A BW vf OL 5 8 BW ' 8 BW ' khz AV 8 Avf 44.7 f 5 BW ' Hence, the correct anwer i Quetion 8 The dependence of drift velocity of electron on electric field in a emiconductor i hown below. The 6 emiconductor ha a uniform electron concentration of n cm and electronic charge 9 q.6 C. If a bia of 5 V i applied acro a m region of thi emiconductor, the reulting current denity in thi region, in ka/cm, i. Drift velocity (cm/) 7 linear contant 5 5 Electric field (V/cm) More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

33 GATE 7 [Forenoon (Set - )] An..6 6 Sol. Given : ND atom/cm, V 5 Electric field E 4 V/cm d Drift velocity i given by, Vd E V d q 9.6 C, V volt, d m cm 7 cm /V-ec [From figure] 5 E 5 Current denity i given by, J ND n qe J.6.6 ka/cm 4 Hence, the correct anwer i.6. Quetion 9 A hown, a uniformly doped Silicon (Si) bar of length L =. m with a donor concentration 6 N D cm i illuminated at x = uch that electron and hole pair are generated at the rate of x GL GL, x L L, where p q 9.6 C, hole diffuion coefficient 7 GL /cm. Hole lifetime i 4, electronic charge Dp cm / and low level injection condition prevail. Auming a linearly decaying teady tate exce hole concentration that goe to at x = L, the magnitude of the diffuion current denity at x = L/, in A/cm, i. Light 6 Si( N D cm ) An. 6 Sol. Given : x GL GL, LP P 4 ec, q 7 GL /cm, 9.6 C Gx ( ) x D P L. m cm /ec G L L P x More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

34 4 GATE 7 [Forenoon (Set - )] So, dp dp GL where, dx L JP qdp dx L D. cm P P P dp 8 P dx J P J P 6 A/cm Hence, the correct anwer i 6. Quetion 4 A three dimenional region R of finite volume i decribed by x y z ; z where x, y, z are real. The volume of R (up to two decimal place). An..785 Sol. Let x y l Here revolution i about z axi Volume of region R ldz Here, l x y Quetion 4 So, volume of region l x y z 4 z R z dz Hence, the correct anwer i.785. Let x(t) be a continuou time periodic ignal with fundamental period T = econd. Let a be the complex Fourier erie coefficient of x(t), where k i integer valued. Conider the following tatement about x(t) : I. The complex Fourier erie coefficient of x(t) are a k where k i integer valued. II. The complex Fourier erie coefficient of x(t) are III. The fundamental angular frequency of x(t) i 6rad/. For the three tatement above, which one of the following i correct? (A) only II and III are true (C) only III i true More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and k a where k i integer valued. (B) only I and III are true (D) only I i true k

35 5 An. Sol. GATE 7 [Forenoon (Set - )] (B) By time caling property for fourier erie If ak i the fourier erie coefficient of x() t then, for ignal x( at ), fourier erie coefficient i ak with time T period a Hence, frequency i a Hence, for ignal x( t ), fourier coefficient i ak and fundamental angular frequency i 6 rad/. Hence, only tatement I and III are true. Hence, the correct option i (B). Quetion 4 A hown, two Silicon (Si) abrupt p-n junction diode are fabricated with uniform donor doping concentration of N D= 4 cm and N D = 6 cm in the n-region of the diode, and uniform acceptor doping concentration of N A 4 cm and N A 6 cm in the p-region of the diode, C repectively. Auming that the revere bia voltage i >> built-in potential of the diode, the ratio C of their revere bia capacitance for the ame applied revere bia, i. p n p n 4 cm 4 cm 6 cm 6 cm An. Sol. p C Diode n p C Diode n 4 cm 4 cm 6 cm 6 cm A Since, C, W N and W ( N) Where, N NA ND qnw Vbi and W C W C Diode C ( N) C Diode More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

36 6 GATE 7 [Forenoon (Set - )] Then, So, 6 6 N NA N 5 /cm D N NA N 5 /cm D 4 C C 5 N 5 N 5 Hence, the correct anwer i. Quetion 4 A half wavelength dipole i kept in the x-y plane and oriented along 45 from the x-axi. Determine the direction of null in the radiation pattern for. Here the angle ( ) i meaured from the z-axi, and the angle ( ) i meaured from the x-axi in the x-y plane. An. Sol. (A) (C) (A) For 9, 45 (B) 9, 5 (D) 9, Applying For E S S jr jie co co along Z-plane r in E ji e E S E S jr co co(9 ) in xy plane r in 9 jrco in ) jie r co co in 9 L ' Hopital rule, E S E S 9 and Hence, the correct option i (A). co 9 45, 9 45, 5 d co in in in co d lim lim 9 d 9 co in d 45 the direction of null will occur. Half wave dipole More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and x z 45 o y

37 7 GATE 7 [Forenoon (Set - )] Quetion 44 In the figure hown, the npn tranitor act a a witch. 5V For the input V ( t ) a hown in the figure, the tranitor witche between the cut-off and aturation in V V () in t V T t (in ec) region of operation, when T i large. Aume collector-to-emitter voltage at aturation VCE ( at).v and bae-to-emitter voltage VBE.7 V. The minimum value of the common-bae current gain () of the tranitor for the witching hould be. An..9 Sol. V () in t 5V k 4.8 k I C 4.8 k V in k B C () I B () E VBE ON.7 V VCEat.V VCC VCE, at 5. IC ma Applying KVL in input loop Vin IB.7 Vin.7 IB. I B.8 ma 8. A IC So, 9.5 I B Hence, the correct anwer i.9. More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

38 8 GATE 7 [Forenoon (Set - )] Quetion 45 Two dicrete-time ignal x[n] and h[n] are both non-zero only for n =,, and are zero otherwie. It i given that x, x, x, h Let yn be the linear convolution of xn and hn. Given that expreion ( y[] y[4]) i. An. Sol. Given : xn ( ) {,, } hn ( ) {, h, h} y and y X ( z) z z H( z) hz hz Y( z) X( z) H( z) Y( z) hz hz z hz hz z hz hz 4 4 Y( z) ( h ) z ( h h ) z ( h h) z h z y( n) {, h, h h, h h, h } y() h h y() h h 4 h y() h h y(4) h y y 4 4, the value of the Hence, the correct anwer i. Quetion 46 Which one of the following option correctly decribe the location of the root of the equation 4 on the complex plane? (A) Four left half plane (LHP) root (B) One right half plane (RHP) root, one LHP root and two root on the imaginary axi (C) Two RHP root and two LHP root (D) All four root are on the imaginary axi More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

39 9 GATE 7 [Forenoon (Set - )] An. (C) Sol. Given : 4 4 Routh Tabulation : 4 Since row of zero appear in the Routh table, we form the auxiliary equation uing the coefficient of row A 4 () A'( ) 4 4 / 6 In the above Routh table, there are two ign change in firt column. So there mut be two pole lie in right half of -plane. Concept : If ROC exit, then pole are ymmetrical about origin. Root of auxiliary equation, Put x, A 4 () 4 Put x x x x j e, e j / j / x ( e ), ( e ) / j / / j / / e, e j / j/ j, j j, j, j, j More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

40 4 GATE 7 [Forenoon (Set - )] The pole-zero diagram i, j j Hence, there are two right hand pole and two left hand pole. Hence, the correct option i (C). Quetion 47 Starting with x =, the olution of the equation x x, after two iteration of Newton-Raphon' method (up to two decimal place) i. An..686 Sol. Let f( x) x x By Newton - Raphon iterative method, f ( xn ) x n x n f '( x ) Given : x f( x) Now, f x x '( ) f '( x) 4 f ( x) So, x x f '( x ) x 4 4 x f( x) f '( x ) f ( x ) x x f '( x ).78 x Hence, the correct anwer i.686. n j More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

41 4 GATE 7 [Forenoon (Set - )] Quetion 48 Let h[n] be the impule repone of a dicrete-time linear time invariant (LTI) filter. The impule repone i given by h, h, h and hn for n < and n >. An..9 Let H ( ) be the dicrete-time Fourier tranform (DTFT) of h[n], where i the normalized angular frequency in radian. Given that H ( ) and, the value of (in radian) i equal to. Sol. Given : hn ( ) H( z) z z H z z z For frequency repone put ( ) [ ] z j e, He e e j j j ( ) [ ] j But He ( ) H( ) j j H( ) [ e e ] H ( ) Let e j Root are So, x j j e e xx x j j e co jin j co.9 Hence, the correct anwer i.9. More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

42 4 GATE 7 [Forenoon (Set - )] Quetion 49 Which one of the following give the implified um of product expreion for the Boolean function F m m m m5 where m, m, m and m 5 are minterm correponding to the input A, B and C with A a the MSB and C a the LSB? An. Sol. (A) AB ABC ABC (B) AC AB ABC (C) AC AB ABC (D) ABC AC ABC (B) Given : A i MSB and C i LSB. So, Function i F( A, B, C ) i.e., F( A, B, C) m m m m 5 F( A, B, C) ABC ABC ABC ABC F( A, B, C) ABC ABC ABC ABC ABC [ X X X ABC ABC ABC ] F( A, B, C) AC( B B) AB( C C) ABC F( A, B, C) AC AB ABC Hence, the correct option i (B). Quetion 5 Which one of the following i the general olution of the firt order differential equation dy ( x y ) dx where x, y are real? (A) y xtan ( x c), where c i a contant (B) y xtan ( x c), where c i a contant An. (C) (D) Sol. Given : y x x c tan ( ), where c i a contant (D) y x tan ( x c) dy, where c i a contant ( x y ) dx (i) Let x y p (ii) dy dp dx dx.(iii) Put equation (ii) and (iii) in (i) dp ( p ) dx dp ( p ) dx dp dx ( p ) More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

43 4 GATE 7 [Forenoon (Set - )] Integrating both ide, dp dx ( p ) tan ( p ) x c p tan( x c) x ytan( x c) [Uing equation (ii)] y xtan( x c) Hence, the correct option i (D). Quetion 5 In the circuit hown, the voltage VIN ( t ) i decribed by :, for t VIN () t 5 Volt, for t where t i in econd. The time (in econd) at which the current I in the circuit will reach the value Ampere i. I VIN () t H H An..4 Sol. I I 5V H H I 5 5 () I I() ( ) 45 () I() More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

44 44 GATE 7 [Forenoon (Set - )] 5 / 5 5 I () it () 5 5e t Let at t t, i( t) become A t 55e Quetion 5 e t t 5 ln 5 t.4 ec Hence, the correct anwer i.4. The figure how an RLC circuit excited by the inuoidal voltage co(t) Volt, where t i in econd. V cot 4 H 5 + V F 6 An..6 Sol. amplitude of V The ratio amplitude of V i Vt ( ) co t rad/ec Z 4 jl4 j Z 5tan 4 V + 4 H I + Z 5 6 F Z V More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

45 45 GATE 7 [Forenoon (Set - )] j j Z 5 5 C 6 Z 5 jtan 5 V I Z, V I Z V Z tan tan V Z V 6 V Hence, the correct anwer i.6. Quetion 5 x x Let f ( x) e for real x. From among the following, chooe the Taylor erie approximation of f ( x ) around x, which include all power of x le than or equal to. 7 (A) x x x (B) x x x (C) x x x 6 (D) x x 7x An. (C) x x Sol. Given : F( x) e Taylor erie i defined a, ( xa) F''( a) ( xa) F'''( a) F( x) F( a) ( xa) f '( a)...!! Taylor erie around x i given by, x x F( x) F() xf'() F''() F'''()...!! x x F( x) e F() e xx F'( x) e.(x ) F '() xx xx F''( x) e. (x) e.(x ) xx ''( ) (4 4 ) F x e x x F ''() xx xx F'''( x) e 8x4 (4x 4x) e.(x ) xx '''( ) 8 4 ( )(4 4 F x e x x x x F '''() 7 More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

46 46 GATE 7 [Forenoon (Set - )] Taylor erie around x x 7x F( x) x 6 Hence, the correct option i (C). Quetion 54 Let I (zdxydy xdz) where x, y, z are real, and let C be the traight line egment from point C A : (,, ) to point B : (4,, ). The value of I i. An. Sol. We need to integrate zdx ydy xdz along path from point A (,,) to B(4,, ) A x (varie) P y (varie) Q z (varie) B y x 4 x 4 z z y Step I : Integrate along line joining A(,,) & P (4,,) Hence dy dz & z I 4 dx8 Step II : Integrate along line joining P(4,,) & Q (4,,) Hence dx dz I ydy y Step III : Integrate along line joining Q(4,,) & B(4,, ) Hence dx dy & x 4 (,,) I xdz 8dz 6 Hence I II I 8 6 (4,,) (4,, ) Hence, the correct anwer i. Quetion 55 j( t kx ky) The expreion for an electric field in free pace i E E ˆ ˆ ˆ ( x y j z) e, where x, y, z repreent the patial coordinate, t repreent time, and, k are contant. Thi electric field (A) doe not repreent a plane wave. (B) repreent a circularly polarized plane wave propagating normal to the z-axi. (C) repreent an elliptically polarized plane wave propagating along the x-y plane. (D) repreent a linearly polarized plane wave. More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and (4,,)

47 47 GATE 7 [Forenoon (Set - )] An. (C) Sol. Given : E ( xˆ yˆ j zˆ )exp j( tkxky) E ( xˆ yˆ)exp j( tkxky) jzˆ exp j( tkxky) firt term E E je E ˆ ˆ ( x y)exp j( tkxky) E ˆ ˆ ( x y)exp j t( kxky) jk r E E e For On comparing equation (i) and (ii), Propagation vector, k kaˆ ˆ x kay Poition vector r xaˆ yaˆ x E ˆ ˆ x y y From above reult we can draw a figure a hown below, y econd term. (i). (ii) E x x k y Here E and k are perpendicular to each other o it i a perpendicular polarization. For E ˆ z exp j( t kx ky) E ˆ zexp j t( kxky) jk r E E e On comparing equation (iii) and (iv), Propagation vector, k kaˆ ˆ x kay r xaˆ yaˆ E x ˆ z y. (iii). (iv) More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

48 48 GATE 7 [Forenoon (Set - )] Again if we draw a figure for above reult a hown below, y x E 45 x k Here reult E and k are perpendicular to ach other becaue plane of incidence i xy and electric field i in z-direction. So, it i alo a perpendicular polarization. Now, E E je E and E both are perpendicular polarized and Hence, the correct option i (C). y E, 9 E. EM wave i elliptically polarized. More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

49 49 GATE 7 [Forenoon (Set - )].General Aptitude. Quetion Some table are helve. Some helve are chair. All chair are benche. Which of the following concluion can be deduced from the preceding entence? (i) At leat one bench i a table (ii) At leat one helf i a bench (iii) At leat one chair i a table (iv) All benche are chair (A) Only (i) (B) Only (ii) (C) Only (ii) and (iii) (D) Only (iv) An. (B) Sol.. Method : T S C B T S B C T S B C Only concluion (ii) follow. Hence, the correct option i (B).. Method : Some table are helve, T S Some helve are chair, S C All chair are benche, C = B Uing equation (ii) and (iii), S B Atleat one helf i a bench. Hence, the correct option i (B).. (i). (ii).. (iii) More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

50 5 GATE 7 [Forenoon (Set - )] Quetion I made arrangement had I informed earlier. (A) could have, been (B) would have, being (C) had, have (D) had been, been An. (A) Sol. Could have, been I could have made arrangement had I been informed earlier. Ue of conditional entence baed on pat participle form. Hence, the correct option i (A). Quetion 4% of death on city road may be attributed to drunken driving. The number of degree needed to repreent thi a a lice of a pie chart i (A) (B) 44 (C) 6 (D) An. (B) 4 Sol. 4% of So, the angle ubtended on pie chart 44. Hence, the correct option i (B). Quetion 4 In the ummer, water conumption i known to decreae overall by 5%. A Water Board official tate that in the ummer houehold conumption decreae by %, while other conumption increae by 7%. Which of the following tatement i correct? (A) The ratio of houehold to other conumption i 8/7 (B) The ratio of houehold to other conumption i /7 (C) The ratio of houehold to other conumption i 7/8 (D) There are error in the official' tatement. An. (D) Sol. Let H be houehold water conumption and P be other water conumption. Total water conumption initially = H + P After 5% decrement in overall water conumption, New overall water conumption =.75 (H + P) After % decrement in houehold water conumption, New houehold water conumption =.8 H After 7% increment in other water conumption, New other water conumption =.7 P More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

51 5 GATE 7 [Forenoon (Set - )] According to given condition New houehold water conumption + New other water conumption = New overall water conumption.8 H +.7 P =.75 (H + P).5 H =.95 P H 9 P Ratio cannot be negative. There are error in the official tatement. Hence, the correct option i (D). Quetion 5 She ha a harp tongue and it can occaionally turn (A) hurtful (B) left (C) methodical (D) vital An. (A) Sol. Hurtful Hurtful mean cauing pain or uffering or omething that i damaging or harmful. The expreion 'harp tongue' define a bitter or critical or rude manner of peaking. Hence, the correct option i (A). Quetion 6 If you are looking for a hitory of India, or for an account of the rie and fall of the Britih Raj, or for the reaon of the cleaving of the ubcontinent into two mutually antagonitic part and the effect thi mutilation will have in the repective ection, and ultimately on Aia, you will not find it in thee page: for though I have pent a lifetime in the country. I lived too near the eat of event, and wa too intimately aociated with the actor, to get the perpective needed for the impartial recording of thee matter. Here, the word antagonitic i cloet in meaning to (A) impartial (B) argumentative (C) eparated (D) hotile An. (C) Quetion 7 A contour line join location having the ame height above the mean ea level. The following i a contour plot of a geographical region. Contour line are hown at 5 m interval in thi plot Q 575 P The path from P to Q i bet decribed by (A) Up-Down-Up-Down (C) Down-Up-Down (B) Down-Up-Down-Up (D) Up-Down-Up More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

52 5 An. Sol. GATE 7 [Forenoon (Set - )] (C) Contour line can be oberved to cro region with height Down-Up-Down atifie. Hence, the correct option i (C). Quetion 8 There are Indian and Chinee in a group of 6 people. How many ubgroup of thi group can we chooe o that every ubgroup ha at leat one Indian? (A) 56 (B) 5 (C) 48 (D) 44 An. (A) Sol. Given : Indian men Total 6 men Chinee men Total number of group N na ( ) 6 64 They will be one empty group hence N = 64 = 6 The number of group having no Indian Down Up Down C C C 7 Number of group having atleat one Indian = Hence, the correct option i (A). Quetion 9 S, T, U, V, W, X, Y and Z are eated around a circular table. T' neighbor are Y and V. Z i eated third to the left of T and econd to the right of S. U' neighbor are S and Y: and T and W are not eated oppoite each other. Who i third to the left of V? (A) X (B) W (C) U (D) T An. (A) Sol. Following circular eating arrangement can be drawn X Z S W U V Only one uch arrangement can be drawn. The peron on third to the left of V i X. Hence, the correct option i (A). T More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and Y

53 5 GATE 7 [Forenoon (Set - )] Quetion Truck ( m long) and car (5 m long) go on a ingle lane bridge. There mut be a gap of at leat m after each truck and a gap of at leat 5 m after each car. Truck and car travel at a peed of 6 km/h. If car and truck go alternately, what i the maximum number of vehicle that can ue the bridge in one hour? (A) 44 (B) (C) 7 (D) 6 An. (A) Sol. Given : Length Gap Truck m m Car 5 m 5 m 5 Speed 6 km/hr m/ 8 Truck m Car m 5 m 5m Ditance 5 Time 5ec Speed If 5 m ditance i covered in 5 econd by two vehicle (one truck and one car) then in one hour maximum number of vehicle paing through the lane = Hence, the correct option i (A). 5 m More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

54 Dear tudent, We have tried to provide the GATE 7 quetion along with their olution. However, if you find any dicrepancy, pleae write to u at info@gateacademy.co.in GATE ACADEMY i not reponible for any error due to data miprint/human error/data inufficiency etc. More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and problem for EC/EE/IN to join For more GATE 7 Subject wie Video Tutorial Subcribe

55 GATE 7 [Afternoon (Set - )].Technical Part. Quetion The maller angle (in degree) between the plane x + y + z = and x y + z = i. An Sol. Given : x yz and x yz For the firt plane A (,,) aˆ ˆ ˆ x ay az For the econd plane B (,,) aˆ aˆ aˆ x y z Angle between two plane can be calculated a, A B co A B Where, AB ( aˆ ˆ ˆ ) (ˆ ˆ ˆ x ay az ax ay az) A B ( ) 9 From equation (i), co co 54.7 Quetion For the circuit hown in the figure, P and Q are the input and Y i the output. PMOS Y..(i) P Q NMOS An. Sol. The logic implemented by the circuit i (A) XNOR (B) XOR (C) NOR (D) OR (B) P i connected to gate terminal of PMOS a well a NMOS. If P i at logic, then PMOS turn ON & NMOS turn OFF. If P i at logic, then PMOS turn OFF & NMOS turn ON. If PMOS i turned ON then output Y i connected to Q. If NMOS i turned ON then output Y i connected to Q. More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

56 GATE 7 [Afternoon (Set - )] P Q PMOS NMOS Output Q (if PMOS i ON) Q (if NMOS i ON) ON OFF (PMOS ON) ON OFF (PMOS ON) OFF ON (NMOS ON) OFF ON (NMOS ON) Hence, the correct option i (B). Quetion Conider an n-channel MOSFET having width W, length L, electron mobility in the channel n and oxide capacitance per unit area C ox. If gate-to-ource voltage VGS.7 V, drain to ource voltage VDS.V W ncox A/V, threhold voltage VTH.V and 5 L, then the tranconductance g m (in ma/v) i. An..5 ma/v Sol. Given : For n-channel MOSFET, VDS. V, VGS.7 V, VTH. V W ncox A/V. ma/v, 5 L VGS VTH and VDS VGS VTH So n-channel MOSFET i in linear region. Tranconductance for n-channel MOSFET i given by, I D ncoxw gm VDS V L GS gm.5..5 ma/v Quetion 4 An n-channel enhancement mode MOSFET i biaed at V GS An. V TH and VDS ( VGS VTH), where V GS i the gate-to-ource voltage, V DS i the drain-to-ource voltage and V TH i the threhold voltage. Conidering channel length modulation effect to be ignificant, the MOSFET behave a a (A) voltage ource with zero output impedance (B) voltage ource with non-zero output impedance (C) current ource with finite output impedance (D) current ource with infinite output impedance (C) More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

57 Sol. GATE 7 [Afternoon (Set - )] The mall ignal equivalent circuit of MOSFET in aturation i a hown in below figure G + D V GS gv r m GS Quetion 5 VGS VTH and VDS VGS VTH Hence MOSFET i in aturation region and therefore it will act a current ource. Becaue channel length modulation effect i coniderable, hence output impedance will exit. Therefor MOSFET behave a a current ource with finite impedance. Hence, the correct option i (C). S The output V of the diode circuit hown in the figure i connected to an averaging DC voltmeter. The reading on the DC voltmeter in volt, neglecting the voltage drop acro the diode, i. in t f 5 Hz k V An. (.8) Sol. Given circuit i hown in below figure, + int f = 5 Hz k V Above circuit i a half wave rectifier and DC voltmeter will read the average value of V, hence average value of V i.e. output of half wave rectifier i given by, Vm Vavg.8 Quetion 6 A two wire tranmiion line terminate in a televiion et. The VSWR meaured on the line i 5.8. The percentage of power that i reflected from the televiion et i. An More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

58 4 GATE 7 [Afternoon (Set - )] Sol. Given : VSWR = 5.8 Reflection coefficient i given by, VSWR 5.8 VSWR Percentage of reflected power i given by, Pr 49.8% P 7 i Hence, the correct anwer i Quetion 7 An n-p-n bipolar junction tranitor (BJT) i operating in the active region. If the revere bia acro the bae collector junction i increaed, then (A) The effective bae width increae and common-emitter current gain increae (B) The effective bae width increae and common-emitter current gain decreae (C) The effective bae width decreae and common-emitter current gain increae (D) The effective bae width decreae and common-emitter current gain decreae An. (C) Sol. If revere bia acro bae collector junction will increae, then (i) Depletion width acro bae collector junction will increae. (ii) Effective bae width will decreae. (iii) Recombination of carrier in bae region will decreae. I C (iv) Common emitter current gain will increae IB Hence, the correct option i (C). Quetion 8 The input x() t and the output yt () of a continuou time ytem are related a, An. The ytem i, (A) Linear and time variant (C) Non-linear and time variant (B) Sol. Given : y() t x( u) du (i) At input x ( u ), t tt y () t x ( u) du tt t t y() t x( u) du tt (B) Linear and time-invariant (D) Non-linear and time invariant More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

59 5 GATE 7 [Afternoon (Set - )] At input x ( u ), At input x ( u ), y () t x ( u) du tt t y () t x ( u) du tt If t x ( u) Ax ( u) Bx ( u) t t y () t A x ( u) dub x ( u) du tt tt y () t Ay () t By () t Hence it i linear. (ii) at input x ( u ), At input x ( u ), t y () t x ( u) du tt tt (i) t y () t x ( u) du x ( u) x ( u t ) t y () t x ( ut ) du tt From equation (i), (ii) tt y ( tt ) x ( u) du (iii) ttt becaue the time hifting in cae of integration will not affect the integrated value. Hence from equation (ii) and (iii) y() t y( t t). Hence it i time invariant. Hence, the correct option i (B). Quetion 9 In the figure, D i a real ilicon p-n junction diode with a drop of.7 V under forward bia condition and D i a Zener diode with breakdown voltage of 6.8V. The input Vin ( t ) i a periodic quare wave of period T, whoe one period i hown in the figure. More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

60 6 GATE 7 [Afternoon (Set - )] 4 V V () in t F T t (ec) D Vout () t 4V D Auming T, where i the time contant of the circuit, the maximum and minimum value of the output waveform are repectively. (A) 7.5 V and.5 V (B) 6. V and.9 V (C) 7.5 V and. V (D) 6. V and.6 V An. (A) T Sol. Cae I : t, Vin( t) 4V D F. B.(.7V) D Break down (6.8 V) Equivalent circuit i hown in below figure, + 4 V V c + F V 6.8V Vout () t Vout 7.5V Apply KVL in the loop hown in figure, 4 V c 7.5 Vc 6.5V T At t capacitor will charge up to 6.5 V due to mall time contant. T Cae II : t T, V ( t) 4V in D RB.. (open circuit) D F. B. More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

61 7 GATE 7 [Afternoon (Set - )] Equivalent circuit i hown in below figure, + Vc 6.5V + F 4 V S.C. Vout () t Apply KVL in the loop hown in figure, 4 V out 6.5 Vout.5V Hence, the correction option i (A). Quetion The rank of the matrix An. 4 Sol. Given : A R4 R4 R5 R R R4 i. More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

62 8 GATE 7 [Afternoon (Set - )] R R R R RR Becaue number of non-zero row in above matrix i 4. Hence, rank of matrix i 4. Quetion Two conducting phere S and S of radii a and b (b a) repectively, are placed far apart and connected by a long, thin conducting wire, a hown in the figure. S S Wire Radiu r a An. Sol. Radiu r b For ome charge placed on thi tructure, the potential and urface electric field on S are V a and that on S are V b and E b, repectively. Then, which of the following i CORRECT? E a, and (A) Va Vb and Ea Eb (B) Va Vb and Ea Eb (C) Va Vb and Ea Eb (D) Va Vb and Ea Eb (C) When two phere are connected through conducting wire, then charge flow from higher potential to lower potential phere until both phere attain ame potential. kqa kqb qa qb So, Va Vb. (i) r r r r Electric field outide phere i given by, kq E r Electric field outide phere A i given by, kqa Ea ra Electric field outide phere B i given by, kqb Eb r b a b a b More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

63 9 GATE 7 [Afternoon (Set - )] Given : r b r a r r.(ii) b a From equation (i) and (ii), kqa kqb ra rb Ea Eb Hence, the correct option i (C). Quetion Conider the random proce X ( t) U Vt, where U i a zero mean Gauian random variable and V i a random variable uniformly ditributed between and. Aume that U and V are tatitically independent. The mean value of the random proce at t i. An. Sol. Given : Xt () U Vt, EU [ ], V ~ UDF [,] f V v () v Otherewie Mean value of V i given by, ab EV [ ] EXt [ ( )] EU [ ] EVt [ ] At t ec, EXt [ ( )] EU [ ] EV [ ] EXt [ ( )] Hence, correct anwer i. Quetion Conider the circuit hown in the figure. fv () v v Y MUX MUX F An. X Z The Boolean expreion F implemented by the circuit i (A) X YZ XY YZ (B) X YZ XZ YZ (C) X YZ XY YZ (D) X YZ XZ YZ (B) More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

64 Sol. GATE 7 [Afternoon (Set - )] Output equation for a multiplexer I I Y Y SI SI S Y MUX P MUX f In the given circuit, output of t MUX, P XY X P XY Output of nd MUX i given by, f Z PZ P f ZXY Z XY f ZXYZ( X Y) f XYZ XZ YZ Hence, the correct option i (B). Quetion 4 Conider the circuit hown in the figure. Aume bae-to-emitter voltage.8v BE current gain ( ) of the tranitor i unity. X Z 8V V and common bae 44 k 4k 6 k k An. 6 The value of the collector-to-emitter voltage V CE (in volt) i. More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

65 GATE 7 [Afternoon (Set - )] Sol. Given : VBE.8V and common bae current gain ( ) Thevenin equivalent circuit of given circuit 8 V 4 k I C R th V th I E k R RR 446.7k th R R 6 VCC 86 Vth R 4.8V R R 6 i.e., I B and hence IC I E Applying KVL in loop, Vth IBRth VBE IERE Vth VBE I E ma IC RE Applying KVL in loop, VCE 8 IC (4 ) V Quetion 5 A connection i made coniting of reitance A in erie with a parallel combination of reitance B and C. Three reitor of value, 5, are provided. Conider all poible permutation of the given reitor into the poition A, B, C and identify the configuration with maximum poible overall reitance, and alo the one with minimum poible overall reitance. The ratio of maximum to minimum value of the reitance (up to econd decimal place) i. An..4 Sol. R B R A R C P Q More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

66 GATE 7 [Afternoon (Set - )] R eq R A R R B B RC R C Becaue parallel combination of two reitance will be approximately to the lower reitance hence R eq will be maximum when RA. 5 8 R eq,max R eq,min 5 5 R R eq,max eq,min Quetion 6 The reidue of a function f( z) ( z4)( z) are (A) and 7 5 (B) (C) and 7 5 (D) An. (B) Sol. Given : f( z) ( z4)( z) Reidue of f ( z ) at z = a f( z)( z a) za and 5 5 and 5 5 So reidue of f( z) ( z4)( z) at z = 4 ( z 4) ( z4)( z) z4 z 5 z4 Reidue of f ( z ) with multiple pole at z = a with order n d ( n )! n n ( z a) f( z) n dz za More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

67 GATE 7 [Afternoon (Set - )] Reidue of ( z4)( z) with pole at z = d ( z )! dz ( z 4)( z ) d dz z 4z d ( z4) ( z4) dz ( z 4) d dz ( z 4) z z ( z 4) ( )( z 4) 4 ( z 4) z ( z 4) z 5 Hence, the correct option i (B). Quetion 7 The general olution of the differential equation, d y dy 5y dx dx in term of arbitrary contant K and K i, z (A) Ke Ke (B) ( 6) x ( 6) x Ke Ke ( 8) x ( 8) x An. (C) (A) Ke Sol. Given : Ke (D) ( 6) x ( 6) x d y dy 5y dx dx ( D D5) y D D5 4 D D 6 4 Ke Ke ( 8) x ( 8) x More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

68 4 GATE 7 [Afternoon (Set - )] M 6 M 6 M x yt () ke ke Mx yt () ke ke Hence, the correct option i (A). Quetion 8 ( 6) x ( 6) x For the ytem hown in the figure, Y () X (). X() + G () + + Y() An. () Sol. X() G () Y() Signal flow graph of above block diagram i hown below X() Y() Forward path gain : P, P Individual loop gain : L Determinate : L Now from Maon gain formula Y() X() k P k k More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

69 5 GATE 7 [Afternoon (Set - )] Y () P P X() L Y () X() Quetion 9 A inuoidal meage ignal i converted to a PCM ignal uing a uniform quantizer. The required ignal to quantization noie ratio (SQNR) at the output of the quantizer i 4 db. The minimum number of bit per ample needed to achieve the deired SQNR i. An. 7 Sol. Given : SQNR = 4dB For a PCM ignal, SQNR(dB) N where, N no. of bit/ample N N 6.5 Minimum number of bit/ample = 7 [to keep SQNR 4 db]. Quetion In the circuit hown V i a inuoidal voltage ource. The current I i in phae with voltage V. The ratio Amplitude of voltage acro the capacitor i. Amplitude of voltage acro the reitor 5 5H I R L V 5F An.. Sol. 5 5H I V + V R V C + 5H Becaue I i in phae with V, hence power factor angle will be zero i.e. circuit i in reonance. X L X C rad/ec LC 5 5 More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

70 6 GATE 7 [Afternoon (Set - )] X C c 5 5 VC IXC I V IR 5I R VC. V R 5 Quetion Which one of the following graph how the Shannon capacity (channel capacity) in bit of a memoryle binary ymmetric channel with croover probability p? (A) (B) Capacity Capacity (C) p (D) p Capacity Capacity An. Sol. p (C) For a memoryle BSC, with cro-over probability p, capacity i given by, p C plog p( p) log ( p) At p, C At p, C At p, C Capacity p Hence, the correct option i (C). Quetion In a DRAM (A) Periodic refrehing i not required (B) Information i tored in a capacitor (C) Information i tored in a latch (D) Both read and write operation can be performed imultaneouly An. (B) More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

71 7 GATE 7 [Afternoon (Set - )] Sol. Inide a dynamic RAM chip, each memory cell hold one bit of information and i made up of two part: a tranitor and a capacitor. Thee are extremely mall tranitor and capacitor o that million of them can fit on a ingle memory chip. The capacitor hold the bit of information - a or a. The tranitor act a a witch that let the control circuitry on the memory chip read the capacitor or change it tate. A capacitor i like a mall bucket that i able to tore electron. To tore a in the memory cell, the bucket i filled with electron. To tore a, it i emptied. Therefore, for dynamic memory to work, either the CPU or the memory controller ha to come along and recharge all of the capacitor holding a before they dicharge. To do thi, the memory controller read the memory and then write it right back. Thi refreh operation happen automatically thouand of time per econd. Thi refreh operation i where dynamic RAM get it name. Dynamic RAM ha to be dynamically refrehed all of the time or it forget what it i holding. The downide of all of thi refrehing i that it take time and low down the memory. Hence, the correct option i (B). Quetion An LTI ytem with unit ample repone hn [ ] 5 [ n] 7 [ n] 7 [ n] 5 [ n 4] i a (A) low-pa filter (B) high-pa filter (C) band-pa filter (D) band-top filter An. (C) Sol. Given : Impule repone hn [ ] 5 [ n] 7 [ n] 7 [ n] 5 [ n 4] Taking Z tranform, 4 Tranfer function, H( z) 57z 7z 5z At low frequency, z H( z) At high frequency, z z H( z) z Since, it i not paing low frequency and high frequency. Hence, it i a band-pa filter. Hence, the correct option i (C). Quetion 4 Which of the following tatement i INCORRECT? (A) Lead compenator i ued to reduce the ettling time. (B) Lag compenator i ued to reduce the teady tate error. (C) Lead compenator may increae the order of a ytem. (D) Lag compenator alway tabilize an untable ytem. An. (D) Sol. Sr. Lag Compenator Lead Compenator. Low pa filter High pa filter. decreae, n decreae increae, n increae More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

72 8 GATE 7 [Afternoon (Set - )] Sr. Lag Compenator Lead Compenator. Gain cro over frequency gc decreae Gain cro over frequency gc increae 4. Bandwidth decreae Bandwidth increae Settling time T increae Settling time T 6. Lag compenator improve the teady tate performance of the ytem (decreae the teady tate error) n More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and n decreae Lead compenator improve the tranient performance of the ytem (decreae the time domain parameter) Hence incorrect tatement i (D) and correct option i (D). Quetion 5 Conider the tate pace realization x () t x () t ut (), x() t 9 x() t 45 with the initial condition x () x() where, ut ( ) denote the unit tep function. The value of lim x( t) x( t) i. t An. (5) Sol. Given : x () t x() t ut () x () t 9 x() t 45...(i) and x () x () State equation i given by, x Ax Bu (ii) On comparing equation (i) and (ii), we get A, B 9 45 Solution of above tate equation in time domain i given by, x() t ZIR ZSR Where, ZIR f [ x()] I / P and ZSR f ( I / P) x () t x() t exp( At) x() exp A( t) Bu() d In cae of zero initial condition we conider zero tate repone. x() t ZSR X () () BU()...(iii) [ I A] 9

73 9 GATE 7 [Afternoon (Set - )] Laplace tranform of tate tranition matrix can be written a, L[ ( t)] ( ) [ I A] 9 BU () U () From equation (iii), we get X () () BU() X() ( 9) 9 Uing partial fraction, X() Taking invere Laplace tranform, we get xt () 9t 5 5e x () t 9t x() t 5 5e x () t, () 5 5 9t x t e The value of x x e 9 ( ), ( ) 55( ) 5 lim x ( t) x ( t) x ( ) x ( ) (5) 5 t Hence, the correct anwer i 5. Quetion 6 The unmodulated carrier power in an AM tranmitter i 5 kw. Thi carrier i modulated by a inuoidal modulating ignal. The maximum percentage of modulation i 5%. If it i reduced to 4%, then the maximum unmodulated carrier power (in kw) that can be ued without overloading the tranmitter i. An. (5. to 5.) Sol. Given : Pc 5kW, ma 5% or.5 More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

74 GATE 7 [Afternoon (Set - )] Maximum modulated power i given by, ma Pt P c.5 5 Pt 5.65kW If the modulation index i ma 4% P c Quetion 7 P 5.8 kw c Figure-I how a 4-bit ripple carry adder realized uing full adder and figure-ii how the circuit of a full adder (FA). The propagation delay of the XOR, AND and OR gate in figure II are n, 5 n and n, repectively. Aume all the input to the 4-bit adder are initially reet to. Y X Y X Y X Y Y Z 4 FA Z FA Z Z FA FA Z S S S S Fig. I X n Y n S n Zn Fig. II At t =, the input to the 4-bit adder are changed to XXXX, YYYY and Z. The output of the ripple carry adder will be table at t (in n) =. An. (5) Sol. Given : Propagation delay of the XOR gate = n, Propagation delay of the AND gate = 5 n, Propagation delay of the OR gate = n Internal diagram of full adder i hown below. Z n More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

75 GATE 7 [Afternoon (Set - )] X n Y n S n Zn Z n The internal diagram of 4 bit ripple carry adder can be form a given below. X Y A S 5 5 C Z B Z X Y A S 5 5 C Z B X Y A S 5 5 C Z B X Y A S 5 5 C Z 4 B More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

76 GATE 7 [Afternoon (Set - )] t (nec) X Y Z A B S C Z X Y A B S C Z X Y A B S C Z 5 X Y A B S 45 C Z 4 5 More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

77 GATE 7 [Afternoon (Set - )] In given 4-bit ripple carry adder, we will get the table output when SSSS and ZZZZ 4 a hown in the above waveform we get SSSS and ZZZZ 4, at t = 5 nec, hence we will get the table output at t = 5 nec. Hence the correct anwer i 5. Quetion 8 A programmable logic array (PLA) i hown in the figure. P P Q Q R R F P Q An. Sol. The Boolean function F implemented i (A) PQR PQR PQR (B) ( P QR)( PQR)( PQ R) (C) PQR PQR PQR (D) ( P QR)( PQR)( PQ R) (C) Function F from the circuit F PQR PQR PQR R Hence, the correction option i (C). Quetion 9 Two n-channel MOSFET, T and T, are identical in all repect except that the width of T i double that of T. Both the tranitor are biaed in the aturation region of operation, but the gate over drive voltage ( VGS VTH ) of T i double that of T, where V GS and V TH are the gate-to-ource voltage and threhold voltage of the tranitor, repectively. If the drain current and tran-conductance of T are I D and g m repectively. The correponding value of thee two parameter for T are (A) 8ID and g m (B) 8ID and 4g m (C) 4ID and 4g m (D) 4ID and g m An. (B) Sol. Given : W W, V V V V D n( GS TN), I K V V g K ( V V ) m n GS TN More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

78 4 GATE 7 [Afternoon (Set - )] VV VGS VTN ncoxw Kn L ncoxw I ( ) D V GS VTN L ncoxw ID 8 ( V ) GS VTN L I D 8I D ncoxw and g m ( V ) GS VTN L ncoxw gm 4 ( V ) GS VTN L g 4g m m Hence, the correct option i (B). Quetion Conider the parallel combination of two LTI ytem hown in the figure. h () t xt () yt () The impule repone of the ytem are h () t ( t ) ( t ) h () t h () t ( t ) If the input x() t i a unit tep ignal, then the energy of yt () i. An. 7 Sol. Given : h () t ( t ) ( t ), h () t ( t ), x () t u () t h () t xt () yt () xt () ht () yt () yt () ht ()* xt () h () t yt () [ ( t) ( t) ( t )]* ut () yt () [ ut ( ) ut ( ) ut ( )] More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

79 5 GATE 7 [Afternoon (Set - )] yt () The energy of output ignal i given by, E yt () dt () dt ( ) dt E 4 7J Quetion An abrupt p-n junction (located at x ) i uniformly doped on both p and n ide. The width of the depletion region i W and the electric field variation in the x-direction i Ex. ( ) Which of the following figure repreent the electric field profile near the p-n junction? (A) Ex () (B) Ex () t n-ide p-ide n-ide p-ide W x W (, ) x (C) n-ide Ex () W p-ide x (D) n-ide W Ex () (, ) p-ide x An. Sol. (A) An electric field i created in the depletion region by the eparation of poitive and negative pace charge denity. Donor ion n ide x n ( x)(c/cm ) qn D x p p ide x qn A Acceptor ion qn D qn A xn x p E( x)(v/cm) x n xp x More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

80 6 GATE 7 [Afternoon (Set - )] From Poion equation dv( x) ( x) dx dv ( x) Ex ( ) dx de( x) ( x).. (i) dx Where V( x) electric potential, Ex ( ) electric field, ( x) volume charge denity From figure, volume charge denity i, qnd, xn x ( x) qn A, x xp From equation (i), ( x) Ex ( ) dx qnd qnd Ex ( ) dx xc.. (ii) At xxn, E( x) qnd xn C qndxn C From equation (ii), electric field Ex ( ) at n-ide i given by, qn D n ( ) D qn x Ex x qnd Ex ( ) ( xxn), xn x.. (iii) In the p-region, electric field i determine from, qn A qn A Ex ( ) dx xc.. (iv) Since Exi ( ) continuou at x =, qnd xn From equation (ii), E() From continuity equation, NA xp ND xn qn Axp So, E() From equation (iv), E() C qn Axp C More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

81 7 GATE 7 [Afternoon (Set - )] From equation (iv), electric field Ex ( ) at p-ide i given by, qn qn Ax A p qn A Ex ( ) x ( xxp ) Hence, the correct option i (A). Quetion Auming that tranitor M and M are identical and have a threhold voltage of V, the tate of tranitor M and M are repectively V.5 V M. V M (A) Saturation, Saturation (C) Linear, Saturation An. (C) Sol. Given : (B) Linear, Linear (D) Saturation, Linear V.5 V M V x. V M For tranitor M : VDS Vx VGS.5 Vx V V V.5 V VGS Vth and VDS VOV So tranitor M i in aturation. Aume M i alo in aturation OV GS th x More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

82 8 GATE 7 [Afternoon (Set - )] So, ID ID aturation aturation k (.5V ) k ( ) n x n.5v x Taking poitive ign.5v x Vx.5 Taking negative ign.5v x Vx.5 For Vx.5, For tranitor M : VGS V Vth VGS So, M i in cutoff region. So Vx.5 [Not poible] For Vx.5, For tranitor M : VGS V Vth VGS andvds VOV So, M i in aturation region. For tranitor M : VGS V VDS Vx.5 V Vth VGS and VDS VOV So, our aumption i wrong and M i in linear region. Hence, the correct option i (C). Quetion Conider the circuit hown in the figure. i i P V The Thevenin equivalent reitance (in ) acro PQ i. Q More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

83 9 GATE 7 [Afternoon (Set - )] An. Sol. Given : i i P V Calculation of R th : Remove V voltage ource by hort circuit i Q i i I dc P V dc R V i i dc th.(i) Idc Idc Idc Applying KCL at node (A), i i Idc Applying KVL in loop (), i ii i ii i Idc.(ii) Applying in loop (), i i.(iv) From equation (ii) and (iii), i Idc i Idc i i Idc Hence from equation (i), R th i A More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and Q

84 GATE 7 [Afternoon (Set - )] Quetion 4 In the circuit hown, tranitor Q and Q are biaed at a collector current of.6 ma. Auming that tranitor current gain are ufficiently large to aume collector current equal to emitter current and V thermal voltage of 6 mv, the magnitude of voltage gain V in the midband frequency range i S (up to econd decimal place). 5V k V V S Q Q R B 5 V An. 5 Sol. Given : (i) Collector current = IC IC.6 ma (ii) i large (iii) Thermal voltage = VT 6 mv. Method : AC equivalent model : Replace all the DC voltage ource by hort circuit. B I B C I C V V r I B I B k E I E E r I B I B I C B C R B More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

85 GATE 7 [Afternoon (Set - )] B I B C I C V V r I B I B k E I E E r I B I B I C B C From the above circuit, output voltage V i given by, V I B.(i) Apply the KVL in the loop hown in above figure. V r I r I.(ii) B B Apply KCL at node E i.e. E, I I I I B B B B ( ) I ( ) I B B I B IB.(iii) From equation (ii) and (iii), V ( r r) IB Where r, r g g m m Hence V IB gm gm Put the value of I B from equation (i) into equation (iv), hence V V gm gm V V g g m m. (iv) More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

86 GATE 7 [Afternoon (Set - )] IC.6 IC Where gm and gm V 6 V T T.6 6 Hence. Method : V 5 V V Voltage gain of thi combination of BJT i given by, k V V RC V r r e e V S Q VT VT 6 Where r e I I.6 e C Q VT VT 6 And r e I I.6 e C R B 5 V V V Hence 5 or 5 V V Key Point : AC equivalent model : Replace all the DC voltage ource by hort circuit. B I B C I C V V r I B I B k E I E E r I B I B I C B C More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

87 GATE 7 [Afternoon (Set - )] V I R.(i) B C Apply KCL at E : I BIB IB IB I B IB.(ii) Apply KVL in loop hown in figure, V r IBrIB.(iii) From equation (ii) and (iii), V r r IB.(iv) From equation (i) and (iv), V RC V r r r re r re V V r R r C e e V RC RC V r r r r e e e e Quetion 5 An integral I over a counterclockwie circle C i given by, z z I edz c z If C i defined a z, then the value of I i An. Sol. (A) (D) iin() (B) iin() (C) iin() (D) 4 iin() Im ( z) i i z i Re( z) i More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

88 4 GATE 7 [Afternoon (Set - )] I Reidue of f ( z ) at z Reidue of Reidue of z z z z e z z a i given by lim ( z a) f( z) z edzi c Re( z i) Re( z i) z at z i za ( z ) z lim ( z i) e i e e zi ( zi )( zi ) i i i e z at z i ( z ) z i lim ( z) e e zi ( zi )( zi ) i z z I edzi Re( z i) Re( z i) c z i I i e e i i i e e I 4i i i i i i e e 4i i i i 4 iin() An. Hence, the correct option i (D). Quetion 6 For a particular intenity of incident light on a ilicon p-n junction olar cell, the photo current denity ( J L ) i.5 ma/cm and the open circuit voltage ( V oc ) i.45 V, conider thermal voltage ( V T ) to be 5 mv. If the intenity of the incident light i increaed by time, auming that the temperature remain unchanged, V oc (in volt) will be. An. (.55) Sol. Open circuit voltage of olar cell i given by, J L Voc VT ln J Where, J L Photo current denity, J Revere aturation current denity, VT Thermal voltage. When intenity of the incident light i increaed by time then JL' J Given : Temperature i contant, o J ' J, V' V JL JL Open circuit voltage Voc.45 V VT ln VT ln J J T T L..(i) More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

89 5 GATE 7 [Afternoon (Set - )] New open circuit voltage, J L' Voc' VT' ln J ' J L' Voc' VT ln J ' J L Voc' VT ln J J L Voc' VT ln VT ln J From equation (i), Voc'.45 (.5ln ).55 V Quetion 7 The value of the integral x y dy dx ( x y) x y dx dy ( x y) (A) ame and equal to.5 (B) ame and equal to.5 (C).5 and.5, repectively (D).5 and.5, repectively An. (C) Sol. Given : Let, And let, x y dy dx ( x y) and I I x y ( x y ) dy dx x ( x y) ( x y) x y dx dy ( x y) are dydx I x dx ( x y) x y I dx.5 ( x) x x y y I dx dy dx dy ( x y) ( x y) ( x y) I I y ( x y) ( x y) dy dy.5 ( y) y Hence, the correct option i (C). More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

90 6 GATE 7 [Afternoon (Set - )] Quetion 8 The tate diagram of a finite tate machine (FSM) deigned to detect an overlapping equence of three bit i hown in the figure. The FSM ha an input In and an output Out. The initial tate of the FSM i S. In = Out = In = Out = S S In = Out = In = Out = In = Out = In = Out = If the input equence i, tarting with the left-mot bit, then the number of time Out will be i. An. (4) Sol. Given : Input equence i Initial tate In Next tate Out S S S S S S S S S S S S S S S S S S S S S S S S S S S S Hence the number of time Out will be i 4. S S In = Out = In = Out = More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

91 7 GATE 7 [Afternoon (Set - )] Quetion 9 A modulating ignal given by phae deviation contant x( t) 5in(4 tco t) V i fed to a phae modulator with K 5rad/V. If the carrier frequency i khz, the intantaneou frequency (in p khz) at t =.5 m ec i. An. 7 Sol. Given : x( t) 5in (4 tco t), k 5rad/v, f khz In PM, frequency deviation i given by, K p d f [ xt ( )] dt At t.5m, p K p d f 5in (4 t co t) dt 5 f 5co 4 tco t) 4 in t 5 co[ ][4 ] f f 5 5 khz Frequency deviation can alo be written a, f f f f f f i i c c fi 5 7 khz Hence, the correct anwer i 7. Quetion 4 Paenger try repeatedly to get a eat reervation in any train running between two tation until they are ucceful. If there i 4% chance of getting reervation in any attempt by a paenger, then the average number of attempt that paenger need to make to get a eat reerved i. An. (.5) Sol. ' X ' Number of attempt X 4 5 P EX ( ) xpx i ( i) i.6.4 c EX ( ).4 (.6).4 (.6).4 (.6).4 (.6).4 EX ( ).4( ) More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

92 8 GATE 7 [Afternoon (Set - )] EX ( ).4 (.6) EX ( ).5.4 Quetion 4 Conider an LTI ytem with magnitude repone An. 8 Sol. ( x) x x... More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and f, f H( f) and phae repone arg { H( f)} f, f If the input to the ytem i xt ( ) 8co t 6in 4 t 4 co 8 t 4 8 6, then the average power of the output ignal y() t i. f f H( f) f f f f H( f) f Otherwie H ( f) f f f f f H( f) f f Otherwie For ( ) 8co X t t 4 ; () 8 X t f Hz 4 () f 8 Y t f 4

93 9 GATE 7 [Afternoon (Set - )] Quetion 4 () 8 f Y t f 4 Y () t 4 4 For () 6in 4 X t t 8 ; f Hz () 6 X t f Hz 8 H( f) for f Hz Y () t 6 8 Y () t For () 4co 8 X t t 6 ; () 4 X t, f 4 6 H( f) for f 4 Hz Y () t 4 6 Y () t Hence Y() t Y() t Y() t Y() t Y() t Yt () 4 4 Note : Power of Yt ( ) 4cot 4 Aco Hence power of Yt ( ) t i A / 4 P Y 8Watt A econd order LTI ytem i decribed by the following tate equation d x () t x () t dt d x () t x () t x () t r () t dt More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

94 4 GATE 7 [Afternoon (Set - )] where () x () t are the two tate variable and () ct () x () t. The ytem i (A) Undamped (ocillatory) (B) Under damped (C) Critically damped (D) Over damped An. (D) Sol. Given : d x () t x () t dx () x () x () r () dx xx x x x r( t) x x rt () x x...(i) State equation i given by, x Ax Bu (ii) On comparing equation (i) and (ii), we get A Characteritic equation i given by, I A Img( j)...(i) ( )( ), Negative unequal real root repreent over-damped ytem. Alo, For a econd order ytem, characteritic equation i given by, nn...(ii) On comparing equation (i) and (ii), we get n and n.6 So, the ytem i over-damped. Hence, the correct option i (D). More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and Real ( )

95 4 GATE 7 [Afternoon (Set - )] Quetion 4 A unity feedback control ytem i characterized by the open-loop tranfer function K ( ) G () The Nyquit path and the correponding Nyquit plot of G () are hown in the figure below, j j j j R j Re -plane K j5.4 K j j jimg G ( )-plane K Re j Nyquit path for G ( ) j5.4 K Nyquit plot for G ( ) If K, then the number of pole of the cloed-loop tranfer function that lie in the right-half of the -plane i (A) (B) (C) (D) An. (C) Sol. Given : K( ) G () and H () Img j5.4 K G ( )-plane j j K K j K Re j5.4 K Number of encirclement about critical point ( j) in anticlockwie i given by, N P Z...(i) Where, P = Number of open loop pole in right half of -plane Z = Number of cloed loop pole in right half of -plane From figure, N = Characteritic equation i given by, More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

96 4 GATE 7 [Afternoon (Set - )] Routh Tabulation : From Routh table, two ign change in firt column i.e., two pole lie in the RHS of plane. P = From equation (i), we get Z Z (ytem i untable) Hence, the correct option i (C). Quetion 44 A unity feedback control ytem i characterized by the open-loop tranfer function ( ) G () K The value of K for which the ytem ocillate at rad/ec i. An. (.75) ( ) Sol. Given : G () K, H () Characteritic equation i given by, GH ( ) ( ) K K 4 Routh Tabulation : 4 K 4K K We will get row of zero when 4 K K.75 K 4 Becaue of row of zero, we form the auxiliary equation uing the coefficient of A () K K( ) row More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

97 4 GATE 7 [Afternoon (Set - )] At rad/ec, K( ).75 4 K Hence, the correct anwer i.75. Alternative method : The ytem i ocillating hence thi i a marginally table ytem. For marginally table ytem : Internal product = External product Internal product = 4K, external product = 4K K.75 4 For a third order ytem only, IP > EP : ytem i table. IP < EP : ytem i untable. IP = EP : ytem i marginal table. Quetion 45 The permittivity of water at optical frequencie i.75. It i found that an iotropic light ource at a ditance d under water form an illuminated circular area of radiu 5 m a hown in the figure. The critical angle i ( C ). Air 5 m Water C d Light ource The value of d (in m) i. An. (4.) Sol. Given : Critical angle i c. The mallet angle of incidence at which a light ray paing from one medium to another le refractive medium can be totally reflected from the boundary between the two. Air Water.75 O c 5m B d Normal A Point ource More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

98 44 GATE 7 [Afternoon (Set - )] The critical angle i given by, n c in n where, n refractive index of air c In Quetion 46 AOB, n refractive index of water c.75 in c c 49. c.75 5 tan c d 5 d 4.m tan 49. In the voltage reference circuit hown in the figure, the op-amp i ideal and the tranitor Q, Q...Q are identical in all repect and have infinitely large value of common-emitter current gain ( ). The collector current ( I C ) of the tranitor i related to their bae-emitter voltage ( V BE ) by the relation I C VBE IS exp V T, where I S i the aturation current. Aume that the voltage V F hown in the figure i.7 V and the thermal voltage VT 6mV. k k 5V 5k + 5V V out V p... Q Q Q Q An..7 The output voltage V out (in volt) i. More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

99 45 Sol. GATE 7 [Afternoon (Set - )] Given circuit i hown below, k k.7 V A 5V.7 V 5k I + 5V V out V.7 p Q Q Q Q Non-inverting voltage V.7 V (from figure) From virtual ground concept, V NI NI V.7 V I VI VP.7.7 Current through 5 kreitor I A 5 5 Apply KVL in loop hown in figure, V out.7 V out.7 V. Quetion 47 A MOS capacitor i fabricated on p-type Si (Silicon) where the metal work function i 4. ev and electron affinity of Si i 4. ev, EC EF.9 ev, where E C and EF are conduction band minimum and the Fermi 4 energy level of Si, repectively. Oxide layer permittivity.9, 8.85 F/cm. Oxide thickne t. m and electronic charge q ox More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and r 9.6 C. If the meaured flat band voltage of thi capacitor i V, then the magnitude of the fixed charge at the oxide-emiconductor interface, in (nc/cm ) i. An Sol. Given : Metal work function em 4. ev EVac Electron affinity e 4eV em EC EFP.9 ev E Fm e e.9 e 4.9 ev Metal m 4. V, 4.9 V Metal emiconductor work function i given by, m m V 4 ox Cox 4.55 F/cm 4 t. ox 9 e e.9 ev Semiconductor EVac E C E FP

100 46 GATE 7 [Afternoon (Set - )] Flat band voltage i given by, V FB Q m C ox Q.8. V C ox Fixed charge at the oxide-emiconductor interface i, Q 9. Cox C/cm Quetion 48 The witch in the circuit hown in the figure, wa open for a long time and i cloed at t. The current it ()(in Ampere) at t.5 ec i. it () A 5 t 5.5 H An. 8.6 Sol. (i) At t : A 5 5 I L ( ) I L ( ) 5 A From property of inductor, I ( ) I ( ) 5 A L L (ii) At t : Inductor behave a a current ource of initial value of 5 A. A A 5 i( ) 5.5 A Apply KCL at node A, i( ) 5 i( ) 5 V More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

101 47 GATE 7 [Afternoon (Set - )] (iii) Calculation of R th : Current ource replaced by O.C R th Time contant, (iv) At t : L.5 ec R 5 th A 5 i( ) A 5 Current it () i given by, t/ it () i( ) i( ) i( ) e /.5 it ( ) (5 ) e t At t =.5 : i(.5) 5e 8.6 A Quetion 49 If the vector function F aˆ ( ˆ ˆ x ykz) ay( kx z) az( kyz) i irrotational, then the value of the contant k, k and k repectively, are (A).,.5,.5 (B).,.,. (C).,.,.5 (D) 4.,.,. An. (B) Sol. Given : F ( yk ˆ ˆ ˆ z) ax ( kx z) ay az( ky) and F i.e. F i irrotational. F aˆ aˆ aˆ x y z x y z ( yk z) ( k x z) ( k y) F aˆ x ( ky) ( kx z) y z aˆ ( ˆ y ky) ( ykz) az ( kx z) ( ykz) x z x z F ( k ) aˆ aˆ ( k ) aˆ ( k ) x y z More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

102 48 GATE 7 [Afternoon (Set - )] For F to be zero, k, k and k Hence, the correct option i (B). Quetion 5 Conider a binary memory le channel characterized by the tranition probability diagram hown in the figure. An. Sol. The channel i (A) Lole (B) Noiele (C) Uele (D) Determinitic (C) Channel tranition matrix i given by,.5.75 PYX /.5.75 Lole channel : A channel decribed by a channel matrix with only one nonzero element in each column i called a lole channel. Example : PYX / 4 4 Determinitic channel : A channel decribed by a channel matrix with only one nonzero element in each row i called a determinitic channel. Example : PYX / Noiele channel : A channel i called noiele if it i both lole and determinitic More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

103 49 GATE 7 [Afternoon (Set - )] Example : PYX / So, the given channel i uele. Hence, the correct option i (C). Quetion 5 Standard air filled rectangular wave guide of dimenion a.9 cm and b. cm are deigned for radar application. It i deired that thee waveguide operate only in the dominant TE mode with operating frequency at leat 5% above the cut-off frequency of the TE mode but not higher than 95% of the next higher cut-off frequency. The range of allowable operating frequency f i (A) 8.9 GHz f. GHz (B) 8.9 GHz f.45 GHz (C) 6.55 GHz f. GHz (D).64 GHz f.4 GHz An. (B) Sol. Given : a.9cm and b.cm Cae : f f c 5% of f c 5 f f..(i) 4 c Cut-off frequency i given by, c m n fc a b For TE mode, m, n 8 c fc 6.55GHz a.9 From equation (i), operating frequency will be, f 8.9GHz 4 Cae : f 95% of f c f.95 f C For TE mode (next higher cut-off frequency), m, n 8 c fc.ghz a.9 9 f.95. f.44 GHz So, the range of operating frequency would be, 8.9GHz f.44ghz Hence, the correct option i (B). More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

104 5 GATE 7 [Afternoon (Set - )] Note : For TE, m, n 8 c fc 4.7 GHz b. For TE, m, n f c c a b Quetion 5 An. 8 f c 6GHz.9. So, next higher cut-off frequency we will get for TE mode. The minimum value of the function Sol. Given : f( x) xx f x in the interval x occur at x =. ( ) x( x ) in [, ] ( ) x x in [, ] f '( x) x x x x x, are the tationary point f ''( x) x f ''( ), (Maxima) f ''(), (Minima) For calculating minimum value of f ( x ) in,, we have to check value of f ( x ) at boundary and tationary point. f f () () (). f ( ) ( ) ( ).666 f () () ().666 Minimum value of function occur at x ( ) ( ) ( ). More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

105 5 GATE 7 [Afternoon (Set - )] Quetion 5 An. (.5) The tranfer function of a caual LTI ytem i H(). If the input to the ytem i where ut ( ) i a unit tep function, the ytem output y( t ) a t i. Sol. Given : in t x() t u() t t in t x() t u() t t, From property of Laplace tranform, LT in tu( t) LT f () t F() f() t LT Fd () t in t t LT ut () in t LT ut () t d d d X () tan X () tan Output of given LTI ytem, yt () xt ()* ht () Taking Laplace tranform, Y () XH () () From final value theorem, Y() tan More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and tan tan Y() lim yt ( ) lim Y( ) t lim.5 tan tan lim

106 5 GATE 7 [Afternoon (Set - )] Quetion 54 The ignal x( t) in(4 t), where t i in econd, i ampled at a rate of 9 ample per econd. The ample ignal i the input to an ideal low pa filter with frequency repone H( f ) a follow :, f khz H( f), f khz What i the number of inuoid in the output and their frequencie in khz? (A) Number of inuoid =, Frequency = 7 (B) Number of inuoid =, Frequencie =, 7, (C) Number of inuoid =, Frequencie =, 7 (D) Number of inuoid =, Frequencie = 7, An. (B) Sol. Given : x( t) in (4 t), f 9 ample/ec Cutoff frequency of LPF = khz. The frequency component preent in ampled ignal i given by, f, f f, f f, f f... m m f 7kHz f m f 9 7 6kHz and khz m f f 8 7 5kHz and khz m f f 7 7 4kHz and khz m m m The number of component preent at the output of LPF will be three and they are 7 khz, khz and khz. Hence, the correct option i (B). Quetion 55 5 An electron ( q ) i moving in free pace with velocity m/ toward a tationary electron ( q) far away. The cloet ditance that thi moving electron get to the tationary electron before the repulive force 8 divert it path i m. Given ma of electron Charge of electron An. 5.6 Sol. Given : e m 9. kg. 9.6 C and permittivity q 6 5 v m/ q 9 F/m Moving Ma of electron m 9. kg 9 Charge of electron e.6 C Velocity of electron of q v 5 () m/ec r Stationary More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

107 5 GATE 7 [Afternoon (Set - )] Initial kinetic energy (KE) initial mv qq Initial potential energy (PE) initial r Final kinetic energy 4 r mv (KE) final v qq Final potential energy (PE) final 4 r From energy conervation theorem, (KE) initial (PE) initial (KE) final (PE) final qq mv 4 r qq r 4 mv 9 q q e.6 9 (.6 ) 8.6 r r m More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

108 54 GATE 7 [Afternoon (Set - )].General Aptitude. Quetion The ninth and the tenth of thi month are Monday and Tueday. (A) Figuratively (B) Retropectively (C) Repectively (D) Rightfully An. (C) Quetion It i to read thi year textbook the lat year. (A) eaier, than (B) mot eay, than (C) eaier, from (D) eaiet, from An. (A) Quetion 5 tudent are taking one or more coure out of Chemitry, Phyic and Mathematic. Regitration record indicate coure enrolment a follow : Chemitry (9), Phyic (86), Mathematic (95), Chemitry and Phyic (8), Chemitry and Mathematic (7), and Phyic and Mathematic (6). How many tudent are taking all ubject? (A) 7 (B) 4 (C) 47 (D) 5 An. (D) Sol. 8 C 9 P M 95 PA ( BC) PA ( ) PB ( ) PC ( ) PA ( B) PA ( C) PB ( C) PA ( B C) PA ( BC) x 5 x Quetion 4 Fatima tart from point P, goe North for km and then Eat for 4 km to reach point Q. She then turn to face point P and goe 5 km in that direction. She then goe North for 6 km. How far i he from point P, and in which direction hould he go to reach point P? (A) 8 km, Eat (B) km, North (C) 6 km, Eat (D) km, North An. (A) Sol. 4 km 8 km km P 5 km 6 km km More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

109 55 GATE 7 [Afternoon (Set - )] Quetion 5 A rule tate that in order to drink beer, one mut be over 8 year old. In a bar, there are 4 people P i 6 year old, Q i 5 year old, R i drinking milkhake and S i drinking a bear. What mut be checked to enure that the rule i being followed? (A) Only P ' drink (B) Only P ' drink and S ' age (C) Only S ' age (D) Only Pdrink, ' Q ' drink and S ' age An. (B) Sol. Since age of P i le than 8 year, o we have to check of P and ince S i drinking a bear o we have to check age of S for Rule i being flowed. Quetion 6 Each of P, Q, R, S, W, X, Y and Z ha been married at mot once. X and Y married and have two children P and Q, Z i grandfather of the daughter S of P. Further Z and W married and are parent of R. Which one of the following mut necearily be FALSE? (A) X i the mother in law to R (B) P and R are not married to each other (C) P i on of X and Y (D) Q cannot be married to R An. (B) Sol. Z X Y R P Q W S = = Marry Son or daughter (i) X i mother in law to R. Therefore option (A) i true. (ii) P and R can marry. Therefore option (B) i fale. (iii) Q and R cannot marry. Therefore option (C) i true. (iv) P i on of X and Y. Therefore option (D) i true. Quetion 7 The number of digit number uch that the digit i never to the immediate right of i (A) 78 (B) 79 (C) 88 (D) 89 An. (C) Sol. Total number of digit number 9 to999 Number which have on immediate right of,,...,9 number,,, number and So, number of digit number uch that digit i never on the immediate right of More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

110 56 GATE 7 [Afternoon (Set - )] Quetion 8 men and 5 women can build a bridge in week. 9 men and 5 women will take week to build the ame bridge. How many men will be needed to build the bridge in one week? An. (A) (B) (C) 6 (D) 9 (C) Sol. Given : men 5 women week 9 men 5 women week Let, x men can finih entire work week y women can finih entire work week 5 x y.(i) Solving equation (i) and (ii), 9 5 x y (ii) x6, y Hence the correct option i (C). Quetion 9 A contour line join location having the ame height above the mean ea level. The following i a contour plot of a geographical region. Contour line are hown at 5 m interval in thi plot. R Q P S T An. km Which of the following i the teepet path leaving from P? (A) P to Q (B) P to R (C) P to S (D) P to T (B) More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

111 57 GATE 7 [Afternoon (Set - )] Quetion If you are looking for a hitory of India, or for an account of the rie and fall of the Britih Raj, or for the reaon of the cleaving of the ubcontinent into two mutually antagonitic part and the effect thi mutilation will have in the repective ection, and ultimately on Aia, you will not find it in thee page for though I have pent a lifetime in the country, I lived too near the eat of event, and wa too intimately aociated with the actor, to get the perpective needed for the impartial recording of thee matter. Which of the following tatement bet reflect the author opinion? (A) An intimate aociation doe not allow for the neceary perpective. (B) Matter are recorded with an impartial perpective. (C) An intimate aociation offer an impartial perpective. (D) Actor are typically aociated with the impartial recording of matter. An. (C) More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and

112 Dear tudent, We have tried to provide the GATE 7 quetion along with their olution. However, if you find any dicrepancy, pleae write to u at info@gateacademy.co.in GATE ACADEMY i not reponible for any error due to data miprint/human error/data inufficiency etc. More quetion and olution of GATE - 7 are available in the Facebook group GATE/ESE/PSU concept and problem for EC/EE/IN to join For more GATE 7 Subject wie Video Tutorial Subcribe