mywbut.com GATE SOLVED PAPER - EC Q.1 to Q.25 carry one mark each.

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1 GATE OLVED PAPE - EC 3 Q. to Q.5 carry one mark each. Q. A bulb in a taircae ha two witche, one witch being at the ground floor and the other one at the firt floor. The bulb can be turned ON and alo can be turned OFF by any one of the witche irrepective of the tate of the other witch. The logic of witching of the bulb reemble (A) and AND gate (B) an O gate (C) an XO gate (D) a NAND gate ol. Let A denote the poition of witch at ground floor and B denote the poition of witch at upper floor. The witch can be either in up poition or down poition. Following are the truth table given for different combination of A and B A B Y(Bulb) up() up() OFF() Down() Down() OFF() up() Down() ON() Down() up() ON() hen the witche A and B are both up or both down, output will be zero (i.e. Bulb will be OFF). Any of the witch change it poition lead to the ON tate of bulb. Hence, from the truth table, we get Y A5 B i.e., the XO gate Hence correct option i (C) Q. Conider a vector field Av ^rv h. The cloed loop line integral # Av : dlv can be expreed a (A) ## ^d # A h: dv over the cloed urface bounded by the loop (B) ### ^d : Adv h over the cloed volume bounded by the loop (C) ### ^d : Adv h over the open volume bounded by the loop (D) ## ^ d # A h: dv over the open urface bounded by the loop ol. toke theorem tate that the circulation a vector field A v around a cloed path l i equal to the urface integral of the curl of A v over the open urface bounded by l. i.e., # Av : dlv ## ^d # Av h: dv Here, line integral i taken acro a cloed path which i denoted by a mall circle on the integral notation where a, the urface integral of ^d # A v h i taken over open urface bounded by the loop. Hence correct option i (D)

2 GATE OLVED PAPE - EC 3 Q. 3 Two ytem with impule repone h^th and h^th are connected in cacade. Then the overall impule repone of the cacaded ytem i given by (A) product of h^th and h^th (B) um of h^th and h^th (C) convolution of h^th and h^th (D) ubtraction of h^th from h^th ol. 3 If the two ytem with impule repone h^th and h^th are connected in cacaded configuration a hown in figure, then the overall repone of the ytem i the convolution of the individual impule repone. Hence correct option i (C) Q. 4 In a forward biaed pn junction diode, the equence of event that bet decribe the mechanim of current flow i (A) injection, and ubequent diffuion and recombination of minority carrier (B) injection, and ubequent drift and generation of minority carrier (C) extraction, and ubequent diffuion and generation of minority carrier (D) extraction, and ubequent drift and recombination of minority carrier ol. 4 The potential barrier of the pn junction i lowered when a forward bia voltage i applied, allowing electron and hole to flow acro the pace charge region (Injection) when hole flow from the p region acro the pace charge region into the n region, they become exce minority carrier hole and are ubject to diffue, drift and recombination procee. Hence correct option i (A) Q. 5 In IC technology, dry oxidation (uing dry oxygen) a compared to wet oxidation (uing team or water vapor) produce (A) uperior quality oxide with a higher growth rate (B) inferior quality oxide with a higher growth rate (C) inferior quality oxide with a lower growth rate (D) uperior quality oxide with a lower growth rate ol. 5 In IC technology, dry oxidation a compared to wet oxidation produce uperior quality oxide with a lower growth rate Hence correct option i (D) Q. 6 The maximum value of q until which the approximation in q. q hold to within % error i (A) c (B) 8c (C) 5c (D) 9c ol. 6 Here, a we know Lim in q. q " but for % error, we can check option (B) firt,

3 GATE OLVED PAPE - EC 3 q 8 c 8 p c#. 34 8c in q in 8c. 39 % error %. %. 39 # 49 Now, we check it for q 5c q 5c 5 p c# c in q in 5c. 77 % error %. 873 o, the error i more than %. Hence, for error le than %, q 8c can have the approximation in q. q Hence correct option i (B) Q. 7 The divergence of the vector field Av xatx+ yaty+ zat z i (A) (B) /3 (C) (D) 3 ol. 7 Given, the vector field A v xavx+ yavy+ zavz o, d $ A v (Divergence of A v ) A A + + A x y z Hence correct option i (D) x y z Q. 8 The impule repone of a ytem i h^th tut ^ h. For an input u^t- h, the output i (A) t ut ^ h (B) tt ^ - h ut- ^ h (C) t ^ - h ut- ^ h (D) t - ut ^ - h ol. 8 Given, the input xt ^ h ut ^ -h It laplace tranform i X ^ h e - The impule repone of ytem i given ht ^ h tu^th It Laplace tranform i H ^ h Hence, the overall repone at the output i Y ^ h XH ^ h ^ h e - 3 it invere laplace tranform i

4 GATE OLVED PAPE - EC 3 ^t yt ^ h - h ut- ^ h Hence correct option i (C) Q. 9 The Bode plot of a tranfer function G^h i hown in the figure below. ol. 9 The gain _ log G ^ h i i 3 db and - 8dB at rad/ and rad/ repectively. The phae i negative for all w. Then G^h i (A) (B) (C) 3 From the given plot, we obtain the lope a log G- log G lope log w - log w (D) 3 From the figure log G -8dB log G 3 db and w rad/ w rad/ o, the lope i lope -8-3 log - log -4 db/ decade Therefore, the tranfer function can be given a G ^ h k at w Gjw ^ h k k w In decibel, log Gj ^ wh log k 3 3 or, k Hence, the Tranfer function i G. ^ h k 39 8 Hence correct option i (B)

5 GATE OLVED PAPE - EC 3 Q. In the circuit hown below what i the output voltage ^V out h if a ilicon tranitor Q and an ideal op-amp are ued? ol. (A) (C) - 5 V (B) -.7 V +.7 V (D) + 5 V For the given ideal op-amp, negative terminal will be alo ground (at zero voltage) and o, the collector terminal of the BJT will be at zero voltage. i.e., V C volt The current in k reitor i given by I 5-5mA k Thi current will flow completely through the BJT ince, no current will flow into the ideal op-amp ( I/ P reitance of ideal op-amp i infinity). o, for BJT we have V C V B I C 5mA i.e.,the bae collector junction i revere biaed (zero voltage) therefore, the collector current (I C ) can have a value only if bae-emitter i forward biaed. Hence, V BE.7 volt & VB- VE 7. & - Vout 7. or, V out -.7 volt Hence correct option i (B) Q. Conider a delta connection of reitor and it equivalent tar connection a hown below. If all element of the delta connection are caled by a factor k, k >, the element of the correponding tar equivalent will be caled by a factor of (A) k (B) k (C) /k (D) k ol. In the equivalent tar connection, the reitance can be given a b a C + + a b c

6 GATE OLVED PAPE - EC 3 a c B a+ b+ c b c A a+ b+ c o, if the delta connection component a, b and c are caled by a factor k then ^kbh^kch A l ka+ kb+ kc k b c k a+ b+ c k A hence, it i alo caled by a factor k Hence correct option i (B) Q. For 885 microproceor, the following program i executed. MVI A, 5H; MVI B, 5H; PT: ADD B; DC B; JNZ PT; ADI 3H; HLT; At the end of program, accumulator contain (A) 7H (B) H (C) 3H (D) 5H ol. The program i being executed a follow MVI A,.5H; A 5H MVI B,.5H; B 5H At the next intruction, a loop i being introduced in which for the intruction DC B if the reult i zero then it exit from loop o, the loop i executed five time a follow : Content in B Output of ADD B (tored value at A) ytem i out of loop i.e., A At thi tage, the 885 microproceor exit from the loop and read the next intruction. i.e., the accumulator i being added to 3 H. Hence, we obtain A A + 3 H H Hence correct option i (A)

7 GATE OLVED PAPE - EC 3 Q. 3 The bit rate of a digital communication ytem i kbit/. The modulation ued i 3-QAM. The minimum bandwidth required for II free tranmiion i (A) / Hz (B) / khz (C) /5 Hz (D) /5 khz ol. 3 In ideal Nyquit Channel, bandwidth required for II (Inter ymbol reference) free tranmiion i b Here, the ued modulation i 3 - QAM (Quantum Amplitude modulation i.e., q 3 or v 3 v 5bit o, the ignaling rate (ampling rate) i b ( " given bit rate) 5 Hence, for II free tranmiion, minimum bandwidth i b khz Hence correct option i (B) Q. 4 For a periodic ignal vt ^ h 3 in t+ co 3t+ 6 in^5t+p/ 4h, the fundamental frequency in rad/ (A) (B) 3 (C) 5 (D) 5 ol. 4 Given, the ignal vt ^ h 3 in t+ co 3t+ 6 in^5t+ p 4 h o we have w rad/ w 3 rad/ w 3 5 rad/ Therefore, the repective time period are T ec w p p T ec w p 3 p T 3 p ec 5 o, the fundamental time period of the ignal i LCM^ p, p,ph L.C.M. ^T, TT3h HCF^, 3, 5h or, T p Hence, the fundamental frequency in rad/ec i w p rad/ Hence correct option i (A)

8 GATE OLVED PAPE - EC 3 Q. 5 In a voltage-voltage feedback a hown below, which one of the following tatement i TUE if the gain k i increaed? (A) The input impedance increae and output impedance decreae (B) The input impedance increae and output impedance alo increae (C) The input impedance decreae and output impedance alo decreae (D) The input impedance decreae and output impedance increae ol. 5 The i/ p voltage of the ytem i given a V in V + V f V + kv out V+ k AV ^Vout AVh V^ + k Ah Therefore, if k i increaed then input voltage i alo increaed o, the input impedance increae. Now, we have V out AV A Vin ^ + kah AV in ^ + ka h ince, V in i independent of k when een from output mode, the output voltage decreae with increae in k that lead to the decreae of output impedance. Thu, input impedance increae and output impedance decreae. Hence correct option i (A) Q. 6 A band-limited ignal with a maximum frequency of 5kHz i to be ampled. According to the ampling theorem, the ampling frequency which i not valid i (A) 5kHz (B) khz (C) 5 khz (D) khz ol. 6 Given, the maximum frequency of the band-limited ignal f m 5kHz According to the Nyquit ampling theorem, the ampling frequency mut be greater than the Nyquit frequency which i given a f N f m # 5 khz o, the ampling frequency f mut atify f $ f N $ khz f

9 GATE OLVED PAPE - EC 3 only the option (A) doe nt atify the condition therefore, 5kHz i not a valid ampling frequency. Hence correct option i (A) Q. 7 In a MOFET operating in the aturation region, the channel length modulation effect caue (A) an increae in the gate-ource capacitance (B) a decreae in the tranconductance (C) a decreae in the unity-gain cutoff frequency (D) a decreae in the output reitance ol. 7 In a MOFET operating in the aturation region, the channel length modulation effect caue a decreae in output reitance. Hence correct option i (D) Q. 8 hich one of the following tatement i NOT TUE for a continuou time caual and table LTI ytem? (A) All the pole of the ytem mut lie on the left ide of the jw axi (B) Zero of the ytem can lie anywhere in the -plane (C) All the pole mut lie within (D) All the root of the characteritic equation mut be located on the left ide of the jw axi. ol. 8 For a ytem to be caual, the.o.c of ytem tranfer function H^h which i rational hould be in the right half plane and to the right of the right mot pole. For the tability of LTI ytem. All pole of the ytem hould lie in the left half of -plane and no repeated pole hould be on imaginary axi. Hence, option (A), (B), (D) atifie an LTI ytem tability and cauality both. But, Option (C) i not true for the table ytem a, have one pole in right hand plane alo. Hence correct option i (C) Q. 9 The minimum eigen value of the following matrix i 3 5 V T X (A) (B) (C) (D) 3 ol. 9 For, a given matrix 6 A@ the eigen value i calculated a A- li where l give the eigen value of matrix. Here, the minimum eigen value among the given option i l e check the characteritic equation of matrix for thi eigen value A- li A (for l ) ^6-49h-5^5-4h+ ^35-4h

10 GATE OLVED PAPE - EC Hence, it atified the characteritic equation and o, the minimum eigen value i l Hence correct option i (A) Q. A polynomial fx () ax ax ax + ax - a with all coefficient poitive ha (A) no real root (B) no negative real root (C) odd number of real root (D) at leat one poitive and one negative real root ol. Given, the polynomial fx ^ h ax ax ax + ax -a ince, all the coefficient are poitive o, the root of equation i given by fx ^ h It will have at leat one pole in right hand plane a there will be leat one ign change from ^a h to ^a h in the outh matrix t column. Alo, there will be a correponding pole in left hand plane i.e.; at leat one poitive root (in.h.p) and at leat one negative root (in L.H.P) et of the root will be either on imaginary axi or in L.H.P Hence correct option i (D) Q. Auming zero initial condition, the repone y^th of the ytem given below to a unit tep input u^th i ol. (A) u^th (B) tu^t h (C) t ut ^ h (D) e - t u^th The Laplace tranform of unit tep fun n i U ^ h o, the O/P of the ytem i given a Y ^ h b lb l For zero initial condition, we check dy t ut ^ h ^ h dt & U ^ h Y^ h-y^h & U ^ h c y m - ^ h or, U ^ h ^y^h h Hence, the O/P i correct which i

11 GATE OLVED PAPE - EC 3 Y ^ h it invere Laplace tranform i given by yt ^ h tu^t h Hence correct option i (B) V ^h Q. The tranfer function of the circuit hown below i V ^ h (A) 5 ṡ + + (B) (C) + + (D) + + ol. For the given capacitance, C mf in the circuit, we have the reactance. X C c -6 # # 4 o, 4 4 V ^h + V ^ 4 4 h Hence correct option i (D) Q. 3 A ource v ^th Vco pt ha an internal impedance of ^4+ j3h. If a purely reitive load connected to thi ource ha to extract the maximum power out of the ource, it value in hould be (A) 3 (B) 4 (C) 5 (D) 7 ol. 3 For the purely reitive load, maximum average power i tranferred when L Th + XTh where Th + jxth i the equivalent thevinin (input) impedance of the circuit. Hence, we obtain L Hence correct option i (C) Q. 4 The return lo of a device i found to be db. The voltage tanding wave ratio (V) and magnitude of reflection coefficient are repectively (A). and. (B).8 and. (C). and. (D).44 and.

12 GATE OLVED PAPE - EC 3 ol. 4 Given, the return lo of device a db i.e., G - db (lo) ^ h in db or, log G - - & G. Therefore, the tanding wave ration i given by + G V - G Hence correct option i (A) Q. 5 Let g^th e -pt, and h^th i a filter matched to g^th. If g^th i applied a input to h^th, then the Fourier tranform of the output i ol. 5 (A) e (C) e -p f -p f (B) e (D) e -pf / The matched filter i characterized by a frequency repone that i given a H^f h G* ^fhexp^-jpfth f where gt ^ h Gf ^ h Now, conider a filter matched to a known ignal g^th. The fourier tranform of the reulting matched filter output g^th will be G^fh Hf ^ hgf ^ h -pf G* ^fhg^fhexp^-jpfth Gf ^ h exp^-jpfth T i duration of g^th Aume exp^- jpfth o, G^fh G_ fi ince, the given Gauian function i gt ^ h e -pt Fourier tranform of thi ignal will be gt ^ h e -pt f e -pf Gf ^ h Therefore, output of the matched filter i G^fh e f -p Hence No Option i correct Q.6 to Q.55 carry two mark each. Q. 6 Let U and V be two independent zero mean Gauain random variable of variance and repectively. The probability P 3V F U 4 9 ^ h i (A) 4/9 (B) / (C) /3 (D) 5/9 ol. 6 Given, random variable U and V with mean zero and variance 4 and 9 i.e., U V u 4

13 GATE OLVED PAPE - EC 3 and v 9 o, PU ^ $ h and PV ^ $ h The ditribution i hown in the figure below fu ^uh e p -u u u fv ^vh -v e v p v e can expre the ditribution in tandard form by auming X u Y u U - u and Y v Y v 3V - v 3 for which we have X U Y V and X 4U alo, Y 9V Therefore, X- Y i alo a normal random variable with X- Y Hence, PX ^ - Y$ h PX Y# ^ - h or, we can ay P^U- 3V # h Thu, P^3V $ Uh Hence correct option i (B) Q. 7 Let A be an m# n matrix and B an n# m matrix. It i given that determinant ^Im + ABh determinant ^In + BAh, where I k i the k# k identity matrix. Uing the above property, the determinant of the matrix given below i V T X (A) (B) 5 (C) 8 (D) 6

14 GATE OLVED PAPE - EC 3 ol. 7 Conider the given matrix be I AB m + T V X where m 4 o, we obtain AB - T T V X V X T V X T V X Hence, we get A T V X, B 8 B Therefore, BA 8 B T V X 4 From the given property Det I AB m + ^ h Det I BA m + ^ h & Det T V X Det +4 T V X Z [ \ ] ] ] ] _ ` a b b b b Note : Determinant of identity matrix i alway. Hence correct option i (B) Q. 8 In the circuit hown below, if the ource voltage 53.3 V V c + then the Thevenin equivalent voltage in Volt a een by the load reitance L i

15 GATE OLVED PAPE - EC 3 (A) + 9c (B) 8+ c (C) 8+ 9c (D) + 6c ol. 8 For evaluating the equivalent thevenin voltage een by the load L, we open the circuit acro it (alo if it conit dependent ource). The equivalent circuit i hown below A the circuit open acro L o I or, j4i i.e., the dependent ource in loop i hort circuited. Therefore, ^j4hv V L j4 + 3 V Th j4 V L 53.3c j c Hence correct option i (C) 4 9c 53.3c c w^ h Q. 9 The open-loop tranfer function of a dc motor i given a. hen V a ^ h + connected in feedback a hown below, the approximate value of K a that will reduce the time contant of the cloed loop ytem by one hundred time a compared to that of the open-loop ytem i (A) (B) 5 (C) (D) ol. 9 Given, open loop tranfer function G ^ h K a K a + + By taking invere Laplace tranform, we have gt ^ h e - t Comparing with tandard form of tranfer function, Ae - t/ t, we get the open loop time contant, t ol

16 GATE OLVED PAPE - EC 3 Now, we obtain the cloed loop tranfer function for the given ytem a G ^ h H Ka ^ h + G ^ h + + Ka Ka + ^ Ka + h By taking invere laplace tranform, we get - ^ka + t h ht ^ h ka. e o, the time contant of cloed loop ytem i obtained a t cl ka + or, t cl (approximately) ka Now, given that k a reduce open loop time contant by a factor of. i.e., t cl t ol or, ka Hence, k a Hence correct option i (C) Q. 3 In the circuit hown below, the knee current of the ideal Zener dioide i ma. To maintain 5V acro L, the minimum value of L in and the minimum power rating of the Zener diode in m, repectively, are ol. 3 (A) 5 and 5 (B) 5 and 5 (C) 5 and 5 (D) 5 and 5 From the circuit, we have I IZ + IL or, I Z I- IL () ince, voltage acro zener diode i 5V o, current through reitor i obtained a

17 GATE OLVED PAPE - EC 3 I 5.5 A - Therefore, the load current i given by I L 5 L ince, for proper operation, we mut have I Z $ I kne o, from Eq. (), we write.5 A - 5 $ ma L 5 ma - 5 $ ma L 4 ma $ 5 L 4 # -3 $ 5 L 4 # - 3 L # # 4 # - L or, 5 # L Therefore, minimum value of L 5 Now, we know that power rating of Zener diode i given by P VI Z Z ^ max h I Z^maxh i maximum current through zener diode in revere bia. Maximum currrent through zener diode flow when load current i zero. i.e., I Z^maxh I Therefore, P 5#.5 5 m Hence correct option i (B) Q. 3 The following arrangement conit of an ideal tranformer and an attenuator which attenuate by a factor of.8. An ac voltage V X V i applied acro X to get an open circuit voltage V YZ acro YZ. Next, an ac voltage V YZ V i applied acro YZ to get an open circuit voltage V X acro X. Then, V / VX, V / VYZ are repectively, YZ X (A) 5/ and 8/ (B) / and 8/ (C) / and / (D) 8/ and 8/ ol. 3 For the given tranformer, we have V V 5. X

18 GATE OLVED PAPE - EC 3 ince, V YZ 8. (attenuation factor) V o, V V YZ ^8. h^5. h X or, V YZ V X at VYZ V X V; V X at VX V Z V; V YZ Hence correct option i (C) Q. 3 Two magnetically uncoupled inductive coil have Q factor q and q at the choen operating frequency. Their repective reitance are and. hen connected in erie, their effective Q factor at the ame operating frequency i (A) q+ q (B) ^/ qh+ ^/ qh (C) ^q + q h/ ^+ h (D) ^q + q h/ ^+ h ol. 3 The quality factor of the inductance are given by q L w and q L w o, in erie circuit, the effective quality factor i given by XLeq Q L L w w eq + + w L L + w + q q + + q + q + Hence correct option i (C) Q. 33 The impule repone of a continuou time ytem i given by h^th d^t- h+ d^t-3h. The value of the tep repone at t i (A) (B) (C) (D) 3 ol. 33 Given, the impule repone of continuou time ytem ht ^ h d^t- h+ d^t-3h From the convolution property, we know xt ^ h* d^t- t h xt ^ -t h o, for the input

19 GATE OLVED PAPE - EC 3 xt ^ h ut ^ h (Unit tep fun n ) The output of the ytem i obtained a yt ^ h ut ^ h* ht ^ h ut ^ h* 6 d^t- h+ d^t-3h@ ut ^ - h+ ut ^ -3h at t y^h u^- h+ u^-3h Hence correct option i (B) Q. 34 The mall-ignal reitance (i.e., dvb/ did) in k offered by the n-channel MOFET M hown in the figure below, at a bia point of V B V i (device data for ' M: device tranconductance parameter kn mncx^/ Lh 4 mav /, threhold voltage V TN V, and neglect body effect and channel length modulation effect) ol. 34 (A).5 (B) 5 (C) 5 (D) Given, V B V V TN V o, we have Drain voltage V D volt V G volt V (Ground) Therefore, V G > V TN and V D > V G -V TN o, the MOFET i in the aturation region. Therefore, drain current i I D kn^vg -VTNh or, I D kn ^V B - h Differentiating both ide with repect to I D k V dvb N^ B-h did ince, V BQ volt (at D.C. Voltage) Hence, we obtain dvb did kn^vb- h -6 # 4# # ^- h.5 # 3.5 k Hence correct option i (A)

20 GATE OLVED PAPE - EC 3 Q. 35 The ac chematic of an NMO common-ource tate i hown in the figure below, where part of the biaing circuit ha been omitted for implicity. For the n -channel MOFET M, the tranconductance g m ma/ V, and body effect and channel length modulation effect are to be neglected. The lower cutoff frequency in HZ of the circuit i approximately at (A) 8 (B) 3 (C) 5 (D) ol. 35 For the given circuit, we obtain the mall ignal model a hown in figure below : e obtain the node voltage at V a V V + + gmvi D L + C & V D -gmvi + L + C Therefore, the output voltage V i obtained a V V L L + C L J -gmvi N K O L + C + D K L + O L C P o, the tranfer function i V -DLCgm Vi + C^D + Lh Then, we have the pole at w C ^ D+ Lh It give the lower cutoff frequency of tranfer function. i.e., w C ^ D+ Lh or, f pc ^ D+ Lh

21 GATE OLVED PAPE - EC 3 - p # # # Hz Hence correct option i (A) 6 3 dy dy Q. 36 A ytem decribed by the differential equation 5 6yt xt dt + + dt ^ h ^ h. Let xt ^ h be a rectangular pule given by < t < xt ^ h * otherwie ol. 36 dy Auming that y^h and - dt (A) e ^ + h^ + 3h - (C) e ^ + h^ + 3h Given, the differential equation at t, the Laplace tranform of y^th i (B) - e - ^ + h^ + 3h (D) dy dy y t dt dt ^ h xt ^ h - - e ^+ h^+ 3h Taking it Laplace tranform with zero initial condition, we have Y ^ h+ 5Y ^ h+ 6Y ^ h X ^ h...() Now, the input ignal i < t < xt ^ h * otherwie i.e., xt ^ h ut ^ h-ut ^ -h Taking it Laplace tranform, we obtain X ^ h - e - e - - ubtituting it in equation (), we get X ^ h Y ^ h e ^ h - e - ^ + h^+ 3h Hence correct option i (B) Q. 37 A ytem decribed by a linear, contant coefficient, ordinary, firt order differential equation ha an exact olution given by y^th for t >, when the forcing function i x^th and the initial condition i y^h. If one wihe to modify the ytem o that the olution become -yt ^ h for t >, we need to (A) change the initial condition to -y^h and the forcing function to xt ^ h (B) change the initial condition to y^h and the forcing function to -xt ^ h (C) change the initial condition to j y^h and the forcing function to j x^th (D) change the initial condition to -y^h and the forcing function to -xt ^ h

22 GATE OLVED PAPE - EC 3 ol. 37 The olution of a ytem decribed by a linear, contant coefficient, ordinary, firt order differential equation with forcing function x^th i y^th o, we can define a function relating x^th and y^th a below dy P + Qy + K xt dt ^ h where P, Q, K are contant. Taking the Laplace tranform both the ide, we get PY^h- Py^h+ QY^h X ^ h...() Now, the olution become y^th -y^th or, Y^h -Y ^ h o, Eq. () change to PY^h- Py^h+ QY^h X^h or, -PY^ h-p y^h-qy^ h X^h...() Comparing Eq. () and (), we conclude that X^h -X ^ h y^h -y^h hich make the two equation to be ame. Hence, we require to change the initial condition to -y^h and the forcing equation to -xt ^ h Hence correct option i (D) Q. 38 Conider two identically ditributed zero-mean random variable U and V. Let the cumulative ditribution function of U and V be F^xh and G^xh repectively. Then, for all value of x (A) F^xh- Gx ^ h # (B) F^xh- Gx ^ h$ (C) ^Fx ()-Gx (). h x# (D) ^Fx ()-Gx (). h x$ ol. 38 The mean of random variable U and V are both zero i.e., U V Alo, the random variable are identical i.e., fu ^uh fv ^vh or, FU ^uh FV ^vh i.e., their cdf are alo ame. o, FU ^uh FV ^vh i.e., the cdf of random variable V will be alo ame but for any intant V $ U Therefore, Gx ^ h Fx ^ h but, xg^xh $ xf^x h or, 6 Fx ^ h- Gx ^ h@ x# Hence correct option i (C) Q. 39 The DFT of a vector 8a b c db i the vector 8a b g db. Conider the product V a b c d a b 8pqrB 8a b c db c d a b c d T The DFT of the vector 8pqrB i a caled verion of d c b a X *hipping Free* Buy Online all GATE Book: hop.nodia.co.in *Maximum Dicount*

23 GATE OLVED PAPE - EC 3 ol. 39 (A) 9a b g dc (B) 9 a b g dc (C) 8a+ b b+ d d+ g g+ ab (D) 8a b g db Given, the DFT of vector 8a b c db a DFT...% 8a b c db/ 8a b g db Alo, we have V a b c d d a b c 8pqrB 8a b c db c d a b...() b c d a T X For matrix circular convolution, we know h h h V x V xn hn h h hx h h hx T XT X where " x, x, x, are three point ignal for x6@ n and imilarly for h6@, n h, h and h are three point ignal. Comparing thi tranformation to Eq(), we get VT a d c b a d 6 pqr@ a b c d c b a 8 B d c b T X T T 6a b c d@ * 6a b c d@ V V a a b b * c c d d T X T X Now, we know that o, x6n@ * x6n@ XDFT 6k@ X, DFT 6k@ V a b c d T X * V V a a b b c g d d T X T X * V a b g d T X 9a b g d C Hence correct option i (A) Y ^ h Q. 4 The ignal flow graph for a ytem i given below. The tranfer function for U ^ h thi ytem i

24 GATE OLVED PAPE - EC 3 ol. 4 (A) + (B) (C) + (D) For the given FG, we have two forward path P k ^h^ h^ h^h - - P k ^h^ h^h^h ince, all the loop are touching to the path P k and P k o, Now, we have Here, the loop are D k Dk D - (um of individual loop) + (um of product of nontouching loop) L ^- 4h^h L ^- 4h^ h L 3 ^- h^ h^ h- - - L 4 ^- h^ h^h- A all the loop L, L, L3 and L 4 are touching to each other o, D - ^L+ L+ L3+ L4h -^ From Maon gain formulae Y ^ h P k D k U ^ h D Hence correct option i (A) Q. 4 In the circuit hown below the op-amp are ideal. Then, V out in Volt i h (A) 4 (B) 6 (C) 8 (D)

25 GATE OLVED PAPE - EC 3 ol. 4 For the given ideal op-amp we can aume - V V V (ideal) + V V V (ideal) o, by voltage diviion Vout # V V out V and, a the I/P current in Op-amp i alway zero therefore, there will be no voltage drop acro K in II op-amp i.e., V V Therefore, V- V V -- ^ h & V - + or, V 4 Hence, V out V 8 volt Hence correct option i (C) Q. 4 In the circuit hown below, Q ha negligible collector-to-emitter aturation voltage and the diode drop negligible voltage acro it under forward bia. If V cc i + 5V, X and Y are digital ignal with V a logic and V cc a logic, then the Boolean expreion for Z i (A) XY (C) XY (B) XY (D) XY

26 GATE OLVED PAPE - EC 3 ol. 4 For the given circuit, we can make the truth table a below X Y Z Logic mean voltage i v volt and logic mean voltage i 5 volt For x, y, Tranitor i at cut off mode and diode i forward biaed. ince, there i no drop acro forward biaed diode. o, Z Y For x, y, Again Tranitor i in cutoff mode, and diode i forward biaed. with no current flowing through reitor. o, Z Y For x, y, Tranitor i in aturation mode and o, z directly connected to ground irrepective of any value of Y. i.e., Z (ground) imilarly for X Y Z (ground) Hence, from the obtained truth table, we get Z XY Hence correct option i (B) Q. 43 A voltage in w t Volt i applied acro YZ. Auming ideal diode, the voltage meaured acro X in Volt, i ol. 43 (A) in w t (B) _ in wt+ in wt i/ (C) ^in wt- in wt h / (D) for all t Given, the input voltage V YZ in wt

27 GATE OLVED PAPE - EC 3 For + ve half cycle V YZ > i.e., V Y i a higher voltage than V Z o, the diode will be in cutoff region. Therefore, there will no voltage difference between X and node. i.e., V X Now, for - ve half cycle all the four diode will active and o, X and terminal i hort circuited i.e., V X Hence, V X for all t Hence correct option i (D) Q. 44 Three capacitor C, C and C 3 whoe value are m F, 5m F, and m F repectively, have breakdown voltage of V, 5V and V repectively. For the interconnection hown below, the maximum afe voltage in Volt that can be applied acro the combination, and the correponding total charge in m C tored in the effective capacitance acro the terminal are repectively, ol. 44 (A).8 and 36 (B) 7 and 9 (C).8 and 3 (D) 7 and 8 Conider that the voltage acro the three capacitor C, C and C 3 are V, V and V 3 repectively. o, we can write V C3...() V3 C ince, Voltage i inverely proportional to capacitance Now, given that C mf ; ^V h max V C 5 mf ; ^V h max 5V C 3 mf ; ^V 3 h max V o, from Eq () we have V V3 5 for ^V 3h max e obtain, V #.8 volt < 5 5

28 GATE OLVED PAPE - EC 3 i.e., V < ^V hmax Hence, thi i the voltage at C. Therefore, V 3 volt V.8 volt and V V+ V3.8 volt Now, equivalent capacitance acro the terminal i C CC 3 eq + C C + C3 5# m F 7 Equivalent voltage i (max. value) V max V 8. o, charge tored in the effective capacitance i Q C eq V max 8 b 8. 7 l # ^ h 3 mc Hence correct option i (C) Q. 45 There are four chip each of 4 byte connected to a 6 bit addre bu a hown in the figure below, AM,, 3 and 4 repectively are mappped to addree (A) CH-FFFH, CH-FFFH, CH-FFFH, 3CH-3FFFH (B) 8H-FFFH, 8H-FFFH, 38H-3FFFH, 48H-4FFFH (C) 5H-8FFH, 5H-8FFH, 35H-38FFH, 55H-58FFH (D) 8H-BFFH, 8H-BFFH, 8H-BFFH, 38H-3BFFH ol. 45 For chip-, we have the following concluion: it i enable when (i)

29 GATE OLVED PAPE - EC 3 and (ii) Input For e have A 3 A and for I/p we obtain ince, A A or A A A4 or A4 A5 or A5 - A can have any value or 9 Therefore, we have the addre range a A 5 A 4 A 3 A A A A 9 A 8 A 7 A 6 A 5 A 4 A 3 A A A From to In Hexadecimal & 8 HtoBFFH imilarly, for chip, we obtain the range a follow E for o, A 3 and A and alo the I/P for A, A, A4, A5 o, the fixed I/p are A 5 A 4 A 3 A A A Therefore, the addre range i A 5 A 4 A 3 A A A A 9 A 8 A 7 A 6 A 5 A 4 A 3 A A A From to In hexadecimal it i from 8 H to BFFH. There i no need to obtain ret of addre ranged a only (D) i matching to two reult. Hence correct option i (D) Q. 46 In the circuit hown below, the ilicon npn tranitor Q ha a very high value of b. The required value of in k to produce I C ma i (A) (B) 3 (C) 4 (D) 5

30 GATE OLVED PAPE - EC 3 ol. 46 The equivalent circuit can be hown a V Th V CC and Th + ince, IC bib ha b. 3 (very high) o, I B i negative in comparion to I C. Therefore, we can write the bae voltage V B o, V I Th C E V Th or, ^ - h^ h or, k+ or, or, 3 ^6 kh^. h+. 8. ^6 k h# ^. h Hence, 6 #. 4 k 8. Hence correct option i (C) Q. 47 Let U and V be two independent and identically ditributed random variable uch that PU PU ^ + h ^ - h. The entropy HU+ V ^ h in bit i (A) 3/4 (B) ol. 47 (C) 3/ (D) log 3 Given, PU ^ + h PU ^ - h where U i a random variable which i identical to V i.e., PV ^ + h PV ^ - h o, random variable U and V can have following value U +, - ; V +, - Therefore the random variable U+ V can have the following value, - henu V - U+ V * henu, V or u -, v henu V Hence, we obtain the probabilitie for U+ V a follow U+ V PU ^ + Vh - # 4

31 GATE OLVED PAPE - EC 3 Therefore, the entropy of the ^U+ Vh i obtained a HU ^ + Vh PU V log / ^ + h ' PU ^ + Vh log 4 log + + log Hence correct option i (C) Common Data Quetion Common Data for Quetion 48 and 49: b # l+ b # l # Bit and are tranmitted with equal probability. At the receiver, the pdf of the repective received ignal for both bit are a hown below. Q. 48 If the detection threhold i, the BE will be (A) (B) 4 ol. 48 (C) 8 (D) 6 For the hown received ignal, we conclude that if i the tranmitted ignal then the received ignal will be alo zero a the threhold i and the pdf of bit i not croing. Again, we can oberve that there i an error when bit i received a it croe the threhold. The probability of error i given by the area encloed by the bit pdf (hown by haded region) 4 P (error when bit received). 5 # # 8

32 GATE OLVED PAPE - EC 3 or P received b tranmitted l 8 ince, the and tranmiion i equiprobable: i.e., P^h P ^ h Hence bit error rate (BE) i BE P received P P received b l + P tranmitted ^ h b tranmitted l ^ h + 8 # 6 Hence correct option i (D) Q. 49 The optimum threhold to achieve minimum bit error rate (BE) i (A) (B) 4 5 ol. 49 (C) (D) 3 The optimum threhold i the threhold value for tranmiion a obtained at the interection of two pdf. From the hown pdf. e obtain at the interection (tranmitted, received) b, l we can obtain the interection by olving the two linear eq x+ y pdf of received bit y 5. x pdf of received bit Hence for threhold 5 4, we have BE P received P P received b l + P tranmitted ^ h b l tranmitted ^ h b # 5 # l # + b # 5 4 # 5 l # <(BE for threhold ) Hence, optimum threhold i 4 5 Hence correct option i (B) Common Data for Quetion 5 and 5: Conider the following figure

33 GATE OLVED PAPE - EC 3 Q. 5 The current I in Amp in the voltage ource, and voltage V in Volt acro the current ource repectively, are (A) 3, - (B) 8, - (C) - 8, (D) -3, ol. 5 At the node, voltage i given a V volt Applying KCL at node I V V I I -3 A Alo, from the circuit, V - 5# V V + V volt Hence correct option i (D) Q. 5 The current in the reitor in Amp i (A) (B) 3.33 (C) (D) ol. 5 Again from the hown circuit, the current in reitor i I V A Hence correct option i (C) Linked Anwer Quetion tatement for Linked Anwer Quetion 5 and 53: A monochromatic plane wave of wavelength l 6 mm i propagating in the direction a hown in the figure below. Ev i, Er v and Ev t denote incident, reflected, and tranmitted electric field vector aociated with the wave.

34 GATE OLVED PAPE - EC 3 Q. 5 The angle of incidence q i and the expreion for Ev i are 4 p# ^x + h (A) 6c and E a x a z e -j ^t - th 3 Vm / ol. 5 (B) 45c and E a a e j z - p# ^tx+ t 4 zh 3 Vm / p# (C) 45c and E 4 ^x+ zh a x a z e -j ^t - th 3 Vm / (D) 6c and E a a e j z - p# ^tx- t 4 zh 3 Vm / For the given incidence of plane wave, we have the tranmitting angle q t 9. c From nell law, we know nin q i nin qt c me in q i c me in qt...() For the given interface, we have m m e, e 45. o, from Eq. () in q i 4. 5 in 9. or, q i. 45c Now, the component of Ev i can be obtained a E v E a E a e -jbk i _ v ox x - v oz zi (oberved from the hown figure) ince, the angle q i 45c o, E ox E Eo oz Therefore, Ev Eo i a a e -jbk _ v x- v zi...() Now, the wavelength of EM wave i l 6 mm o, b p p 3 # l 4 Alo, direction of propagation i av ax az k v + v o, k x + z

35 GATE OLVED PAPE - EC 3 ubtituting it in equation (), we get Ev Eo i a a e -j _ v x- v zi Hence correct option i (C) Q. 53 The expreion for Ev r i p# (A).3 E 4 ^x- zh a x a z e -j ^t + th 3 Vm / (B) E a a e j z p# - ^tx+ t 4 zh 3 Vm / p# 4 ^x+ zh 3 ol. 53 p# (C).44 E 4 ^x- zh a x a z e -j ^t + th 3 Vm / p# (D) E 4 ^x+ zh a x a z e -j ^t + th 3 Vm / e obtain the reflection coefficient for parallel polarized wave (ince, electric field i in the plane of wave propagation) a hco qt- hco qi G z...() h co q + h co q A we have already obtained t q i 45c, q t 9. c m Alo, h 45 h h e. 45. m and h h h e ubtituting thee in eq. () we get G z co 9. c co 45c co 9. c+ 45. co 45c Therefore, the reflected field ha the magnitude given by Ero T E ' io or E ro G z Eio -.3 Eio Hence, the expreion of reflected electric field i Ev r. Eo -j p# k -3 _-a v x-a v zi e 4 3 () Again, we have the propagation vector of reflected wave a av ax az k v - v or, k x - z ubtituting it in Eq. (), we get Ev r.3 E 4 o -j p# x-z - _-a v x-a v zie 3 b l Ev r.3 E jp# 4 ^x- zh o - V _ av x+ av zie 3 m Hence correct option i (A) i

36 GATE OLVED PAPE - EC 3 tatement for Linked Anwer Quetion 54 and 55: The tate diagram of a ytem i hown below. A ytem i decribed by the tatevariable equation X o AX+ Bu ; y CX + D u Q. 54 The tate-variable equation of the ytem hown in the figure above are Xo X u (A) > H > H Xo X u (B) > H > H ol. 54 y 6 X + u y - 6 X + u Xo X u (C) > H > H Xo X u (D) > H > H y - 6 -@ X -u y 6 -@ X -u For the hown tate diagram we can denote the tate x, x a below o, from the tate diagram, we obtain xo -x-u xo - x+ ^h^-h^h^- hu+ ^-h^h^- hx o - x + x + u x and y ^- h^hx+ ^-h^h^- hx+ ^h^-h^h^-h^hu x - x + u Hence, in matrix form we can write the tate variable equation xo x > xo H > - u - H> + - x H > H x and y 8 - B > + u x H which can be written in more general form a X o > - X H > H y 8 - BX+ u Hence correct option i (A) Q. 55 The tate tranition matrix e At of the ytem hown in the figure above i (A) e -t -t e > -t -th (B) > -t -th te e -te e ol. 55 (C) e -t > -t -th (D) e -t te -t - > -t H e e e From the obtained tate-variable equation

37 GATE OLVED PAPE - EC 3 e have A > - - H o, I - A > H and ^I - Ah - + ^ + h > + H V + ^ + h + T X Hence, the tate tranition matrix i obtained a e At - - L ^I-Ah Z V_ ] + b - L [ ` ] ^ + h + b \ T Xa - e > -t -th te e Hence correct option i (A) General Aptitude (GA) Quetion Q.56 to Q.6 carry one mark each. Q. 56 Chooe the grammatically COECT entence: (A) Two and two add four (B) Two and two become four (C) Two and two are four (D) Two and two make four ol. 56 Two and two make four Hence correct option i (D) Q. 57 tatement: You can alway give me a ring whenever you need. hich one of the following i the bet inference from the above tatement? (A) Becaue I have a nice caller tune. (B) Becaue I have a better telephone facility (C) Becaue a friend in need i a friend indeed (D) Becaue you need not pay toward the telephone bill when you give me a ring ol. 57 You can alway given me a ring whenever you need. Becaue a friend i need i a friend indeed Hence correct option i (C) Q. 58 In the ummer of, in New Delhi, the mean temperature of Monday to

38 GATE OLVED PAPE - EC 3 ol. 58 edneday wa 4 C and of Tueday to Thurday wa 43c C. If the temperature on Thurday wa 5% higher than that of Monday, then the temperature in c C on Thurday wa (A) 4 (B) 43 (C) 46 (D) 49 Let the temperature on Monday, Tueday, edneday and Thurday be repectively a TM, TTU, T, TTH o, from the given data we have TH + TTU + T 4...() 3 and TTU + T + TTH 43...() 3 alo, a the temperature on Thurday wa 5 % higher than that of Monday i.e. T TH 5. T M...(3) olving eq (), () and (3), we obtain T TH 46cC Hence correct option i (C) Q. 59 Complete the entence: Dare... mitake. (A) commit (B) to commit (C) committed (D) committing ol. 59 Dare to commit mitake Hence correct option i (B) Q. 6 They were requeted not to quarrel with other. hich one of the following option i the cloet in meaning to the word quarrel? (A) make out (B) call out (C) dig out (D) fall out ol. 6 They were requeted not to quarrel with other. Quarrel ha a imilar meaning to fall out Hence correct option i (D) Q. No Carry Two Mark Each Q. 6 A car travel 8km in the firt quarter of an hour, 6km in the econd quarter and 6 km in the third quarter. The average peed of the car in km per hour over the entire journey i (A) 3 (B) 36 (C) 4 (D) 4 ol. 6 Given, the ditance travelled by the car in each quarter interval a Ditance Time Duration 8 km 4 hr 6 km 4 hr 6 km 4 hr Therefore, the total time taken hr

39 GATE OLVED PAPE - EC 3 Hence, Hence correct option i (C) Total ditance travelled km average peed Total ditance travelled Total time taken 3 34 / 4 km/ hr Q. 6 Find the um to n term of the erie ^ n + h 99 ^ n - h (A) + (B) + 8 ol n n ^ - h 99 ^ - h (C) + n (D) + n 8 8 It will be eay to check the option for given erie. From the given erie e get um of term um of term and um of 3 term Checking all the option one by one, we oberve that only (D) option atifie a n 99 ^ - h n + n 8 99 ^ - h o, ^ - h ^ - h Hence correct option i (D) Q. 63 tatement: There were different tream of freedom movement in colonial India carried out by the moderate, liberal, radical, ocialit, and o on. hich one of the following i the bet inference from the above tatement? (A) The emergence of nationalim in colonial India led to our Independence (B) Nationalim in India emerged in the context of colonialim (C) Nationalim in India i homogeneou (D) Nationalim in India i heterogeneou ol. 63 Nationalim in India i heterogeneou Hence correct option i (D) Q. 64 The et of value of p for which the root of the equation 3x + x+ p^p- h are of oppoite ign i (A) ^-3,h (B) ^, h (C) ^, 3h (D) ^, 3h ol. 64 Given, the quadratic equation 3x + x + P^P - h It will have the root with oppoite ign if *hipping Free* Buy Online all GATE Book: hop.nodia.co.in *Maximum Dicount*

40 GATE OLVED PAPE - EC 3 PP ^ - h < o it can be poible only when P < and P - > or P > and P - < The t condition tend to no olution for P. Hence, from the econd condition, we obtain < P < i.e., P i in the range ^, h Hence correct option i (B) Q. 65 hat i the chance that a leap year, elected at random, will contain 53 unday? (A) /7 (B) 3/7 ol. 65 (C) /7 (D) 5/7 In a leap year, there are 366 day o, 5 week will have 5 aturday and for remaining two day ^366-5 # 7 h. e can have the following combination aturday, unday unday, Monday Monday, Tueday Tueday, edneday edneday, Thurday Thurday, Friday Friday, aturday Out of thee even poibilitie, only two conit a aturday. Therefore, the probability of aturday i given a P 7 Hence correct option i (A)

41 GATE OLVED PAPE - EC 3 Anwer heet. (C) 3. (B) 5. (*) 37. (D) 49. (B) 6. (C). (D) 4. (A) 6. (B) 38. (C) 5. (D) 6. (D) 3. (C) 5. (A) 7. (B) 39. (A) 5. (C) 63. (D) 4. (A) 6. (A) 8. (C) 4. (A) 5. (C) 64. (B) 5. (D) 7. (D) 9. (C) 4. (C) 53. (A) 65. (A) 6. (B) 8. (C) 3. (B) 4. (B) 54. (A) 7. (D) 9. (A) 3. (C) 43. (D) 55. (A) 8. (C). (D) 3. (C) 44. (C) 56. (D) 9. (B). (B) 33. (B) 45. (D) 57. (C). (B). (D) 34. (A) 46. (C) 58. (C). (B) 3. (C) 35. (A) 47. (C) 59. (B). (A) 4. (A) 36. (B) 48. (D) 6. (D) **********