GATE EC Q.1-30 Carry One Mark Each. Consider the network graph shown in the figure. Which one of the following is NOT a tree of this graph?

Transcription

1 GATE EC 004 Q. - 0 Carry One Mark Each MCQ. Consider the network graph shown in the figure. Which one of the following is NOT a tree of this graph? (A) a (C) c (B) b (D) d SOL. MCQ. For a tree there must not be any loop. So a, c, and d don t have any loop. Only b has loop. Hence (B) is correct option. The equivalent inductance measured between the terminals and for the circuit shown in the figure is

2 Page GATE EC (A) L+ L+ M (B) L+ L M (C) L + L + M (D)L + L M SOL. The sign of M is as per sign of L If current enters or exit the dotted terminals of both coil. The sign of M is opposite of L If current enters in dotted terminal of a coil and exit from the dotted terminal of other coil. Thus L eq L+ L M Hence (D) is correct option. MCQ. The circuit shown in the figure, with R Ω, L 4 H and C F has input voltage vt ( ) sin t. The resulting current it () is (A) 5 sin( t + 5. c) (B) 5 sin( t 5. c) (C) 5 sin( t + 5. c) (D) 5 sin( t 5. c) SOL. Here ω and V + 0c Y + R j ω C + jωl + j# + j # 4 + j4-5+ tan c I V* Y ( + 0c)( c) c Thus it () 5 sin( t + 5. c) Hence (A) is correct option. MCQ.4 For the circuit shown in the figure, the time constant RC ms. The input voltage is v ( t) sin 0 t. The output voltage vo () t is equal to i

3 Page GATE EC (A) sin( 0 t 45c) (B) sin( 0 t + 45c) (C) sin( 0 t 5c) (D) sin( 0 t + 5c) SOL.4 Hence (A) is correct option. vi () t sin 0 t Here ω 0 rad and V i + 0c jωc Now V 0. V t R j CR V i + + ω j ω C + 0c + - j # 0 # 0 45c v0 () t sin( 0 t 45c) MCQ.5 For the R L circuit shown in the figure, the input voltage vi () t u() t. The current it () is SOL.5 Hence (C) is correct option. Input voltage vi () t ut () Taking laplace transform Vi () s s

4 Page 4 GATE EC Impedance Zs () s + Vi () s Is () s + ss ( + ) or Is () ; s s + E Taking inverse laplace transform it () ( e ) u( t) t At t 0, it () 0 At t, it () 0. At t, it () 05. Graph (C) satisfies all these conditions. MCQ.6 The impurity commonly used for realizing the base region of a silicon n p n transistor is (A) Gallium (B) Indium (C) Boron (D) Phosphorus SOL.6 Trivalent impurities are used for making p type semiconductor. Boron is trivalent. Hence option (C) is correct MCQ.7 If for a silicon npn transistor, the base-to-emitter voltage ( V BE ) is 0.7 V and the collector-to-base voltage ( V CB ) is 0. V, then the transistor is operating in the (A) normal active mode (B) saturation mode (C) inverse active mode (D) cutoff mode SOL.7 Here emitter base junction is forward biased and base collector junction is reversed biased. Thus transistor is operating in normal active region. Hence option (A) is correct. MCQ.8 Consider the following statements S and S. S : The β of a bipolar transistor reduces if the base width is increased. S : The β of a bipolar transistor increases if the dopoing concentration in the base is increased. Which remarks of the following is correct? (A) S is FALSE and S is TRUE (B) Both S and S are TRUE (C) Both S and S are FALSE (D) S is TRUE and S is FALSE SOL.8 Hence option (D) is correct. We have β α α

5 Page 5 GATE EC Thus α -" β - α." β. If the base width increases, recombination of carrier in base region increases and α decreases & hence β decreases. If doping in base region increases, recombination of carrier in base increases and α decreases thereby decreasing β. Thus S is true and S is false. MCQ.9 An ideal op-amp is an ideal (A) voltage controlled current source (C) current controlled current source (B) voltage controlled voltage source (D) current controlled voltage source SOL.9 MCQ.0 An ideal OPAMP is an ideal voltage controlled voltage source. Hence (B) is correct option. Voltage series feedback (also called series-shunt feedback) results in (A) increase in both input and output impedances (B) decrease in both input and output impedances (C) increase in input impedance and decrease in output impedance (D) decrease in input impedance and increase in output impedance SOL.0 In voltage series feed back amplifier, input impedance increases by factor ( + Aβ) and output impedance decreases by the factor ( + Aβ). R if Ri ( + Aβ) R of Ro ( + A β ) Hence (C) is correct option. MCQ. The circuit in the figure is a (A) low-pass filter (C) band-pass filter (B) high-pass filter (D) band-reject filter SOL. This is a Low pass filter, because At ω V0 0 Vin and at ω 0 V0 V Hence (A) is correct option. in

6 Page 6 GATE EC MCQ. Assuming V CEsat 0. V and β 50, the minimum base current ( I B ) required to drive the transistor in the figure to saturation is (A) 56 μa (C) 60 ma (B) 40 ma (D) ma SOL. MCQ. SOL. Applying KVL we get VCC IC RC VCE 0 or I VCC VCE C 0..8 ma RC k Now I IC.8 B m 56 μa β 50 Hence option (A) is correct. A master - slave flip flop has the characteristic that (A) change in the output immediately reflected in the output (B) change in the output occurs when the state of the master is affected (C) change in the output occurs when the state of the slave is affected (D) both the master and the slave states are affected at the same time A master slave D-flip flop is shown in the figure. In the circuit we can see that output of flip-flop call be triggered only by transition of clock from to 0 or when state of slave latch is affected. Hence (C) is correct answer. MCQ.4 The range of signed decimal numbers that can be represented by 6-bits s complement number is (A) - to + (B) -6 to +6 (C) -64 to +6 (D) - to +

7 Page 7 GATE EC SOL.4 The range of signed decimal numbers that can be represented by n bits s n - n - complement number is ( ) to + ( ). Thus for n 6 we have 6-6- Range ( ) to + ( ) to + Hence (A) is correct answer. MCQ.5 SOL.5 A digital system is required to amplify a binary-encoded audio signal. The user should be able to control the gain of the amplifier from minimum to a maximum in 00 increments. The minimum number of bits required to encode, in straight binary, is (A) 8 (B) 6 (C) 5 (D) 7 The minimum number of bit require to encode 00 increment is n $ 00 or n $ 7 Hence (D) is correct answer. MCQ.6 Choose the correct one from among the alternatives ABCD,,, after matching an item from Group most appropriate item in Group. Group Group P. Shift register. Frequency division Q. Counter. Addressing in memory chips R. Decoder. Serial to parallel data conversion (A) P, Q, R (B) P, Q, R (C) P, Q, R (D) P, Q, R SOL.6 MCQ.7 Shift Register " Serial to parallel data conversion Counter " Frequency division Decoder " Addressing in memory chips. Hence (B) is correct answer. The figure the internal schematic of a TTL AND-OR-OR-Invert (AOI) gate. For the inputs shown in the figure, the output Y is (A) 0 (B) (C) AB (D) AB SOL.7 For the TTL family if terminal is floating, then it is at logic. Thus Y ( AB + ) AB. 0 0

8 Page 8 GATE EC Hence (A) is correct answer. MCQ.8 Given figure is the voltage transfer characteristic of (A) an NOMS inverter with enhancement mode transistor as load (B) an NMOS inverter with depletion mode transistor as load (C) a CMOS inverter (D) a BJT inverter SOL.8 Hence option (C) is correct MCQ.9 The impulse response hn [ ] of a linear time-invariant system is given by hn [ ] un [ + ] + un [ ) nn [ 7] where un [ ] is the unit step sequence. The above system is (A) stable but not causal (B) stable and causal (C) causal but unstable (D) unstable and not causal SOL.9 A system is stable if / hn ( ) <. The plot of given hn ( ) is n 6 Thus / hn ( ) / hn ( ) n n < Hence system is stable but hn ( )! 0 for n < 0. Thus it is not causal. Hence (A) is correct answer. MCQ.0 The distribution function Fx () x of a random variable x is shown in the figure. The probability that X is

9 Page 9 GATE EC SOL.0 (A) zero (B) 0.5 (C) 0.55 (D) 0.0 Hence (D) is correct option. Fx ( # X< x) px ( x) PX ( x) + - or PX ( ) PX ( ) PX ( ) z MCQ. The z -transform of a system is Hz () z 0.. If the ROC is z < 0., then the impulse response of the system is (A) ( 0.) n n un [] (B) ( 0.) u[ n ] (C) (.) 0 n un [] (D) (.) 0 u[ n ] n SOL. MCQ. SOL. MCQ. SOL. Hence (D) is correct answer. Hz () z z 0. We know that z < 0. n au[ n ] * az z < a Thus hn [ ] (0.) u[ n ] n The Fourier transform of a conjugate symmetric function is always (A) imaginary (B) conjugate anti-symmetric (C) real (D) conjugate symmetric The Fourier transform of a conjugate symmetrical function is always real. Hence (C) is correct answer. The gain margin for the system with open-loop transfer function ( + s) GsHs () (), is s (A) (B) 0 (C) (D) The open loop transfer function is ( + s) GsHs () () s Substituting s jω we have ( + jω) Gj ( ω) Hj ( ω ) ω...()

10 Page 0 GATE EC Gj ( ω) Hj ( ω ) 80c + tan ω The frequency at which phase becomes 80c, is called phase crossover frequency. Thus 80 80c + tan ωφ or tan ω φ 0 or ω φ 0 The gain at ω φ 0 is Gj ( ω) Hj ( ω ) + ω ω Thus gain margin is 0 and in db this is. Hence (D) is correct option MCQ.4 Given GsHs () () K.The point of intersection of the asymptotes of ss ( + )( s+ ) the root loci with the real axis is (A) 4 (B). (C). (D) 4 SOL.4 MCQ.5 Centroid is the point where all asymptotes intersects. ΣReal of Open Loop Pole ΣReal Part of Open Loop Pole σ ΣNo.of Open Loop Pole ΣNo.of Open Loop zero. Hence (C) is correct option. In a PCM system, if the code word length is increased from 6 to 8 bits, the signal to quantization noise ratio improves by the factor (A) 6 8 (B) (C) 6 (D) 8 SOL.5 When word length is 6 S # 6 `N j N 6 When word length is 8 S # 8 6 `N j N 8 S ^ N h 6 N 8 Now 4 S 6 ^ N hn 6 Thus it improves by a factor of 6. Hence (C) is correct option. MCQ.6 An AM signal is detected using an envelop detector. The carrier frequency and modulating signal frequency are MHz and khz respectively. An appropriate value for the time constant of the envelop detector is (A) 500μ sec (B) 0μ sec

11 Page GATE EC (C) 0μ. sec (D) μ sec SOL.6 MCQ.7 SOL.7 MCQ.8 Hence (B) is correct option. Carrier frequency f c # 0 6 Hz Modulating frequency f m # 0 Hz For an envelope detector π f c > > πf Rc m < RC < πf c πf m < RC < πf c πf m < RC < 6 π0 0 # 59. # 0-7 < RC < so, 0 μsec sec best lies in this interval. # -5 An AM signal and a narrow-band FM signal with identical carriers, modulating signals and modulation indices of 0. are added together. The resultant signal can be closely approximated by (A) broadband FM (B) SSB with carrier (C) DSB-SC (D) SSB without carrier Hence (B) is correct option. SAM () t Ac[ + 0. cos ωmt] cos ωmt snbfm () t Accos[ ωct+ 0. sin ωmt] st () SAM () t + SNBfm () t Ac[ + 0. cos ωmt] cos ωct+ Accos( ωct+ 0. sin ωmt) Accos ωct+ Ac0. cos ωmtcos ωct + Accos ωctcos(0. sin ωmt) Acsin ωct. sin( 0. sin ωmt) As 0. sin ω mt,+ 0. to 0. so cos(. 0sin ω mt). As when θ is small cos θ. and sin θ, θ, thus sin(. 0sin ω mt) 0. sincos ωctcos ωmt+ Ac cos ωct Ac0. sin ωmtsin ωct Accos ωct+ 0. Accos( ωc+ ωm) t Thus it is SSB with carrier. cosec In the output of a DM speech encoder, the consecutive pulses are of opposite polarity during time interval t# t # t. This indicates that during this interval (A) the input to the modulator is essentially constant (B) the modulator is going through slope overload USB

12 Page GATE EC (C) the accumulator is in saturation (D) the speech signal is being sampled at the Nyquist rate SOL.8 MCQ.9 Consecutive pulses are of same polarity when modulator is in slope overload. Consecutive pulses are of opposite polarity when the input is constant. Hence (A) is correct option. The phase velocity of an electromagnetic wave propagating in a hollow metallic rectangular waveguide in the TE 0 mode is (A) equal to its group velocity (B) less than the velocity of light in free space (C) equal to the velocity of light in free space (D) greater than the velocity of light in free space SOL.9 We know that vp > c > vg. Hence (D) is correct option. MCQ.0 Consider a lossless antenna with a directive gain of + 6 db. If mw of power is fed to it the total power radiated by the antenna will be (A) 4 mw (B) mw (C) 7 mw (D) /4 mw SOL.0 Hence (A) is correct option. We have 4πU(, θ φ) G D (, θφ ) Prad For lossless antenna P rad P in Here we have P rad Pin mw and 0 log GD (, θφ ) 6 db or G D (, θφ ) 98. Thus the total power radiated by antenna is 4πU( θ, φ ) P G (, ) m # mw rad D θφ Q Carry Two Marks Each MCQ. For the lattice shown in the figure, Za j Ω and Zb Ω. The values of the open z z circuit impedance parameters 6 z z G are

13 Page GATE EC SOL. j + j j + j (A) + j + j G (B) + j j G + j + j + j + j (C) j j G (D) + j + j G We know that V zi + zi V zi + zi where z V I I 0 z V I I 0 Consider the given lattice network, when I 0. There is two similar path in the circuit for the current I. So I I For z applying KVL at input port we get V IZ ( a+ Zb) Thus V I( Za+ Zb) z ( Za+ Zb) For Z applying KVL at output port we get V Z I Z I a b Thus V I( Za Zb) z ( Za Zb) For this circuit z z and z z. Thus R V SZa+ Zb Za ZbW z z z z G S W Z Z Z Z S a b a+ bw S W T X Here Za j and Zb Ω Thus z z j j z z G + j + j G Hence (D) is correct option.

14 Page 4 GATE EC MCQ. The circuit shown in the figure has initial current il ( 0 ) A through the inductor and an initial voltage vc ( 0 ) V across the capacitor. For input vt () ut (), the Laplace transform of the current it () for t $ 0 is SOL. (A) s (B) s + s + s + s + s + (C) s (D) s + s + s + s + Applying KVL, Ldi() t vt () Ri() t + + # itdt () dt C 0 Taking L.T. on both sides, + + Is () vc ( 0 ) Vs () RI() s + LsI() s Li( 0 ) + + sc sc vt () ut () thus Vs () s Hence Is () I() s + si() s + s s s Is () + s 6 + s s s or Is () s + s + s + Hence (B) is correct option. MCQ. Consider the Bode magnitude plot shown in the fig. The transfer function Hs () is ( s + 0) (A) ( s+ )( s+ 00 ) 0 ( s + ) (C) ( s+ 0 )( s+ 00 ) 0( s + ) (B) ( s+ 0 )( s+ 00 ) 0 ( s + 00) (D) ( s+ )( s+ 0 )

15 Page 5 GATE EC SOL. The given bode plot is shown below At ω change in slope is +0 db " zero at ω At ω 0 change in slope is 0 db " poles at ω 0 At ω 00 change in slope is 0 db " poles at ω 00 Ks ( + ) Thus Ts () s s ( + )( + ) 0 00 Now 0 log0 K 0 " K ( s + ) 00( s + ) Thus Ts () s s ( + )( + ) ( s + 0)( s + 00) 0 00 Hence (C) is correct option. Vo () s MCQ.4 The transfer function Hs () of an RLC circuit is given by V() s 6 Hs () 0 s + 6 0s + 0 The Quality factor (Q-factor) of this circuit is (A) 5 (B) 50 (C) 00 (D) 5000 i SOL.4 MCQ.5 Characteristics equation is s + 0s+ 0 0 Comparing with s + ξωns+ ωn 0 we have ω n ξω 0 Thus ξ Now Q 50 ξ 00. Hence (B) is correct option. For the circuit shown in the figure, the initial conditions are zero. Its transfer Vc () s function Hs () is V() s i

16 Page 6 GATE EC (A) s (C) s + 0 s s (B) s (D) s s s SOL.5 Hence (D) is correct option. V0 () s Hs () Vi () s sc R sl + + s LC + scr + sc s ( 0 # 0 ) + s( 0 # 0 ) s + s+ s + 0 s+ 0 MCQ.6 d y dy A system described by the following differential equation + + y x() t is dt dt initially at rest. For input xt () ut (), the output yt () is (A) ( e + e ) u( t) (B) ( e t + e ) u( t) (C) ( 05. e t e ) u() t (D) ( 05. e t + + e ) u() t SOL.6 MCQ.7 Hence Correct Option is (A) dy dy Given, + + y xt dt dt ^ h Taking Laplace Transformation both sides, we have 6 s + Y^sh Xs ^ h s or Ys ^ h + ss ^ + h^s+ h s s + s + Increasing Laplace transformation gives, yt ^ h e t ^ + e t hu^th Consider the following statements S and S S : At the resonant frequency the impedance of a series RLC circuit is zero. S : In a parallel GLC circuit, increasing the conductance G results in increase in its Q factor.

17 Page 7 GATE EC Which one of the following is correct? (A) S is FALSE and S is TRUE (B) Both S and S are TRUE (C) S is TRUE and S is FALSE (D) Both S and S are FALSE SOL.7 Impedance of series RLC circuit at resonant frequency is minimum, not zero. Actually imaginary part is zero. Z R+ j ωl ` ωc j At resonance ωl 0 and Z R that is purely resistive. Thus S ωc is false Now quality factor Q R C L Since G R, Q G If G - then Q. provided C and L are constant. Thus S is also false. Hence (D) is correct option. MCQ.8 In an abrupt p n junction, the doping concentrations on the p side and n-side 6 are NA 9# 0 /cm respectively. The p n junction is reverse biased and the total depletion width is μm. The depletion width on the p side is (A).7 μm (B) 0. μm (C).5 μm (D) 0.75 μm SOL.8 We know that WN p A WN n D or W Wn# ND μ # 0 p 0. μm N 6 A 9# 0 Hence option (B) is correct. MCQ.9 The resistivity of a uniformly doped n type silicon sample is 05Ω. - mc. If the electron mobility ( μ n ) is 50 cm /V-sec and the charge of an electron is 6. # 0-9 Coulomb, the donor impurity concentration ( N D ) in the sample is (A) # 0 6 /cm (B) # 0 6 /cm (C) 5. # 0 5 /cm (D) 5# 0 5 /cm SOL.9 Hence option (B) is correct. Conductivity σ nqu n or resistivity ρ σ nqμ n Thus n qρμ n. 6 # 0 # 0. 5 # 50 /cm - For n type semiconductor n N D C L 6

18 Page 8 GATE EC MCQ.40 Consider an abrupt p n junction. Let V bi be the built-in potential of this junction and V R be the applied reverse bias. If the junction capacitance ( C j ) is pf for Vbi + VR V, then for Vbi + VR 4 V, C j will be (A) 4 pf (B) pf (C) 0.5 pf (D) 0.5 pf SOL.40 We know that C e SNAND j ε ; ( Vbi + VR)( NA + ND) E Thus C j \ ( V + V ) Now Cj ( Vbi + VR) Cj ( Vbi + VR) 4 or Cj C j 0.5 pf Hence option (D) is correct. bi R MCQ.4 Consider the following statements Sq and S. S : The threshold voltage ( V T ) of MOS capacitor decreases with increase in gate oxide thickness. S : The threshold voltage ( V T ) of a MOS capacitor decreases with increase in substrate doping concentration. Which Marks of the following is correct? (A) S is FALSE and S is TRUE (B) Both S and S are TRUE (C) Both S and S are FALSE (D) S is TRUE and S is FALSE SOL.4 Increase in gate oxide thickness makes difficult to induce charges in channel. Thus V T increases if we increases gate oxide thickness. Hence S is false. Increase in substrate doping concentration require more gate voltage because initially induce charges will get combine in substrate. Thus V T increases if we increase substrate doping concentration. Hence S is false. Hence option (C) is correct. MCQ.4 The drain of an n-channel MOSFET is shorted to the gate so that VGS VDS. The threshold voltage ( V T ) of the MOSFET is V. If the drain current ( I D ) is ma for VGS V, then for V GS V, I D is (A) ma (B) ma (C) 9 ma (D) 4 ma

19 Page 9 GATE EC SOL.4 We know that I D KV ( V) GS T Thus IDS ( VGS VT ) I DI ( VGS VT ) Substituting the values we have ID ( ) 4 I D ( ) or I D 4IDI 4 ma Hence option (D) is correct. MCQ.4 SOL.4 The longest wavelength that can be absorbed by silicon, which has the bandgap of. ev, is. μm. If the longest wavelength that can be absorbed by another material is 0.87 μm, then bandgap of this material is (A). 46 A/cm (B) ev (C) ev (D) ev Hence option (A) is correct. E g \ λ Thus or E E g g E g λ #.. 46 ev 087. λ MCQ.44 SOL.44 The neutral base width of a bipolar transistor, biased in the active region, is 0.5 μ m. The maximum electron concentration and the diffusion constant in the base are 0 4 / cm and Dn 5 cm /sec respectively. Assuming negligible recombination in the base, the collector current density is (the electron charge is 6. # 0-9 Coulomb) (A) 800 A/cm (B) 8 A/cm (C) 00 A/cm (D) A/cm Concentration gradient 4 dn dx - # # 0 - q 6. # 0 9 C D n 5 4 dn 0 dx # 0 J C qd dn n dx # 0 # 5 # # 0 Hence option (B) is correct. 8 A/cm

20 Page 0 GATE EC MCQ.45 Assume that the β of transistor is extremely large and VBE 07. V, IC and V CE in the circuit shown in the figure SOL.45 (A) IC ma, VCE 4.7 V (B) IC 05. ma, VCE 75. V (C) IC ma, VCE 5. V (D) IC 05. ma, VCE 9. V The thevenin equivalent is shown below V T R V R + R C # 5 V 4 + Since β is large is large, IC. IE, IB. 0 and I VT VBE E 0. 7 ma R 00 E Now V CE 5.kIC 00IE 5.k# m 00 # m 5. V Hence (C) is correct option MCQ.46 A bipolar transistor is operating in the active region with a collector current of ma. Assuming that the β of the transistor is 00 and the thermal voltage ( V T ) is 5 mv, the transconductance ( g m ) and the input resistance ( r π ) of the transistor in the common emitter configuration, are (A) gm 5 ma/v and rπ 5.65 kω (B) gm 40 ma/v and rπ 40. kω (C) gm 5 ma/v and rπ 5. k Ω (D) gm 40 ma/v and rπ 5. kω

21 Page GATE EC SOL.46 When IC >> ICO IC g m ma mA/V VT 5mV β r π kω g - m 40 # 0 Hence (D) is correct option. MCQ.47 The value of C required for sinusoidal oscillations of frequency khz in the circuit of the figure is (A) π μf (C) π 6 (B) π μf μf (D) π 6 μf] SOL.47 MCQ.48 The given circuit is wein bridge oscillator. The frequency of oscillation is π f RC or C πrf π # 0 # 0 π μ Hence (A) is correct option. In the op-amp circuit given in the figure, the load current i L is (A) V s (B) V s R R

22 Page GATE EC SOL.48 (C) Vs (D) V s RL R The circuit is as shown below We know that for ideal OPAMP V - V + Applying KCL at inverting terminal V- Vs + V- V0 0 R R or V- Vo V s...() Applying KCL at non-inverting terminal V+ V+ Vo + IL + 0 R R or V+ Vo+ ILR 0...() Since V- V+, from () and () we have Vs+ ILR 0 or I L V s R Hence (A) is correct option. MCQ.49 In the voltage regulator shown in the figure, the load current can vary from 00 ma to 500 ma. Assuming that the Zener diode is ideal (i.e., the Zener knee current is negligibly small and Zener resistance is zero in the breakdown region), the value of R is (A) 7 Ω (B) 70 Ω (C) 70 Ω (D) 4 Ω SOL.49 If I Z is negligible the load current is

23 Page GATE EC Vz R I L as per given condition 00 ma VZ # # 500 ma R At IL 00 ma 5 00 ma V Z 5 V R or R 70Ω At IL 500 ma ma V Z 5 V R or R 4 Ω Thus taking minimum we get R 4 Ω Hence (D) is correct option. MCQ.50 SOL.50 In a full-wave rectifier using two ideal diodes, V dc and V m are the dc and peak values of the voltage respectively across a resistive load. If PIV is the peak inverse voltage of the diode, then the appropriate relationships for this rectifier are (A) V Vm dc, PIV V m (B) I Vm dc, PIV V m π π (C) V Vm dc, PIV V m (D) V V m dc, PIV V π π Hence (B) is correct option. m MCQ.5 The minimum number of - to - multiplexers required to realize a 4- to - multiplexers is (A) (B) (C) (D) 4 SOL.5 Number of MUX is 4 required. Hence (C) is correct answer. and. Thus the total number multiplexers is MCQ.5 The Boolean expression AC + BC is equivalent to (A) AC + BC + AC (B) BC + AC + BC + ACB (C) AC + BC + BC + ABC (D) ABC + ABC + ABC + ABC SOL.5 MCQ.5 Hence (D) is correct answer. AC + BC AC+ BC AC( B + B) + BC( A + A) ACB + ACB + BC A + BC A 00, 00, 00 correspond to the s complement representation of which one of the following sets of number (A) 5,9, and 57 respectively (B) -6, -6, and -6 respectively

24 Page 4 GATE EC (C) -7, -7 and -7 respectively (D) -5, -9 and -57 respectively SOL.5 Hence (C) is correct answer Thus s complement of 00, 00 and 00 is 7. So the number given in the question are s complement correspond to -7. MCQ.54 SOL.54 MCQ.55 The 855 Programmable Peripheral Interface is used as described below. (i) An AD / converter is interface to a microprocessor through an 855. The conversion is initiated by a signal from the 855 on Port C. A signal on Port C causes data to be stobed into Port A. (ii) Two computers exchange data using a pair of 855s. Port A works as a bidirectional data port supported by appropriate handshaking signals. The appropriate modes of operation of the 855 for (i) and (ii) would be (A) Mode 0 for (i) and Mode for (ii) (B) Mode for (i) and Mode for (ii) (C) Mode for (i) and Mode 0 for (ii) (D) Mode for (i) and Mode for (ii) For 855, various modes are described as following. Mode : Input or output with hand shake In this mode following actions are executed. Two port (A & B) function as 8 - bit input output ports.. Each port uses three lines from C as a hand shake signal. Input & output data are latched. Form (ii) the mode is. Mode : Bi-directional data transfer This mode is used to transfer data between two computer. In this mode port A can be configured as bidirectional port. Port A uses five signal from port C as hand shake signal. For (), mode is Hence (D) is correct answer. The number of memory cycles required to execute the following 8085 instructions (i) LDA 000 H (ii) LXI D, FOFH would be

25 Page 5 GATE EC (A) for (i) and for (ii) (C) for (i) and for (ii) (B) 4 for (i) and for (ii) (D) for (i) and 4 for (ii) SOL.55 MCQ.56 LDA 6 bit & Load accumulator directly this instruction copies data byte from memory location (specified within the instruction) the accumulator. It takes 4 memory cycle-as following.. in instruction fetch. in reading 6 bit address. in copying data from memory to accumulator LXI D, ( F0F) 4 & It copies 6 bit data into register pair D and E. It takes memory cycles. Hence (B) is correct answer. In the modulo-6 ripple counter shown in figure, the output of the - input gate is used to clear the J-K flip-flop The -input gate is (A) a NAND gate (C) an OR gate (B) a NOR gate (D) a AND gare SOL.56 In the modulo - 6 ripple counter at the end of sixth pulse (i.e. after 0 or at 0) all states must be cleared. Thus when CB is the all states must be cleared. The input to -input gate is C and B and the desired output should be low since the CLEAR is active low Thus when C and B are 0, 0, then output must be 0. In all other case the output must be. OR gate can implement this functions. Hence (C) is correct answer. MCQ.57 Consider the sequence of 8085 instructions given below LXI H, 958 MOV A, M CMA MOV M, A Which one of the following is performed by this sequence? (A) Contents of location 958 are moved to the accumulator

26 Page 6 GATE EC (B) Contents of location 958 are compared with the contents of the accumulator (C) Contents of location 859 are complemented and stored in location 859 (D) Contents of location 589 are complemented and stored in location 589 SOL.57 Hence (A) is correct answer. LXI H, 958H ; 958H " HL MOV A, M ; (958H) " A CMa ; A " A MOV M, A ; A " M This program complement the data of memory location 958H. MCQ.58 A Boolean function f of two variables x and y is defined as follows : f(,) 00 f(,) 0 f(,) ;(,) f 0 0 Assuming complements of x and y are not available, a minimum cost solution for realizing f using only -input NOR gates and - input OR gates (each having unit cost) would have a total cost of (A) unit (B) 4 unit (C) unit (D) unit SOL.58 MCQ.59 SOL.59 Hence (D) is correct answer. We have fxy (,) xy+ xy+ xy xy ( + y) + xy x+ xy or fxy (,) x+ y Here compliments are not available, so to get x we use NOR gate. Thus desired circuit require unit OR and unit NOR gate giving total cost unit. It is desired to multiply the numbers 0AH by 0BH and store the result in the accumulator. The numbers are available in registers B and C respectively. A part of the 8085 program for this purpose is given below : MVI A, 00H LOOP HLT END The sequence of instructions to complete the program would be (A) JNX LOOP, ADD B, DCR C (B) ADD B, JNZ LOOP, DCR C (C) DCR C, JNZ LOOP, ADD B (D) ADD B, DCR C, JNZ LOOP Hence (D) is correct answer. MVI A, 00H ; Clear accumulator LOOP ADD B ; Add the contents of B to A

27 Page 7 GATE EC DCR C ; Decrement C JNZ LOOP ; If C is not zero jump to loop HLT END This instruction set add the contents of B to accumulator to contents of C times. Hence (D) is correct answer. MCQ.60 A khz sinusoidal signal is ideally sampled at 500 samples/sec and the sampled signal is passed through an ideal low-pass filter with cut-off frequency 800 Hz. The output signal has the frequency. (A) zero Hz (B) 0.75 khz (C) 0.5 khz (D) 0.5 khz SOL.60 Hence Correct Option is (C) Here f s 500 samples/sec, f m khz The sampled frequency are.5 khz, 0.5 khz, Since LPF has cut-off frequency 800 Hz, then only output signal of frequency 0.5 khz would pass through it MCQ.6 A rectangular pulse train st () as shown in the figure is convolved with the signal cos ( 4p# 0 t). The convolved signal will be a (A) DC (C) 8 khz sinusoid (B) khz sinusoid (D) 4 khz sinusoid SOL.6 Hence Correct Option is (D) St ^ h + cos ωst+ cos ωst+... T cos 4π # 0 t ω s s cos 8π # 0 t ^ + h π π # 0# 0 0. # 0 # St ^ h* xt ^ h S^τh# ^τ thdτ # 0 # cos ωst+ cos ωst+...@ dt cos 8π # 0 t # 6

28 Page 8 GATE EC So, frequencies present will be f! f,f! f! f ; f 0 khz f m 8π # 0 4kHz π s m s s m s Hence 4 khz sinusoidal signal will be present MCQ.6 Consider the sequence xn [ ] [ 4 j5+ j5]. The conjugate anti-symmetric part - of the sequence is (A) [ 4 j.5, j, 4 j.5] (B) [ j.5,, j.5] (C) [ j.5, j, 0] (D) [ 4,, 4] SOL.6 MCQ.6 SOL.6 Hence (A) is correct answer. We have xn ( ) [ 4 j5, + j, 4] x*( n ) [ 4 + j5, j, 4] x*( n) [4, j, 4 + j5] x cas - * xn ( ) x( n) ( n) [ 4 j, j 4 j ] A causal LTI system is described by the difference equation yn [ ] αyn [ ] xn [ ] + βxn [ ] The system is stable only if (A) α, β < (B) α >, β > (C) α <, any value of β (D) β <, any value of α Hence (C) is correct answer. We have yn ( ) αyn ( ) xn ( ) + βxn ( ) Taking z transform we get Yz () Yzz () α Xz () + βxzz () Yz () βz or Xz () c m...(i) αz β z( z) or Hz () α ( z ) It has poles at! α / and zero at 0 and β /. For a stable system poles must lie inside the unit circle of z plane. Thus α < or α < But zero can lie anywhere in plane. Thus, β can be of any value.

29 Page 9 GATE EC MCQ.64 A causal system having the transfer function Hs ( ) /( s+ ) is excited with 0 ut (). The time at which the output reaches 99% of its steady state value is (A).7 sec (B).5 sec (C). sec (D). sec SOL.64 Hence (C) is correct option. We have rt () 0u() t or Rs () 0 s Now Hs () s + Cs () Hs () $ Rs () $ 0 0 s + s s( s + ) or Cs () 5 5 s s + ct () 5[ e t ] The steady state value of ct () is 5. It will reach 99% of steady state value reaches at t, where 5[ e t ] 099. # 5 or e t 099. e t 0. or t ln 0. or t. sec MCQ.65 The impulse response hn [ ] of a linear time invariant system is given as n, hn [ ] * 4 n, 0 otherwise jπn/ 4 If the input to the above system is the sequence e, then the output is (A) 4 jπn/ 4 e (B) 4 jπn/ 4 e (C) 4e jπn/ 4 (D) 4e j π n/ 4 SOL.65 Hence (D) is correct answer. j n/ 4 We have xn ( ) e π and hn ( ) 4 δ( n+ ) δ( n+ ) δ( n ) + 4 δ( n ) Now yn ( ) xn ( )* hn ( ) / xn ( khk ) ( ) / xn ( khk ) ( ) k k or yn ( ) xn ( + ) h( ) + xn ( + ) h( ) + xn ( ) h() + xn ( ) h() π 4 e ( ) π e ( ) π e ( ) π j n+ j n+ j n j 4 e ( n )

30 Page 0 GATE EC e + 6e + e 4 e e e e e e π 6 π 4 π π j n j n 4 4 π e [0] e [ cos 4] or yn ( ) 4e j r n π ( ) π ( ) π j n j n j ( n ) π + + j ( n ) MCQ.66 Let xt () and yt () with Fourier transforms Ff () and Yf () respectively be related as shown in Fig. Then Yf () j π f (A) Xf ( /) e (C) Xf (/ ) e jπf (B) Xf (/ ) e jπf (D) Xf ( /) e j π f SOL.66 From given graph the relation in xt () and yt () is yt () x[ ( t+ )] xt () F Xf () Using scaling we have xat ( ) F a X f c a m F Thus x( t) X f c m Using shifting property we ge jπft0 xt ( t 0 ) e X() f Thus x[ ( t+ )] e X f jπf F jπf( ) e X f bl bl x[ ( t+ )] Hence (B) is correct answer. F j f e π X f c m MCQ.67 SOL.67 A system has poles at 0. Hz, Hz and 80 Hz; zeros at 5 Hz, 00 Hz and 00 Hz. The approximate phase of the system response at 0 Hz is (A) 90c (B) 0c (C) 90c (D) 80c Approximate (comparable to 90c) phase shift are Due to pole at 00. Hz " 90c Due to pole at 80 Hz " 90c Due to pole at 80 Hz " 0 Due to zero at 5 Hz " 90c

31 Page GATE EC MCQ.68 Due to zero at 00 Hz " 0 Due to zero at 00 Hz " 0 Thus approximate total 90c phase shift is provided. Hence (A) is correct option. Consider the signal flow graph shown in Fig. The gain x5 is x (A) ( be + cf + dg) abcd (B) bedg ( be + cf + dg) (C) abcd ( be + cf + dg) + bedg (D) ( be + cf + dg) + bedg abcd SOL.68 MCQ.69 Mason Gain Formula Σp Ts () k k In given SFG there is only one forward path and possible loop. p abcd (sum of indivudual loops) - (Sum of two non touching loops) ( L+ L+ L) + ( LL) Non touching loop are L and L where LL bedg Thus Cs () Rs () Hence (C) is correct option p ( be + cf + dg) + bedg abcd ( be + cf + dg) + bedg If A G, then sin At is ( ) ( ) (A) sin 4t + sin t sin( 4t) + sin( t) sin( 4t) + sin( t) sin( 4t) + sin( t) G sin( t) sin( t) (B) sin() t sin( t) G ( ) ( ) (C) sin 4t + sin t sin( 4t) sin( t) sin( 4t) + sin( t) sin( 4t) + sin( t) G ( ) ( ) (D) cos t + cos t cos( 4t) + cos( t) cos( 4t) + cos( t) cos( 4t) + cos( t) G

32 Page GATE EC SOL.69 Hence (A) is correct option We have A G Characteristic equation is [ λi A] 0 or λ + λ + 0 or ( λ+ )( λ+ ) 0 or λ + 5λ+ 4 0 Thus λ 4 and λ Eigen values are 4 and. Eigen vectors for λ 4 ( λ I A) X 0 or λ + x λ + G x G 0 x G x G 0 or x x 0 or x + x 0 We have only one independent equation x x. Let x K, then x K, the Eigen vector will be x K x G K K G G Now Eigen vector for λ ( λ I A) X 0 or or λ + x λ + G x G 0 G x x G 0 We have only one independent equation x x Let x K, then x K. Thus Eigen vector will be x K x G K G K G Digonalizing matrix x x M x x G G Now M ` j G

33 Page GATE EC Now Diagonal matrix of sin At is D where sin( λt) 0 sin( 4t) 0 D 0 sin( λ t) G 0 sin( t) G Now matrix B sin At MDM sin( 4t) 0 `j G 0 sin( t) G G sin( 4 t) sin( t) sin( 4t) sin( t) `j sin( 4t) + sin( t) sin( 4t) sin( t) G sin( 4t) sin( t) sin( 4t) sin( t) `j sin( 4t) sin( t) sin( 4t) + sin( t) G sin( 4t) + sin( t) sin( 4t) + sin( t) ` s j sin( 4t+ sin( t) sin( 4t) + sin( t) G λ MCQ.70 SOL.70 The open-loop transfer function of a unity feedback system is Gs () K ss ( + s+ )( s+ ) The range of K for which the system is stable is (A) > K > 0 (B) > K > 0 4 (C) < K < (D) 6 < K < 4 For ufb system the characteristic equation is + Gs () 0 + Gs () + K 0 ss ( + s+ )( s+ ) 4 s + 4s + 5s + 6s+ K 0 The routh table is shown below. For system to be stable, ( 4K) 0 < K and 0 < 7 / This gives 0 < K < 4 s 4 5 K s s 7 K s 4K 0 7 / s 0 K

34 Page 4 GATE EC Hence (A) is correct option 4 MCQ.7 For the polynomial Ps () s+ s+ s+ s+ s+ 5 the number of roots which lie in the right half of the s plane is (A) 4 (B) (C) (D) SOL.7 Hence (B) is correct option. 5 4 We have Ps () s + s + s + s+ 5 The routh table is shown below. If ε " 0 + then ε + ε 5ε 4 44 is positive and ε ε+ is negative. Thus there are two sign change in first column. Hence system has root on RHS of plane. s 5 s 4 5 s ε 0 s ε + ε 5 0 s 5ε 4ε 44 ε+ s 0 0 MCQ.7 The state variable equations of a system are : xo x x u, xo x and y x + u. The system is (A) controllable but not observable (B) observable but not controllable (C) neither controllable nor observable (D) controllable and observable SOL.7 Hence (D) is correct option. x x We have x G u 0 G x G+ 0 G and x Y [ 0] u x Here A 0 G, B 0 G and C [ 0] The controllability matrix is Q C [ B AB] 0 G det Q C! 0 Thus controllable The observability matrix is

35 Page 5 GATE EC T T T Q 0 [ C A C ]! 0 0 G det Q 0! 0 Thus observable MCQ.7 SOL.7 Given A 0 0 G, the state transition matrix e At is given by t 0 e (A) > e t H (B) e t 0 tg 0 0 e (C) e t t 0 0 e > th (D) 0 e e t G 0 Hence (B) is correct option. s 0 0 s ( si A) 0 s G 0 G 0 0 s G ( si A) ( s ) 0 s 0 ( s ) 0 ( s ) G > 0 H s e At L [( si A)] t e 0 tg 0 e MCQ.74 Consider the signal xt () shown in Fig. Let ht () denote the impulse response of the filter matched to xt, () with ht () being non-zero only in the interval 0 to 4 sec. The slope of ht () in the interval < t < 4 sec is (A) - sec (B) sec - SOL (C) sec (D) sec The impulse response of matched filter is ht () xt ( t) Since here T 4, thus ht () x( 4 t) The graph of ht () is as shown below.

36 Page 6 GATE EC From graph it may be easily seen that slope between Hence (B) is correct option. < t < 4 is. MCQ.75 SOL.75 MCQ.76 A mw video signal having a bandwidth of 00 MHz is transmitted to a receiver through cable that has 40 db loss. If the effective one-side noise spectral density at the receiver is 0-0 Watt/Hz, then the signal-to-noise ratio at the receiver is (A) 50 db (B) 0 db (C) 40 db (D) 60 db The SNR at transmitter is SNR tr P tr N B # 00 # 0 9 In db SNR tr 0 log 0 90 db Cable Loss 40 db At receiver after cable loss we have SNR Rc db Hence (A) is correct option. A 00 MHz carrier of V amplitude and a MHz modulating signal of V amplitude are fed to a balanced modulator. The ourput of the modulator is passed through an ideal high-pass filter with cut-off frequency of 00 MHz. The output of the filter is added with 00 MHz signal of V amplitude and 90c phase shift as shown in the figure. The envelope of the resultant signal is (A) constant (B) + sin( π # 0 6 t) SOL.76 (C) 5 sin( π 0 6 t) (D) 5 + cos( π # 0 6 t) 4 4 Hence (C) is correct option. We have f c 00 MHz 00 # 0 6 and fm MHz # 0 6

37 Page 7 GATE EC The output of balanced modulator is VBM () t [ cos ωct][ cos ωct] [ cos( ωc+ ωm) t+ cos( ωc ωm) t] If VBM () t is passed through HPF of cut off frequency fh 00 # 0, then only ( ωc+ ωm) passes and output of HPF is VHP () t cos( ωc+ ωm) t Now V0 () t VHP () t + sin( π # 00 # 0 ) t cos[π00 # 0 + π# # 0 t] + sin(π# 00 # 0 ) t cos[ π0 8 + π0 6 t] + sin( π0 8 ) t [ cos(π0 tt ) cos(π0 t)] sin[π0 tsin(π0 t) + sin π0 t] 6 8 cos( 0 t) cos 0 t 6 8 π π + sin π0 t sin π0 t ` j This signal is in form Acos π0 8 t+ Bsin π0 8 t The envelope of this signal is A + B 6 cos( 0 t) 6 ` π + sin( π0 t j ` j 6 cos ( 0 t) 6 6 π + + sin ( π0 t) sin( π0 t) sin( π0 6 t) 4 5 sin( π0 6 t) MCQ.77 SOL.77 Two sinusoidal signals of same amplitude and frequencies 0 khz and 0. khz are added together. The combined signal is given to an ideal frequency detector. The output of the detector is (A) 0. khz sinusoid (B) 0. khz sinusoid (C) a linear function of time (D) a constant Hence (A) is correct option. st () Acos[ π0 # 0 t] + Acos[ π0. # 0 t] Here T 00μ sec 0 # 0 and T 99μ sec 0. # 0 Period of added signal will be LCM [ T, T]

38 Page 8 GATE EC Thus T LCM[ 00, 99] 9900μ sec Thus frequency f 0. khz 9900μ MCQ.78 Consider a binary digital communication system with equally likely 0 s and s. When binary 0 is transmitted the detector input can lie between the levels 05. V and V with equl probability : when binary is transmitted, the voltage at the detector can have any value between 0 and V with equal probability. If the detector has a threshold of 0. V (i.e., if the received signal is greater than 0. V, the bit is taken as ), the average bit error probability is (A) 0.5 (B) 0. (C) 0.05 (D) 0.5 SOL.78 The pdf of transmission of 0 and will be as shown below : Probability of error of P( 0 # X # 0. ) 0. Probability of error of 0 : P(. 0# X # 05. ) 005. # 0. Average error Hence (A) is correct option. P( 0 # X # 0. ) + P( 0. # X # 0. 5) MCQ.79 SOL.79 A random variable X with uniform density in the interval 0 to is quantized as follows : If 0 # X # 0., xq 0 If 0. < X #, xq 07. where x q is the quantized value of X. The root-mean square value of the quantization noise is (A) 0.57 (B) 0.98 (C). 05 (D) Hence (B) is correct option. The square mean value is σ ( x x ) f( x) dx -# q

39 Page 9 GATE EC or # ( x x ) f( x) dx 0 # 0. q 0. ( x 0) f( x) dx+ ( x 0. 7) f( x) dx 0 x x. x x ; E + ; + E 0 # σ RMS σ MCQ.80 Choose the current one from among the alternative ABCD,,, after matching an item from Group with the most appropriate item in Group. Group Group. FM P. Slope overload. DM Q. μ-law. PSK R. Envelope detector 4. PCM S. Hilbert transform T. Hilbert transform U. Matched filter (A) - T, - P, - U, 4 - S (B) - S, - U, - P, 4 - T (C) - S, - P, - U, 4 - Q (D) - U, - R, - S, 4 - Q 0. SOL.80 MCQ.8 Hence (C) is correct option. FM $ Capture effect DM $ Slope over load PSK $ Matched filter PCM $ μ law Three analog signals, having bandwidths 00 Hz, 600 Hz and 600 Hz, are sampled at their respective Nyquist rates, encoded with bit words, and time division multiplexed. The bit rate for the multiplexed. The bit rate for the multiplexed signal is (A) 5. kbps (B) 8.8 kbps (C) 57.6 kbps (D) 8.4 kbps SOL.8 Since fs fm, the signal frequency and sampling frequency are as follows f m 00 Hz $ 400 samples per sec f m 600 Hz $ 00 samples per sec f m 600 Hz $ 00 samples per sec Thus by time division multiplexing total 4800 samples per second will be sent. Since each sample require bit, total 4800 # bits per second will be sent Thus bit rate R b 4800 # kbps Hence (C) is correct option. MCQ.8 Consider a system shown in the figure. Let Xf () and Yf () and denote the Fourier

40 Page 40 GATE EC transforms of xt () and yt () respectively. The ideal HPF has the cutoff frequency 0 khz. The positive frequencies where Yf () has spectral peaks are (A) khz and 4 khz (B) khz and 44 khz (C) khz and 4 khz (D) khz and 4 khz SOL.8 The input signal Xf () has the peak at khz and khz. After balanced modulator the output will have peak at fc! khz i.e. : 0! $ and 9 khz 0! ( ) $ 9 and khz 9 khz will be filtered out by HPF of 0 khz. Thus khz will remain. After passing through khz balanced modulator signal will have! khz signal i.e. and 4 khz. Thus peak of Yf () are at khz and 4 khz. Hence (B) is correct option. MCQ.8 A parallel plate air-filled capacitor has plate area of 0 m and plate separation of 0 m. It is connect - ed to a 0.5 V,.6 GHz source. The magnitude of the displacement current is ( 6 - ε 0 π F/m) (A) 0 ma (B) 00 ma (C) 0 A (D).59 ma SOL.8 The capacitance is - -4 C εo A 885. # 0 # d - # 0 The charge on capacitor is - - Q CV # # 0 Displacement current in one cycle Q I fq # # #. 59 T Hence (D) is correct option. ma

41 Page 4 GATE EC MCQ.84 A source produces binary data at the rate of 0 kbps. The binary symbols are represented as shown in the figure given below. The source output is transmitted using two modulation schemes, namely Binary PSK (BPSK) and Quadrature PSK (QPSK). Let B and B be the bandwidth requirements of BPSK and QPSK respectively. Assume that the bandwidth of he above rectangular pulses is 0 khz, B and B are (A) B 0 khz, B 0 khz (B) B 0 khz, B 0 khz (C) B 0 khz, B 0 khz (D) B 0 khz, B 0 khz SOL.84 The required bandwidth of M array PSK is BW R b n where n M and R b is bit rate For BPSK, M n $ n Thus B Rb # 0 0 khz n $ For QPSK, M 4 n MCQ.85 Thus B Rb 0 khz Hence (C) is correct option. Consider a 00 Ω, quarter - wave long (at GHz) transmission line as shown in Fig. It is connected to a 0 V, 50 Ω source at one end and is left open circuited at the other end. The magnitude of the voltage at the open circuit end of the line is (A) 0 V (C) 60 V (B) 5 V (D) 60/7 V SOL.85 Hence (C) is correct option. VL ZO V Z in in

42 Page 4 GATE EC or V L ZO V Z in 0 # V 50 in MCQ.86 In a microwave test bench, why is the microwave signal amplitude modulated at khz (A) To increase the sensitivity of measurement (B) To transmit the signal to a far-off place (C) To study amplitude modulations (D) Because crystal detector fails at microwave frequencies SOL.86 Hence (D) is correct option. MCQ.87 jkz - k t If E ( atx + jat ω jkz - j t y) e and H (/ k ωμ)( aty+ kat ω x) e, the time-averaged Poynting vector is (A) null vector (B) ( k/ ωμ) at z (C) ( k/ ωμ) t (D) ( k/ ωμ) t a z a z SOL.87 Hence (A) is correct option. R avg Re[ E# H * ] E# H * jkz jωt ( at ja ) e k x+ ty # ( jatx+ aty) e ωμ at k ( j)( j) k z ; 0 ωμ ωμ E Thus R avg Re[ E# H * ] 0 jkz+ jωt MCQ.88 Consider an impedance Z R+ jx marked with point P in an impedance Smith chart as shown in Fig. The movement from point P along a constant resistance circle in the clockwise direction by an angle 45c is equivalent to (A) adding an inductance in series with Z (B) adding a capacitance in series with Z (C) adding an inductance in shunt across Z (D) adding a capacitance in shunt across Z

43 Page 4 GATE EC SOL.88 MCQ.89 SOL.89 MCQ.90 SOL.90 Suppose at point P impedance is Z r+ j( ) If we move in constant resistance circle from point P in clockwise direction by an angle 45c, the reactance magnitude increase. Let us consider a point Q at 45c from point P in clockwise direction. It s impedance is Z r 05. j or Z Z+05. j Thus movement on constant r - circle by an + 45c in CW direction is the addition of inductance in series with Z. Hence (A) is correct option. A plane electromagnetic wave propagating in free space is incident normally on a large slab of loss-less, non-magnetic, dielectric material with ε > ε 0. Maxima and minima are observed when the electric field is measured in front of the slab. The maximum electric field is found to be 5 times the minimum field. The intrinsic impedance of the medium should be (A) 0π Ω (B) 60π Ω (C) 600π Ω Hence (D) is correct option. We have VSWR E max 5 Emin + or Γ Thus Γ η η Now Γ η + η or η 0π η + 0π or η 4π Γ Γ (D) 4π Ω A lossless transmission line is terminated in a load which reflects a part of the incident power. The measured VSWR is. The percentage of the power that is reflected back is (A) 57.7 (B). (C) 0. (D). Hence (D) is correct option. Γ The VSWR + Γ or Γ

44 Page 44 GATE EC Thus or P P ref inc P ref Γ P 9 inc 9 i.e..% of incident power is reflected. Answer Sheet. (B) 9. (A) 7. (D) 55. (B) 7. (B). (D) 0. (D) 8. (B) 56. (C) 74. (B). (A). (D) 9. (B) 57. (A) 75. (A) 4. (A). (C) 40. (D) 58. (D) 76. (C) 5. (C). (D) 4. (C) 59. (D) 77. (A) 6. (C) 4. (C) 4. (D) 60. (C) 78. (A) 7. (A) 5. (C) 4. (A) 6. (D) 79. (B) 8. (D) 6. (B) 44. (B) 6. (A) 80. (C) 9. (B) 7. (B) 45. (C) 6. (C) 8. (C) 0. (C) 8. (A) 46. (D) 64. (C) 8. (B). (A) 9. (D) 47. (A) 65. (D) 8. (D). (A) 0. (A) 48. (A) 66. (B) 84. (C). (C). (D) 49. (D) 67. (A) 85. (C) 4. (A). (B) 50. (B) 68. (C) 86. (D) 5. (D). (C) 5. (C) 69. (A) 87. (A) 6. (B) 4. (B) 5. (D) 70 (A) 88. (A) 7. (A) 5. (D) 5. (C) 7 (B) 89. (D) 8. (C) 6. (A) 54. (D) 7 (D) 90. (D)

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