2 π Hz. 2π and 1 π Hz

Transcription

1 GATE/EC-009 EC : ELECTRONICS AND COMMUNICATION ENGINEERING Page GATE EC 009 Q. - Q.0 carry one mark each. MCQ. SOL. MCQ. The order of the differential equation 3 dy dy 4 y e dt dt t + c m + is (A) (B) (C) 3 (D) 4 The highest derivative terms present in DE is of nd order. Hence (B) is correct answer. The Fourier series of a real periodic function has only (P) cosine terms if it is even (Q) sine terms if it is even (R) cosine terms if it is odd (S) sine terms if it is odd Which of the above statements are correct? (A) P and S (B) P and R (C) Q and S (D) Q and R SOL. The Fourier series of a real periodic function has only cosine terms if it is even and sine terms if it is odd. Hence (A) is correct answer. MCQ.3 A function is given by ft ( ) sin t+ cos t. Which of the following is true? (A) f has frequency components at 0 and π Hz (B) f has frequency components at 0 and π Hz (C) f has frequency components at π and π Hz (D) f has frequency components at 0. π and π Hz SCO 8-9, 4th Floor, Sector 34-A, Chandigarh, 600 Ph.: Mobile:

2 GATE/EC-009 EC : ELECTRONICS AND COMMUNICATION ENGINEERING Page SOL.3 MCQ.4 Given function is ft () sin t+ cos t cos t + cos t + cos t The function has a DC term and a cosine function. The frequency of cosine terms is ω πf " f Hz π The given function has frequency component at 0 and π Hz. Hence (B) is correct answer. A fully charged mobile phone with a V battery is good for a 0 minute talk-time. Assume that, during the talk-time the battery delivers a constant current of A and its voltage drops linearly from V to 0 V as shown in the figure. How much energy does the battery deliver during this talk-time? (A) 0 J (C) 3. kj (B) kj (D) 4.4 J SOL.4 MCQ.5 The energy delivered in 0 minutes is t t E # VIdt I # Vdt I# Area 0 0 # ( 0 + ) # kj Hence (C) is correct option. In an n-type silicon crystal at room temperature, which of the following can have a concentration of 4# 0 9 cm - 3? (A) Silicon atoms (B) Holes (C) Dopant atoms (D) Valence electrons SOL.5 Only dopant atoms can have concentration of 4# 0 9 cm - 3 in n type silicon at room temperature. Hence option (C) is correct. MCQ.6 The full form of the abbreviations TTL and CMOS in reference to logic families are (A) Triple Transistor Logic and Chip Metal Oxide Semiconductor (B) Tristate Transistor Logic and Chip Metal Oxide Semiconductor (C) Transistor Transistor Logic and Complementary Metal Oxide Semiconductor (D) Tristate Transistor Logic and Complementary Metal Oxide Silicon SCO 8-9, 4th Floor, Sector 34-A, Chandigarh, 600 Ph.: Mobile:

3 GATE/EC-009 EC : ELECTRONICS AND COMMUNICATION ENGINEERING Page 3 SOL.6 MCQ.7 SOL.7 TTL " Transistor - Transistor logic CMOS " Complementary Metal Oxide Semi-conductor Hence (C) is correct answer. The ROC of z -transform of the discrete time sequence n n xn ( ) un ( ) b u( n ) 3 l b l is (A) 3 < z < (B) z > (C) z < 3 (D) < z < 3 Hence (A) is correct answer n n xn [ ] un ( ) b u( n ) 3 l b l Taking z transform we have n 3 n n n Xz () n z n / b z 3 l / b l n 0 n 3 n 3 n n n z / b z 3 l / b l n 0 n 3 First term gives z 3 < " < 3 z Second term gives z > " > z Thus its ROC is the common ROC of both terms. that is < z < 3 MCQ.8 The magnitude plot of a rational transfer function Gs () with real coefficients is shown below. Which of the following compensators has such a magnitude plot? (A) Lead compensator (C) PID compensator (B) Lag compensator (D) Lead-lag compensator SOL.8 This compensator is roughly equivalent to combining lead and lad compensators in the same design and it is referred also as PID compensator. Hence (C) is correct option MCQ.9 A white noise process Xt () with two-sided power spectral density 0 W/ Hz # 0 SCO 8-9, 4th Floor, Sector 34-A, Chandigarh, 600 Ph.: Mobile:

4 GATE/EC-009 EC : ELECTRONICS AND COMMUNICATION ENGINEERING Page 4 is input to a filter whose magnitude squared response is shown below. The power of the output process Yt () is given by (A) 5# 0 7 W (B) # 0 6 W (C) # 0 6 W (D) # 0 5 W SOL.9 Correct Option is ( ) MCQ.0 Which of the following statements is true regarding the fundamental mode of the metallic waveguides shown? (A) Only P has no cutoff-frequency (B) Only Q has no cutoff-frequency (C) Only R has no cutoff-frequency (D) All three have cutoff-frequencies SOL.0 MCQ. Rectangular and cylindrical waveguide doesn t support TEM modes and have cut off frequency. Coaxial cable support TEM wave and doesn t have cut off frequency. Hence (A) is correct option. A fair coin is tossed 0 times. What is the probability that only the first two tosses will yield heads? (A) 0 c m (B) C b l SOL. MCQ. (C) 0 0 c m (D) C b l Number of elements in sample space is 0. Only one element " HHTTTTTTTT,,,,,,,,,, is event. Thus probability is 0 Hence (C) is correct answer. If the power spectral density of stationary random process is a sine-squared function of frequency, the shape of its autocorrelation is SCO 8-9, 4th Floor, Sector 34-A, Chandigarh, 600 Ph.: Mobile:

5 GATE/EC-009 EC : ELECTRONICS AND COMMUNICATION ENGINEERING Page 5 SOL. Correct Option is ( ) MCQ.3 If fz () c+ cz, then (A) (C) 0 unit circle + fz () # dz is given by z π c (B) π ( + c0) π jc (D) π ( + c0) SOL.3 Hence (C) is correct answer We have fz () c0+ cz - () f () z fz c c z z 0 z z( + c0) + c z Since f () z has double pole at z 0, the residue at z 0 is z( + c0) + c Res f () z z 0 lim z. f ( z) lim z. z " 0 z " 0 c m c z Hence [ + fz ( )] # f () z dz # dz πj [Residue at z 0] z unit circle unit circle πjc MCQ.4 In the interconnection of ideal sources shown in the figure, it is known that the 60 V source is absorbing power. SCO 8-9, 4th Floor, Sector 34-A, Chandigarh, 600 Ph.: Mobile:

6 GATE/EC-009 EC : ELECTRONICS AND COMMUNICATION ENGINEERING Page 6 Which of the following can be the value of the current source I? (A) 0 A (B) 3 A (C) 5 A (D) 8 A SOL.4 Circuit is as shown below Since 60 V source is absorbing power. So, in 60 V source current flows from + to - ve direction So, I+ I I I I is always less then A So, only option (A) satisfies this conditions. Hence (A) is correct option. MCQ.5 SOL.5 MCQ.6 The ratio of the mobility to the diffusion coefficient in a semiconductor has the units (A) V - (B) cm.v (C) V. cm - (D) Vs. Hence option (A) is correct. Unit of mobility μ n is Unit of diffusion current D n is μn Thus unit of is / Dn cm V. sec cm sec cm cm V V$ sec sec V In a microprocessor, the service routine for a certain interrupt starts from a fixed location of memory which cannot be externally set, but the interrupt can be delayed or rejected Such an interrupt is (A) non-maskable and non-vectored (B) maskable and non-vectored (C) non-maskable and vectored (D) maskable and vectored SOL.6 Vectored interrupts : Vectored interrupts are those interrupts in which program control transferred to a fixed memory location. Maskable interrupts : Maskable interrupts are those interrupts which can be rejected SCO 8-9, 4th Floor, Sector 34-A, Chandigarh, 600 Ph.: Mobile:

7 GATE/EC-009 EC : ELECTRONICS AND COMMUNICATION ENGINEERING Page 7 or delayed by microprocessor if it is performing some critical task. Hence (D) is correct answer. MCQ.7 If the transfer function of the following network is Vo () s V() s + scr i The value of the load resistance R L is SOL.7 (A) R 4 (C) R For given network we have ( RL XC) Vi V 0 R + ( RL XC) RL V0 () s srl + C Vi () s R RL + + sr C L (B) R (D) R RL R + RR sc + R L L MCQ.8 RL R + RRLsC + RL + R + RsC RL But we have been given V0 () s TF.. Vi () s + scr Comparing, we get + R & R R L R L Hence (C) is correct option. Consider the system dx Ax + Bu with A dt 0 0 G and B p q G where p and q are arbitrary real numbers. Which of the following statements about the controllability of the system is true? (A) The system is completely state controllable for any nonzero values of p and q (B) Only p 0 and q 0 result in controllability (C) The system is uncontrollable for all values of p and q SCO 8-9, 4th Floor, Sector 34-A, Chandigarh, 600 Ph.: Mobile:

8 GATE/EC-009 EC : ELECTRONICS AND COMMUNICATION ENGINEERING Page 8 (D) We cannot conclude about controllability from the given data SOL.8 Hence (C) is correct option. Here A 0 0 G and B p q G 0 p p AB 0 G q G q G p q S 8 B ABB q p G S pq pq 0 Since S is singular, system is completely uncontrollable for all values of p and q. MCQ.9 For a message signal mt () cos( πft m ) and carrier of frequency f c, which of the following represents a single side-band (SSB) signal? (A) cos( πf t) cos( π f t) (B) cos( πft) m c (C) cos[ π ( f + f ) t] (D) [ + cos( πf t) cos( πf t) c m c m c SOL.9 MCQ.0 Hence (C) is correct option. cos( πfmt) cos( π fct) $ DSB suppressed carrier cos( π ft c ) $ Carrier Only cos[ π ( fc+ fm) t] $ USB Only [ + cos( πf t) cos( πf t)] $ USB with carrier m c Two infinitely long wires carrying current are as shown in the figure below. One wire is in the y z plane and parallel to the y axis. The other wire is in the x y plane and parallel to the x axis. Which components of the resulting magnetic field are non-zero at the origin? (A) xyz,, components (B) xy, components (C) yz, components (D) xz, components SOL.0 Due to A current wire in x y plane, magnetic field be at origin will be in x direction. Due to A current wire in y z plane, magnetic field be at origin will be in z SCO 8-9, 4th Floor, Sector 34-A, Chandigarh, 600 Ph.: Mobile:

9 GATE/EC-009 EC : ELECTRONICS AND COMMUNICATION ENGINEERING Page 9 direction. Thus x and z component is non-zero at origin. Hence (D) is correct option. Q. to Q.60 carry two marks each. MCQ. SOL. Consider two independent random variables X and Y with identical distributions. The variables X and Y take values 0, and with probabilities, 4 and 4 respectively. What is the conditional probability PX ( + Y X Y 0)? (A) 0 (B) /6 (C) /6 (D) Hence (C) is correct option. We have px ( 0) py ( 0) px ( ) py ( ) 4 px ( ) py ( ) 4 Let X+ Y $ A and X Y 0 $ B Now PA ( + B) PX ( + Y X Y 0) PB ( ) Event PA ( + B) happen when X+ Y and X Y 0. It is only the case when X and Y. Thus PA ( + B) # Now event PB ( ) happen when X Y 0 It occurs when X Y, i.e. X 0 and Y 0 or X and Y or X and Y Thus PB ( ) # + # + # PA ( + B) 6 / Now PB ( ) 66 / 6 MCQ. The Taylor series expansion of sin x x π at x π is given by ( x π) (A) + ( x π) +... (B) ! 3! SCO 8-9, 4th Floor, Sector 34-A, Chandigarh, 600 Ph.: Mobile:

10 GATE/EC-009 EC : ELECTRONICS AND COMMUNICATION ENGINEERING Page 0 SOL. ( x π) (C) ( x π) +... (D) ! 3! Hence (D) is correct answer. We have fx () sin x x π Substituting x π y,we get sin( y + π) sin y fy ( + π) ( sin y ) y y y... y y 3 5 y y c + m 3! 5! 4 y y or fy ( + π) ! 5! Substituting x π y we get 4 ( x π) ( x π) fx () ! 5! MCQ.3 If a vector field V is related to another vector field A through V 4# A, which of the following is true? (Note : C and S C refer to any closed contour and any surface whose boundary is C. ) (A) # V$ dl # # A$ ds (B) # A$ dl # # V$ ds C S C (C) # Δ# V$ dl # # Δ# A$ ds (D) # Δ # V$ dl # # V$ ds C S C C S C SOL.3 Hence (B) is correct option. We have V 4# A...() By Stokes theorem # A$ dl ## ( 4# A) $ ds...() From () and () we get # A$ dl ## V$ ds MCQ.4 Given that Fs () is the one-side Laplace transform of ft, () the Laplace transform of t # f() τ dτ is SOL.4 0 (A) sf() s f() 0 (B) () s Fs s (C) # F() τ dτ (D) [ ( ) ( )] 0 s Fs f 0 By property of unilateral laplace transform t L Fs () f() d 0 # τ τ + f() τ dτ 3 s s # 3 Here function is defined for 0 < τ < t, Thus t L Fs () # f() τ s 0 SCO 8-9, 4th Floor, Sector 34-A, Chandigarh, 600 Ph.: Mobile: C S C

11 GATE/EC-009 EC : ELECTRONICS AND COMMUNICATION ENGINEERING Page Hence (B) is correct answer. MCQ.5 SOL.5 Match each differential equation in Group I to its family of solution curves from Group II Group I Group II dy y A. dx x. Circles dy y B. dx x. Straight lines dy C. x dx y 3. Hyperbolas dy D. dx x y (A) A,B 3,C 3,D (B) A, B 3, C, D (C) A,B,C 3,D 3 (D) A 3,B,C,D Hence (A) is correct answer dy y (A) dx x or dy # y dx # x MCQ.6 or log y log x+ log c or y cx Straight Line Thus option (A) and (C) may be correct. (B) dy y dx x or dy # y dx # x or log y log x+ log c or log y log + log c x or y c Hyperbola x The Eigen values of following matrix are R V S 3 5W S 3 6W S W (A) 3, j, T6 j X (B) 6+ 5,3 j + j,3 j SCO 8-9, 4th Floor, Sector 34-A, Chandigarh, 600 Ph.: Mobile:

12 GATE/EC-009 EC : ELECTRONICS AND COMMUNICATION ENGINEERING Page SOL.6 (C) 3 + j, 3 j, 5+ j (D) 3, + 3 j, 3j Sum of the principal diagonal element of matrix is equal to the sum of Eigen values. Sum of the diagonal element is + 3.In only option (D), the sum of Eigen values is. Hence (D) is correct answer. MCQ.7 An AC source of RMS voltage 0 V with internal impedance Zs ( + j) Ω feeds a load of impedance ZL ( 7+ 4j) Ω in the figure below. The reactive power consumed by the load is (A) 8 VAR (C) 8 VAR (B) 6 VAR (D) 3 VAR SOL.7 From given circuit the load current is I L V 0+ 0c 0+ 0c Zs + ZL ( + j ) + ( 7+ 4j ) j ( 8 6 j) 0+ 0c - + φ where φ tan φ 4 The voltage across load is V L IZ L L The reactive power consumed by load is P r VI * * IZ# I Z I L L L L L L L (7 # 4 j) 0+ 0c 8+ 6j Thus average power is 8 and reactive power is 6. Hence (B) is correct option. ( 7 + 4j) 8 + 6j MCQ.8 The switch in the circuit shown was on position a for a long time, and is move to position b at time t 0. The current it () for t > 0 is given by SCO 8-9, 4th Floor, Sector 34-A, Chandigarh, 600 Ph.: Mobile:

13 GATE/EC-009 EC : ELECTRONICS AND COMMUNICATION ENGINEERING Page 3 5t 50t (A) 0. e u( t) ma (B) 0 e u( t) ma 50t 000t (C) 0. e u( t) ma (D) 0 e u( t) ma SOL.8 At t 0, the circuit is as shown in fig below : V (0 ) 00 V Thus V (0 + ) 00 V At t 0 +, the circuit is as shown below + I( 0 ) 00 0 ma 5k At steady state i.e. at t 3 is I( 3) 0 t + RCeq Now it () I(0 ) e u( t) (. 05μ+ 03. μ). 0μ C eq 0.6 μf 05. μ+ 03. μ+ 0. μ RCeq 5 # 0 # 0. 6 # 0 it () 0 e 50t u( t) ma Hence (B) is correct option. MCQ.9 In the circuit shown, what value of R L maximizes the power delivered to R L? (A) 4Ω. (B) 3 8 Ω SCO 8-9, 4th Floor, Sector 34-A, Chandigarh, 600 Ph.: Mobile:

14 GATE/EC-009 EC : ELECTRONICS AND COMMUNICATION ENGINEERING Page 4 (C) 4 Ω (D) 6 Ω SOL.9 For P max the load resistance R L must be equal to thevenin resistance R eq i.e. R L R eq. The open circuit and short circuit is as shown below The open circuit voltage is V oc 00 V From fig I A 8 V x 4# V I 00 Vx A 4 4 I sc I+ I 5 A R Voc th 00 4 Ω Isc 5 Thus for maximum power transfer R L R eq 4 Ω Hence (C) is correct option. MCQ.30 The time domain behavior of an RL circuit is represented by L di / + Ri V ( Be Rt L sin t) u( t) dt 0 +. SOL.30 For an initial current of i() V0 0, the steady state value of the current is given by R (A) it () " V 0 (B) it ()" V 0 R R (C) it () V0 " ( + B) (D) it () V0 " ( + B) R R Steady state all transient effect die out and inductor act as short circuits and forced response acts only. It doesn t depend on initial current state. From the given time domain behavior we get that circuit has only R and L in series with V 0. Thus at steady state it ()" i( 3 ) V 0 R Hence (A) is correct option. SCO 8-9, 4th Floor, Sector 34-A, Chandigarh, 600 Ph.: Mobile:

15 GATE/EC-009 EC : ELECTRONICS AND COMMUNICATION ENGINEERING Page 5 MCQ.3 In the circuit below, the diode is ideal. The voltage V is given by (A) min ( Vi, ) (B) max ( Vi, ) (C) min ( V, ) (D) max ( V, ) i i SOL.3 MCQ.3 Let diode be OFF. In this case A current will flow in resistor and voltage across resistor will be V.V Diode is off, it must be in reverse biased, therefore Vi > 0 " Vi > Thus for Vi > diode is off and V V Option (B) and (C) doesn t satisfy this condition. Let Vi <. In this case diode will be on and voltage across diode will be zero and V V i Thus V min( Vi, ) Hence (A) is correct option. Consider the following two statements about the internal conditions in a n channel MOSFET operating in the active region. S : The inversion charge decreases from source to drain S : The channel potential increases from source to drain. Which of the following is correct? (A) Only S is true (B) Both S and S are false (C) Both S and S are true, but S is not a reason for S (D) Both S and S are true, and S is a reason for S SOL.3 Both S and S are true and S is a reason for S. Hence option (D) is correct. MCQ.33 In the following a stable multivibrator circuit, which properties of v0 () t depend on R? SCO 8-9, 4th Floor, Sector 34-A, Chandigarh, 600 Ph.: Mobile:

16 GATE/EC-009 EC : ELECTRONICS AND COMMUNICATION ENGINEERING Page 6 SOL.33 (A) Only the frequency (B) Only the amplitude (C) Both the amplitude and the frequency (D) Neither the amplitude nor the frequency The R decide only the frequency. Hence (A) is correct option MCQ.34 In the circuit shown below, the op-amp is ideal, the transistor has VBE 06. V and β 50. Decide whether the feedback in the circuit is positive or negative and determine the voltage V at the output of the op-amp. SOL.34 (A) Positive feedback, V 0 V. (B) Positive feedback, V 0 V (C) Negative feedback, V 5 V (D) Negative feedback, V V The circuit is shown in fig below SCO 8-9, 4th Floor, Sector 34-A, Chandigarh, 600 Ph.: Mobile:

17 GATE/EC-009 EC : ELECTRONICS AND COMMUNICATION ENGINEERING Page 7 The voltage at non inverting terminal is 5 V because OP AMP is ideal and inverting terminal is at 5 V. Thus I C 0 5 ma 5k V E I E R E m#. 4k. 4V IE IC V Thus the feedback is negative and output voltage is V V. Hence (D) is correct option. MCQ.35 A small signal source Vi ( t) Acos 0t+ Bsin 0 t is applied to a transistor amplifier as shown below. The transistor has β 50 and hie 3Ω. Which expression best approximate V () t 0 6 (A) V () t 500( Acos 0t+ Bsin 0 t) 0 (B) V ( t) 500( Acos 0t+ Bsin 0 t) 0 (C) V ( t) 500Bsin 0 t 0 (D) V ( t) 50Bsin 0 t SOL.35 The output voltage is hferc V 0 AV r i. V i hie Here R C 3 Ω and h ie 3kΩ Thus V # 3k Vi 3k. 50( Acos 0t+ Bsin 0 6 t) Since coupling capacitor is large so low frequency signal will be filtered out, and best approximation is V 0. 50Bsin 0 6 t Hence (D) is correct option. MCQ.36 If X in logic equation 6 X+ Z{ Y+ ( Z+ X+ X( X+ Y)}, then (A) Y Z (B) Y Z (C) Z (D) Z 0 SCO 8-9, 4th Floor, Sector 34-A, Chandigarh, 600 Ph.: Mobile:

18 GATE/EC-009 EC : ELECTRONICS AND COMMUNICATION ENGINEERING Page 8 SOL.36 We have 6 X+ Z{ Y+ ( Z+ X+ Z( X+ Y)] Substituting X and X 0 we get [ + Z{ Y+ ( Z+ Y)}][ 0+ Z( + Y)] or [ ][ Z( )] + A and 0 + A A or Z ) Z 0 Hence (D) is correct answer MCQ.37 What are the minimum number of - to - multiplexers required to generate a - input AND gate and a - input Ex-OR gate (A) and (B) and 3 (C) and (D) and SOL.37 The AND gate implementation by : mux is as follows Y AI0+ AI AB The EX OR gate implementation by : mux is as follows Y BI0+ BI AB + BA Hence (A) is correct answer. MCQ.38 Refer to the NAND and NOR latches shown in the figure. The inputs ( P, P ) for both latches are first made (0, ) and then, after a few seconds, made (, ). The corresponding stable outputs ( Q, Q ) are (A) NAND: first (0, ) then (0, ) NOR: first (, 0) then (0, 0) (B) NAND : first (, 0) then (, 0) NOR : first (, 0) then (, 0) (C) NAND : first (, 0) then (, 0) NOR : first (, 0) then (0, 0) (D) NAND : first (, 0) then (, ) NOR : first (0, ) then (0, ) SCO 8-9, 4th Floor, Sector 34-A, Chandigarh, 600 Ph.: Mobile:

19 GATE/EC-009 EC : ELECTRONICS AND COMMUNICATION ENGINEERING Page 9 SOL.38 For the NAND latche the stable states are as follows For the NOR latche the stable states are as follows Hence (C) is correct answer. MCQ.39 What are the counting states ( Q, Q ) for the counter shown in the figure below (A), 0, 00,, 0,... (B) 0, 0,, 00, 0... (C) 00,, 0, 0, (D) 0, 0, 00, 0, 0... SOL.39 The given circuit is as follows. The truth table is as shown below. Sequence is 00,, 0, CLK J K Q J K Q SCO 8-9, 4th Floor, Sector 34-A, Chandigarh, 600 Ph.: Mobile:

20 GATE/EC-009 EC : ELECTRONICS AND COMMUNICATION ENGINEERING Page 0 Hence (A) is correct answer. MCQ.40 A system with transfer function Hz () has impulse response h (.) defined as h(), h() 3 and hk () 0 otherwise. Consider the following statements. S : Hz () is a low-pass filter. S : Hz () is an FIR filter. Which of the following is correct? (A) Only S is true (B) Both S and S are false (C) Both S and S are true, and S is a reason for S (D) Both S and S are true, but S is not a reason for S SOL.40 We have h(), h(3) otherwise hk () 0. The diagram of response is as follows : It has the finite magnitude values. So it is a finite impulse response filter. Thus S is true but it is not a low pass filter. So S is false. Hence (A) is correct answer. MCQ.4 Consider a system whose input x and output y are related by the equation 3 yt () xt ( τ) g( τ) dτ where ht () is shown in the graph. # 3 Which of the following four properties are possessed by the system? BIBO : Bounded input gives a bounded output. Causal : The system is causal, LP : The system is low pass. LTI : The system is linear and time-invariant. (A) Causal, LP (B) BIBO, LTI (C) BIBO, Causal, LTI (D) LP, LTI SOL.4 Here ht ()! 0 for t < 0. Thus system is non causal. Again any bounded input xt () gives bounded output yt. () Thus it is BIBO stable. SCO 8-9, 4th Floor, Sector 34-A, Chandigarh, 600 Ph.: Mobile:

21 GATE/EC-009 EC : ELECTRONICS AND COMMUNICATION ENGINEERING Page Here we can conclude that option (B) is correct. Hence (B) is correct answer. MCQ.4 The 4-point Discrete Fourier Transform (DFT) of a discrete time sequence {,0,,3} is (A) [0, + j,, j ] (B) [, + j, 6, j ] (C) [6, 3j,, + 3j ] (D) [6, + 3j, 0, 3j ] SOL.4 MCQ.43 Hence (D) is correct answer We have xn [ ] {,,,) 03 and N 4 For N 4, [] N jπnk/ N Xk [] / xne [ ] k 0,... N n 0 3 jπnk/ 4 Xk / xne [ ] k 0,,... 3 n 0 3 / Now X[ 0 ] xn [ ] n 0 x[ 0] + x[ ] + x[ ] + x[ 3] / n 0 x[ ] xne [ ] jπn/ x[0] + x[] e + x[] e + x[3] e j3 + j3 3 / n 0 X[ ] xne [ ] j π n jπ / jπ jπ3 / x[0] + x[] e + x[] e + x[3] e / n 0 X[ 3 ] xne [ ] j3πn/ j π j π j π3 x[0] + x[] e + x[] e + x[3] e + 0 j3 j3 Thus [ 6, + j3, 0, j3] j3π / j3π j9π / The feedback configuration and the pole-zero locations of Gs () s s+ s + s + are shown below. The root locus for negative values of k, i.e. for 3 < k < 0, has breakaway/break-in points and angle of departure at pole P (with respect to the positive real axis) equal to SCO 8-9, 4th Floor, Sector 34-A, Chandigarh, 600 Ph.: Mobile:

22 GATE/EC-009 EC : ELECTRONICS AND COMMUNICATION ENGINEERING Page (A)! and 0c (B)! and 45c (C)! 3 and 0c (D)! 3 and 45c SOL.43 The characteristic equation is + GsHs () () 0 or Ks ( s+ ) + 0 s + s+ or s + s+ + K( s s+ ) 0 or K s + s+ s s For break away & break in point differentiating above w.r.t. s we have dk ( s s+ )( s+ ) ( s + s+ )( s ) 0 ds ( s s+ ) Thus ( s s+ )( s+ ) ( s + s+ )( s ) 0 or s! Let θ d be the angle of departure at pole P, then θd θp+ θz+ θz 80c θ d 80c ( θp+ θz+ θ) 80c ( 90c c) 45c Hence (B) is correct option. s MCQ.44 An LTI system having transfer function + and input xt () sin ( t+ ) is in s + s+ steady state. The output is sampled at a rate ω s rad/s to obtain the final output { xk ( )}. Which of the following is true? (A) y (.) is zero for all sampling frequencies ω s (B) y (.) is nonzero for all sampling frequencies ω s (C) y (.) is nonzero for ω >, but zero for < s (D) y (.) is zero for ω >, but nonzero for < s SCO 8-9, 4th Floor, Sector 34-A, Chandigarh, 600 Ph.: Mobile: ω s ω

23 GATE/EC-009 EC : ELECTRONICS AND COMMUNICATION ENGINEERING Page 3 SOL.44 MCQ.45 SOL.45 MCQ.46 Hence (A) is correct answer. The unit step response of an under-damped second order system has steady state value of -. Which one of the following transfer functions has theses properties? (A) 4. (B) 38. s s+. s + 9. s+ 9. (C) 4. (D) 38 s 59. s+. s 9. s+ 9. For under-damped second order response Ts () k n ω where ξ < s + ξωns + ωn Thus (A) or (B) may be correct For option (A) ω n. and ξωn. 59 " ξ. For option (B) ω n 9. and ξωn. 5 " ξ Hence (B) is correct option. A discrete random variable X takes values from to 5 with probabilities as shown in the table. A student calculates the mean X as 3.5 and her teacher calculates the variance of X as.5. Which of the following statements is true? SOL.46 k PX ( k) (A) Both the student and the teacher are right (B) Both the student and the teacher are wrong (C) The student is wrong but the teacher is right (D) The student is right but the teacher is wrong Hence (B) is correct option. The mean is X Σxp i i() x # 0. + # # # # X Σxp i i() x # # # # # Variance σ x X ^Xh 0. ( 3). MCQ.47 A message signal given by mt ( ) ( ) cos ωt ( ) sin ωt amplitude - modulated with a carrier of frequency ω C to generator st ( )[ + mt ( )]cos ωc t. What is the power efficiency achieved by this modulation scheme? (A) 833. % (B). % (C) 0 % (D) 5% SCO 8-9, 4th Floor, Sector 34-A, Chandigarh, 600 Ph.: Mobile:

24 GATE/EC-009 EC : ELECTRONICS AND COMMUNICATION ENGINEERING Page 4 SOL.47 MCQ.48 Hence (C) is correct option. mt () cos ω t sin ωt sam () t [ + mt ( )]cos ωc t Modulation index mt () max Vc m `j + ` j η m ` 00% # m + ` j j # 00% 0% + A communication channel with AWGN operating at a signal to noise ration SNR >> and bandwidth B has capacity C. If the SNR is doubled keeping constant, the resulting capacity C is given by (A) C. C (B) C. C+ B (C) C. C + B (D) C. C + 0.3B SOL.48 Hence (B) is correct option. We have C B log S ` + N j. B log S `N j As S N If we double the S ratio then N >> MCQ.49 SOL.49 C. B log S ` N j. Blog + Blog. B+ C S N A magnetic field in air is measured to be B B x y 0 c yt xt x y m + x + y What current distribution leads to this field? [Hint : The algebra is trivial in cylindrical coordinates.] (A) J Bz 0 t, r! 0 μ 0 c x + y m (B) J Bz 0 t, r! 0 μ 0 c x + y m (C) J 0, r! 0 (D) J Bz 0 t, r! 0 μ 0 c x + y m Hence (C) is correct option. We have B v B x y 0 c ay ax x y + x + y m...() To convert in cylindrical substituting SCO 8-9, 4th Floor, Sector 34-A, Chandigarh, 600 Ph.: Mobile:

25 GATE/EC-009 EC : ELECTRONICS AND COMMUNICATION ENGINEERING Page 5 x r cos φ and y rsin φ a x cos φar sin φaφ and a y sin φar + cos φaφ In () we have B v Ba v 0 φ Now H v Bv Ba v 0 φ constant μ0 μ0 J v 4# H v 0 since H is constant MCQ.50 A transmission line terminates in two branches, each of length 4 λ, as shown. The branches are terminated by 50Ω loads. The lines are lossless and have the characteristic impedances shown. Determine the impedance Z i as seen by the source. SOL.50 (A) 00Ω (B) 00Ω (C) 50Ω (D) 5Ω The transmission line are as shown below. Length of all line is λ 4 Z i Z i Ω Z Z 50 L Ω Z Z 50 L Z L3 Zi Zi 00Ω 00Ω 00Ω Z i Z 50 5 Z 00 0 Ω L3 Hence (D) is correct option. Common Date for Question 5 and 5 : SCO 8-9, 4th Floor, Sector 34-A, Chandigarh, 600 Ph.: Mobile:

26 GATE/EC-009 EC : ELECTRONICS AND COMMUNICATION ENGINEERING Page 6 Consider a silicon p n junction at room temperature having the following parameters: Doping on the n-side # 0 7 cm -3 Depletion width on the n-side 0μ. m Depletion width on the p side 0μ. m Intrinsic carrier concentration 4. # 0 0 cm -3 Thermal voltage 6 mv - Permittivity of free space 885. # 0 4 F.cm - Dielectric constant of silicon MCQ.5 SOL.5 MCQ.5 The built-in potential of the junction (A) is 0.70 V (B) is 0.76 V (C) is 0.8 V (D) Cannot be estimated from the data given Hence option (B) is correct. We know that N A W P N D W N 7 6 or N NDWN A # 0 # 0. # 0 6 # 0 WP # 0 The built-in potential is V bi V T n NAND c n m i ln 0 0 # # # # e 0 o (. 4# 0 ) The peak electric field in the device is (A) 0.5 MV. cm -, directed from p region to n region (B) 0.5 MV. cm -, directed from n region to p region (C) 80. MV. cm -, directed from p-retion to n region (D).80 MV. cm -, directed from n region to p region 6 SOL.5 The peak electric field in device is directed from p to n and is E endxn from p to n εs endxn from n to p εs # 0 # # 0 # # MV/cm 885. # 0 # Hence option (B) is correct. SCO 8-9, 4th Floor, Sector 34-A, Chandigarh, 600 Ph.: Mobile:

27 GATE/EC-009 EC : ELECTRONICS AND COMMUNICATION ENGINEERING Page 7 Common Data for Questions 53 and 54 : The Nyquist plot of a stable transfer function Gs () is shown in the figure are interested in the stability of the closed loop system in the feedback configuration shown. MCQ.53 Which of the following statements is true? (A) Gs () is an all-pass filter (B) Gs () has a zero in the right-half plane (C) Gs () is the impedance of a passive network (D) Gs () is marginally stable SOL.53 The plot has one encirclement of origin in clockwise direction. Thus Gs () has a zero is in RHP. Hence (B) is correct option. MCQ.54 The gain and phase margins of Gs () for closed loop stability are (A) 6 db and 80c (B) 3 db and 80c (C) 6 db and 90c (D) 3 db and 90c SOL.54 The Nyzuist plot intersect the real axis ate Thus G. M. 0 log x 0 log db And its phase margin is 90c. Hence (C) is correct option. Common data for Questions 55 & 56 : The amplitude of a random signal is uniformly distributed between -5 V and 5 V. MCQ.55 SOL.55 If the signal to quantization noise ratio required in uniformly quantizing the signal is 43.5 db, the step of the quantization is approximately (A) V (B) 0.05 V (C) V (D) 0.0 V Hence (C) is correct option. We have SNR n SCO 8-9, 4th Floor, Sector 34-A, Chandigarh, 600 Ph.: Mobile:

28 GATE/EC-009 EC : ELECTRONICS AND COMMUNICATION ENGINEERING Page 8 or n 6 n n $ n. 7 No. of quantization level is 7 8 Step size required is VH VL 5 ( 5) MCQ.56 If the positive values of the signal are uniformly quantized with a step size of 0.05 V, and the negative values are uniformly quantized with a step size of 0. V, the resulting signal to quantization noise ration is approximately (A) 46 db (B) 43.8 db (C) 4 db (D) 40 db 0 8 SOL.56 For positive values step size s V For negative value step size s - 0. V No. of quantization in + ive is s 005. Thus n + 00 $ n 7 No. of quantization in ve Q s Thus n- 50 $ n 6 S `N j n db + S `N j n db - Best S `N j db 0 Hence (B) is correct option. Statement for Linked Answer Question 57 and 58 : Consider for CMOS circuit shown, where the gate voltage v 0 of the n-mosfet is increased from zero, while the gate voltage of the p MOSFET is kept constant at 3 V. Assume, that, for both transistors, the magnitude of the threshold voltage is V and the product of the trans-conductance parameter is ma. V - SCO 8-9, 4th Floor, Sector 34-A, Chandigarh, 600 Ph.: Mobile:

29 GATE/EC-009 EC : ELECTRONICS AND COMMUNICATION ENGINEERING Page 9 MCQ.57 SOL.57 For small increase in V G beyond V, which of the following gives the correct description of the region of operation of each MOSFET (A) Both the MOSFETs are in saturation region (B) Both the MOSFETs are in triode region (C) n-mosfets is in triode and p MOSFET is in saturation region (D) n- MOSFET is in saturation and p MOSFET is in triode region For small increase in V G beyond V the n channel MOSFET goes into saturation as VGS "+ ive and p MOSFET is always in active region or triode region. Hence (D) is correct option. MCQ.58 Estimate the output voltage V 0 for VG 5. V. [Hints : Use the appropriate currentvoltage equation for each MOSFET, based on the answer to Q.57] (A) 4 (B) 4 + (C) 4 3 (D) SOL.58 Hence (C) is correct option. Statement for Linked Answer Question 59 & 60 : Two products are sold from a vending machine, which has two push buttons P and P. When a buttons is pressed, the price of the corresponding product is displayed in a 7 - segment display. If no buttons are pressed, ' 0 ' is displayed signifying Rs 0. If only P is pressed, is displayed, signifying Rs. If only P is pressed 5 is displayed, signifying Rs. 5 If both P and P are pressed, ' E ' is displayed, signifying Error The names of the segments in the 7 - segment display, and the glow of the display for 0,, 5 and E are shown below. SCO 8-9, 4th Floor, Sector 34-A, Chandigarh, 600 Ph.: Mobile:

30 GATE/EC-009 EC : ELECTRONICS AND COMMUNICATION ENGINEERING Page 30 Consider () push buttons pressed/not pressed in equivalent to logic /0 respectively. () a segment glowing/not glowing in the display is equivalent to logic /0 respectively. MCQ.59 SOL.59 If segments a to g are considered as functions of P and P, then which of the following is correct (A) g P + P, d c+ e (B) g P + P, d c+ e (C) g P + P, e b+ c (D) g P + P, e b+ c The given situation is as follows The truth table is as shown below P P a b c d e f g From truth table we can write a b PP + PP P NOT Gate c PP + PP P NOT Gate d c+ e and c PP P+ P OR GATE f PP P+ P OR GATE g PP P+ P OR GATE Thus we have g P+ P and d c+ e. It may be observed easily from figure that Led g does not glow only when both P and P are 0. Thus SCO 8-9, 4th Floor, Sector 34-A, Chandigarh, 600 Ph.: Mobile:

31 GATE/EC-009 EC : ELECTRONICS AND COMMUNICATION ENGINEERING Page 3 g P+ P LED d is all condition and also it depends on d c+ e Hence (B) is correct answer. MCQ.60 SOL.60 What are the minimum numbers of NOT gates and - input OR gates required to design the logic of the driver for this 7 - Segment display (A) 3 NOT and 4 OR (B) NOT and 4 OR (C) NOT and 3 OR (D) NOT and 3 OR As shown in previous solution NOT gates and 3-OR gates are required. Hence (D) is correct answer. Answer Sheet. (B) 3. (C) 5. (A) 37. (A) 49. (C). (A) 4. (A) 6. (D) 38. (C) 50. (D) 3. (B) 5. (A) 7. (B) 39. (A) 5. (B) 4. (C) 6. (D) 8. (B) 40. (A) 5. (B) 5. (C) 7. (C) 9. (C) 4. (B) 53. (B) 6. (C) 8. (C) 30. (A) 4. (D) 54. (C) 7. (A) 9. (C) 3. (A) 43. (B) 55. (C) 8. (C) 0. (D) 3. (D) 44. (A) 56. (B) 9. (*). (C) 33. (A) 45. (B) 57. (D) 0. (A). (D) 34. (D) 46. (B) 58. (C). (C) 3. (B) 35. (D) 47. (C) 59. (B). (*) 4. (B) 36. (D) 48. (B) 60. (D) SCO 8-9, 4th Floor, Sector 34-A, Chandigarh, 600 Ph.: Mobile: