Q. No. 1 5 Carry One Mark Each. 1. Choose the most appropriate word from the options given below to complete the following sentence.

Transcription

1 EC-GATE-4 PAPER- Q. No. 5 Carry One Mark Each. Choose the most appropriate word from the options given below to complete the following sentence. Answer: (B) Communication and interpersonal skills are important in their own ways. (A) each (B) both (C) all (D) either. Which of the options given below best completes the following sentence? She will feel much better if she. (A) will get some rest (B) gets some rest (C) will be getting some rest (D) is getting some rest Answer: (B) 3. Choose the most appropriate pair of words from the options given below to complete the following sentence. Answer: (C) She could not the thought of the election to her bitter rival. (A) bear, loosing (B) bare, loosing (C) bear, losing (D) bare, losing 4. A regular die has si sides with numbers to 6 marked on its sides. If a very large number of throws show the following frequencies of occurrence:.67;.67; 3.5; 4.66; 5.68; 6.8. We call this die Answer: (B) (A) irregular (B) biased (C) Gaussian (D) insufficient For a very large number of throws, the frequency should be same for unbiased throw. As it not same, then the die is baised. 5. Fill in the missing number in the series Answer: nd number is in increa sin g order as shown above st number

2 EC-GATE-4 PAPER- Q. No. 6 Carry One Mark Each 6. Find the odd one in the following group Q,W,Z,B B,H,K,M W,C,G,J M,S,V,X (A) Q,W,Z,B (B) B,H,K,M (C) W,C,G,J (D) M,S,V,X Answer: (C) a W Z B B H K N W C G J M S V X Lights of four colors (red, blue, green, yellow) are hung on a ladder. On every step of the ladder there are two lights. If one of the lights is red, the other light on that step will always be blue. If one of the lights on a step is green, the other light on that step will always be yellow. Which of the following statements is not necessarily correct? Answer: (D) (A) The number of red lights is equal to the number of blue lights (B) The number of green lights is equal to the number of yellow lights (C) The sum of the red and green lights is equal to the sum of the yellow and blue lights (D) The sum of the red and blue lights is equal to the sum of the green and yellow lights 8. The sum of eight consecutive odd numbers is 656. The average of four consecutive even numbers is 87. What is the sum of the smallest odd number and second largest even number? Answer: 63 Eight consecutive odd number =656 a-6, a-, a-, a,a+,a+4, a+6 a+8=656 a=8 Smallest m=75 () Average consecutive even numbers a + a + a + + a + 4 = 87 4 a = 86 Second largest number =88 +=63 9. The total eports and revenues from the eports of a country are given in the two charts shown below. The pie chart for eports shows the quantity of each item eported as a percentage of the total quantity of eports. The pie chart for the revenues shows the percentage of the total revenue generated through eport of each item. The total quantity of eports of all the items is 5 thousand tonnes and the total revenues are 5 crore rupees. Which item among the following has generated the maimum revenue per kg?

3 EC-GATE-4 PAPER- Eports Revenues Item5 % Item 6 6% Item4 Item % Item % Item3 Item 5 % Item 6 9% Item % Item % % 9% Item 4 6% Item3 3% (A) Item (B) Item 3 (C) Item 6 (D) Item 5 Answer: (D) Item: Item: = 5 = Item = Item 3 Item: 6 Item:5 9.8 Item 6 6 = = 5 = =.6.6 = Item 5 3. It takes 3 minutes to empty a half-full tank by draining it at a constant rate. It is decided to simultaneously pump water into the half-full tank while draining it. What is the rate at which water has to be pumped in so that it gets fully filled in minutes? (A) 4 times the draining rate (B) 3 times the draining rate (C).5 times the draining rate (D) times the draining rate Answer: (A) V = 3(s) drawing rate = s half Total volume =6 S tank (s )() (s) = 3s s (s) s = 3s s = 4s s = 4drawing rate 7 3 3

4 EC-GATE-4 PAPER- Q. No. 5 Carry One Mark Each. The determinant of matri A is 5 and the determinant of matri B is 4. The determinant of matri AB is. Answer: AB = A. B = ( 5 ).( 4) =. Let X be a random variable which is uniformly chosen from the set of positive odd numbers less than. The epectation E[X]is. Answer:5 X =,3,5,...,99 n = 5 ( number of observations) n i [ ] E = = = 5 = 5 n 5 5 i= 3. For t, t t < the maimum value of the function f ( t) e e = occurs at (A) t = log e 4 (B) t = log e (C) t = (D) t = log e 8 Answer: (A) ' t t f t e = + 4e = e 4e e = t = log 4 and f t < at t = log t t t 4 e '' 4 e 4. The value of lim + is (A) ln (B). (C) e (D) Answer: (C) lim + = e ( standard limit ) 5. If the characteristic equation of the differential equation Answer: (A) d y dy + α + y = d d has two equal roots, then the values of α are ( A) ± B, ( C) For equal roots, Discriminant B 4AC = α = 4 4 α = ± ± j ( D) ± / 4

5 EC-GATE-4 PAPER Norton s theorem states that a comple network connected to a load can be replaced with an equivalent impedance Answer: (D) (A) in series with a current source (C) in series with a voltage source Norton s theorem (B) in parallel with a voltage source (D) in parallel with a current source I N Zequ Load 7. In the figure shown, the ideal switch has been open for a long time. If it is closed at t=, then the magnitude of the current (in ma) through the 4kΩ resistor at t = + is. 5kΩ 4 kω kω V + i µ F t = mh Answer:. ma For t = o+ i( o + ) =.ma 9K i o +.ma 5 kω 4 kω V + i ( + ) 8. A silicon bar is doped with donor impurities N D =.5 5 atoms / cm 3. Given the intrinsic carrier concentration of silicon at T = 3 K is n i =.5 cm -3. Assuming complete impurity ionization, the equilibrium electron and hole concentrations are (A) n =.5 6 cm -3, p =.5 5 cm -3 (B) n =.5 cm -3, p =.5 5 cm -3 (C) n =.5 5 cm -3, p =.5 cm -3 (D) n =.5 5 cm -3, p = 5 cm -3 Answer: (D) N =.5 Atom / cm D h =.5 / cm i Since complete ionization taken place, h = N =.5 / cm P D (.5 ) ni 5 n = = = / cm 5 3 5

6 EC-GATE-4 PAPER An increase in the base recombination of a BJT will increase (A) the common emitter dc current gain β (B) the breakdown voltage BV CEO (C) the unity-gain cut-off frequency f T (D) the transconductance g m Answer: (B). In CMOS technology, shallow P-well or N-well regions can be formed using (A) low pressure chemical vapour deposition (B) low energy sputtering (C) low temperature dry oidation (D) low energy ion-implantation Answer: (D). The feedback topology in the amplifier circuit (the base bias circuit is not shown for simplicity) in the figure is V CC (A) Voltage shunt feedback R c I o V o (B) Current series feedback (C) Current shunt feedback (D) Voltage series feedback Answer: (B) By opening the output feed back signed becomes zero. Hence it is current sampling. As the feedback signal v f is subtracted from the signal same v s it is series miing.. In the differential amplifier shown in the figure, the magnitudes of the common-mode and Answer: (B) differential-mode gains are A cm and A d, respectively. If the resistance R E is increased, then (A) A cm increases (B) common-mode rejection ratio increases (C) A d increases (D) common-mode rejection ratio decreases A d does not depend on R E A cm decreases as R E is increased A A d CMRR = = cm Increases + V i R S V S ~ R C R E + V I V EE R E V CC R C 6

7 EC-GATE-4 PAPER A cascade connection of two voltage amplifiers A and A is shown in the figure. The openloop gain A v, input resistance R in, and output resistance R O for A and A are as follows: A:A =,R = k Ω,R = kω v in A : A = 5,R = 5k Ω, R = Ω v in The approimate overall voltage gain V out / V in is. + V A in A R L kω + V out Answer: 34.7 V Z i R L Overall voltage gain, Av = = AV A V Vi Zi + Z R L + Z 5k k = 5 5k k k + + A = 34.7 V 4. For an n-variable Boolean function, the maimum number of prime implicants is Answer: (D) (A) (n-) (B) n/ (C) n (D) (n-) For an n-variable Boolean function, the maimum number of prime implicants ( n ) = 5. The number of bytes required to represent the decimal number in packed BCD (Binary Coded Decimal) form is. Answer: 4 In packed BCD (Binary Coded Decimal) typically encoded two decimal digits within a single byte by taking advantage of the fact that four bits are enough to represent the range to 9. So, is required 4-bytes to stored these BCD digits 6. In a half-subtractor circuit with X and Y as inputs, the Borrow (M) and Difference (N = X - Y) are given by Answer: (C) (A) M = X, Y, N = XY (B) M = XY, N = X Y (C) M = X Y, N = X Y (D) M = XY N = X Y 7

8 Function Table for Half-subtractor is X Y Difference (N) Borrow (M) Hence, N = X Y and m = XY Hence, N = X Y and m = XY EC-GATE-4 PAPER An FIR system is described by the system function Answer: (C) 7 3 H ( z) = + z + z The system is (A) maimum phase (B) minimum phase (C) mied phase (D) zero phase Minimum phase system has all zeros inside unit circle maimum phase system has all zeros outside unit circle mied phase system has some zero outside unit circle and some zeros inside unit circle. 7 3 for H s z z = + + One zero is inside and one zero outside unit circle hence mied phase system 8. Let [n] = [-n]. Let X(z) be the z-transform of [n]. If.5 + j.5 is a zero o X(z), which one of the following must also be a zero of X(z). (A).5 - j.5 (B) /(.5 + j.5) (C) /(.5 - j.5) (D) + j4 Answer: (B) Given [ n] = [ n] [ ] z = z Time reversal property in z transform if one zero is.5 + j.5 then other zero will be.5 + j.5 9. Consider the periodic square wave in the figure shown. 3 4 t 8

9 EC-GATE-4 PAPER- The ratio of the power in the 7 th harmonic to the power in the 5 th harmonic for this waveform is closest in value to. Answer:.5 For a periodic sequence wave, nth harmonic component is α n power in nth harmonic component is α n Ratio of the power in 7 th harmonic to power in 5 th harmonic for given waveform is 7 5 = The natural frequency of an undamped second-order system is 4 rad/s. If the system is damped with a damping ratio.3, the damped natural frequency in rad/s is. Answer: 38.5 r / sec Given ω = 4 r / sec ξ =.3 ω = ω d d d n n ξ ω = 4.3 ω = 38.5 r / sec. For the following sytem, ( s ) + s s + ( s) + + s Y ( s) When X ( s) =, the transfer function ( s) y s is Answer: (D) s + ( A) s If X ( s) = Y( s) X ( s) B s ; The block diagram becomes + C s + s s X ( s) ( + ) + D s + s s s ( + ) Y ( s) Y( s) s = s = s X ( s) s +. ( s + ) / s + s s + s s ( + ) ( + ) + S ( S + ) 9

10 EC-GATE-4 PAPER- The capacity of a band-limited additive white Gaussian noise (AWGN) channel is given by P C = Wlog + bits per second (bps), where W is the channel bandwidth, P is the σ w average power received and fied P σ =, approimately (A).44 (B).8 (C).7 (D).36 Answer: (A) σ is the one-sided power spectral density of the AWGN. For a, the channel capacity (in kbps) with infinite bandwidth ( W ) is P ω ln + P σ ω C = lim ω log lim w + = ω σ ω ln P P ln ln + + σ ω P P σ ω = lim. = lim ln ω P σ σ ln ω P σ ω σ ω [ + ].ln σ This limit is equivalent to ln P P lim = = = ln e =.44 KGpa ω σ 3. Consider sinusoidal modulation in an AM system. Assuming no overmodulation, the µ when the maimum and minimum values of the envelope, modulation inde respectively, are 3 V and V, is. Answer:.5 A t µ = A t ma + ma 3 µ = = = A t min A t min 4. To maimize power transfer, a lossless transmission line is to be matched to a resistive load impedance via a / 4 λ transformer as shown. lossless transmission line λ / 4 transformer ZL = 5Ω ZL = Ω Answer: 7.7Ω The characteristic impedance inω of the λ / 4 transformer is.

11 Here impedance is matched by using QWT ( λ 4) ' Z = ' Z Z L in = 5 = 5 = Z = 7.7Ω EC-GATE-4 PAPER Which one of the following field patterns represents a TEM wave travelling in the positive direction? Answer: (B) ( A) E = + 8y, ˆ H = 4zˆ ( B) E = y, ˆ H = 3zˆ ( C) E + z, ˆ H = + yˆ ( D) E = 3y, ˆ H = + 4zˆ For TEM wave Electric field (E), Magnetic field (H) and Direction of propagation (P) are orthogonal to each other. Here P = + a By verification E = a, H = 3a y E H = a a = + a P y z z Q. No Carry Two Marks Each 6. The system of linear equations 3 a 5 3 b = 4 has 5 c 4 (A) a unique solution (B) infinitely many solutions (C) no solution (D) eactly two solutions Answer: (B) [ ] 3 5 A / B = R R 3R R3 + R R 3 R 3 R Since, rank A = rank A / B < number of unknowns Equations have infinitely many solutions.

12 EC-GATE-4 PAPER The real part of an analytic function f(z) where z =. + jy is given by e -y cos(). The imaginary part of f(z) is Answer: (B) (A) e y cos() (B) e -y sin() (C) -e y sin() (D) e -y sin() y real part u = e cos and V =? v v dv = d + dy y u u y y y = d + dy( U sin g C R equations) = e cos d e sin dy = d e sin y y Integrating, we get V = e sin 8. The maimum value of the determinant among all real symmetric matrices with trace 4 is. Answer: 49 y General real symmetric matri is z det = yz and trace is y + z = 4 given z = 4 y... * Let f = yz det = y + 4y u sin g * Using maima and minima of a function of two variables, we have f is maimum at =, y = 7 and therefore, maimum value of the determinant is If r = aˆ + yaˆ + zaˆ and r = r, then div r ( ln r) y z Answer: 3 r ( ln r) = div ( r ( ln r) ) = div( r ) = 3 r r ˆ ˆ ln r = a ln r = a ˆ = a = r r r r =. 3. A series LCR circuit is operated at a frequency different from its resonant frequency. The operating frequency is such that the current leads the supply voltage. The magnitude of current is half the value at resonance. If the values of L, C and R are H, F and Ω, respectively, the operating angular frequency (in rad/s) is. Answer:.45 r/sec The operating frequency (w ), at which current leads the supply. i.e., ω <ω r again magnitude of current is half the value at resonance

13 EC-GATE-4 PAPER- i,e.,. at ω = ω I = V z V at ω=ω Iresonance = R Iresonance I = V V i.e., = = Z = R z R Given R = Ω; L=H; C=F ωc Z = R + ω L = = + ω = ωc R L 4 By substituting R, L & C values,, ω = ω = = ω ω Assume ω =, then, + = = = 4.79,.8 if = 4.79 ω=.8 r sec if =.8 ω=.45r sec But ω <ω r So, operating frequency ω =.45 r sec 3. In the h-parameter model of the -port network given in the figure shown, the value of h (in S) is. 3Ω 3Ω 3Ω ' Ω Ω Ω ' Answer:.4 If two, π n ws are connected in parallel, The y-parameter are added 3

14 EC-GATE-4 PAPER- i.e., y = y + y equ 3 3 y = y = yequ = y y y h = y y y y where y = y y y y The value of h = y = y =.833 y 5 = h = In the figure shown, the capacitor is initially uncharged. Which one of the following epressions describes the current I(t) (in ma) for t >? R 5V + kω R kω I C µ F Answer: (A) 5 t / τ t / τ ( A) I( t) = ( e ), τ = msec B I t = e, τ = msec 3 5 t / τ 5 t / τ ( C) I( t) = ( e ), τ = 3msec ν = = + [ ] t V t V V V e c R final initial final τ = R equ.cequ 3 R equ = K K KΩ 3 C = µ F equ 3 6 t τ D I t = e, τ = 3 msec 4

15 EC-GATE-4 PAPER- τ = V initial msec 3 = volts Vfinal = Vs.s = 5. = volts 3 3 t ν R ( t) = e τ 3 3 t ν t R t 5 τ τ ν R ( t) = e volts i R (t) = = e ma 3 K In the magnetically coupled circuit shown in the figure, 56 % of the total flu emanating from one coil links the other coil. The value of the mutual inductance (in H) is. Ω M ( + ) 6cos 4t 3 V ~ 4 H 5H ( /6) F Answer:.49 Henry Given 56% of the total flu emanating from one coil links to other coil. i.e, K = 56%.56 We have, K = M L L L = 4H; L = 5H M =.56 m =.5H 34. Assume electronic charge q =.6-9 C, kt/q = 5 mv and electron mobility µ n = cm /V-s. If the concentration gradient of electrons injected into a P-type silicon sample is /cm 4, the magnitude of electron diffusion current density (in A/cm ) is. Answer: 4 9 kj Given q =.6 ; =.5 mv, µ n = cm / v s q Dn kj From Einstein relation, = µ n q Dn = 5mV cm / v S 5cm / s Diffuion current Density J = =.6 5 = 4A / cm 9 dn q Dn d 5

16 EC-GATE-4 PAPER Consider an abrupt PN junction (at T = 3 K) shown in the figure. The depletion region ε is width X n on the N-side of the junction is. µm and the permittivity of silicon si.44 - F/cm. At the junction, the approimate value of the peak electric field (in kv/cm) is. P + region N region X N A >> N 6 3 D N D = / cm Answer: 3.66 Given =.µ m, =.44 F / µ N = / cm D n Si n 6 3 q ND Peak Electric field, E = = = 3.66KV / cm.44 n 36. When a silicon diode having a doping concentration of N A = 9 6 cm -3 on p-side and N D = 6 cm -3 on n-side is reverse biased, the total depletion width is found to be 3µ m. Given that the permittivity of silicon is.4 F/cm, the depletion width on the p-side and the maimum electric field in the depletion region, respectively, are Answer: (B) (A).7 µ m and.3 5 V/cm (B).3 µ m and V/cm (C).3 µ m and.4 5 V/cm (D). µ m and.4 5 V/cm Given N = 9 / cm ; N = / cm A.4 F / cm n A = = p D n p D Total depletion width = + = 3µ m = N 9 N = Total Depletion width, + = 3µ m n p n 9 + = 3µ m p p p p =.3µ m 9 6 qna p.6 9.3µ m Ma. Electric field, E = =.4 5 = 4.5 V / cm 6

17 EC-GATE-4 PAPER The diode in the circuit shown has V on =.7 Volts but is ideal otherwise. If V i = 5sin ( ωt) Volts, the minimum and maimum values of V O (in Volts) are, respectively, kω V i R R kω V o + V (A) -5 and.7 (B).7 and 5 (C) -5 and 3.85 (D).3 and 5 Answer: (C) When Vi makes Diode 'D' OFF, V = Vi V min = 5V When V makes diode 'D' ON, i ( V.7 ) V = + V + V i on R + R ( ) 5.7 k V ( ma) = V k + k = 3.85V 38. For the n-channel MOS transistor shown in the figure, the threshold voltage V Th is.8 V. Neglect channel length modulation effects. When the drain voltage V D =.6 V, the drain current I D was found to be.5 ma. If V D is adjusted to be V by changing the values of R and V DD, the new value of I D (in ma) is V DD R D G S (A).65 (B).75 (C).5 (D).5 Answer: (C) 7

18 EC-GATE-4 PAPER- Given VTh =.8V When V =.6V, I =.5mA = µ cos w V V L D D n DS Th ω µ n cos =.785 A / V L When V = V D ω I = µ cos V V L = D n DS Th [ Device is in sat] mA 39. For the MOSFETs shown in the figure, the threshold voltage Vt = V and W K = µ C =. ma / V. The value of ID (in ma) is. L VDD = + V R kω R kω I D Answer:.9 W Given Vt = V, K = µ cos =.A / V L W ID = I D = µ n cos VGs V t L.mA / V 5 = =.9mA VDD = 5V K K V I D 5V 8

19 EC-GATE-4 PAPER In the circuit shown, choose the correct timing diagram of the output (y) from the given waveforms W, W, W3 and W4. X D Q Clk FF > Q output ( y) X D Q > Q ClK X X W W W3 W4 (A) W (B) W (C) W3 (D) W4 Answer: (C) This circuit has used negative edge triggered, so output of the D-flip flop will changed only when CLK signal is going from HIGH to LOW ( to ) C LK X X Y( w 3 ) 9

20 EC-GATE-4 PAPER- This is a synchronous circuit, so both the flip flops will trigger at the same time and will respond on falling edge of the Clock. So, the correct output (Y) waveform is associated to w 3 waveform. 4. The outputs of the two flip-flops Q, Q in the figure shown are initialized to,. The sequence generated at Q upon application of clock signal is J Q > K Q Q J Q > KQ (A) (B) (C) (D) Answer: (D) Clock Initial st CP nd CP 3 rd CP 4 th CP So, the output sequence generated at Q is. 4. For the 885 microprocessor, the interfacing circuit to input 8-bit digital data (DI DI 7 ) from an eternal device is shown in the figure. The instruction for correct data transfer is (A) MVI A, F8H (B) IN F8H (C) OUT F8H (D) LDA F8F8H CLK ( ) J Q K ( Q ) J Q ( ) K Q Q Q A A A 3 to 8 Decoder C B A / M RD A 3 A 4 A 5 A 6 A 7 GA GB G Digital Inputs A 8 A 9 A A A A 3 A 4 A 5 I / O Device 7 7 DI DI DO D DS DS Data Bus ( D D ) 7

21 Answer: (D) EC-GATE-4 PAPER- This circuit diagram indicating that it is memory mapped I/O because to enable the 3-to-8 decoder G A is required active low signal through ( Io m ) and G B is required active low through ( R D ) it means I/o device read the status of device LDA instruction is appropriate with device address. Again to enable the decoder o/p of AND gate must be and Ds signal required is which is the o/p of multi-i/p AND gate to enable I/O device. So, A A A A A A A A A A A A A A A A F 8 F 8 Device address = F8F8H The correct instruction used LDA F8F8H 43. Consider a discrete-time signal n for n [ n] = otherwise If y[n] is the convolution of [n] with itself, the value of y[4] is. Answer: Given [ n] [ ] = [ n ]* [ n] y n n n for n = elsewhere [ ] = [ ] [ ] y n k. n k k= 4 [ ] [ ] [ ] y 4 = k. G k k= = = = 44. The input-output relationship of a causal stable LTI system is given as y[ n] = αy[ n ] + β [ n] If the impulse response h[n] of this system satisfies the condition = [ ] relationship between α and β is ( A) α = β / ( B) α = + β / ( C) α = β ( D) α = β Σ h n, n = the

22 EC-GATE-4 PAPER- Answer: (A) Given system equation as [ ] = α [ ] + β [ ] y n y n n y z β = z α z β H( z) = α z h [ ] = β( α) [ ] [ ] h n u n causal system h= [ ] Also given that h n = β = α β α = β α = 45. The value of the integral sin c ( 5t) Answer:. dt is. We can use pasrevalis theorem sin 5πt Let ( t) sin ( 5t) = 5πt in frequency domain X ( f 5 ).5.5 f Now, t dt t df.5 = = 5.5 = 5 = =. 5 5

23 EC-GATE-4 PAPER An unforced liner time invariant (LTI) system is represented by Answer: (C) = If the initial conditions are () = and () = -, the solution of the state equation is ( A) ( t) =, ( t) = t B t = e, t = e t t ( C) ( t) = e, ( t) = e t Solution of state equation of X( t) = L SI A.X ( ) X( ) = A = S + = S + [ SI A] [ SI A] L SI A L SI A S + = ( S + )( S + ) S + S + = S + L S + ( ) = L S + t e = t e t X t e = t X t e t X t e X t = e = t X X t e t = e t t t D t = e, t = e t 3

24 EC-GATE-4 PAPER The Bode asymptotic magnitude plot of a minimum phase system is shown in the figure. G ( jω) ( db) ω ( rad / s) in log scale Answer:.5 If the system is connected in a unity negative feedback configuration, the steady state error of the closed loop system, to a unit ramp input, is. G ( jw) db ( db/dec) ( db ( dec) ) wr/sec Due to initial slope, it is a type- system, and it has non zero velocity error coefficient K V The magnitude plot is giving db at r/sec. Which gives k v k = v The steady state error e ss A = k given unit ramp input; A = v e ss ss = e =.5 4

25 EC-GATE-4 PAPER Consider the state space system epressed by the signal flow diagram shown in the figure. C 3 S u 3 a 3 S S c y a a The corresponding system is (A) always controllable (B) always observable (C) always stable (D) always unstable Answer: (A) From the given signal flow graph, the state model is X X X X = + u X a3 a a X 3 3 X Y = [ C C C 3 ] X X 3 A = ;B = ;C = C C C a a a 3 Controllability: Qc B AB A B QC = a a a a + Q = C = [ ] 3 Observability C C C C3 QO = CA a3c3 c + a c3 c + ac3 CA c a c a a a c c a a a c a c c a a + ( + ) + + ( + ) Q depends on a,a,a 3 & c & c & c3. It is always controllable 5

26 EC-GATE-4 PAPER The input to a -bit quantizer is a random variable X with pdf f ( ) = e for and f ( ) = for <, for < For outputs to be of equal probability, the quantizer threshold should be. Answer:.35 X one bit Quantizer Q ( ) One bit quantizer will give two levels. Both levels have probability of Pd of input X is f ( π) T Let T be the thsuhold Q T = < T Where and are two levels { } P Q r.e d T = = = e e. = e T T + = e T T ln T T = = =.693 =.35 6

27 EC-GATE-4 PAPER Coherent orthogonal binary FSK modulation is used to transmit two equiprobable symbol s t cos f t s t = cosπ f t, where α = 4mV. Assume an AWGN Answer: (C) waveforms = α π and N =.5 W / Hz. Using an channel with two-sided noise power spectral density u / optimal receiver and the relation Q( v) = e du π the bit error probability for a data v rate of 5 kbps is ( A) Q( ) ( B) Q( ) ( C) Q( 4 ) ( D) Q( 4 ) For Binary F SK E Bit error probability = Q N O E Energy per bit [No. of symbols = No. of bits] A T 3 [ ] E =,A = 4,T = inverse of data rate E = = 6 N = 6 Pe = Q = Q 4 5. The power spectral density of a real stationary random process.x(t) is given by Answer: 4 S =, w f w ( f ), f > w The value of the epectation π Given S ( f ) w jπft R ( τ ) =.e df w w, f w = w, f w E X t t 4w is. jπwt jπwt e e sin πwt = = w jπt w πt sin πw. 4w 4 Now, E ( t ). t R.. π 4w = π π = 4w w π. 4w 7

28 EC-GATE-4 PAPER In the figure, M(f) is the Fourier transform of the message signal.m(t) where A = Hz and B = 4 Hz. Given v(t) = cos ( π fct) and w( t) = cos π ( fc + A) t, where fc > A The cutoff frequencies of both the filters are f C M ( f ) m( t) v ( t) A High Pass Filter B B A f w ( t) Low Pass Filter s( t) Answer: 6 The bandwidth of the signal at the output of the modulator (in Hz) is. m( t) M( f ) M( f ) After multiplication with V( t) = cos( π f t) = Let w t m t.v t f A B B A W f specturm of w t is c f A f B f + B f + A f A f B f + B f + A c c c c c c c c f c f c After high pass filter After multiplication with cos( ( fc A) t) f A f B f f + B f + A C C π + and low pass filter of cut off f c 8 C C C

29 EC-GATE-4 PAPER- Bandwidth = A B = 4 = 6 ( A B ) ( A B) O f 53. If the electric field of a plane wave is O O E Z, t = 3cos ωt kz + 3 y4sin ωt kz + 45 mv / m, the polarization state of the plane wave is (A) left elliptical (C) right elliptical Answer: (A) (B) left circular (D) right circular = ( + ) ( ω + ) E = 3cos( ωt kz + 3 ) E = 4cos( ωt kz + 45 ) E z t 3cos cot kz 3o a 4 sin t kz 45 a y y y y o o At z = E = 3cos ω t + 3 E = 4sin ω t + 45 E E so Elliptical polarization Q= 3 35 = 5 left hand elliptical (LEP) 54. In the transmission line shown, the impedance Z in (in ohms) between node A and the ground is. A Z in =? Ω Z = 5 Ω, L =.5λ 5Ω Answer: 33.33Ω λ Here l = ( l λ ) Zin = = ZL = 5Ω Zin = ( 5) = = 33.33Ω 3 9

30 EC-GATE-4 PAPER For a rectangular waveguide of internal dimensions a b( a > b), the cut-off frequency for the TE mode is the arithmetic mean of the cut-off frequencies for TE mode and TE Answer: mode. If a = 5 cm, the value of b (in cm) is. t c C = a tc = K ; tc = K a a tc = K + a b Given t c f = c c K + = a b + a a a b a + f K 3 + = = + = 5 b b.+.45 = b = b = cm b 3