GATE EC : ELECTRONICS & COMMUNICATION ENGINEERING Set - 3 N o. of Questi ons : 65 Maxi mum Mar k s : 100

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1 GATE 06 EC : ELECTRONICS & COMMUNICATION ENGINEERING Set - N o. of Questi ons : 65 Maxi mum Mar k s : 00 I NSTRUCTI ONS. Total of 65 questions carrying 00 marks, out of which 0 questions carrying a total of 5 marks are in General Aptitude (GA). The Engineering Mathematics will carry around 5% of t h e t ot al m ar k s, the General Aptitude section will carry 5% of t h e t ot al m ar k s and the r em ai n i n g 70% of t h e t ot al m ar k s.. Types of Qu est i on s (a) M u l t i pl e Ch oi ce Qu est i on s (M CQ) carrying or marks each in all papers and sections. These questions are objective in nature, and each will have a choice of four options, out of which the candidate has to mark the correct answer(s). (b) Nu m er i cal An sw er Qu est i on s of or marks each in all papers and sections. For these questions the answer is a real number, to be entered by the candidate using the virtual keypad. No choices will be shown for these type of questions. 4. For -m ar k multiple-choice questions, / m ar k s will be deducted for a wrong answer. Likewise, for -m ar k s multiple-choice questions, / marks will be deducted for a wrong answer. There is no negative marking for numerical answer type questions. GENERAL APTI TUDE (GA) (Q. 5) : Car r y On e M ar k Each. An apple costs Rs. 0. An onion costs Rs. 8. Select the most suitable sentence with respect to gr ammar and usage. (a) The price of an apple is greater than an onion. (b) The price of an apple is more than onion. (c) The price of an apple is greater than that of an onion. (d) Apples are more costlier than onions.. The Buddha said, Holding on to anger is like grasping a hot coal with the intent of throwing it at someone else; you are the one who gets burnt. Select the wor d below which is closest in meaning to the word underlined above. (a) bur ning (b) ignit ing (c) clutching (d) flinging. M has a son Q and a daughter R. He has no other childr en. E is the mother of P and daughter-inlaw of M. How is P related to M? (a) P is the son-in-law of M. (b) P is the grandchild of M. (c) P is the daughter-in-law of M. (d) P is the grandfather of M. 4. The number that least fits this set: (4, 44, 97 and 64) is. (a) 4 (b) 44 (c) 97 (d) I t takes 0 s and 5 s, respectively, for two trains t r avel l i n g at di ffer ent const ant speeds t o completely pass a telegraph post. The length of the first train is 0 m and that of the second train is 50 m. The magnitude of the difference i n t he speeds of t he t wo t r ai ns (i n m/s) i s. (a).0 (b) 0.0 (c).0 (d).0 (Q.6 0) : Car r y Tw o M ar k s Each 6. The velocity V of a vehicle along a straight line is measur ed in m/s and plot t ed as shown wit h r espect to time in seconds. At t he end of the 7 seconds, how much will the odometer reading increase by (in m)? (a) 0 (b) (c) 4 (d) 5

2 SOLVED PAPER 06 (SET-) 7. The over whelming number of people infected with rabies in India has been flagged by the World Health Organization as a source of concern. I t is estimated that inoculating 70% of pets and stray dogs against r abies can lead t o a significant reduction in the number of people infected with rabies. Which of the following can be logically inferred from the above sentences? (a) The number of people in I ndia infected with rabies is high (b) The number of people in other parts of the world who are infected with rabies is low (c) Rabi es can be er adi cat ed i n I ndi a by vaccinating 70% of stray dogs (d) St r ay dogs ar e t he main sour ce of r abies wor ldwide 8. A flat is shared by four first year undergraduate students. They agreed to allow the oldest of them to enjoy some extra space in the flat. Manu is t wo months older than Sr avan, who is t hr ee mont hs younger t han Tr ideep. Pavan is one month older than Sravan. Who should occupy the extra space in the flat? (a) Manu (c) Tr ideep (b) Sravan (d) Pavan 9. Find the area bounded by the lines x + y 4, x y 5 in the first quadrant (a) 4.95 (b) 5.5 (c) 5.70 (d) A straight line is fit to a data set (ln x, y). This line intercepts the abscissa at ln x 0. and has a slope of 0.0. What is the value of y at x 5 from the fit? (a) 0.00 (b) 0.04 (c) 0.04 (d) 0.00 TECH NI CAL SECTI ON (Q. 5) : Car r y On e M ar k Each. Consider a square matrix x A, where x is unknown. I f the eigenvalues of the matrix A are ( + j) and ( j), then x is equal to (a) + j (c) + sin( z). For f(z), z is. (b) j (d) the residue of the pole at z 0. The probability of getting a "head" in a single toss of a bi ased coi n i s 0.. The coi n i s t ossed repeatedly till a "head" is obtained. I f the tosses are independent, then the probability of getting "head" for the first time in the fifth toss is 4. The integral dx is equal to 0 ( x) 5. Consider the first order initial value problem y y + x x, y(0), (0 x < ) with exact solution y(x) x + e x. For x 0., the percentage difference between the exact solution and the solution obtained using a single iteration of the second-order Runge-Kutta method with step-size h 0. is. 6. Consider the signal x(t) cos(6t) + sin(8t), where t is in seconds. The Nyquist sampling rate (in samples/second) for the signal y(t) x(t + 5) is (a) 8 (b) (c) 6 (d) sin( t) sin( t) 7. I f the signal x(t) * with * denoting t t the convolution operation, then x(t) is equal to (a) sin( t) t (b) sin( t) t sin( t) sin( t) (c) (d) t t 8. A discrete-time signal x[n] [n ] + [n 5] has z-t r ansfor m X(z). I f Y(z) X( z) i s t he z-transform of another signal y[n], then (a) y[n] x[n] (b) y[n] x[ n] (c) y[n] x[n] (d) y[n] x[ n] 9. I n the RLC circuit shown in the figure, the input voltage is given by v i (t) cos(00t) + 4 sin(500t). The output voltage v o (t) is (a) cos(00t) + sin(500t) (b) cos(00t) + 4 sin(500t) (c) sin(00t) + cos(500t) (d) sin(00t) + 4 cos(500t)

3 SOLVED PAPER 06 (SET-) 0. The I -V characteristics of three types of diodes at the r oom temper atur e, made of semiconductor s X, Y and Z, are shown in the figure. Assume that the diodes are uniformly doped and identical in all respects except their materials. I f E gx, E gy and E gz are the band gaps of X, Y and Z, respectively, then 4. Consi der t he ci r cui t shown i n t h e fi gur e. Assuming V BE V EB 0.7 volt, the value of the dc voltage V C (in volt) is (a) E gx > E gy > E gz (b) E gx E gy E gz (c) E gx < E gy < E gz (d) no relationship among these band gaps exists.. The figure shows the band diagram of a Metal Oxide Semiconductor (MOS). The surface region of this MOS is in 5. I n the astable multivibrator circuit shown in the figure, the frequency of oscillation (in khz) at the output pin is (a) inver sion (c) deplet ion (b) accumulat ion (d) flat band. The figure shows the I-V characteristics of a solar cell illuminated unifor mly with solar light of power 00 mw/cm. The solar cell has an area of cm and a fill fact or of 0.7. The maximum efficiency (in %) of the device is 6. I n an 8085 microprocessor, the contents of the accumulator and the carry flag are A7 (in hex) and 0, r espectively. I f the instr uction RLC is executed, then the contents of the accumulator (in hex) and the carry flag, respectively, will be (a) 4E and 0 (b) 4E and (c) 4F and 0 (d) 4F and 7. The logic functionality r ealized by the circuit shown below is (a) OR (b) XOR. The diodes D and D in the figure are ideal and the capacitors are identical. The product RC is very large compared to the time period of the ac vol t age. A ssumi ng t hat t he di odes do not breakdown in the reverse bias, the output voltage V o (in volt) at the steady state is (c) NAND (d) AND 8. The minimum number of -input NAND gat es required to implement a -input XOR gate is (a) 4 (b) 5 (c) 6 (d) 7

4 4 SOLVED PAPER 06 (SET-) 9. The block diagram of a feedback control system is shown in the figure. The overall closed-loop gain G of the system is (a) G G G G H GG (b) G G G G H (c) (d) G G G G G H G G G G G G G H 0. For me unity feedback control system shown in the figure, the open-loop transfer function G(s) is given as G(s). s( s ) The steady state error e ss due to a unit step input is. A binar y baseband digital communicat ion syst em employs t he signal, 0 t T S p(t) TS 0, ot her wise f or t r an sm i ssi on of bi t s. T he gr aph i cal representation of the matched filter output y(t) for this signal will be (a) (b) (c) (d) (a) 0 (b) 0.5 (c).0 (d). For a superheterodyne receiver, the intermediate frequency is 5 MH z and t he local oscillator frequency is.5 GHz. I f the fr equency of the received signal is greater than the local oscillator frequency, then the image frequency (in MHz) is.. An analog baseband signal, bandlimited to 00 Hz, is sampled at the Nyquist rate. The samples are quantized into four message symbols t hat occur independently with probabilities p p and p p. The information rate (bits/sec) of the message source is. 4. I f a r ight-handed cir cular ly polarized wave is incident normally on a plane perfect conductor, then the reflected wave will be (a) right-handed circularly polarized (b) left-handed circularly polarized (c) elliptically polarized with a tilt angle of 45 (d) horizontally polarized 5. Far aday's law of electr omagnetic induction is mathematically described by which one of the following equations? (a) B 0 (b) D V B (c) E t D (d) H E t

5 SOLVED PAPER 06 (SET-) 5 (Q.6 55) : Car r y Tw o M ar k s Each 6. The par t i cul ar sol ut i on of t he i ni t i al val ue problem given below is d y dy 6y dx dx 0 with y(0) and dy dx x 0 6 (a) ( 8x) e 6x (c) ( + 0x) e 6x (b) ( + 5x) e 6x (d) ( x) e 6x 7. I f the vector s e (, 0, ), e (0,, 0) and e (, 0, ) form an orthogonal basis of the threedi mensional r eal space, t hen t he vect or u (4,,-) u e e e 5 5 (a) u e e e 5 5 (b) u e e e 5 5 (c) can be expressed as (d) u e e e A triangle in t he xy-plane is bounded by t he straight lines x y, y 0 and x. The volume above the triangle and under the plane x + y + z 6 is. 9. The values of the integral z e dz along a j z closed contour c in anti-clockwise direction for (i) the point z 0 inside the contour c, and (ii) the point z 0 outside the contour c, respectively, are (a) (i).7, (ii) 0 (b) (i) 7.9, (ii) 0 (c) (i) 0, (ii).7 (d) (i) 0, (ii) A signal cos cos( ) t t is the input to an LTI system with the transfer function H(s) e s + e s. I f C k denotes the k th coefficient in the exponential Fourier series of the output signal, then C is equal to (a) 0 (b) (c) (d) c. T h e ROC (r egi on of con ver gen ce) of t h e z-t r ansf or m of a di scr et e-t i m e si gn al i s represented by the shaded region in the z-plane. I f the signal x n (.0) n, < n < +, then the ROC of its z-transform is represented by (a) (b) (c) (d). Assume that the circuit in the figure has reached the steady state before time t 0 when the resistor suddenly burns out, resulting in an open cir cuit. The cur r ent i (t) (in amper e) at t 0 + is.

6 6 SOLVED PAPER 06 (SET-). I n the figure shown, the current i (in ampere) is. 7. The inject ed excess electr on concentr ation pr ofile in the base region of an npn BJT, biased in the active region, is linear, as shown in the figure. If the area of the emitter-base junction is 0.00 cm, n 800 cm /(V-s) in the base region and depletion layer widths ar e negligible, then the collector current Ic (in ma) at room temperature is. (Given: ther mal voltage V T 6 mv at room temperature, electronic charge q C) 4. The z-parameter matrix for the twoport network shown is z z z z (a) (b) 9 9 (c) 6 9 (d) A continuous-time speech signal x a (t) is sampled at a r at e of 8 k H z an d t h e sam pl es ar e subsequently grouped in blocks, each of size N. The DFT of each block is to be computed in real time using the radix- decimation-in-frequency FFT algor ithm. I f the pr ocessor per for ms all oper ations sequentially, and takes 0 (s for com pu t i n g each compl ex m u l t i pl i cat i on (including multiplications by and -) and the t i m e r equ i r ed f or addi t i on/su bt r act i on i s negl i gi bl e, t hen t he maxi mum val ue of N is. 6. The di r ect for m st r uct ur e of an FI R (fi ni t e impulse response) filter is shown in the figure. 8. Figures I and I I show two MOS capacitors of unit ar ea. The capacitor in Figur e I has insulator materials X (of thickness t nm and dielectric constant 4) and Y (of thickness t nm and dielect r ic const ant 0). The capacit or in Fi gur e I I has onl y i nsul at or mat er i al X of t hi ck ness t Eq. I f t he capaci t or s ar e of equal capacitance, then the value of t Eq (in nm) is 9. The I-V characteristics of the zener diodes Dl and D are shown in Figure I. These diodes are used in the circuit given in Figure I I. I f the supply voltage is varied from 0 to 00 V, then breakdown occurs in The filter can be used to approximate a (a) low-pass filt er (b) high-pass filt er (c) band-pass filt er (d) band-st op filt er (a) D only (b) D only (c) both D and D (d) none of D and D

7 SOLVED PAPER 06 (SET-) For the circuit shown in the figure, R R R, L H an d C F. I f t h e i npu t V in cos(0 6 t), then the overall voltage gain (V out /V in ) of the circuit is. 4. Following is the K-map of a Boolean function of five variables P, Q, R, S and X. The minimum sumof-pr oduct (SOP) expr ession for the function is 4. I n the circuit shown in the figure, the channel length modulation of all transistors is non-zero ( 0). Also, all transistors operate in saturation and have negligible body effect. The ac small signal voltage gain (V O /V in ) of the circuit is (a) PQSX PQSX QRSX QRSX (b) QSX QSX (c) QSX + QSX (d) QS + QS 44. For the circuit shown in the figure, the delays of NOR gates, multiplexers and inverters are ns,.5 ns and ns, respectively. I f all the inputs P, Q, R, S and T are applied at the same time instant, the maximum propagation delay (in ns) of the circuit is. g ( r r r ) (a) m o o o g ( r r ) (b) m o o gm g ( r ( ( r ) r ) (c) m o o o gm (d) g m( ro ( ro ) ro ) g m 4. I n the circuit shown in the figure, transistor M i s i n sat ur at i on and has t r ansconduct ance g m 0.0 siemens. I gnor ing inter nal par asitic capacitances and assuming the channel length modulation to be zero, the small signal input pole frequency (in khz) is. 45. For the circuit shown in the figure, the delay of the bubbled NAND gate is ns and that of the counter is assumed to be zero. I f the clock (Clk) frequency is GHz, then the counter behaves as a (a) mod-5 counter (b) mod-6 counter (c) mod-7 counter (d) mod-8 counter

8 8 SOLVED PAPER 06 (SET-) 46. The first two r ows in the Routh table for the characteristic equation of a certain closed-loop control system are given as The range of K for which the system is stable is (a).0 < K < 0.5 (b) 0 < K < 0.5 (c) 0 < K < (d) 0.5 < K < 47. A second-order linear time-invariant system is described by the following state equations d x ( t ) x ( t ) u ( t ) dt d x ( t ) x ( t ) u ( t ) dt where x (t) and x (t) are the two state variables an d u(t ) den ot es t he i nput. I f t he out pu t c(t) x (t), then the system is (a) controllable but not observable (b) observable but not controllable (c) both controllable and observable (d) neither controllable nor observable 48. The for war d-pat h t r ansfer funct i on and t he feedback-path transfer function of a single loop negative feedback control system are given as K ( s ) G(s) and H(s), s s respectively. I f the variable parameter K is real posit ive, t hen the location of the br eakaway point on the root locus diagram of the system is. 49. A wide sense st at ionar y r andom pr ocess X(t) passes t hr ough the LTI system shown in t he figure. I f the autocorrelation function of X(t) is R X (), then the autocorrelation function R Y () of the output Y(t) is equal to 50. A voice-grade AWGN (additive white Gaussian noise) telephone channel has a bandwidt h of 4.0 k H z and t wo-si ded noi se power spect r al density Watt per Hz. I f information at the rate of 5 kbps is to be transmitted over this channel with arbitrarily small bit error rate, t hen t he mi nimum bi t -ener gy E b (in mj/bi t ) necessary is. 5. The bit error probability of a memoryless binary symmetric channel is 0 5. If 0 5 bits are sent over this channel, then the probability that not more than one bit will be in error is. 5. Consider an air-filled rectangular waveguide with di mensions a.86 cm and b.06 cm. At 0 GHz operating frequency, the value of the pr opagat i on const ant (per m et er ) of t h e corresponding propagating mode is 5. Consider an air-filled rectangular waveguide with dimensions a.86 cm and b.06 cm. The incr easing order of the cut-off frequencies for differ ent modes is (a) TE 0 < TE 0 < TE < TE 0 (b) TE 0 < TE < TE 0 < TE 0 (c) TE 0 < TE 0 < TE 0 < TE (d) TE 0 < TE < TE 0 < TE A r adar oper at ing at 5 GH z uses a common ant enna for tr ansmission and r ecept ion. The antenna has a gain of 50 and is aligned for maximum directional radiation and reception to a target km away having radar cross-section of m. I f it transmits 00 kw, then the received power (in W) is. 55. Consider the charge profile shown in the figure. The r esul t ant pot ent i al di st r i but i on i s best descr ibed by (a) (b) (a) R X () + R X ( T 0 ) + R X ( + T 0 ) (b) R X () R X ( T 0 ) R X ( + T 0 ) (c) R X () + R X ( 70) (d) R X () R X ( 70) (c) (d)

9 SOLVED PAPER 06 (SET-) 9 GENERAL APTI TUDE AN SWERS. (c). (c). (b) 4. (c) 5. (a) 6. (d) 7. (a) 8. (c) 9. (b) 0. (a) TECH NI CAL SECTI ON. (d). (). (0.070) 4. () 5. (0.06) 6. (c) 7. (a) 8. (c) 9. (b) 0. (c). (a). (). (0) 4. (0.5) 5. (5.65) 6. (d) 7. (d) 8. (a) 9. (b) 0. (a). (485). (6.55). (c) 4. (b) 5. (c) 6. (a) 7. (d) 8. (0) 9. (b) 0. (b). (d). ( ). ( ) 4. (a) 5. (8) 6. (c) 7. (6.656) 8. (.6) 9. (a) 40. ( ) 4. (c) 4. ( ) 4. (b) 44. (7) 45. (d) 46. (d) 47. (a) 48. (.4) 49. (b) 50. (.50) 5. (0.75) 5. (58.07) 5. (c) 54. (0.0) 55. (c) GENERAL APTI TUDE. The meaning of underlined word grasping means clut ching (or holding somet hing t ight ly).. Daughter (+ or ) M R Son Daughter-in-law (+) Q Husband Grand Child E EXPL ANATI ONS P Mother (+) Male ( ) Female (+ or ) So, from the above relation diagram it is clear that P is the grandchild of M. 5. Here the speed of first train (S ) 0 m/s 0 and the speed of second train (S ) 50 0 m/s 5 The magnitude of difference in the speeds of two trains (S S ) ( 0) m/s m/s 6. From the given figure. The odometer r eading increases fr om star ting point to end point. Odometer reading Area of the given diagram Area of the velocity and time graph per second st sec triangle nd sec square rd sec square + triangle 4 th sec triangle 5 th sec straight line 0 6 th sec triangle 7 th sec triangle Total Odometer reading at 7 seconds 0 m 5 m So, the odometer reading is increased by 5 m. 8. Manu age sravan age + months 9. Manu age Trideep age months Pavan age Sravan s age + month From the above statement Trideep age > Man > Pavan > Sravan Hence, Trideep can occupy the extra space in the flat.

10 0 SOLVED PAPER 06 (SET-) 4 A,0 B [0, 7], 5 C,0 5 D 0, E [4, ], F [0, ] Requi r ed ar ea Ar ea of BFE + Ar ea of quadrilateral FEOC 4 6 (4.5) sq. units So, t he ar ea bounded by t he gi ven l i nes i s 5.5 sq. units. 0. The equation of a straight line is y mx + c m slope 0.0 set (log x, y) Now, the line intercepts the abscissa at log x 0. So, at abscissa y 0 y mx + c c c 0.00 y mx + c y 0.0 og x + c At x 5 y 0.0 log So, the value of y is TECH NI CAL SECTI ON. From the given square matrix det (A) x ( + j) ( j) x () (j) x (j ) ( ) x + x (since j ) x x sin z. Given f(z) z 5 z z z... z! 5! z z... z! 5! Residue of f(z) (at z 0). The probability of getting head for the first time in fifth loss 4. Let I I (P) (0.7) 4 (0.) dx x x [(0) ] 5. Given the first order initial value problem is dy dx y x x y(0) 0 x < Given f(x, y) y + x x, x o 0, y o, h 0. k hf (x o, y o ) 0. ( + (0) 0 ) 0. k hg (x 0 + h, y 0 + k ) 0 0. ((y 0 + k ) + (x 0 + h) (x 0 + h) ) (since h 0.) 0. (( + 0.) + (0.) (0.) ) 0. ( ) 0.9 y y k k For exact solut ion, (since k 0. and k 0.9) y(x) x + e x y(0.) (0.) + e error Relative error error y( x) Percentage Error ( )% 0.06%

11 SOLVED PAPER 06 (SET-) 6. Given signal x(t) cos (6t) + sin (8t) and y(t) x(t + 5) Where t is in seconds y(t) cos ( t + 0) + sin (6 t + 40) f m 6 f m 8 f m 8 Hz (f S ) min f m 6 Hz Thus the Nyquist sampling rate is 6 Hz. 7. H er e t he convol ut i on of t wo si nc pul ses i s sinc pulse. So, V 0( t) cos 00 t Now consider only 4 sin 500t, then cir cuit becomes So, again V n 0 ( t ) 4 sin 500 t Finally (according to superposition theorem) V 0 (t) V 0( t) V 0 ( t) V 0 (t) cos (00t) + 4 sin (500 t) 0. From the given fig : n So, Now x (t) sin t t x(t) x (t) * x (t) X() X () X () X () x(t) x (t) sin t t So, the value of x(t) is sin t t 8. Here (a) n x(n) X(z/a) a ( ) n x(n) X( z) but x(n) [n ] + [n 5] y(n) ( ) n x (n) ( ) n [(n ) + [n 5] y(n) (n ) (n 5) x(n) Hence, the value of signal y(n) is x(n). 9. The input voltage is given by V i (t) cos 00t + 4 sin 500t Let us apply Super position theor em only consider cos 00t, then circuit becomes V r V r > V r > V r E g Where V r is cut-in voltage So, E gz > E gy > E gx. Fill factor where P max P T Pmax Pmax P I V T SC OC Maximum power Theoretical power I SC V OC Pmax P max 6 0 W Now M aximum efficiency ( max ) P P max in 6 0 W W 00 0 cm cm 00 % % Thus the maximum efficiency (in %) of the device is %.

12 SOLVED PAPER 06 (SET-). Total period T T + T m sec Fr equency of oscillat ions (f) T 5.65 khz Diodes are ideal therefore during Positive cycle of output voltage (V 0 ) V During Negative cycle, the diodes are Reverse biased so output (V 0 ) 0 V V 0 0 V (always) So, the output voltage V 0 (in volt) at steady state is 0 V. 4. Here V E (.5 0.7) 5. V B I B.8 V V E V EB (.8 0.7). V V B 0k. 0. 0k 0k 0. I C I 50 0k V C I C (K) 50(0.) ( k) 0k 0.5 V Hence, the value of the dc voltage V C is 0.5V. 6. Given After executing instruction 7. From given fig. it is a AND gate I nput Output A B Y The minimum no. of NAND gates required for -input EX-OR gate 4 9. From the given block diagram, we have Y( s) X( s) 0. Given G(s) G(s) GG G H G G and H( s) s( s ) Type- System, to the unit step input the e ss 0 So, the steady state error e ss due to a unit step input is 0.. f If 5 MHz f Lo 500 MHz From the above figure T T 0.69 (R A + R B ) C 0.69 (.k + 4.7k) m sec 0.69 R B C 0.69 (4.7 k) m sec f s f Lo f If f s f Lo + f If 55 MHz f si image frequency f s f If MHz So, the image frequency is 485 MHz.. Here the samples are quantized into four message symbols that occur independently with probabilities are given by

13 SOLVED PAPER 06 (SET-) P 0.5 P P 0.75 P 0.75 H 0.5 log 0.5 log l og 0.75 log log log log I nformation rate H log. The gr aphical r epr esentation of the matched filter output y(t) for this signal is given by 4. Here RHCP right handed circularly polarized LHCP Left-handed circularly polarized I f the wave is incident on perfect conductor then reflection coefficient is given by E r0 E E r0 i0 E i0 80 I f incident wave is traveling along + Z direction then the reflected wave will be traveling along Z direction. Thus, the reflected wave is left hand circularly polarized (LHCP). 5. Faraday s law of electr omagnetic induction is mathematically described by the equation is given by E E B t or H t 6. The differential equation is given by d y dy 6y dx dx 0 D + D (D + 6) 0 D 6, 6 The solution is given by 6x 6 y C e C xe y(0) C C (i) y 6 x 6x e C xe x...(i) dy dx 6 x 8 C 6 x 6 x e 6xe e dy 8 + C dx x C C 8 The solution of y e 6x 8 x e 6x 7. The equation is given by (Since C & C 8) u x e + x e + x e...(i) Putting the value of u, e, e & e in the given equation (i), we get (4,, ) x (, 0, ) + x (0,, 0) x x x 4 (a) (b) x + x (c) On solving these equations, we get x, 5 x, x 5 u e e e Volume zdx dy x 6 x x0 y0 + x (, 0, ) y dx dy 0 since x y, y x & x x + y + z 6 z 6 x y

14 4 SOLVED PAPER 06 (SET-) 9. (i) When the point z 0 inside the contour c. z e dz j ( z ) c jf () j e 7.9 (ii) When the point z 0 outside the contour c. z e dz j ( z ) 0 ( z lies out side c) c 0. The transfer function H(e j ) e j + e j cos. Cir cuit at t 0, the direction of current and correct component was not given. Not t 0 + Here x(t) H(j 0 ) 0 y(t) cos t cos cos 80 t x(t) cos t 0 H(j 0 ) cos () y(t) cos(t + 80) y(t) cos t cos(t + ) T T T T 0 y(t) cos( 0 t + ) cos( 0 t + ) y(t) 0 0 e e j ( t ) j ( t ) e y(t) 0 0 e e C Thus the value of C is.. x(y) j ( t) j ( t) () n u( n) u( n ) ROC ( z ) ( z ½) n e j ( 0t ) j ( 0t ) e So, the ROC of z-tansform is null. e j ( 0t) j ( 0t). 4 So, i(0 + ) A 4 The magnit ude of he cur r ent is ampere (negat ive). From above fig. Nodal Analysis (V 8) V (V 8) V 0 Now from KCL 4V 6 V 4 Volts i i A H ence t he cur r ent (i ) in t he given cir cui t is Amp. 4. Given network is in lattice form Where Z a Z b 0 Z c 0 Z d 6 But it is not symmetrical & balanced

15 SOLVED PAPER 06 (SET-) 5 So, Z Z V I I 0 V I I 0 KVL V I V I Z Also V Z I KVL Z V I 0 I 0 V I I 0 V I Z // 6 // 6 So, t he Z-par amet er mat r ix for the t wo por t network is given by Z 5. The number of complex multiplications required N for DI F-FFT log N N log N (0 sec) 5 sec Solving the above equation, we get N 8 6. From the given figure y(n) 5[x(n) x(n )] Y(e j ) 5[ e j ] X(e j ) H(e j ) 5[ e j ] Thus, it is a Band pass filter 7. We know that dn Collector current (I C ) AeD n dx Ae n V Where A Area for the emitter-base function e Electronic charge V T Thermal voltage I C I C ma Hence the collector current (I C ) is ma. 8. We known that 9. Now C A d Where C Capacitance A Area of a capacitor dielectric constant CC Now C C C r 0 C t eq t eq r m.6 nm T So, the value of thickness (t eq ) is.6 nm. dn dx From the above fig., it is clear that both zener diodes are in reverse biased V BZ 80 V V BZ 70 V D have list saturation current. When we will vary the voltage above 80V D get breaks down & will replaced by 80V. & through it current can flow through it. But because of D we will take minimum current i.e. net cur r ent equals t o r ever se sat ur at ion current of D as we know. 4

16 6 SOLVED PAPER 06 (SET-) 40. The diode have least saturation will break down first and it will replaced by its break down voltage and t he net cur r ent equal upt o ot her di ode reverse saturation current. From above fig. Vx 0 s 0 6 V 0 s s s in 6 6 V V 0 0 s Vx 0 Vx s 0 s V 0 V in V0 Vx s s 0. 6 Vx V in s 0 s overall voltage gain V 0 Vin 4. The ac small signal voltage gain V 0 V gm r0 r0 r0 in gm 4. C M 50PF [ A V ] A V g m R D 0.0 A V 0 C Mi 50PF [ + 0] F 0.55 nf f p f p R C i mi 5K 0.55 mf khz So, t he smal l signal i nput pol e fr equency is kh z. 4. The minimum sum-of-pr oduct (SOP) expr ession for the function Q.S.X Q.S.X 44. The maximum propagation delay of the circuit t pd, NOR + t pd, mux + t pd, NOR + t pd,mux ( ) ns 7 ns 45. At 6 th Clock pulse 46. Q Q Q 0 0 I t makes NAND gate output 0 at 8 ns due to its delay. At that time counter receives 7th, 8th Clock pulse and counts, 000. Hence, it is mod 8 count er. s s s s 0 k k 4 k(k ) 4 k 4 4k + 6k 4 > 0 4k + 8k k 4 > 0 4k(k + ) (k + ) > 0 (4k ) (k + ) > 0 (4k ) > 0, k + > 0 k >, k > < k < 0 for stability So the range of k for which the system is stable is 0.5 < k < 47. Given d x ( t ) x ( t ) u ( t ) dt x x + U and d x ( t ) x ( t ) u ( t ) dt x x + U c x

17 SOLVED PAPER 06 (SET-) 7 x x 0 U 0 x [c] 0 x x Hence by applying Gilbert s test, the system is controllable but not observable. 48. Given transfer function k( s ) G(s), H(s) ( s s ) At break away point d ds s s s 0 x dk 0 ds d ds ( s s ) ( s s ) ( s ) ( s ) ( s s ) d ds ( s s ) ( s )(s ) ( s s ) 0 s 4s ,.4 Valid BAP is Y(t) X(t) X(t T o ) Autocorrelation function for o/p R y () E[y(t) Y(t + )] R y () E [(X(t) X (t T o )] [X (t + ) X (t + T o )] R y () E [(X(t) X (t + ) X(t) X (t + T o ) X (t T o ) X(t + ) + X (t T o ) X (t + T o )] R y () [R x () R x ( T o ) R x ( + T o ) + R x ()] R y () R x () R x ( T o ) R x ( + T o ) So, the autocor r ection function of R y () of the output Y(t) is R x () R x ( T o ) R x ( + T o ). 50. From given data Channel transmission rate (C) 5 kbps Channel band width B 4 khz N S C B log N S 68. E b S J / sec.50 R bits / sec b C B log ( + S/N) log ( + S/N) C B ( + S/N) C/B 5/4 89 S/N 89 S 89 N S E b 89. R b.50 So the minimum bit energy (E b ) is.50 mj/bit. 5. P 0 5 N 0 5 Given question can be solved by two methods. M et h od : Binomial : n C x p x q n x P[x 0] + P[x ] M et h od : Poisson C 0 (0 ) ( 0 ) C (0 ) ( 0 ) () () e x! e ()! e 0.75 x np (since n 0 5, p 0 5 ) e So, the probability that not more than one bit will be in error is Given Air filled r ect angular waveguide a.86 cm b.06 cm f 0 GHz Assume dominant mode (TE 0 ) is propagating in t he waveguide, So cut-off frequency of TE 0 mode is given by 0 c f c (TE 0 ) a 0.86 f c 6.56 GHz

18 8 SOLVED PAPER 06 (SET-) propagation constant is given by j i f 0 0 c j f j58.07 m Therefore the value of propagation constant is given by 5. Given m a.86 cm b.06 cm air filled r ect angular waveguide c f c(te ) (Since m, n ) a b 0 0 (.6).06 The received power, P R is given by (0.06) P R 4 (4 ) (0 ) W Hence, the received power is 0.0 W. 55. Let us consider a and b For line () : Now the equation of the line to point (, 0) and (0, ) is given by t y 0 ( ) t ( t ) dt For line () : y t ( t ) t 0 The equation of the line to the point (0, ) and (, 0) is given by f c(te ) 6.5 GHz 0 c 0 f c(te 0) 4.76 GHz b.06 0 c 0 f c(te 0 ). GHz a.86 0 c 0 f c(te 0 ) 6.56 GHz a.86 I ncreasing order of the cut-off frequency is given by TE 0 < TE 0 < TE 0 < TE 54. Given frequency, f 5 GHz Hz 8 c 0 wave length, m f 5 0 gain of antenna, G 50 Range of target, R max km 0 m, radar cross-section, m, transmitted power, P t 00 kw Radar range equation is given by R max 4 Pt G G 4 Since A G e (4 ) P 4 R y t t ( t ) dt 0 At t 0 : y ( ) ( 0) t y y At t 0 + : y y + t ( t ) ( t ) y ( t ) 0 t