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BSNL JTO 9 DETAILED SOLUTION 1 BSNL Junior Telecom Officers-JTO 9 SOLUTIONS Engineers Institute of India-Eii offers exclusive coaching program for JTO preparations under team of JTO and working

1 BSNL JTO 9 DETAILED SOLUTION 1 BSNL Junior Telecom Officers-JTO 9 SOLUTIONS Engineers Institute of India-Eii offers exclusive coaching program for JTO preparations under team of JTO and working professional from BSNL. JTO coaching program for BSNL JTO 16 Vacancy JTO classroom coaching JTO Postal correspondence coaching JTO All India Test series Highest selections in JTO with 8 rank under All India 5 Rank To buy these books online your Name, Phone at eiidelhi@gmail.com Ph ENGINEERS INSTITUTE OF INDIA 16 All Rights Reserved Address : 8-B/7, Jia Sarai, Near IIT, Hauz Khas, New Delhi Ph

2 BSNL JTO 9 DETAILED SOLUTION Junior Telecom Officers-JTO 9 SOLUTIONS Technical Section - I 4. Pinch off voltage is define as the minimum drain to source voltage where I D enters into saturation. 7. GaAs is the direct bandgap material in which most of the energy will be released in the form of light during recombination. 9. Schottky diode contains a metal-semiconductor junction. 1. In n type semiconductor, concentration of holes is equal to n Pn n 1. Given: V 1 = AV BI, I 1 = CV DI I1 CV DI Now the admittance looking into the port-1 is, Y V AV BI Now V = I R L CIRL DI CRL D Y AI R BI AR B L L i n 1 1. LED is device which converts electrical energy into optical energy, this process is called electorluminescence I k 1 RK R1 R RN I 5º(5 ) J I1 5 J 5 J ENGINEERS INSTITUTE OF INDIA 16 All Rights Reserved Address : 8-B/7, Jia Sarai, Near IIT, Hauz Khas, New Delhi Ph

3 BSNL JTO 9 DETAILED SOLUTION 14º(5 ) J I1 1º VAB 5 J (5 I 1 9) 17 º 16. Given: l, we know that, for lossless transmission line Zin = Z L 18. We know that, for a rectangular waveguide cut off frequency is given by; m n C fc a b For m = 1, n = or n = 1, m = i..e, TE 1 and TE 1 mode have the least f c. If a > b, then TE 1 is dominant mode, having least f c. 19. Radiation resistance of a quarter wave monopole is R r = 6.5 Rm 1. We know, shunt resistance Rsh m 1 Here, R m = 1 I 1 m 1 I.1 m 1 R sh We know, Diode current I e 1/( I Amp ID I e. Given: D = D 1 and l = 4l 1 We know that, Resistance of wire is given by, l R resistivity A l R D /4 l1 R1 D /4 1 l 4l R /4 4 /4 1 D D1 4l R R 1 1 D1 V / V T 1. Given: L = H, A = 1 mm, D = 1m di Now voltage, V L dt ENGINEERS INSTITUTE OF INDIA 16 All Rights Reserved Address : 8-B/7, Jia Sarai, Near IIT, Hauz Khas, New Delhi Ph

4 BSNL JTO 9 DETAILED SOLUTION 4 4. Given: R H = m /c = 9 1 m We know that, Mobility = R H m V S 91 and carrier density = 1 qr /m The conductivity of a semiconductor is = nq n + pq p (1) By mass action law, n i P=...() n From (1) and (), ni nqn q p n d 1 Now qn n 1 q p dn n d Now for minimum conductivity, dn n ni 6. We know that, 7. We know that, p n 1.4 () m E g H m m Z For lossless transmission line with L JZ ZL =, the equivalent impedance, Zin Z 8 Z JZ L Zin Zin JZ 5J 9 j 1 L 5J 5 L nh 8. Given: VBE.7V, 99 and Vc 7.5V Apply KVL in base emitter loop, 9 + R BI B +.7 = R BI B = 8. (1) Apply KVL in collector emitter loop, ENGINEERS INSTITUTE OF INDIA 16 All Rights Reserved Address : 8-B/7, Jia Sarai, Near IIT, Hauz Khas, New Delhi Ph

5 BSNL JTO 9 DETAILED SOLUTION I c = I c = 9.5 ma From (1), R B = 88.8 k Here, Z ' Z Z Z ' For maximum power transfer, R = 1 and 1 Pmax.5 watt 41 5% of P max =.15 watt 1.15 R 1 R R 8 4. By applying reciprocity theorem, we get, L I = 6A 4. By source transformation ENGINEERS INSTITUTE OF INDIA 16 All Rights Reserved Address : 8-B/7, Jia Sarai, Near IIT, Hauz Khas, New Delhi Ph

6 BSNL JTO 9 DETAILED SOLUTION 6 By KVL I 1 7I = (1) 6I 7I 1 51V x = 6I 7I 1 51 I = 76I 7I 1 = From (1) and () 44. Calculation of I SC: () I = 94.4 ma 4 ISC.8mA 5 Calculation of V th: 4 Vth Vth By Nodal analysis V th = 8V 4 Vth 8 Rth 1k I At t = SC I L( ) = 1.6 Amp At t > ENGINEERS INSTITUTE OF INDIA 16 All Rights Reserved Address : 8-B/7, Jia Sarai, Near IIT, Hauz Khas, New Delhi Ph

7 BSNL JTO 9 DETAILED SOLUTION 7 Now L.4. R / 5 i()() t i e1.6 t e t L i(.15).755amp 47. Here D 1 will be short-circuited and D will be open circuited. 5. 4sint 4.4 I ma ma 11 1 By nodal analysis, 1 V1 V1 V 1 V 6 V V V V...(1) 1 Now V 1 = = 6V From (1) V = V = 8V ENGINEERS INSTITUTE OF INDIA 16 All Rights Reserved Address : 8-B/7, Jia Sarai, Near IIT, Hauz Khas, New Delhi Ph

8 BSNL JTO 9 DETAILED SOLUTION 8 Technical Section - II 1. TWT is a specialized vacuum tube in which the radio waves interact with the electron beam while traveling down a wire helix which surrounds the beam. These have wide bandwidth, but output is limited to a few hundred watts.. Given:- F(A, B, C) = M (,, ) m(1, 4, 5, 6, 7) ABC ABC ABC ABC ABC BC()() A A ABC AB C C BC ABC AB BC A() B BC BC A[()()] B B B C BC AB AC A BC. Number of address lines required to address 8K bytes of memory is 1. As, 1 = 819 = 8K bytes 4. ASCII is a character encoding scheme. ASCII code represent text in computer, communication equipment and other devices that use text. 5. Binary number = s complement = 111 s complement = = 11 = 14 in BCD 6. Vi V 1 V V Va 1 i ENGINEERS INSTITUTE OF INDIA 16 All Rights Reserved Address : 8-B/7, Jia Sarai, Near IIT, Hauz Khas, New Delhi Ph

9 BSNL JTO 9 DETAILED SOLUTION 9 Vi 1 Vi Vb 11 Due to virtual short circuit. V V a b V i V Vi 4V V V i V Vi V.5 V i i 7. If V = +1V, then D ON and D 1 OFF. 1 VUT 8V.5 If V = 1V, then D 1 ON and D OFF. 1 VLT 5V 8. (i) x y x y (Demorgan's theorem) (ii) x y x y x y (iii) x y xy x y (iv) x y x. y xy x y 9. A Boolean function can be expressed as sum of products (SOP) i.e., sum of minterms and product of sums i.e., product of maxterm. 1. CMOS logic family having the lowest power dissipation.1 mw and highest noise margin as (V OH) min is closer to the power supply voltage and (V OH) max is closer to zero. 11. T Flip flop: Characteristic table: T Q Q(n+1) Q( n 1) T Q T Q 1. The nyquist plot of G(J)H(J) of a closed loop control system encloses the point ( 1, J) Gain Margin < 1 ENGINEERS INSTITUTE OF INDIA 16 All Rights Reserved Address : 8-B/7, Jia Sarai, Near IIT, Hauz Khas, New Delhi Ph

10 BSNL JTO 9 DETAILED SOLUTION 1 Gain Margin = ive (db) 1. Given:- () d y () t dt dy t 4()() y t ( r t 1) r t dt () s Y () s 4()() sy s () Y s R s R s e s Y () s 1 s e R() s s s We know that, DSB-SC signal is given by:- s()()cos t A m t f t. Where, A cos f t is carrier signal. c Here, () 1 cos cos c s() t 1 a cos f t cos f tcos f t m a m c m t a f t f t m a m c c 15. FM signal is given by, s() t Acos(()) t c Where, () t i i t t i dt c f () f k m t dt t s() t Accos fct () k f m t dt fct () k m t dt t f 16. Given:- P()() t F. T P f Condition for zero inter symbol interference in the absence of noise is n P() f 1 nr b n P f 1 n Tb 17. The magnitude response of an ideal equalizer for rectifying a distortion characterized by J T sin() c ft e ft is 1 1 f J ft T sin() c ft e T sin() ft J ft e sin() ft ft 18. We know that, Minimum sampling frequency f s = f m is also known as Nyquist rate. 8kHz = f m f m = 4 khz ENGINEERS INSTITUTE OF INDIA 16 All Rights Reserved Address : 8-B/7, Jia Sarai, Near IIT, Hauz Khas, New Delhi Ph

11 BSNL JTO 9 DETAILED SOLUTION A klystron is a linear beam vacuum tube, in which bunching of electrons is caused by velocity modulation.. Program counter points to the address location from where the next byte i.e., next instruction is to be fetched. 1. Given: f = 1GHz d = km P t = 1 Watt P r = dbm We know that, P G G PG P G G 4 d 4 d t t r t r t r 1G P G r 4. In memory mapped I/o, the processor can manipulate I/o data residing in interface register with the same instructions that are used to manipulate memory location.. When i = 1. dx =. sum =. x = 4. {sum < x} when i = dx = 4. sum = 4. x = 8. {sum < x} when i = dx = 16. sum =. x = 16. {sum > x} sum =. 4. Total number of comparison of element is section sort is (n ) = () = 4 comparison 4, comparisons takes ms. Now, 4 elements takes (4) comparisons So, 4 comparisons required ms 1,,,, 1 s 4, ENGINEERS INSTITUTE OF INDIA 16 All Rights Reserved Address : 8-B/7, Jia Sarai, Near IIT, Hauz Khas, New Delhi Ph

12 BSNL JTO 9 DETAILED SOLUTION 1 (4),,,, 1 1 s 1.sec 1 s 4 4 4, 5. Post fix expression is ABC*/D EF/+ Given: A = 6, B =, C =, D =, E = 4 and F = So, ABC*/D EF/+ = 66/ + = 1 + = + = *operator encounter Pop top element and * = 6 Again encounter / operator pop top two element and find the value operator encounter pop top two element and find the value 1 = / operator encounter pop top two element and find the value 4 ENGINEERS INSTITUTE OF INDIA 16 All Rights Reserved Address : 8-B/7, Jia Sarai, Near IIT, Hauz Khas, New Delhi Ph

13 BSNL JTO 9 DETAILED SOLUTION 1 + operator encounter pop top two element and find the value + = Value printed 6. Address range : CC to CFFF So, first 4 bits are 11 So, address lines are A 15, A 14, A 1, A 1. Next byte is changing from C to F. i.e., 11 to So next two address lines are A 11, A ALE Address Latch Enable PSW Program Status Word CMA Complement Accumulator RLC Rotate Accumulator Left 8. LXIB 7 H B = H, C = 7H MVI A, 8FH MVI C, 68H A = 8FH C = 68 H SUB C A = A C = 7H ANI F H A AND F = 7 H STAX B Content of memory location 7H is 7H. 9. There are total 9 jump instruction. 1. JC, Jump if carry flag is set bytes. JNC, Jump if carry flag is reset bytes. JZ, Jump if zero flag is set bytes 4. JNZ, Jump if zero flag is reset bytes 5. JP, Jump if positive bytes 6. JM, Jump if minus bytes 7. JPE, Jump if parity even bytes 8. JPO, Jump if parity odd bytes 9. JMP, Unconditional Jump bytes 7 bytes ENGINEERS INSTITUTE OF INDIA 16 All Rights Reserved Address : 8-B/7, Jia Sarai, Near IIT, Hauz Khas, New Delhi Ph

14 BSNL JTO 9 DETAILED SOLUTION LC rad/sec 1. Minimum size of ROM = n n = 4. V V I I e V / V T s I I e s V / VT I V I ln V VT ln I s VT I s Now from the figure, V I V V V i i T ln Is 1 Case 1: Vi V V1 VT ln...(1) I s Case : Vi 4V V VT ln...() I Now () (1) 1 1 I s V V1 VT ln VT ln VT ln I s I s I s 1 ln V V1 V T V V1 V T ln io io I1 io. g.. g g V I V I 4. mo m1 m1 i 1 i G()() s H s k( s 4) s( s 1) q() s 1()() G s H s s ENGINEERS INSTITUTE OF INDIA 16 All Rights Reserved Address : 8-B/7, Jia Sarai, Near IIT, Hauz Khas, New Delhi Ph

15 BSNL JTO 9 DETAILED SOLUTION 15 s s ks k 4 s s k s 4 dk Now, ds ( 1)( 4)() s s s s s 8s s 4 s s s 8s 4 s.5, P x y z x y z x y z x y z P z ()() x y z x y P z()() x y z x y P x y z 7. Given: q() s 4 s s s s RH Table: 4 s 1 s 1 s a { a } a- a 1 s s When, a =, then a. So, there are sign changes in the first column. So, there are a roots in the right half of the s plane. 8. There are variables, out of which are required for select lines and 1 will be used for input. Let A be the input variable. Let A be the input variable. ENGINEERS INSTITUTE OF INDIA 16 All Rights Reserved Address : 8-B/7, Jia Sarai, Near IIT, Hauz Khas, New Delhi Ph

16 BSNL JTO 9 DETAILED SOLUTION Clk CLR L C 1 1 X Load inputs 1 1 Count next binary state X X X Clear outputs X 1 No change 4. D flip flop is set initially So, Sum = 1111, carry = Here, P1 ( s 1)( s 4) 1 L1, L s 1( 1)( s 4) s From the Mason s gain formula, Y () s P k k R() s 1 ( s 1)( s 4) 1 1 s 1( s1)( s4) s s s s s 4. G()() s H s ( k.66) s s( s 1) k.66 J G()()()() J H J1 G Jgc H Jgc J ( J 1) k (.66) gc gc gc k (.66) 1(1 1) k Given:- k = 96, f s = 8kHz, n = 8, s = 1 We know that, for TDM, data rate = (nk+s)f s = 615 kbps 6.15 Mbps ENGINEERS INSTITUTE OF INDIA 16 All Rights Reserved Address : 8-B/7, Jia Sarai, Near IIT, Hauz Khas, New Delhi Ph

17 BSNL JTO 9 DETAILED SOLUTION Given: d = 6 km, f d = 1 GHz We know that, Downlink loss = 4 d So, uplink loss = = 5.15 db /(1 1) 45. In uniform quantizer, the quantization noise is given by; N q 1 V Where, is step size and it is given by V Nq 1 Peak to peak n Peak to peak n 46. For Matched filter, impulse response h()()(1) t P T t P t dB Now output y()() t *()()() P t h t p h...(1) t d Now, Put t = 1 ENGINEERS INSTITUTE OF INDIA 16 All Rights Reserved Address : 8-B/7, Jia Sarai, Near IIT, Hauz Khas, New Delhi Ph

18 BSNL JTO 9 DETAILED SOLUTION 18 1 y(1). d Given: Responsivity R =.5 A/W I p 51 p p p 5 1 Watt 1 Power loss = L = 1 = db = So, the required transmitted power is = = 1 4 watt = 1 dbm 48. Given: Coupling C = db and directivity D = db P db C 1log1 P1 P C 1log 1 db 1 P db 4 D 1log1 P 1log P D 1 P4 P.1watt.1 1 P P 4.1mwatt 4 db 5. Given: D =, f = MHz We know that, 4 Directivity D A e 4 A [ 1 8 /( 1)] 6 e Ae 1 m For further Query contact us at eiidelhi@gmail.com Ph ENGINEERS INSTITUTE OF INDIA 16 All Rights Reserved Address : 8-B/7, Jia Sarai, Near IIT, Hauz Khas, New Delhi Ph

ONE MARK QUESTIONS. 1. Consider the network graph shown in the figure. Which one of the following is NOT a tree of this graph?

ONE MARK QUESTIONS. 1. Consider the network graph shown in the figure. Which one of the following is NOT a tree of this graph? ELECTRONICS & COMMUNICATION ENGINEERING ONE MARK QUESTIONS 1. Consider the network graph shown in the figure. Which one of the following is NOT a tree of this graph? (a.) (b.) (c.) (d.). The equivalent