1 | P a g e ENGINEERS CAREER GROUPwww.engineerscareergroup.com S.C.O-61, SEC-40/C, CHANDIGARHengineerscareergroup@live.in ONE MARK QUESTIONS 1.Two independent random variable X & Y are uniformly distributed in the interval (-1,1) the probability that max [X,Y] is less than is (a) (b) 9/16(c) (d) 2/3 Solution: (b) )21) 4 , x (max( 1214 ,211 p21y 1 ,21x p=1691694343dy21dx212 / 112 / 11 2. If1 xthen value of xxis (a) 2 /e (b) 2 /e (c) x(d) 1 Solution: (a) 1 xsin i cos ei 2 / 2 / 2 / isin i cos e2 / e ) e ( ) i (21 i 2 / i i 2 /e 3. 3 z 21 z1) z ( fif C = close Counter clockwise path in z plan such that 1 1 zthen value ofdz ) z ( fj 21c is Solution: (c) ] dz3 z z[j 21d ) t ( fj 212 2 | P a g e ENGINEERS CAREER GROUPwww.engineerscareergroup.com S.C.O-61, SEC-40/C, CHANDIGARHengineerscareergroup@live.in By Cauchy Integral 1 1 z1 1 iy x1 ) 1 y ( x2 2 Readius = 1 centre = (0,-1) Ft at1 zlies inside= -3 outside 03 z1i 21c z = -3 outside z = 1 Case : 21 dz1 z1I1 1 0 1 I I2 1 Total 1c1 dz ) z (j 21 4. In the circuit shown (a) C B A Y (b)c ) B A ( Y (c)C ) B A ( Y (d) Y = AB + C Solution: (a) The circuit is equivalent to inverter as below C ) B A ( YC ) B A ( YC B . A Y(using De-Morgan) 5. The impedance looking into notes 1 and 2 in the given circuit is 1 k99ib100 k129 k 5 VoltsABCABYC3 | P a g e ENGINEERS CAREER GROUPwww.engineerscareergroup.com S.C.O-61, SEC-40/C, CHANDIGARHengineerscareergroup@live.in (a) 50(b) 100 (c) 5 K (d 10.1 kSolution: (a) Apply KVL 10000 (-ib) = 100 (I+100 ib) -10000 ib = 100 I + 100 x 100 is -- 10000 ib = 100 I + 10000 ib -- 20000 is = 100 I -Ib =I20000100 -Ib = 200I We knowOCV =V V = 100 ( I + 100 ib) V = 100 2IIV = 50 I 50IV 50 Rin Solution : (a) 6.A system with transfer from with ) 4 s )( 3 s )( 1 s () 2 s )( 9 s ( ) s ( G2 Is excited byt sinthe steady state output of system zero at(a) =1 rad/sec.(b) = 2 rad/sec. (c)= 3rad /sec.(d)= 4 rad/sec. Solution: (c) ) t ( fFor y(t) = 0 4 | P a g e ENGINEERS CAREER GROUPwww.engineerscareergroup.com S.C.O-61, SEC-40/C, CHANDIGARHengineerscareergroup@live.in | ) j ( G | =zero 16 ( 9 ( 14 )( 9 () j ( G2 2 22 2 Atw= 3 , read/sec. zero ) j ( GThus y(t) = 0 Option (c) 7.In the sum of products function f (X,Y,A) =) 5 , 4 , 3 , 2 ( ,the prime implications are(a)Y X , Y X (b)Z Y X . Z Y X , Y X (c)XY YZ X , Z Y X (d)Z Y X , Z Y X , YZ X , Z Y X 8.If x[n] =] n [ u )21( )31(n | n |, then the region of convergence (ROC) of its Z transform in the Z-plane will be (a) 3 | z |31(b) 21| z |31 (c) 3 | z |21(d) | z |31 Solution: (c) For ROC is3 | z |31 For) n ( u )21(nROC is 21| z |3 | z |21 ROC doesnot include any role 9. The directivity of the antenna is (a) 10 dB(b) 12.6 dB(c) 11.5 dB(d) 18 dB 5 | P a g e ENGINEERS CAREER GROUPwww.engineerscareergroup.com S.C.O-61, SEC-40/C, CHANDIGARHengineerscareergroup@live.in 10. A coaxial cable with an inner diameter of 1 mm and outer diameter of 2.4 mm is filled with a dielectric of relative permittivity10.89. Given Mo =H/m 10 47. m / F3610Eo9 theZo(Characteristic impedance) of the cable is (a) 330 (b) 100 (c) 143.3 (d) 43.4Solution: Answer is NOT in the Options rrablin 12021Zo rr)ab( In 60 Zo89 . 10114 . 2ln 60 ZoOptism is B ( ) (x) not any valid answer If100 Zoif abln Zo used Q. 11 A source alphabet consist of N symbols was the probabilities of the Dust 2 symbol same a source encoder increases the probability of first symbol by small amount and 2nd by after encoding the entropy(a) increase(b) remains the same(c) increases only if N = 2(d) decreases Solution: (d) ) ( Decreases Because the entropy is maximum of the message are equiprobable if probability9 / the entropy ( ) 12. The diodes and capacitors in the circuit shown are ideal. The voltage v(t) across the diode D1 isD2C2C1V(t)cos( t)+_D1 6 | P a g e ENGINEERS CAREER GROUPwww.engineerscareergroup.com S.C.O-61, SEC-40/C, CHANDIGARHengineerscareergroup@live.in (a)1 ) t cos( (b)) t sin( (c) 1-cos( ) t (d) 1-sin( ) t13. In the circuit shown -j1 1J1++11