Detailed solution of IES 2014 (ECE) Conventional Paper II. solve I 0 and use same formula again. Saturation region

Transcription

1 etailed olution of IS 4 (C) Conventional Pape II qv qv Sol. (a) IC I e Ie K K 4 I =.7 Fo I C = m olve I and ue ame fomula again K IC V ln 5ln 4 q I.7 =.8576 Volt Sol. (b) VGS VS Vupply 5V N MOS channel, V 5V, S GS V V 4V So VS VGS V Satuation egion V V w GS I ncox L I L C V V n ox GS w Now c / V n ox w m Sol.(c) Y C F by CMOS logic o, V dd C F V dd V o C F V C F

2 Sol.(d) Sol.(e) R e lim G H G H Fo unit tep k k 3 R 4 e Fo unit amp input ignal R e V b m Cutoff a b Cutoff 8V me e a b b / Sol.(f) V in 6 8 t 5in5t Sol.(g) V V in f t in f t fc fm m c m 95.5MHz 5 99.Hz f f m. 995Hz Powe diipated in eito e kk Vm P 7.W R he bu inteface unit (IU) end out addee, fetche intuction fom memoy, ead data fom pot and memoy, and wite data to pot and memoy. IU ha the following component. (i) Queue : he IU fetche up to ix intuction byte. he IU toe thee pe-fetched byte in a fit in fit out egite et called a queue. hi pe-fetch and queue cheme geatly peed up poceing. Fetching the next intuction while the cuent intuction execute i called pipelining. (ii) Segment Regite : Fou egment eite in the IU ae ued to hold the uppe 6 bit of the tating addee of fou memoy egment that the 886 i woking with at a paticula time. he fou egment egite ae the code egment (CS) egite, the tack egment (SS) egite, the exta egment (S) egite, and the data egment (S) egite. (iii) Intuction Pointe : he intuction pointe egite hold the 6-bit adde, o offet, of the next code byte within thi code egment. he value contained in the IP in efeed to a an offet becaue thi value mut be offet fom (added to) the egment bae adde in CS to poduce the equied -bit phyical adde ent out by the IU.

3 Sol.(h) Pointe liteally mean one that point to omething and pointe in compute language ae ued to point infomation. C language alo uppot pointe it can be able to declae and manipulate pointe. he pointe in C ae flexible and can point to anything fom imple data type to function including ue defined data type. C alo povide facility fo conveting a pointe of one type to anothe. In C a pointe vaiable i declaed by uing the * pointe. int *pi; pi i an intege pointe/pointe to an intege float *pf; pf i a float pointe cha *pc; pf i a chaacte pointe Sol.(a) N W N L L W Fo elative ize of bae egion W W. L L Fo elative ize of emitte egion Calculate W L value whee N pe cc N pe cc cm 6 8,, 6.8 / ec.3, W cm, L 3.4 cm 4 3 Sol.(b) In ve half cycle diode will be ON o capacito will be chaged upto peak value and it i paed to high pa RC cicuit but hee RC>> o no diffeentiation and output will be ame a input with vey mall tilt. V C R R V o In ve half diode will be off but o/p will be aco R, o again paed though high pa cicuit but again RC>> o output will be ame a input with vey mall tilt. V C R V o Since RC>> he output will be ame a input in cae of High pa cicuit but ince it i paed though capacito o output will be quae wave with mall tilt uch that C value of output will be zeo So hee output will be quae wave only in which aea i zeo ove one time peiod. 3

4 Sol.(c) V V i = V d=(v V ) R V i = i o (V V ) R L V o Ouput V =-V Sol. (d) I x V R x unknown eitance M I I V M V g V.V K.5m / V, V V Vg Ix I I I V, a not cuent though gate. M So I & I x I I V x R uming R x V k I k hen, V GS V & K V g V V 3 V V 4.58V, 6.58V (not poible) V 4.58V t 4

5 Sol.3(a) C adde (Page-57 Moi-Meno book of digital) Sol.3(c) XNOR Gate Sol. 3(d)(i) So when bit ae equal, It detect. V Y = Opeation : When = and =, NMOS i in cut off o Y = and if any of the input i high, NMOS goe in atuation egion and Y = (ii) V uth able 5

6 Sol.4(a) Sol.4(b) G Y K hen it chaacteitic equation i G H k k 4 69 k k b k k b 5.5 Fo ocillation, b = k k 4.6ad / ec G k p p k chaactetic equation p n k, k t 4 4 ec hu, p = Mp % n. e Sol.4(c) (i) G /.6 k.77 k / k / chaacteitic equation 6

7 k n, k. fo k.8 fo k k k 6 (ii) (iii) 3 k.5 4k 5 Sol.4(d) 3 4k k.5 5 a 5 a k.5 k.5 4k 5 k.5 4k k 5 k 3.94 k o k and k 3.94 o G fo table ytem k 3.94, but k 3.94 k chaacteiticequation k 3 3 k a k a k Fo ytem to be table k 7

8 So k and k >,, Sol.5(a)(i) S e(t) m(t) m(t) he eo diffeence i in the ange of S and it maximum value i S S o and it i known a Quantization noie eo. ddition to coect tacking eo can alo take when S i too lage in thi cae input ignal i malle than tep ize and econtuction of ignal m(t) i no longe valid o in thi cae: S S m(t) (ii) (iii) f. S o avoid lope oveload eo m t dm t dt dm t dt max max f S co t m S Sf f m Fo avoiding lope ove load poblem Sol.5(b) S Fo avoiding tep ize limitation So Hee S f f f 3f hen Pin 3 W.8 m 4d km f b Mbp P f b b 8

9 nhc, n = incident photon nhc P.987 Watt So P fb hen P P e in L P in L log P P d P d d.l hen P d 35.8d P d 95.55d So 6.73 d d.l Sol.5(c) Lmax ntopy i given by 5.68km H f xlog f xdx f x H' x f ' xlog f x f ' xdx f x Fo f x H, H' x max f ' x contant, Condition fo unifom ditibution Now, given that it vay fom M to M; the aea unde pdf i unity M f(x) M M X Sol.6(a) M H log dx M max M M H log M max Rectangula waveguide dimenion =..3 cm (i) If guide i ai filled cut off fequency 3.3 f c c a 6.5GHz 9

10 f c c 3 3.4GHz a.3 So at GHz, only mode will popagate. So fo dominant mode popagation fequency ange = 6.5 GHz < f < 3.4 GHz Popagation contant f f c (ii) if filled with dielectic of efactive index =.5 o.5 f ' c fc GHz.5 f ' f ' c c fc GHz.5 c 3 5 GHz b..5.5 = GHz Sol.6(b) Given f ' c c.9ghz.3. So at GHz,,, and mode will popagate becaue f > f c fo thee mode. Fequency ange of opeation fo dominant mode 4.35GHz f 8.7GHz Popagation contant at 6. GHz 6.GHz 4.35GHz So wave will popagate and j f f c j9.83 d, / 4 dco co Nomalized eultant field fo -element aay = co Fo maximum adiation co º,

11 Fo half powe 6º, º Sol.6(c): S Sol.7(c) Z Z Z Z Z Z Z Z ZZ Z Z Z Z Z Z # include <tdio.h> # include <conio.h> main ( ) { int i, j, n, k; i =, j = pintf( %d %d,i j) pintf( nte value of n ); canf( %d, &n); fo(k=; k<n; k) { j = ij; pintf( %d, j); } } Refe page no.63 a Micowave engineeing book