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                                                     >^M
>          THIS IS TI 1999 jadavpur for ECE students.for cs another paper is ^M
>given^M
>^M
>1.two transistors are connected Vbe is 0.7volts .this is simple ckt.one ^M
>transistor is diode equivalent. & asked the o/p across the 2 nd transistor.^M
>2.simple k map ans is Bbar.^M
>3.^M
>^M
>                               Emitter^M
>---R-------transistorbase| --^M
>                          | ---^M
>                                   collector^M
>             in above capacitor is connected parallel with resistance ^M
>r.capacitor is not shown^M
>             in fig.capacitor is used for in this ckt:^M
>^M
>^M
>             ans:a.speedupb.active bypass  c.decoupling^M
>   4.^M
>^M
>   -----R------I----------o/p^M
>           |___R____ |^M
>                             in above r is resistence.I is cmos inverter.^M
>                             then ckt is used for:^M
>^M
>^M
>                             a.schmitt trigger b.latch  c.inverter  ^M
>d.amplifier^M
>^M
>^M
>      5.simple amplifier ckt  openloop gain of amplifier is 4.V in ^M
>=1v.asked for V x?^M
>      amplifdier + is connected to base. - is connected to i/p in between ^M
>5k is connected.^M
>      from o/p feedback connected to - of amplifier with 15k.this is ckt.^M
>^M
>^M
>      6.resistence inductot cap are serially connected to ac voltage 5 ^M
>volts.voltage across^M
>      inductor is given.R I C values are given & asked for^M
>      voltages across resistence & capacitor.^M
>      7.^M
>              ___  R_____^M
>             |            |^M
>      ---R------OPAMP ----------^M

                 >             |---^M
>             R1        R1 is for wjhat i mean what is the purpose of R1.^M
>             |^M
>^M
>             ground^M
>^M
>^M
>     8.asked for Vo at the o/p.it is like simple cmos realization that is n ^M
>block is above^M
>     & p block is below.Vdd is 3 volts at supply.V threshold 5 volts.^M
>     9.2 d ffs are connected in asyncro manner .clock 10 MEGAHZ.gate delay ^M
>is 1 nanosec.^M
>     A B are the two given D FFs.asked for AB output is:^M
>^M
>^M
>     a.updown^M
>     b.up c. updown glitching like that (take care abt glitching word)^M
>^M
>     10.^M
>^M
>^M
>     ----------------| subtractor|---------o/p^M
>         |___HPF____|^M

                         >^M
>                     the ckt is LPF ,HPF or APF ?^M
>^M
>   11.in a queue at the no of elements removed is proportional to no of ^M
>elements in^M
>   the queue.then no of elements in the queue:^M
>   a.increases decreases exp or linearly(so these are the 4 options given ^M
>choose 1 option)^M
>   12.with 2 i/p AND gates u have to form a 8 i/p AND gate.which is the ^M
>fastest in the^M
>   following implementations.^M
>   ans we think ((AB)(CD))((EF)(GH))^M
>   13.with howmany 2:1 MUX u can for   8:1 MUX.answer is 7.^M
>   14. there are n states then ffs used are log n.^M
>   15.cube each side has r units resistence then the resistence across ^M
>diagonal of cube.^M
>   16.op amp connections asked for o/p^M
>   the answer is (1+1/n)(v2-v1).check it out.practise this type of model.^M
>   17.^M
>       _____________ supply^M
>   ---|__           ___|^M
>  Ii     >________ |___    Tranistot^M
>                       > _______Vo^M
>                       > _______Vo^M
>                        |^M
>                        |^M
>                        R       |^M
>                        |       |  Io^M
>                        ground.^M
>^M
>^M
>^M
>^M
>    asked for Io/Ii=? transistor gain is beta.^M
>^M
>^M
>    a.(1+beta)square b.1+beta  c. beta^M
>^M
>^M
>    18.y=kxsquare. this is transfer function of a block with i/p x & o/p ^M
>y.if i/p is^M
>    sum of a & b then o/p is :--^M
>^M
>    a. AM b.FM  c. PM^M
>    19.^M
>                 ------MULTIPLIER--- |^M

                       >                |                    |^M
>        _____R__|__OPAMP______________________Vo^M
>                 ---^M
>                 |^M
>                ground.^M
>                v in = -Ez then o/p Vo =?^M
>                answer is squareroot of -Ez.multiplier i/ps are a & b then ^M
>its o/p^M
>                is a.b;^M


