Texas Instruments Paper

texas instruments paper .

in this paper there was 20 questions as follows in 60 minutes .
second part consists of 36 que. in 30 minutes all questions are
diagramatical.(figurs)..

 

1. if a 5-stage pipe-line is flushed and then we have to execute 5 and 12
instructions respectively then no. of  cycles will be
a. 5 and 12
b. 6 and 13
c. 9 and 16
d.none 


2. k-map


ab
----------
c 1   x  0  0
 1   x  0  x

solve it 

a. A.B
B. ~A
C. ~B
D. A+B


3.CHAR  A[10][15] AND INT B[10][15] IS DEFINED
WHAT'S THE ADDRESS OF A[3][4] AND B[3][4]
IF ADDRESS OD A IS OX1000 AND B IS 0X2000

A. 0X1030 AND 0X20C3
B. OX1031 AND OX20C4
AND SOME OTHERS..

4. int f(int *a)
{
 int b=5;
a=&b;
}

main()
{
int i;
printf("\n %d",i);
f(&i);
printf("\n %d",i);
}

what's the output .

1.10,5
2,10,10
c.5,5
d. none

5. main()
{
int i;
fork();
fork();
fork();
printf("----");
}

how many times the printf will be executed .
a.3 
b. 6
c.5
d. 8


6. 
void f(int i)
{
int j;
for (j=0;j<16;j++)
{
if (i & (0x8000>>j))
printf("1");
else
printf("0");
}
}
what's the purpose of the program 

a. its output is hex representation of i
b. bcd
c. binary
d. decimal

 
7.#define f(a,b) a+b
#define g(a,b) a*b


main()
{
int m;
m=2*f(3,g(4,5));
printf("\n m is %d",m);
}

what's the value of m
a.70
b.50
c.26
d. 69

8.

main()
{
char a[10];
strcpy(a,"\0");
if (a==NULL)
printf("\a is null");
else
printf("\n a is not null");}

what happens with it .
a. compile time error.
b. run-time error.
c. a is  null
d. a is not  null.

9. char a[5]="hello"

a. in array we can't do the operation .
b. size of a is too large
c. size of a is too small
d. nothing wrong with it .

10. local variables can be store by compiler
a. in register or heap
b. in register or stack
c .in stack or heap .
d. global memory.

11. average and worst time complexity in a sorted binary tree is

12. a tree is given and ask to find its meaning (parse-tree)
  (expression tree)
ans. ((a+b)-(c*d))  ( not confirmed)
13. convert 40.xxxx into binary .

14. global variable conflicts due to multiple file occurance 
is resolved during 
a. compile-time
b. run-time 
c. link-time
d. load-time

15. 
two program is given of factorial.
one  with  recursion and one without recursion .
question was which program won't run for very big no. input because of
stack overfow .
a. i only  (ans.)
b. ii only
c. i& ii both .
c. none 

16. 
struct a
{
int a;
char b;
int c;
}

union b
{
char a;
int b;
int c;
};
which is correct .
a. size of a is always diff. form size of b.(ans.)
b. size of a is always same  form size of b.
c. we can't say anything because of not-homogeneous (not in ordered)
d. size of a can be same if ...





>
>
>1.two transistors are connected Vbe is 0.7volts .this is simple ckt.one 
>transistor is diode equivalent. & asked the o/p across the 2 nd transistor.
>2.simple k map ans is Bbar.
>3.
>
>                               Emitter
>---R-------transistorbase| --
>                          | ---
>                                   collector
>             in above capacitor is connected parallel with resistance 
>r.capacitor is not shown
>             in fig.capacitor is used for in this ckt:
>
>
>             ans:a.speedupb.active bypass  c.decoupling
>   4.
>
>   -----R------I----------o/p
>           |___R____ |
>                             in above r is resistence.I is cmos inverter.
>                             then ckt is used for:
>
>
>                             a.schmitt trigger b.latch  c.inverter  
>d.amplifier
>
>
>      5.simple amplifier ckt  openloop gain of amplifier is 4.V in 
>=1v.asked for V x?
>      amplifdier + is connected to base. - is connected to i/p in between 
>5k is connected.
>      from o/p feedback connected to - of amplifier with 15k.this is ckt.
>
>
>      6.resistence inductot cap are serially connected to ac voltage 5 
>volts.voltage across
>      inductor is given.R I C values are given & asked for
>      voltages across resistence & capacitor.
>      7.
>              ___  R_____
>             |            |
>      ---R------OPAMP ----------
>             |---
>             R1        R1 is for wjhat i mean what is the purpose of R1.
>             |
>
>             ground
>
>
>     8.asked for Vo at the o/p.it is like simple cmos realization that is n 
>block is above
>     & p block is below.Vdd is 3 volts at supply.V threshold 5 volts.
>     9.2 d ffs are connected in asyncro manner .clock 10 MEGAHZ.gate delay 
>is 1 nanosec.
>     A B are the two given D FFs.asked for AB output is:
>
>
>     a.updown
>     b.up c. updown glitching like that (take care abt glitching word)
>
>     10.
>
>
>     ----------------| subtractor|---------o/p
>         |___HPF____|
>
>                     the ckt is LPF ,HPF or APF ?
>
>   11.in a queue at the no of elements removed is proportional to no of 
>elements in
>   the queue.then no of elements in the queue:
>   a.increases decreases exp or linearly(so these are the 4 options given 
>choose 1 option)
>   12.with 2 i/p AND gates u have to form a 8 i/p AND gate.which is the 
>fastest in the
>   following implementations.
>   ans we think ((AB)(CD))((EF)(GH))
>   13.with howmany 2:1 MUX u can for   8:1 MUX.answer is 7.
>   14. there are n states then ffs used are log n.
>   15.cube each side has r units resistence then the resistence across 
>diagonal of cube.
>   16.op amp connections asked for o/p
>   the answer is (1+1/n)(v2-v1).check it out.practise this type of model.
>   17.
>       _____________ supply
>   ---|__           ___|
>  Ii     >________ |___    Tranistot
>                       > _______Vo
>                        |
>                        |
>                        R       |
>                        |       |  Io
>                        ground.
>
>
>
>
>    asked for Io/Ii=? transistor gain is beta.
>
>
>    a.(1+beta)square b.1+beta  c. beta
>
>
>    18.y=kxsquare. this is transfer function of a block with i/p x & o/p 
>y.if i/p is
>    sum of a & b then o/p is :--
>
>    a. AM b.FM  c. PM
>    19.
>                 ------MULTIPLIER--- |
>                |                    |
>        _____R__|__OPAMP______________________Vo
>                 ---
>                 |
>                ground.
>                v in = -Ez then o/p Vo =?
>                answer is squareroot of -Ez.multiplier i/ps are a & b then 
>its o/p
>                is a.b;
>