Design a circuit to detect when 3 and only 3 bits are set
out of 8 bits.(eg. o0101100)
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Answer / xyz
You fools...the question is to design a circuit
(hardware)..not writing a program.
| Is This Answer Correct ? | 31 Yes | 0 No |
Answer / esther
Add the digits and check the sum. It would imply a 3-bit
adder (max. value = 8 can be represented with three bits)
and compare the result to 011.
| Is This Answer Correct ? | 31 Yes | 2 No |
Answer / jayasuriya
Use a 8 bit parallel in serial out shift register. Use the
serial out bit to increment a counter(a 3 bit register), if
its one. After 8 shifts, compare the counter with 011b to
detect 3.
| Is This Answer Correct ? | 21 Yes | 1 No |
Answer / priya
keep the number in A reg and ratate it 8 times if carry
generates more than 3 it will indicate a non desired numer
| Is This Answer Correct ? | 12 Yes | 5 No |
Answer / aiyush
The first guy's got it write, every other answer is more
complicated than needed. Also, we cannot use a 3-input AND
like the second guy said b/c we are given 8 bits, not 3. A
3-bit adder and a comparator is all you need.
| Is This Answer Correct ? | 5 Yes | 2 No |
Answer / ravichandra
step1: start
step2: mov the byte to accumulator
step3: intialize length = 0 , count = 0
step4: rotate acc. to right through carry.
step5: if carry = 1 , increment count
step6: increment length
step7: if length < 8 , goto step3
step8: if count = 3 , (do required action)
step9: end
| Is This Answer Correct ? | 6 Yes | 4 No |
Answer / arnab kumar biswas
2 bit adder will do the trick.give 0 to one i/p.and transmit
the bits serially one by one to the other i/p.o/p of the
adder goes to latch,where it will stay until 8 bits have
come.then use comparator to compare if answer is 3 or not.
| Is This Answer Correct ? | 1 Yes | 0 No |
Answer / kalyanpa
for serially arriving stream:
use a 2 bit accumulator( S1 and S2) with outputs O1 and O2.
Final Output will be Y = ~O1 . O2
For parallel,
use combinational logic , probably priority encoder to
reduce delay.
| Is This Answer Correct ? | 0 Yes | 2 No |
Answer / gautam bhattacharya
Step 1: Store the 8 bit value in a accumulator
Step 2: Store 0x1 in a register0, initialize two counter
with 0 i.e. store zero in a reg1 and reg2.
LOOP:
Step 3: Check if AND operation between the value in
register0 and accumulator is set i.e. 1
if yes, increment reg1 and reg2
If no, increment only reg2
step 4: Left shift the value of register0 by 1
step 5: if ( reg2 >=8), exit LOOP
if ( reg1 >= 3), show that 3 bit is set
Else Go To LOOP
MOV XAR1, #Data
MOV XAR0, #0
MOV XAR2, #0
Loop:
TBIT *XAR1, #Count
BF Loop1, NTC
INR *XAR0
Loop1:
INR *XAR2
MOV AL, *XAR1
CMP AL, #0x03
BF Loop3, EQ
MOV AL, *XAR2
CMP AL, #0x80
BF Loop, NEQ
Loop3:
EXIT
| Is This Answer Correct ? | 0 Yes | 3 No |
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