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Specify the size of a ROM (number of words and number of bits per word) that will accommodate the truth table for the following combinational circuit :

a code converter from a 4-digit BCD number to a binary number.
largest number is 9999(1001100110011001)in 4 digit BCD which needs 15 bits in binary representation ,so total 10^4 combination ,hence 10^4 *15 bits ?

what you want to say is:

if 4 digit bcd num given then the o/p is corresponding binary number

total no of possible 4 digit binary number =104

and output is the binary number largest 4 digit BCD is 1001100110011001 ie. 9999

corresponding binary number needs 15 bits to represent 9999

so total 104 combination and each combination correspond to a 15 bit binary number.

Toal size of ROM required (104   *  15 )

yes Sir.. exactly

+1 vote

in the question code converter from a 4-digit BCD number is mention ,

Now , the table for code converter have 16 rows , hence total 24 = 16 combinations are there and each combination correspond to a 14 bit binary number. The size for converter table is 216

Total size of ROM required (216   *  14 ) as each combination is 14 bits long.

Same type of question  -->  http://gateoverflow.in/26836/rom-size-to-build-adder-substractor

answered by Veteran (62k points) 14 138 556

http://gateoverflow.in/?qa=blob&qa_blobid=6434520157519357056

Sir please see their explanation. so confused :(

This question  is taken from mano&Kime book .

The above link is the solution to Mano & kime  book.

but the thing is in that snap they clearly mention a code converter hence 216 is taken for 16 inputs.

Sir see the qs in 1.c. the explanation is highlighted for c part

yes, code converter have 16 inputs,

for 4 bits , it is 2= 16 ,  as it is binary number . see the table for code -converter ( both binary to gray & gray to binary )

http://digitalbyte.weebly.com/code-converters.html

as it has 16 inputs hence 216  number of words posible.

+1 vote

Refresher:

Binary-coded decimal (BCD) is a class of binary encodings of decimal numbers.

Eg: 1001 in BCD representation represents decimal no. 9

1001 1001 in BCD representation represents decimal no. 99 and so on.

Doubt Clarification:

//Note: I/P is in BCD, Output is in Binary.

1 digit BCD input, binary output:

Max possible I/p: 1001 (9)     O/P: = 1001 (9)

Input lines eqd.     = 4

Output lines reqd. = 4

2 digit BCD input, binary output:

Max possible I/p: 1001 1001 (99)  O/P: 1100011(99)

Input lines reqd.     = 8     //count the no. of input bits

Output lines reqd. =  7     // ceil (log 99) = ceil (6.6293) = 7

//or simply count the no. of output bits

3 digit BCD input, binary output:

Max possible I/p: 1001 1001 1001 (999)  O/P: 1111100111(999)

Input lines eqd.     = 12

Output lines reqd. =  10     // ceil (log 999) = ceil (9.964) = 10

4 digit BCD input, binary output:

Max possible I/p: 1001 1001 1001 1001 (9999)  O/P: 10011100001111(9999)

Input lines eqd.     = 16

Output lines reqd. =  14     // ceil (log 9999) = ceil (13.28) = 14

Hence, Size of ROM with 16 input lines and 14 output lines is: 2^16 * 14 bits

answered by Junior (551 points) 9