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Binary number problem

Hello,

I have a 14 bit number in the RAM. The RAM has 19 bit address bus. I generate a 19 bit address and retrieve that 14 bit number and then concatenate it with 5 bit number to get the data from other part of the memory. So, for example the number is (011F) hex at memory location

0000000000000000000. I get that number by generating memory address 0000000000000000000. Then I concatenate (011F) hex with "00000" to make 19 bit memory address again to reach somewhere in the memory and pick another data. The binay equivalent of (011F) hex is "0000000100011111". I put this number on the address bus bits 5 to 18 and put the "00000" on the address bus LSB bits from 0 to 4. So, the 14 bit number becomes (23E0)hex which is the base address of my first LUT value. My memory is 512K x 18 = 9M bits.18 bit is the data bus. I partioned the memory between the pointers and Lookup table values. The pointers are 257 x 64 = 16448 decimal and rest of the memory is occupied b y the LUT My questions are as follows
  1. When I concatenate the 14 bit number with the 5 bit number and make the address change what am I actually doing to the number am i multiplying the 14 bit number by 32 or what am i doing exactly?
  2. what is the actual maximum number I can use to point to a LUT.

Please Advice! John.

Reply to
john
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You are multiplying the 14-bit number by 2`5 (32) then adding the 5-bit number to it

Someone else can do that one.

--
Bill Sloman, Nijmegen
Reply to
bill.sloman

Hi,

Thanks for the reply! Would you please advice me that how many address locations I can access in the RAM or after multiplying the number with

32 and adding 5 bits to it what will be the range of the memory access I will have?

Thanks

snipped-for-privacy@ieee.org wrote:

Reply to
john

On Fri, 12 May 2006 09:13:14 -0700, john top-posted:

0000000000000000000-1111111111111111111

That'd be table 0, item 0 through table 16,383, item 31.

Hope This Helps! Rich

Reply to
Rich Grise

Hi,

i am sorry. I did not understand it. would you please explain in little more detail..

Thanks John

Reply to
john

john wrote:

you have 19 bit address bus so you have 2^19 = 524288 address

0000000000000000000 = 0 0000000000000000001 = 1 0000000000000000010 = 2 0000000000000000011 = 3 .... .... 1111111111111111100 = 524284 1111111111111111101 = 524285 1111111111111111110 = 524286 1111111111111111111 = 524287

Ahmed

Reply to
Ahmed Samieh

I don't know how much simpler I can make it. You've already stated the problem _and_ the solution. You have a 14-bit address bus that you want to shift left by 5 bits, and append an addional 5-bit number, that indicates which table element, so that you get 2^14 tables:

Address: xx xxxx xxxx xxxx (conveniently grouped in hexadecimal nybbles) range: 00 0000 0000 0000 = 0 decimal 11 1111 1111 1111 = 16,383 decimal So then, you shift that address left by 5 bits, and append 00000 binary:

Address: xxx xxxx xxxx xxx0 0000 range: 000 0000 0000 0000 0000 = 0 decimal 111 1111 1111 1110 0000 = 524256 decimal, but remember, this is the address of element 0 of table 16,393.

Then, you use the lower 5 bits for the table element, and you get:

ttt tttt tttt ttte eeeee to address a specific LUT element.

So, it's up to you to figure out how to arrange your memory, because you can access it all.

Good Luck! Rich

Reply to
Rich Grise

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