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Given the following in 8-bit 2s complement numbers:

11000011 = -61 (decimal)
00011111 = +31 (decimal)

I am required to obtain a boolean expression of a logic circuit whose output out goes high when its 8-bit input in (also in 2s complement representation) is in the following range:

-61 < in < 31

Number line for 8 bit numbers (2s complement):

10000000 (most negative) ..... 11000011 (-61) ..... 00000000 ..... 00011111 (31) ..... 01111111 (most positive)

Is there any way of solving this problem besides brute-forcing and comparing bit-by-bit?

Edit: The following statement is not allowed

out = ((in < 11000011 && in > 10000000) || (in > 00011111 && in < 01111111)) ? 1'b0 : 1'b1;
JTJM
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  • Are you allow to use comparison `<` and `>`? Are you allow to use operator `?`? That is a if-then-else... maybe they are asking you to use only logic ports, like `AND`, `XOR`, `NOT`, etc... – GabrielOshiro Nov 22 '15 at 04:09
  • Your answer seems correct. What about `out = (in >= 11000011 && in <= 00011111) ? 1'b1 : 1'b0;` ? assuming you can use `>=` and `<=` operators. – GabrielOshiro Nov 22 '15 at 04:09
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    can you access each bit of the input individually like `in[7]` for bit 7 of the input, `in[6]` for bit 6 of the input, etc? If so, you could use `(~in[7] & ~in[6] & ~in[5]) | ((in[7] & in[6] & in[1] & in[0])` – GabrielOshiro Nov 22 '15 at 04:27
  • @GabrielOshiro I have clarified it with the assignment setter and apparently, I'm not allowed to access the entire data stream. I only have access to the individual bits (something like your comment with in[7]-in[1]) so I went ahead and listed all the numbers out to find a pattern manually. Not pretty though. Thanks for pointing me in the right direction. – JTJM Dec 02 '15 at 15:20

1 Answers1

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I'm not sure if there is a faster way to do this. But what I did was to list the numbers out in 2s complement format before trying to find a pattern. The following chunks of numbers are sorted in numerical order (from 00000000 to 11111111 so that the pattern can be more clearly seen).

Let the MSB be A and LSB be H. The equation is: A B C + A B D + A B E + A B F + A' B' C' D' + A' B' C' E' + A' B' C' F' + A' B' C' G' + A' B' C' H'

A' B' C' D' (easiest to observe):

00000000 (<- min) 00000001 00000010 00000011 00000100 00000101 00000110 00000111 00001000 00001001 00001010 00001011 00001100 00001101 00001110 00001111

A' B' C' E' + A' B' C' F' + A' B' C' G' + A' B' C' H':

00010000 00010001 00010010 00010011 00010100 00010101 00010110 00010111 00011000 00011001 00011010 00011011 00011100 00011101 00011110

A B D + A B E + A B F:

11000100 11000101 11000110 11000111 11001000 11001001 11001010 11001011 11001100 11001101 11001110 11001111 11010000 11010001 11010010 11010011 11010100 11010101 11010110 11010111 11011000 11011001 11011010 11011011 11011100 11011101 11011110 11011111

A B C (easiest to observe):

11100000 11100001 11100010 11100011 11100100 11100101 11100110 11100111 11101000 11101001 11101010 11101011 11101100 11101101 11101110 11101111 11110000 11110001 11110010 11110011 11110100 11110101 11110110 11110111 11111000 11111001 11111010 11111011 11111100 11111101 11111110 11111111 (<-max)

JTJM
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