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physics=world
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1. The original expression is:
F(A,B,C,D) = Σm(0,1,2,5,8,9,10)
If I changed the order of ABCD to:
F(C,B,A,D)
What would be the minterms now?2. Use the truth table for 0 to 15
I used the truth table in the regular order
ABCD and for CBAD.ABCD | F
0000 | 1
0001 | 1
0010 | 1
0011 | 0
0100 | 0
0101 | 1
0110 | 0
0111 | 0
1000 | 1
1001 | 1
1010 | 1
1011 | 0
1100 | 0
1101 | 0
1110 | 0
1111 | 0
So, here I basically thought that if C was 1 in the regular column then in the new truth table I just look for 1 under the C column. Is this right or do I need to redo the columns so that everything under the A column from the regular truth table is shifted over to 2 columns over?
CBAD | F
0000 | 1
0001 | 1
0010 | 1
0011 | 1
0100 | 0
0101 | 1
0110 | 0
0111 | 0
1000 | 1
1001 | 0
1010 | 1
1011 | 0
1100 | 0
1101 | 0
1110 | 0
1111 | 0
I got F(C,B,A,D) = Σm(0,1,2,3,5,8,10)
is this correct? I thought that if you changed the order than the minterms should change and only one minterm changed.
F(A,B,C,D) = Σm(0,1,2,5,8,9,10)
If I changed the order of ABCD to:
F(C,B,A,D)
What would be the minterms now?2. Use the truth table for 0 to 15
The Attempt at a Solution
I used the truth table in the regular order
ABCD and for CBAD.ABCD | F
0000 | 1
0001 | 1
0010 | 1
0011 | 0
0100 | 0
0101 | 1
0110 | 0
0111 | 0
1000 | 1
1001 | 1
1010 | 1
1011 | 0
1100 | 0
1101 | 0
1110 | 0
1111 | 0
So, here I basically thought that if C was 1 in the regular column then in the new truth table I just look for 1 under the C column. Is this right or do I need to redo the columns so that everything under the A column from the regular truth table is shifted over to 2 columns over?
CBAD | F
0000 | 1
0001 | 1
0010 | 1
0011 | 1
0100 | 0
0101 | 1
0110 | 0
0111 | 0
1000 | 1
1001 | 0
1010 | 1
1011 | 0
1100 | 0
1101 | 0
1110 | 0
1111 | 0
I got F(C,B,A,D) = Σm(0,1,2,3,5,8,10)
is this correct? I thought that if you changed the order than the minterms should change and only one minterm changed.