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Solving digital system problems using MATLAB Write a MATLAB script CTask2p3.m to: a) Generate the truth table for the following Boolean expressions and display it in the MATLAB command window (Use logical functions: and, not, or)

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Answer #1

Solution: Part a: Code: octave:59> fprintf(" fprintf(" AA BB CC DD XX\n"); fprintf("- -In") for AA = [0 1] for BB = [O 1] for CC- [0 1] for DD - [0 1] xXor ((and (AA, BB)), (and (CC,DD))); fprintf("%d %d %d %d %d\n", AA, BB, CC, DD, XX); end end end end fprintf(" fprintf(" AA YY\n"); fprintf("- for AA - [0 1] YY and ( AA , and ( AA , AA )); fprintf("%d %d\n", AA, end YY); Output:

Copyable code:

%Print statements

fprintf("-----------\n")

fprintf(" AA BB CC DD XX\n");

fprintf("-----------\n")

%Loops

for AA = [0 1]

for BB = [0 1]

for CC = [0 1]

for DD = [0 1]

%Expression

XX = or ((and(AA,BB)),(and(CC,DD)));

fprintf(" %d %d %d %d %d\n", AA, BB, CC, DD, XX);

end

end

end

end

%Second expression

fprintf("-----------\n")

fprintf(" AA YY\n");

fprintf("-----------\n")

for AA = [0 1]

YY = and(AA,and(AA,AA));

fprintf(" %d %d\n", AA, YY);

end

Part b: Code: octave: 142> %Part 1 bin num = 10011011'; dec_num-bin2dec(bin num); fprintf("\n\nConverted decimal number:%d" , dec-num); %Part 2 dec num137: bin num-dec2bin (dec num); fprintf("\n\nConverted binary number:%s ", bin num); %Pa rt 3 hex numA4C3'; decnun-hex2dec (hexnum) ; fprintf("\n\nConverted decimal number:%d" , dec-num); &Part 4 dec num52710: hex num-dec2hex(dec num); fprintf("\n\nConverted hexadecimal number:%s",hex-num); - - Output: Converted decimal number 155 Converted binary number: 10001001 Converted decimal number:42179 Converted hexadecimal number: CDE6

Copyable code:

%Part 1

bin_num = '10011011';

dec_num=bin2dec(bin_num);

fprintf("\n\nConverted decimal number:%d",dec_num);

%Part 2

dec_num = 137;

bin_num=dec2bin(dec_num);

fprintf("\n\nConverted binary number:%s",bin_num);

%Part 3

hex_num = 'A4C3';

dec_num=hex2dec(hex_num);

fprintf("\n\nConverted decimal number:%d",dec_num);

%Part 4

dec_num = 52710;

hex_num=dec2hex(dec_num);

fprintf("\n\nConverted hexadecimal number:%s",hex_num);

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Answer 1

Solution: Part a: Code: octave:59> fprintf(" fprintf(" AA BB CC DD XX\n"); fprintf("- -In") for AA = [0 1] for BB = [O 1] for CC- [0 1] for DD - [0 1] xXor ((and (AA, BB)), (and (CC,DD))); fprintf("%d %d %d %d %d\n", AA, BB, CC, DD, XX); end end end end fprintf(" fprintf(" AA YY\n"); fprintf("- for AA - [0 1] YY and ( AA , and ( AA , AA )); fprintf("%d %d\n", AA, end YY); Output:

Copyable code:

%Print statements

fprintf("-----------\n")

fprintf(" AA BB CC DD XX\n");

fprintf("-----------\n")

%Loops

for AA = [0 1]

for BB = [0 1]

for CC = [0 1]

for DD = [0 1]

%Expression

XX = or ((and(AA,BB)),(and(CC,DD)));

fprintf(" %d %d %d %d %d\n", AA, BB, CC, DD, XX);

end

end

end

end

%Second expression

fprintf("-----------\n")

fprintf(" AA YY\n");

fprintf("-----------\n")

for AA = [0 1]

YY = and(AA,and(AA,AA));

fprintf(" %d %d\n", AA, YY);

end

Part b: Code: octave: 142> %Part 1 bin num = 10011011'; dec_num-bin2dec(bin num); fprintf("\n\nConverted decimal number:%d" , dec-num); %Part 2 dec num137: bin num-dec2bin (dec num); fprintf("\n\nConverted binary number:%s ", bin num); %Pa rt 3 hex numA4C3'; decnun-hex2dec (hexnum) ; fprintf("\n\nConverted decimal number:%d" , dec-num); &Part 4 dec num52710: hex num-dec2hex(dec num); fprintf("\n\nConverted hexadecimal number:%s",hex-num); - - Output: Converted decimal number 155 Converted binary number: 10001001 Converted decimal number:42179 Converted hexadecimal number: CDE6

Copyable code:

%Part 1

bin_num = '10011011';

dec_num=bin2dec(bin_num);

fprintf("\n\nConverted decimal number:%d",dec_num);

%Part 2

dec_num = 137;

bin_num=dec2bin(dec_num);

fprintf("\n\nConverted binary number:%s",bin_num);

%Part 3

hex_num = 'A4C3';

dec_num=hex2dec(hex_num);

fprintf("\n\nConverted decimal number:%d",dec_num);

%Part 4

dec_num = 52710;

hex_num=dec2hex(dec_num);

fprintf("\n\nConverted hexadecimal number:%s",hex_num);