(a) Convert the following numbers to be stated bases: (i) 125.625, to binary (ii) 111011.10011, to octal (b) Given that X = 11011011 and Y= 11101101, (i) co...
Question 1 Report
(a) Convert the following numbers to be stated bases:
(i) 125.625, to binary
(ii) 111011.10011, to octal
(b) Given that X = 11011011 and Y= 11101101,
(i) construct the truth table for X, Y Z;
(ii) draw the logic gate of X, Y = Z
(iii) state the value of Z.
Answer Details
(a) Number-base conversions
(i) Convert \(125.625_{10}\) to base 2.
For the integer part, successive division by 2 gives:
| Division | Quotient | Remainder |
|---|---|---|
| 125 ÷ 2 | 62 | 1 |
| 62 ÷ 2 | 31 | 0 |
| 31 ÷ 2 | 15 | 1 |
| 15 ÷ 2 | 7 | 1 |
| 7 ÷ 2 | 3 | 1 |
| 3 ÷ 2 | 1 | 1 |
| 1 ÷ 2 | 0 | 1 |
Reading the remainders from bottom to top gives \(125_{10}=1111101_2\).
For the fractional part:
\(0.625\times2=1.25\), giving 1;
\(0.25\times2=0.5\), giving 0;
\(0.5\times2=1.0\), giving 1.
Therefore, \(0.625_{10}=0.101_2\), and hence
\[\boxed{125.625_{10}=1111101.101_2}\]
(ii) Convert \(111011.10011_2\) to base 8. Group the binary digits in threes from the binary point:
\[111\;011.100\;110_2=73.46_8\]
\[\boxed{111011.10011_2=73.46_8}\]
(b) \(X=11011011\), \(Y=11101101\), and \(Z=X\mathbin{\text{ AND }}Y\).
(i) Truth table for the AND operation
| X | Y | Z = X AND Y |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
(ii) Logic-gate diagram
(iii) Value of \(Z\)
| Bit position | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|
| X | 1 | 1 | 0 | 1 | 1 | 0 | 1 | 1 |
| Y | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 1 |
| Z | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 1 |
\[\boxed{Z=11001001_2}\]
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