(a) Perform the following conversions. 1. 38F 16 to octal 2. 153 8 to binary 3. 456 10 to hexadecimal 4. 1001 0111 2 in to decimal (b) Draw the logic circuit for the following Boolean expression expressed in two forms. Also write the truth table. Y = AC + BC (i) Y = C(A + B) (ii)

Question

(a) Perform the following conversions.

[latex]38F_{16}[/latex] to octal
[latex]153_{8}[/latex] to binary
[latex]456_{10}[/latex] to hexadecimal
1001 [latex]0111_{2}[/latex] in to decimal

(b) Draw the logic circuit for the following Boolean expression expressed in two forms. Also write the truth table.

[latex]Y=\overline{A} \overline{C}+ B \overline{C}[/latex] (i)

[latex]Y=\overline{C}(\overline{A}+B)[/latex] (ii)

Accepted Answer

(a) 1. Writing the binary equivalent of each hexdigit of [latex]38F_{16}[/latex], we get 0011 1000 1111. We get 001 110 001 111 after arranging this in groups of 3 bits. Writing the octal digit for each group, we get the desired octal number, i.e. [latex]1617_{8}[/latex].

Writing the binary equivalent for each octal digit of [latex]153_{8}[/latex], we get 001 101 011, which is the answer.
Succesive division of 456 by 16 gives:

Divisor	Number	Remainder
16	456
16	28	8
16	1	C
16	0	1
	The answer is 1C8.

By providing the weightage of each bit, we get the following:

[latex]1 imes 2^{7}+0 imes 2^{6}+0 imes 2^{5}+1 imes 2^{4}+0 imes 2^{3}+1 imes 2^{2}+1 imes 2^{1}+1 imes 2^{0} [/latex]

= 1 × 128 + 0 × 64 + 0 × 32 + 1 × 16 + 0 × 8 + 1 × 4 + 1 × 2 + 1 × 1

= 128 + 0 + 0 + 16 + 0 + 4 + 2 + 1

= 151

(b) The logic circuits are shown in Figures 31 and 32 and the truth table is provided in Table 2.

Form (i)

Form (ii)

Table 2 Truth Table
Input Variables			Form 1 Intermediate Variables		Form 1 Output	Form 2 Intermediate Variables		Form 2 Output
A	B	C	A′·C′	B·C′	Y = A′·C′ + B·C′	[latex]\underline{A^{\prime}+B}[/latex]	[latex]\underline{C^{\prime}}[/latex]	[latex]\underline{Y=C^{\prime} .\left(A^{\prime}+B ight)}[/latex]
0	0	0	1	0	1	[latex]\underline{1}[/latex]	[latex]\underline{1}[/latex]	[latex]\underline{1}[/latex]
0	0	1	0	0	0	[latex]\underline{1}[/latex]	[latex]\underline{0}[/latex]	[latex]\underline{0}[/latex]
0	1	0	1	1	1	[latex]\underline{1}[/latex]	[latex]\underline{1}[/latex]	[latex]\underline{1}[/latex]
0	1	1	0	0	0	[latex]\underline{1}[/latex]	[latex]\underline{0}[/latex]	[latex]\underline{0}[/latex]
1	0	0	0	0	0	[latex]\underline{0}[/latex]	[latex]\underline{1}[/latex]	[latex]\underline{0}[/latex]
1	0	1	0	0	0	[latex]\underline{0}[/latex]	[latex]\underline{0}[/latex]	[latex]\underline{0}[/latex]
1	1	0	0	1	1	[latex]\underline{1}[/latex]	[latex]\underline{1}[/latex]	[latex]\underline{1}[/latex]
1	1	1	0	0	0	[latex]\underline{1}[/latex]	[latex]\underline{0}[/latex]	[latex]\underline{0}[/latex]

Table 2 Truth Table
Input Variables			Form 1 Intermediate Variables		Form 1 Output	Form 2 Intermediate Variables		Form 2 Output
A	B	C	A′·C′	B·C′	Y = A′·C′ + B·C′	$\underline{A^{\prime}+B}$	$\underline{C^{\prime}}$	$\underline{Y=C^{\prime} .\left(A^{\prime}+B\right)}$
0	0	0	1	0	1	$\underline{1}$	$\underline{1}$	$\underline{1}$
0	0	1	0	0	0	$\underline{1}$	$\underline{0}$	$\underline{0}$
0	1	0	1	1	1	$\underline{1}$	$\underline{1}$	$\underline{1}$
0	1	1	0	0	0	$\underline{1}$	$\underline{0}$	$\underline{0}$
1	0	0	0	0	0	$\underline{0}$	$\underline{1}$	$\underline{0}$
1	0	1	0	0	0	$\underline{0}$	$\underline{0}$	$\underline{0}$
1	1	0	0	1	1	$\underline{1}$	$\underline{1}$	$\underline{1}$
1	1	1	0	0	0	$\underline{1}$	$\underline{0}$	$\underline{0}$

Question III.8: (a) Perform the following conversions. 1. 38F 16 to octal 2....

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