Derived gates are the gates which are derived from the fundamental logical gates

The NAND is the combination of NOT and AND. The NAND is generated by inverting the output of an AND operator. The algebraic expression of the NAND function is : Y = A . B

(i) BCD - Binary Coded Decimal.
(ii) EBCDIC - Extended Binary Coded Decimal Interchange Code.
(iii) ASCII - American Standard Code for Information Interchange,
(iv) Unicode.
(v) ISCII - Indian Standard Code for Information Interchange

46₁₀ - 101110₂

Step 1: Convert given Decimal number into Binary.
Step 2: Check if the binary number contains 8 bits. if less add 0 at the left most bit, to make it as 8 bits.
Step 3: Invert all bits (i.e. Change 1 as 0 and 0 as 1).

(i) The octal number system is playing a vital role in digital computer work.
(ii) The octal number system has a base of 8: It means that it has eight unique symbols such as 0,1,2,3,4,5,6 and 7.
(iii) Thus, each digit of an octal number can have any value from 0 to 7

a) AND gate truth table

b) OR Gate Truth table

c) Not Gate Truth Table

(i) ISCII is the system of handling the character of Indian local languages. This as a 8-bit coding system.
(ii) Therefore it can handle 256 (28) characters. This system is formulated by the Department of Electronics in India in the year 1986-88 and recognized by Bureau of Indian Standards (BIS) 
(iii) Now this coding system is integrated with Unicode.

(i) There are only two digits in the Binary system, namely, 0 and 1. The numbers in the binary system are represented to the base 2 and the positional multipliers are the powers of 2.
(ii) The left most bit in the binary number is called as the Most Significant Bit (MSB) and it has the largest positional weight.
(iii) The right most bit is the Least Significant Bit (LSB) and has the smallest positional weight.

(A) A6₁₆
\(\begin{matrix} A & 6 \\ \downarrow & \downarrow \\ \begin{matrix} 10 \\ \begin{matrix} \downarrow \\ 1010 \end{matrix} \end{matrix} & \begin{matrix} 6 \\ \begin{matrix} \downarrow \\ 0110 \end{matrix} \end{matrix} \end{matrix}\)
A6₁₆=10100110₂
(B) BE₁₆
\(\begin{matrix} B & E \\ \downarrow & \downarrow \\ \begin{matrix} 11 \\ \begin{matrix} \downarrow \\ 1010 \end{matrix} \end{matrix} & \begin{matrix} 14 \\ \begin{matrix} \downarrow \\ 1110 \end{matrix} \end{matrix} \end{matrix}\)
BE16=10111110₂
(C) 9BC8₁₆
\(\begin{matrix} 9 & B \\ \downarrow & \downarrow \\ \begin{matrix} 9 \\ \begin{matrix} \downarrow \\ 1001 \end{matrix} \end{matrix} & \begin{matrix} 11 \\ \begin{matrix} \downarrow \\ 1011 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} C & 8 \\ \downarrow & \downarrow \\ \begin{matrix} 12 \\ \begin{matrix} \downarrow \\ 1100 \end{matrix} \end{matrix} & \begin{matrix} 8 \\ \begin{matrix} \downarrow \\ 1000 \end{matrix} \end{matrix} \end{matrix}\)
9BC8₁₆=10001101111001000₂
(D) BC9₁₆
\(\begin{matrix} B & C & 9 \\ \downarrow & \downarrow & \downarrow \\ \begin{matrix} 11 \\ \downarrow \\ 1011 \end{matrix} & \begin{matrix} 12 \\ \downarrow \\ 1100 \end{matrix} & \begin{matrix} 9 \\ \downarrow \\ 1001 \end{matrix} \end{matrix}\)
BC9₁₆=101111001001₂

(3674)₈ in binary From
      3     6      7     4
      011  110   111  100
(3674)₈ = (011110111100)₂
(3674)₈ in decimal Form 
(3674)₈ = 3 x 8³ + x 8² + 7 x 8¹ + 4 x 8⁰
           = 1536 + 384 + 56 + 4 = (1980)₁₀
Thus (3674)₈ \(\rightarrow \) (0111110111100)₂ = (1980)₁₀

1101010₂ + 101101₂
1101010
+101101
________
10010111
________
= 10010111₂

=98₁₀ = 110010
8bit format of 98₁₀ =     0 1 1 0 0 0 1 0
1's complement     =     1 0 0 1 1 1 0 1
Add 1 Bit                =                     +1
                                       _____________
2's complement      =    1 0 0 1 1 1 0  1
                                       _____________