(D) NAND gate and NOT gateĪns: B DeMorgan’s first theorem shows the equivalence of NOR gate and Bubbled AND gate (Logic diagrams for De Morgan’s First Theorem is shown in fig.1(a) Aįig.1(a) Logic Diagrams for De Morgan’s First Theorem i.e., - 8 = 1000 1000 -Sign Magnitude -ĭeMorgan’s first theorem shows the equivalence of (A) OR gate and Exclusive OR gate. The MSB is the sign bit followed by the magnitude bits. 8 is equal to signed binary number (A) 10001000 (C) 10000000Īns: A - 8 is equal to signed binary number 10001000 6ĭIGITALS ELECTRONICS (To represent negative numbers in the binary system, Digit 0 is used for the positive sign and 1 for the negative sign. Which of the memory is volatile memory (A) ROM (B) RAM (C) PROM (D) EEPROMĪns: B RAM is a volatile memory (Volatile memory means the contents of the RAM get erased as soon as the power goes off.) Q.25 The Boolean expression A.B + A.B + A.B is equivalent to (A) A + B (B) A.B (C) A + B (D) A.B Ans: A The Boolean expression A.
Decimal to bcd priority encoder k map code#
The Gray code for decimal number 6 is equivalent to (A) 1100 (B) 1001 (C) 0101 (D) 0110 Ans: C The Gray code for decimal number 6 is equivalent to 0101 (Decimal number 6 is equivalent to binary number 0110) + The hexadecimal number ‘A0’ has the decimal value equivalent to (A) 80 (B) 256 (C) 100 (D) 160 Ans: D The hexadecimal number ‘A0’ has the decimal value equivalent to 160 ( A 0 161 160 = 10X161 + 0X160 = 160) Ans: A EPROM contents can be erased by exposing it to Ultraviolet rays (The Ultraviolet light passes through a window in the IC package to the EPROM chip where it releases stored charges.
In this case 24 greater than or equal to 1)ĮPROM contents can be erased by exposing it to (A) Ultraviolet rays. 2ĭIGITALS ELECTRONICS (For Mod-m Counter, we need N flip-flops where N is chosen to be the smallest number for which 2N is greater than or equal to m. How many Flip-Flops are required for mod–16 counter? (A) 5 (B) 6 (C) 3 (D) 4 Ans: D The number of flip-flops is required for Mod-16 Counter is 4.
The number of control lines for a 8 – to – 1 multiplexer is (A) 2 (B) 3 (C) 4 (D) 5 Ans: B The number of control lines for an 8 to 1 Multiplexer is 3 (The control signals are used to steer any one of the 8 inputs to the output) The Boolean expression is ABC + A BC is equivalent to 1 The simplification of the Boolean expression ABC + A BC is (A) 0 (B) 1 (C) A (D) BC Ans: B The NAND gate output will be low if the two inputs are (A) 00 (B) 01 (C) 10 (D) 11 Ans: D The NAND gate output will be low if the two inputs are 11 (The Truth Table of NAND gate is shown in Table.1.1) X(Input) Y(Input) F(Output) 0 0 1 0 1 1 1 0 1 1 1 0 Table 1.1 Truth Table for NAND Gate Choose correct or the best alternative in the following: Q.1 OBJECTIVE TYPE QUESTIONS Each Question carries 2 marks.
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