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For the logic diagram shown below, the output f is
From given logic diagram, output is:
= A.(A + B) + B.(B + C)
= A.A + A B + BB + BC
= A + AB + B + BC
= A(1 + B) + B(1 + C)
= A + B
The minimum number of NOR gates requires to implement the expression
f(A, B, C) = πM(0, 1, 2, 3, 4, 7) is
Given expressio n is f = πM( 0, 1, 2, 3, 4, 7)
The Kmap for this expression is shown below.
The minimized expression in POS form is
The minimized expression 'f' can be implemented using two input NOR gate as shown below.
Thus, we require six NOR gates.
Consider the following statements associated with logic gates:
1. An OR gate in the positive logic system becomes an AND gate in the negative logic system and viceversa.
2. The dual of an expression is obtained by changing ANDs to ORs, ORs to ANDs.Os to 1s, to Os and by complementing the variables.
3. NAND and NOR are the only two available universal logic gates.
4. Universal gates do not follow commutative law
5. On logic diagrams, an inversion bubble at the point where the input is connected to a gate, indicates an activeHIGH input.
Which of the statements given above is/are not correct?
Hence, statements 1 and 3 are correct while statements 2, 4 and 5 are false.
The number of two input NAND gate required to implement an OR gate and an EXNOR gate are respectively
OR gate using NAND gate:
EXNOR gate using NAND gate:
Thus, we require 3 NAND gates for an OR gate while 5 NAND gates for an EXNOR gate.
Three set of simultaneous equations are given by
(i)
(ii) AB = AC
(iii)
The solution of the above three set of simultaneous equations will be
Given,
or,
i.e. must be 0 and B must be 0 Hence, A = 1 and B = 0
Also, AB = AC or 1.B  1.C
or B = C
Hence, C = 0 (∵ B = 0)
Also,
or, 1.0 + 1.1 + 0.D = 1.D
or D = 1
Thus, the values of A, 6, Cand Dare given by
A = 1, B = 0, C = 0 and D = 1.
If one of the input to an EXNOR gate is fixed to logic ‘1’, then output of this gate will be
Output = X.1 + = X = Input
The truth table for above circuit is shown below.
Hence, the output y is equivalent to output of an AND gate having two inputs A and 6. Thus, an AND gate is equivalent to a series switching circuit.
An XOR gate produces output 1 only when two inputs are
For same inputs, A A = 0
For different inputs,
Consider the truth table shown beiow:
The logic gate represented by the above truth table is
For same inputs, output (f) = 0
For different inputs, output (f) = 1
Hence, given truth table represents an EXOR gate.
A logic gate is an electronic circuit which
The output of a logic gate is ‘0’ when both its inputs are different. The gate is
Truth tables
Thus, the gate can be either an EXNOR or a NOR gate.
For the diode circuit shown below, both the diodes D_{1}, and D_{2 }are ideal.
If V_{R}= +5 V, then match ListI (Values of V_{1} and V_{2}) with ListIl (Output voltage V_{0}) and select the correct answer using the codes given below the lists:
ListI
V_{1} = 0 and V_{2} = 5 volt
V_{1 }= V_{2} = 5 volt
V_{1} = V_{2} = 0 volt
Listll
1 . 5 volt
2. 0.12 volt
3. 0.24 volt
4. 2.5 volt
Codes:
A B C
(a) 3 1 2
(b) 4 1 3
(c) 2 3 4
(d) 3 4 2
Thus,
= 0.12 V
A circuit has three inputs and one output. The output is 1 if at least two of the three input variables are 1, otherwise it is zero. The minimum number of basic gates required to implement the output (Y) are
The truth table for the given condition is shown below:
The Kmap for above truth table is shown below.
Thus, Y = AB + BC + CA which can be implemented using 3 AND gates and 1 OR gate (total 4 basic gates).
The boolean expression for the output f of the digital circuit shown below is
Let the output of second last NAND gate be X.
Then,
Thus,
Given circuit represents EXNOR gate using NAND gate
In the truth table of an Ninput OR gate, in the column for the output of the gate,
A two input OR gate truth table is shown below:
From the truth table of OR gate it is clear tha output is 0’s only when all the inputs are 0’s otherwise for all other input combination outpu is always 1's. Hence, in the column for the outpu of the OR gate, the number of zeros is always 1's.
Consider the following statements associated with logic gates:
1. Logic circuit of any complexity can be realised by using only the three basic gates namely AND, GR and NOT.
2. AND, OR and NOT .gates are called universal building blocks.
3. AND/OR/INVERT logic (AO! logic) can be converted to NAND logic or NOR logic.
4. A NAND gate can be used as an inverter by connecting all its input terminals except one, to logic 1 and applying the signal to be inverted to the remaining terminal.
Q. Which of the statements given above is/are correct?
Assertion (A): When a number of variables are to be XNORed, a number of twoinput XNOR gates can be used.
Reason (R): Three or more variable XNOR gates will make the circuit costly.
Since three or more variable XNOR gates do not exist, therefore when a number of variables are to be XNORed, a number of two input XNOR gases have to be used. Thus, reason is not a correct statement.
When n is even,
When n is odd,
The simplified version of the logic circuit shown below consists of
From the given circuit, output is
(∴ (X + Y) (X + Z) = X + YZ)
Therefore, the given logic circuit can be simplified as Do shown below.
Thus, it require a AND gate and a NOT gate.
The circuit shown in figure is a ECL ORAND INVERTER circuit. The output Z is
In case of ECL circuit, the floating inputs are considered as logic ‘0’. Hence output
22 docs274 tests

22 docs274 tests
