Unit - 1
Boolean Algebra, Types of Number Systems, Types of Codes and Forms of Expression
Q1) Explain universal gates in detail.
A1)
NAND gate
It is a digital circuit which has two or more inputs and single output and it is the inversion of logical AND gate.
The truth table of 2-input NAND gate is:
A | B | Y = (A.B)’ |
0 | 0 | 1 |
0 | 1 | 1 |
1 | 0 | 1 |
1 | 1 | 0 |
Here A, B are the inputs and Y is the output of two input NAND gate. When both inputs are ‘1’, then the output, Y is ‘0’. If at least one of the input is zero, then the output, Y is ‘1’. This is just the inverse of AND operation.
The image shows the symbol of NAND gate:
Fig.: NAND gate
NAND gate works same as AND gate followed by an inverter.
Timing Diagram:
NOR gate
It is a digital circuit that has two or more inputs and a single output which is the inversion of logical OR of all inputs.
The truth table of 2-input NOR gate is:
A | B | Y = (A+B)’ |
0 | 0 | 1 |
0 | 1 | 0 |
1 | 0 | 0 |
1 | 1 | 0 |
Here A and B are the two inputs and Y is the output. If both inputs are ‘0’, then the output is ‘1’. If any one of the input is ‘1’, then the output is ‘0’. This is exactly opposite to two input OR gate operation.
The symbol of NOR gate is:
Fig.: NOR gate
NOR gate works exactly same as that of OR gate followed by an inverter.
Timing Diagram:
Q2) Explain EX-OR gate and Ex-NOR gate.
A2)
Ex-OR gate
It stands for Exclusive-OR gate. Its function varies when the inputs have even number of ones.
The truth table of 2-input Ex-OR gate is:
A | B | Y = A⊕B |
0 | 0 | 0 |
0 | 1 | 1 |
1 | 0 | 1 |
1 | 1 | 0 |
Here A, B are the inputs and Y is the output of two input Ex-OR gate. The output (Y) is zero instead of one when both the inputs are one.
Therefore, the output of Ex-OR gate is ‘1’, when only one of the two inputs is ‘1’. And it is zero, when both inputs are same.
The symbol of Ex-OR gate is as follows:
Fig.: XOR gate
It is similar to that of OR gate with an exception for few combination(s) of inputs. Hence, the output is also known as an odd function.
Timing Diagram:
Ex-NOR gate
It stands for Exclusive-NOR gate. Its function is same as that of NOR gate except when the inputs having even number of ones.
The truth table of 2-input Ex-NOR gate is:
A | B | Y = A⊙B |
0 | 0 | 1 |
0 | 1 | 0 |
1 | 0 | 0 |
1 | 1 | 1 |
Here A, B are the inputs and Y is the output. It is same as Ex-NOR gate with the only modification in the fourth row. The output is 1 instead of 0, when both the inputs are one.
Hence the output of Ex-NOR gate is ‘1’, when both inputs are same and 0, when both the inputs are different.
The symbol of Ex-NOR gate is:
Fig.: XNOR gate
It is similar to NOR gate except for few combination(s) of inputs. Here the output is ‘1’, when even number of 1 is present at the inputs. Hence is also called as an even function.
Timing Diagram:
Q3) Write analog vs digital system.
A3)
Analog | Digital |
An analog signal is a continuous signal that represents physical measurements. | Digital signals are time separated signals which are generated using digital modulation. |
It is denoted by sine waves | It is denoted by square waves |
It uses a continuous range of values that help you to represent information. | Digital signal uses discrete 0 and 1 to represent information. |
Temperature sensors, FM radio signals, Photocells, Light sensor, Resistive touch screen are examples of Analog signals. | Computers, CDs, DVDs are some examples of Digital signal. |
The analog signal bandwidth is low | The digital signal bandwidth is high. |
Analog signals are deteriorated by noise throughout transmission as well as write/read cycle. | Relatively a noise-immune system without deterioration during the transmission process and write/read cycle. |
Analog hardware never offers flexible implementation. | Digital hardware offers flexibility in implementation. |
It is suited for audio and video transmission. | It is suited for Computing and digital electronics. |
Processing can be done in real-time and consumes lesser bandwidth compared to a digital signal. | It never gives a guarantee that digital signal processing can be performed in real time. |
Analog instruments usually have s scale which is cramped at lower end and gives considerable observational errors. | Digital instruments never cause any kind of observational errors. |
Analog signal doesn't offer any fixed range. | Digital signal has a finite number, i.e., 0 and 1. |
Q4) What are the 3 basic boolean laws?
A4)
The three basic laws of Boolean Algebra are:
- Commutative law
- Associative law
- Distributive law
Commutative Law
- The logical operation carried between two Boolean variables when gives the same result irrespective of the order the two variables, then that operation is said to be Commutative. The logical OR & logical AND operations between x & y are shown below
x + y = y + x
x.y = y.x
- The symbol ‘+’ and ‘.’ indicates logical OR operation and logical AND operation.
- Commutative law holds good for logical OR & logical AND operations.
Associative Law
- If a logical OR operation of any two Boolean variables is performed first and then the same operation is performed with the remaining variable providing the same result, then that operation is said to be Associative. The logical OR & logical AND operations of x, y & z are:
x + (y + z) = (x + y) + z
x.(y.z) = (x.y).z
- Associative law holds good for logical OR & logical AND operations.
Distributive Law
- If a logical OR operation of any two Boolean variables is performed first and then AND operation is performed with the remaining variable, then that logical operation is said to be Distributive. The distribution of logical OR & logical AND operations between variables x, y & z are :
x.(y + z) = x.y + x.z
x + (y.z) = (x + y).(x + z)
- Distributive law holds good for logical OR and logical AND operations.
These are the Basic laws of Boolean algebra and we can verify them by substituting the Boolean variables with ‘0’ or ‘1’.
Q5) Convert (11101)2 to BCD.
A5)
Step 1 − Convert to Decimal
Binary Number − 111012
Its Decimal Equivalent −
Step | Binary Number | Decimal Number |
Step 1 | 111002 | ((1 × 24) + (1 × 23) + (1 × 22) + (0 × 21) + (0 × 20))10 |
Step 2 | 111002 | (16 + 8 + 4 + 0 + 0)10 |
Step 3 | 111002 | 2810 |
Binary Number − 111002 = Decimal Number − 2810
Step 2 -- Convert to BCD
Decimal Number − 2810
Its BCD Equivalent:
Step | Decimal Number | Conversion |
Step 1 | 2810 | 00102 10002 |
Step 2 | 2810 | 00101000BCD |
Result
(11100)2 = (00101000)BCD
Q6) Convert (1000)BCD to Excess-3.
A6)
Step 1 − Convert to decimal
(1000)BCD = 810
Step 2 − Add 3 to decimal
(8)10 + (3)10 = (11)10
Step 3 − Convert to Excess-3
(11)10 = (1011)2
Result
(1000)BCD = (1011)XS-3
Q7) What are alpha numeric codes? Explain.
A7)
A binary digit or bit can be represented by two symbols as '0' or '1'.
This is not sufficient for communication between two computers as we need many more symbols for communication.
These symbols are representing 26 alphabetic characters with capital and small letters, numbers from 0 to 9, punctuation marks and other symbols.
These alphanumeric codes represent numbers and alphabetic characters.
Most of them also represent other characters like symbol and various instructions necessary for conveying information.
The code at least represents 10 digits and 26 letters of alphabet i.e. total 36 items.
The following three types of alphanumeric codes are commonly used for the data representation.
- American Standard Code for Information Interchange (ASCII).
- Extended Binary Coded Decimal Interchange Code (EBCDIC).
- Five bit Baudot Code.
Q8) Convert the Boolean function into Standard SoP form.
f = p’qr + pq’r + pqr’ + pqr
A8)
Step 1 – By using the Boolean postulate, x + x = x and also writing the last term pqr two more times we get
⇒ f = p’qr + pq’r + pqr’ + pqr + pqr + pqr
Step 2 – By Using Distributive law for 1st and 4th terms, 2nd and 5th terms, 3rdand 6th terms.
⇒ f = qr(p’ + p) + pr(q’ + q) + pq(r’ + r)
Step 3 – Then Using Boolean postulate, x + x’ = 1 we get
⇒ f = qr(1) + pr(1) + pq(1)
Step 4 – hence using Boolean postulate, x.1 = x we get
⇒ f = qr + pr + pq
⇒ f = pq + qr + pr
This is the required Boolean function.
Q9) Convert the Boolean function into Standard PoS form.
f = (p + q + r).(p + q + r’).(p + q’ + r).(p’ + q + r)
A9)
Step 1 – By using the Boolean postulate, x.x = x and writing the first term p+q+r two more times we get
⇒ f = (p + q + r).(p + q + r).(p + q + r).(p + q + r’).(p +q’ + r).(p’ + q + r)
Step 2 – Now by using Distributive law, x + (y.z) = (x + y).(x + z) for 1st and 4thparenthesis, 2nd and 5th parenthesis, 3rd and 6th parenthesis.
⇒ f = (p + q + rr’).(p + r + qq’).(q + r + pp’)
Step 3 − Applying Boolean postulate, x.x’=0 for simplifying of the terms present in each parenthesis.
⇒ f = (p + q + 0).(p + r + 0).(q + r + 0)
Step 4 − Using Boolean postulate, x + 0 = x we get
⇒ f = (p + q).(p + r).(q + r)
⇒ f = (p + q).(q + r).(p + r)
This is the simplified Boolean function.
Hence, both Standard SoP and Standard PoS forms are Dual to one another.
Q10) Simplify the Boolean function,
f = p’qr + pq’r + pqr’ + pqr
A10)
Method 1
Given
f = p’qr + pq’r + pqr’ +pqr.
In first and second term r is common and in third and fourth terms pq is common.
So, taking out the common terms by using Distributive law we get,
⇒ f = (p’q + pq’)r + pq(r’ + r)
The terms present in first parenthesis can be simplified by using Ex-OR operation.
The terms present in second parenthesis is equal to ‘1’ using Boolean postulate we get
⇒ f = (p ⊕q)r + pq(1)
The first term can’t be simplified further.
But, the second term is equal to pq using Boolean postulate.
⇒ f = (p ⊕q)r + pq
Therefore, the simplified Boolean function is f = (p⊕q)r + pq
Method 2
Given f = p’qr + pq’r + pqr’ + pqr.
Using the Boolean postulate, x + x = x.
Hence we can write the last term pqr two more times.
⇒ f = p’qr + pq’r + pqr’ + pqr + pqr + pqr
Now using the Distributive law for 1st and 4th terms, 2nd and 5th terms, 3rdand 6th terms we get.
⇒ f = qr(p’ + p) + pr(q’ + q) + pq(r’ + r)
Using Boolean postulate, x + x’ = 1 and x.1 = x for further simplification .
⇒ f = qr(1) + pr(1) + pq(1)
⇒ f = qr + pr + pq
⇒ f = pq + qr + pr
Therefore, the simplified Boolean function is f = pq + qr + pr.
Hence we got two different Boolean functions after simplification of the given Boolean function. Functionally, these two functions are same. As per requirement, we can choose one of them.
Q11) Find the complement of the Boolean function,
f = p’q + pq’.
A11)
Using DeMorgan’s theorem, (x + y)’ = x’.y’ we get
⇒ f’ = (p’q)’.(pq’)’
Then by second law, (x.y)’ = x’ + y’ we get
⇒ f’ = {(p’)’ + q’}.{p’ + (q’)’}
Then by using, (x’)’=x we get
⇒ f’ = {p + q’}.{p’ + q}
⇒ f’ = pp’ + pq + p’q’ + qq’
Using x.x’=0 we get
⇒ f = 0 + pq + p’q’ + 0
⇒ f = pq + p’q’
Therefore, the complement of Boolean function, p’q + pq’ is pq + p’q’.
Q12) Simplify f(X,Y,Z)=∏M(0,1,2,4)f(X,Y,Z)=∏M(0,1,2,4)using K-map.
A12)
Therefore, the simplified Boolean function is
f = (X + Y).(Y + Z).(Z + X)
Q13) Simplify:
F(P,Q,R,S)=∑(0,2,5,7,8,10,13,15)
A13)
F = P’Q’R’S’ + PQ’R’S’ + P’Q’RS’ +PQ’RS’ + QS
F = P’Q’S’ + PQ’S’ + QS
F = Q’S’ +QS
Q14) Simplify:
F(A,B,C)=π(0,3,6,7)
A14)
F = A’BC +ABC +A’B’C’ +ABC’
F = BC + C’ ( A’B’ + AB )