Logical Expressions & Their Equivalent Forms

 All of the Boolean Expressions can be represented in either canonical form or standard form. In this tutorial, I am going to explain both forms with the help of solved examples.

Learning Objectives:

• Determine the canonical forms of a Boolean expression
• Determine the standard forms of a Boolean expression
• Minterms and maxterms explanation with the help of two-variable and three-variable functions
• What is meant by the sum of products (SOP) and product of sum (POS) expressions?
• Convert an SOP expression into a standard SOP and a POS expression into a standard POS
• Non-standard SOP expression to standard SOP expression
• Non-standard POS expression to standard POS expression
• Evaluate the complement of a function from canonical form
• Conversion between standard forms 
• standard SOP to standard POS
• standard POS to standard SOP
• Examples of all the topics mentioned above

The canonical forms of a Boolean expression || Standard Form of a Boolean expression

There are two canonical forms: Sum of minterms and product of maxterms.

Similarly, there are two standard forms: Sum of products and product of sums.

Minterms:

There is a method of expressing Boolean expression algebraically from the truth table with the help of the following steps.

• Draw truth table
• Assign minterms to all possible combinations of variables. For three variables there are 8 possible combinations (2³) and there are 8 minterms as well.
• Pick only those minterms that produce 1 at the output.
• spao="font-size: medium;">'OR' all the minterms

ANY BOOLEAN FUNCTION CAN EXPRESS IN THE SUM OF MINTERMS.

Example: Express AND gate expression in the sum of minterms

A	B	Output	Minterms
0	0	0	A’.B’  or  m0
0	1	0	A’.B  or  m1
1	0	0	A.B’  or  m2
1	1	1	A.B  or  m3

F = A.B = m3

F(A,B) = ∑ (3 )

It is a short notation for the sum of minterms.

Example: Express OR gate expression in the sum of minterms

A	B	Output	Minterms
0	0	0	A’.B’  or  m0
0	1	1	A’.B  or  m1
1	0	1	A.B’  or  m2
1	1	1	A.B  or  m3

F = A’.B + A.B’ + A.B = m1 + m2 +  m3

F(A,B) = ∑ ( 1,2,3)

It is a short notation for the sum of minterms.

From the above two examples of AND and OR gates, notice that we only pick those minterms or combinations at which the function produces a "HIGH" output.

Maxterms

There is another method of representing Boolean expressions algebraically from the truth table with the help of the following steps.

• Draw truth table
• Assign maxterms to all possible combinations of variables for 3 variables there are 8 possible combinations 
• Pick only those maxterms that produce 0
• 'AND' all the maxterms

Example: Express AND gate expression in the product of maxterms

A	B	Output	Maxterms
0	0	0	A + B  or  M0
0	1	0	A + B’  or  M1
1	0	0	A’ + B  or  M2
1	1	1	A’ + B'  or  M3

F = M0 . M1 .  M2 = (A + B)(A + B')(A' + B)

F(A,B) = π (0,1,2)

It is a short notation for the product of maxterms.

Example: Express OR gate expression in the product of maxterms

A	B	Output	Maxterms
0	0	0	A + B  or  M0
0	1	1	A + B’  or  M1
1	0	1	A’ + B  or  M2
1	1	1	A’ + B’  or  M3

F = M0 = (A + B)

F(A,B) = π (0)

It is a short notation for the product of maxterms.

ANY BOOLEAN FUNCTION CAN EXPRESS IN THE PRODUCT OF MAXTERMS.

Example: The truth table is given. Express the function in the sum of minterms and the product of maxterms. 

F = A + B'C

A	B	C	Output	Minterms	Maxterms
0	0	0	0	A’.B’.C’ (m0)	A+B+C  (M0)
0	0	1	1	A’.B’.C  (m1)	A+B+C’  (M1)
0	1	0	0	A’.B.C’  (m2)	A+B’+C  (M2)
0	1	1	0	A’.B.C  (m3)	A+B’+C’  (M3)
1	0	0	1	A.B’.C’  (m4)	A’+B+C  (M4)
1	0	1	1	A.B’.C  (m5)	A’+B+C’  (M5)
1	1	0	1	A.B.C’  (m6)	A’+B’+C  (M6)
1	1	1	1	A.B.C  (m7)	A’+B’+C’  (M7)

Sum Of Minterms:

F = A’.B’.C + A.B’.C’ + A.B’.C + A.B.C’ + A.B.C

F = m0 + m4 + m5 + m6 + m7

F(A,B,C) = ∑ (1,4,5,6,7 )

Product Of Maxterms:

F = (A+B+C)(A+B’+C)(A+B’+C’)

F = M0 M2 M3

F(A,B,C) = π (0,2,3 )

Convert The Boolean Expression In Standard POS | Standard SOP:

Standard SOP:

When two or more product terms are summed by Boolean addition, the resulting sum is a Sum of Products (SOP).

 Examples:

AB + CD

AB' + CD'

A standard SOP expression is one in which all the variables in the domain apear in each product term in the expression.

 

Examples:

An example of a standard SOP expression is given below:

AB'C + AB'C' + ABC +A'BC + A'B'C

Domain: Set of variables A, B, C

Each product term contains all three variables.

ANY BOOLEAN FUNCTION CAN BE EXPRESSED IN THE SUM OF PRODUCTS.

Convert an ordinary SOP into a standard SOP:

To convert an ordinary SOP into a standard SOP, we use the following laws of Boolean algebra:

1. A variable added to its complement is equal to '1'
2. Any variable multiplied by 1 remains the same
3. Distributive law

These points will help in the conversion process

1. Check out for non-standard terms
2. Find out the missing variable
3. Multiply each term by a term containing a variable added to its complement. For example A = A + A' and AB = AB(B + B'). Since B + B' = 1

Examples of standard SOP form:

Similarly, the expression AB' + C is also the Sum Of Product SOP, but it is not in standard form. In standard for each product, the term contains all sets of variables either complimented or uncomplimented in the domain.

Now, let's convert it into standard form.

AB'(C + C') + C Point 1 and Point 2

AB'C + AB'C' + C Point 3

AB'C + AB'C' + C(A + A') Repeat point 1 and point 2

AB'C + AB'C' + AC +AC' Repeat point 3

AB'C + AB'C' + AC(B + B') +AC'(B + B') Repeat point 1 and point 2

AB'C + AB'C' + ACB + ACB' + AC'B + AC'B' Repeat point 3

AB'C + AB'C' + ACB + AC'B + AC'B' term AB;C appears twice

This expression has three variables in its domain (A, B, C), and each product term has all three variables and now it is in standard form.

Standard POS:

When two or more sum terms are multiplied, the resulting expression is a product of sum (POS).

 Examples of standard POS form:

(A + B)(C + D)

(X + Y)(Z + X)

An important point to remember is that in the POS expression no two variables are under the same bar. Or you can say, the over-bar is restricted to single or individual variables only.

A' + B'+ C'        allowed 

(A + B + C)'           not allowed  

A standard POS expression is one in which all the variables in the domain appear in each sum term in the expression.

 

 Examples:

 (A + B' + C)(A + B' + C')(A + B + C)(A' + B +C)(A' + B' + C)
It is a standard POS expression.

Convert an ordinary POS into a standard POS:

Any Boolean expression can be expressed in POS form. To convert an ordinary POS into a standard POS, we use these laws:

1. A variable multiplied by its complement is equal to 0
2. If '0' is added to any term, the result will remain unchanged (additive identity)
3. Apply distributive law

These points will help in the conversion process

1. Check out for non-standard terms
2. Find out the missing variable
3. Add a term that is equivalent to '0'. For example, X.X' = 0
4. Apply this rule (A+B)(A+C) = A + BC. Learn more about these rules in my article laws of Boolean algebra.

Examples:

(X + Y)(Z + X)

This example is in POS form, but not in standard form. Because each product term does not contain all the set of variables either complimented or uncomplimented in the domain. the domain of this expression is the set of variables X, Y, Z. Now, convert it into standard form.

= (X + Y)(Z + X)

The missing variable is Z in the first term and Y in the second term.

= (X + Y + Z.Z')(Y.Y' + Z + X)

Apply the rule stated in point 4.

= (X + Y + Z)(X + Y +Z')(X + Y + Z)(X + Y' + Z)

= (X + Y +Z')(X + Y + Z)(X + Y' + Z)

The term X + Y + Z repeat twice. 

Complement Of a Function Evaluated With The Help Of Canonical Form:

Recall the truth table for the OR gate. in the table below, OR gate is expressed in the sum of minterms and the product of maxterms. From the truth table, we can conclude some important points.

A	B	Output (OR)	Minterms	Maxterms
0	0	0	A’.B’  or  m0	A + B  or  M0
0	1	1	A’.B  or  m1	A + B’  or  M1
1	0	1	A.B’  or  m2	A’ + B  or  M2
1	1	1	A.B  or  m3	A’ + B’  or  M3
			F(A,B) = ∑ ( 1,2,3)	F(A,B) = π (0)

1. The complement of a function is evaluated from the truth table. The complement of a function is equal to the sum of minterms missing in the original function
2. Another important property that is clear from the truth table is  Mj ' = mj. It means any minterm with subscript j is equal to the complement of the maxterm with the same subscript j.
3. To explain my point, I will proceed with the help of the simplest example of the OR gate. The complement of the OR gate is the NOR gate.

F = m1 + m2 +  m3

F' = ( m1 + m2 +  m3)'

Apply DeMorgan's Law

F' = m1'. m2' .  m3'

F' = M1 . M2 . M3 

F'(A,B) = π (1, 2, 3)

F'(A,B) = ∑ (0) = π (1, 2, 3)

A	B	Output (NOR)	Minterms	Maxterms
0	0	1	A’.B’  or  m0	A + B  or  M0
0	1	0	A’.B  or  m1	A + B’  or  M1
1	0	0	A.B’  or  m2	A’ + B  or  M2
1	1	0	A.B  or  m3	A’ + B’  or  M3
			F(A,B) = ∑ (0)	F(A,B) = π (1, 2, 3)

The complemented function that is NOR has minterms that are not in the original OR function. In other words NOR function has maxterms that are missing in the original OR function.

Examples:

Evaluate the complement of the following Boolean function/ logic expression with the help of canonical forms. F = XY' + X'Y

F = XY' + X'Y

F' = (XY' + X'Y)'

De Morgan's law

F' = (XY')' (X'Y)'

again apply De Morgan's Law

F' = (X' + Y)(X + Y')

Apply distributive law

F' = XX' + X'Y' + XY + YY'

XX' = 0

F' = X'Y' + XY

The above expression is the complement of the given Boolean function. It is obtained for simplification of laws theorems. For more examples please check out my article here on Boolean expression simplification exercises.

Now our aim is to evaluate with the help of canonical forms.

Step 1: Draw truth tables along with minterms and maxterms

X	Y	Output	Minterms	Maxterms
0	0	0	X’.Y’  or  m0	X + Y  or  M0
0	1	1	X’.Y  or  m1	X + Y’  or  M1
1	0	1	X.Y’  or  m2	X’ + Y  or  M2
1	1	0	X.Y  or  m3	X’ + Y’  or  M3
			F(X,Y) = ∑ ( 1,2)	F(X,Y) = π (0,3)

F = M0 . M3

F = m1 + m2

F' = (m1 + m2)'

DeMorgan's law

F' = m1' . m2'

F' = M1 . M2

This expression F' = M1. M2 is the maxterm notation for the complement of the given function. now draw its truth table. Recall that maxterms are those where the function produces '0' at the output.

X	Y	Output	Minterms	Maxterms
0	0	1	X’.Y’  or  m0	X + Y  or  M0
0	1	0	X’.Y  or  m1	X + Y’  or  M1
1	0	0	X.Y’  or  m2	X’ + Y  or  M2
1	1	1	X.Y  or  m3	X’ + Y’  or  M3
			F(X,Y) = ∑ (0 , 3)	F(X,Y) = π (1, 2)

F' = m0 + m3

F' = X'Y' + XY

F'(X , Y) = ∑ (0 , 3 )

Conversion Between Standard Forms: SOP To POS | POS To SOP:

F = AB + CD SOP form

F = (A +B)(C +D) POS form

F = (AB + C)(CD + A) Neither POS nor SOP

I have already discussed conversion from ordinary SOP/POS to standard SOP/POS. For more examples please check out my next article for solved examples.

To convert standard POS to SOP or vice versa please do these simple steps.

1. Convert the given expression into standard form. If it is given in ordinary SOP, convert it into standard SOP, if it is given in ordinary POS, convert it into standard POS. Standard SOP expression contains all the minterms (combination of variables that produce 1) and standard POS expression contains all the maxterms (combination of variables that produce 0).
2. Write binary equivalent for each term in standard SOP or POS expression. for example, 1+0+0 is the binary equivalent value for (A'+B+C)
3. Evaluate the number of possible combinations with the help of several variables (for 2 variables, the number of possible combinations is 22 )
4. If the POS expression contains 2 product terms (for 2 variable expressions) the rest of the 2 terms are SOP terms. And if the POS expression contains 5 product terms (for 3 variable expressions) the rest of the 3 terms are SOP terms.
5. Draw the truth table for ease.

Example: (A + B' +C)(A + B +C') (POS To SOP)

Step 1:

The expression is already in standard form.

Step 2:

Write binary equivalent for each combination of variables.

A + B' + C = 0 + 1 + 0

A + B + C' = 0 + 0 + 1

Step 3:

The total number of possible combinations is 23  = 8.

Step 4:

The POS expression contains only 2 combinations of variables. The rest of the 6 terms are SOP.

Step 5:

Have a look at the truth table.

A	B	C	Output	Minterms	Maxterms
0	0	0	0		A+B+C  (M0)
0	0	1	1	A’.B’.C  (m1)
0	1	0	0		A+B’+C  (M2)
0	1	1	1	A’.B.C  (m3)
1	0	0	1	A.B’.C’  (m4)
1	0	1	1	A.B’.C  (m5)
1	1	0	1	A.B.C’  (m6)
1	1	1	1	A.B.C  (m7)

F = A’.B’.C + A’.B.C  + A.B’.C’ + A.B’.C + A.B.C’ + A.B.C

Example: A'B'C + AB'C' +ABC (SOP To POS)

Step 1:

The expression is already in standard form.

Step 2:

Write binary equivalent for each combination of variables.

A'.B'.C = 0.0.1

A.B'.C' = 1.0.0

A.B.C = 1.1.1

Step 3:

The total number of possible combinations is 23  = 8.

Step 4:

The SOP expression contains only 3 combinations of variables. The rest of the 6 terms are POS.

Step 5:

Have a look at the truth table.

A	B	C	Output	Minterms	Maxterms
0	0	0	0		A+B+C  (M0)
0	0	1	1	A’.B’.C  (m1)
0	1	0	0		A+B’+C  (M2)
0	1	1	0		A+B’+C’  (M3)
1	0	0	1	A.B’.C’  (m4)
1	0	1	0		A’+B+C’  (M5)
1	1	0	0		A’+B’+C  (M6)
1	1	1	1	A.B.C  (m7)

F = (A+B+C)(A+B’+C)(A+B’+C’ )(A’+B+C’)(A’+B’+C)

Check out my next article for solved examples here.

Classification Of Amplifiers Based On Q-Point | Power Amplifiers

 Another article on the classification of amplifiers. This type of classification is based on the position of the Q point on the load line.

Learning Objectives:

• Learn about the difference between classes of power amplifiers

• FAQs

• Why is power dissipation considered a big issue in power amplifiers?

• Is a large-signal operation different from a small-signal operation?

• How to eliminate crossover distortion?

Introduction 

Large signal amplifiers or power amplifiers provide a large amount of power to the load. This amplifier category has different classes. Every class has a different region of operation on the load line. The types of amplifiers are usually at the output stage of the system to drive or power the device connected as a load. Some important features are 

• Power efficiency

• Maximum power 

• The power dissipated should be as low as possible

• Impedance matching to the output device. Low output resistance so that the power loss is minimum.

This classification can also be viewed as the output stage of an amplifier. It means different classes of amplifiers can be used as an output stage of a multistage amplifier system. Each class has a different shape of the output waveform. The following figure compares the output waveforms obtained from each class.

How does a power amplifier look like?

A power amplifier can either be a simple power transistor (not a common transistor but designed with some extra thermal stability properties) or an integrated circuit along with special thermal properties. Power transistors have a larger surface area for efficient heat sinking.

Power amplifiers process large signals, and a larger amount of power is required. So, all of the power devices must dissipate the internally generated heat. For power, the BJT collector is a critical junction. a heat sink is necessary to attach with it. Also, the case of power transistors has a large contact area. Internally generated heat flows from the case then heat sinks and then dissipates in the surrounding air.

Why is power dissipation considered a big issue in power amplifiers?

Power is defined as the rate at which energy is consumed or dissipated in a device. Power generated must be dissipated fast enough to avoid temperature buildup. The heat generated is equal to heat dissipated or consumed. 

Large signal operation is also applied to digital switching circuits. As you know switching devices operate either in cut-off or saturation modes. But its operation is limited to either of the two states only. So, heat dissipation is not an issue in a digital switching system.

Is large-signal operation different from small-signal operation:

For signals that have a power greater than 1W are considered large signals. In large signal operations, approximation and models are not applicable. In a small signal, it is assumed that the incoming signal is small enough and can not be able to change device parameters that are beta and re`.

A small-signal swings over a small part of the AC load line. While a large signal approaches the limits of the AC load line.

Class A Amplifiers - A brief overview:

It necessarily operates in a linear region for 360 degrees of the input cycle and cut off at 0 degrees. Following is the simple BJT class A amplifier circuit and its load line.

The maximum possible signal swing is achieved when the Q point is centred at the AC load line. With the help of a load line, we can conclude the following points.

• Collector current is maximum (peak value of current or peak of the sinusoid) at IC(sat) and minimum at its cut off value.

• Similarly, VCE collector to emitter voltage is maximum (peak voltage or peak of the sinusoid) is at Vce(sat) and minimum at its cut off value.

Class B Amplifiers - A brief overview:

Class B amplifiers are biased such that they operate in linear region for 180 degrees of the input signal half cycle of the sinusoid) and then cut off for the next 180 degrees. Q point is at cut off or it is biased at cut off. 

Since half of the input cycle is clipped, the output of the class B amplifier is highly distorted. As I discussed earlier, they are linear amplifiers, the shape of the output waveform is the same as that of the input wave. To get the same output waveform similar to the input waveform (similar in shape) there are two complementary transistors. This class of amplifiers use two complementary transistors. One of the transistors amplifies the positive part of the signal and the other transistor amplifies the negative part of the signal.

A simple class B amplifier with the help of two complementary BJTs (one is NPN and the other is PNP) and the load line is given below.

During the positive half cycle, Q1 is ON Vce varies from its Q point value (that is 0V) to its maximum value (that is Vcc). During the negative half-cycle, Q1 remains turned off. Similarly, during the positive half cycle, Q2 remains turned off. During the negative half-cycle, it produces the negative half of the Vce.

During the positive half cycle, Q1 is ON Ic varies from its Q point value (that is 0V) to its maximum value (that is IC(sat)). During the negative half-cycle, Q1 remains turned off. Similarly, during the positive half cycle, Q2 remains turned off. During the negative half-cycle, it produces the negative half of the Vce.

Why does cross over distortion occur?

One of the most simple classes B circuits is shown in the figure above.

To turn on the transistor VBE ≥ 0.7V. When the base voltage is 0 (VBE = 0 or VBE < 0.7V) neither of the transistors conducts. So there is no waveform at the output. Because of this output waveform is not the same in shape as that of an input waveform. This type of distortion is known as cross over distortion.

Class AB Amplifiers - A brief overview:

This class of amplifiers are biased such that they operate in a linear region and conduct slightly more than 180 degrees. The amplifier is then cut off for the rest of the cycle. The Q point is slightly above the cut off point.

How to eliminate crossover distortion?

This class is specifically designed to eliminate crossover distortion. Slight changes made in the class B circuit will eliminate the distortion.

Class C Amplifiers - A brief overview:

The class C amplifiers are biased such that they operate less than 180 degrees of the input signal. The output is highly distorted. They are also known as tuned amplifiers and are used at radio frequencies.

Class C Amplifier

The figure above shows a simple CE amplifier. The idea of biasing is noticeable. The base is negatively biased with the voltage source VBB. The signal applied must have an amplitude (peak value) greater than VBB+VBE. The transistor is ON for a very short interval of time and produces a pulse (pulse of Ic) at the output. The output is highly distorted. To get the output voltage similar to the input voltage in shape and tank circuit is used.