 एस्लाम टेल्याONArikum students the things which we had done in the last lecture. In fact, take many table the Khanatha, just may we compare different CMOS and TTL gates and we look at the actual values. So let us start our discussion by revising the contents which we had studied in the last lecture. The operational characteristics were four or five important parameters. The first parameter was DC supply voltage. So every circuit needs a DC supply voltage so that it works. So we usually use 5 volts for TTL for 5 volts supply voltage. Similarly, we need 5 volts for CMOS. There is another series of CMOS in which we need 3.3 volts supply voltage. Then we talked about the noise margin. Noise margin was that every logic level, high or low, it is represented by certain voltage levels. So if that signal is within those voltage levels, then the whole circuit will work well. But due to some external noise, some other factors, the logic 1 or logic 0 representation comes out of that voltage range. Due to that, the circuit might not work. So we had calculated the noise margins. We saw that the 5 volt version of CMOS has more noise margins. That means it can work very well in a noisy environment. The third parameter which we looked at was the power dissipation. Every circuit uses power. Of course, current flows in it. It needs energy. How many watts are consumed in it. So we looked for TTL and CMOS. The power requirement for TTL is more or less fixed. Power requirement for CMOS varies with respect to frequency. Because there is a capacitive load in CMOS circuits. When there is a capacitor, there will be a charge and a discharge. The discharge depends on the frequency it operates. When the frequency of a CMOS circuit increases, the power dissipation also increases. So the variable power dissipation of CMOS circuits will be there. When we measure it, there will be a static power dissipation and a dynamic power dissipation. We will see that in the table. We saw another parameter, the propagation delay. We said that if there is a variable on the input, then after a special delay, you will get a value on the output. So the delay would determine how fast a signal can change at the input of any gate. If the delay factor is changing at a high frequency, then it will not work. We talked about a product of the speed and dissipation. In both the products, there will be a number. You can compare it with different gates. Finally, we saw the fan out. Fan out is how many gates can we connect to the output of another gate. We said it depends on the current which the gate can supply to the inputs of different gates. Again, there is a restriction. There are fixed number of fan outs in TTL. CMOS depends on the frequency. If you want to operate the circuit at a high frequency, then your fan out will be less. That means you will have to connect less number of gates. If you want to operate at a low frequency, then you can connect more number of gates to the output of a gate. These were the five important parameters. Let us have a look at the table where we compare different gates and see the actual numbers. After that, we will start the Boolean algebra. Let us first consider the TTL series of gates. We have the 7-4 standard gate, the 7-4S short key gate, the 7-4LS the low power short key, 7-4AS the advanced short key, and the 7-4ALS the advanced low power short key, and the 7-4F the fast version of TTL. If we look at the propagation delays, the standard version has a propagation delay of 9 nanoseconds, which is almost the highest propagation delay. 7-4LS propagation delay is 9.5 nanoseconds. Fast and ALS are very low. AS is also very low, it is 1.7. If we look at the power dissipation, let us measure in milliwatts, the 7-4S short key power dissipation is more than 20. Low power short key power dissipation is 2, the advanced low power short key power dissipation is 1.2. An important parameter is the speed power product. Let us measure in pico joules. Let us look at these. 4.8 is the lowest of ALS. So, this gate is better than all, in terms of power dissipation and propagation delay. The standard version is the highest, it is 90. Maximum clock rate on which you can operate is also given. Standard TTL gate operates at 35 MHz. The advanced short key operates up to 200 MHz. The advanced low power short key operates up to 70 MHz. Fan out, again 7-4 standard 10 gates can be connected. The short key 20 gates can be connected. The advanced short key 40 gates can be connected. Let us now look at the CMOS series. First, let us look at the 5V series, and then look at the 3.3V series. In the 5V series, you have the 7-4AC and the 7-4AHC. Propagation delay is less compared to TTL logic. Power dissipation is very less. 0.00275 is for 7-4HC. Similarly, 0.0055 is for 7-4AC. CMOS series tells us that power dissipation varies on low frequency and high frequency. Power dissipation static that means no operations or low frequencies. Power dissipation as I have said before is 0.00275. Same gate, if it operates at 100 kHz, its power dissipation increases to 0.0625. Similarly, for 7-4AC, the power dissipation increases to 0.08 at 100 kHz. Speed power product is at 100 kHz. 1.125 is for 7-4AC. 0.4 is for 7-4AC. 0.23 is for 7-4AHC. If we compare the speed power product with TTL, it is very less. Maximum clock rate is 50 for AC. For AC, it is 160 MHz. And for AHC, it is 170 MHz. Now, let us have a look at the 3.3V CMOS series, the 7-4LV, the 7-4LVC and the 7-4ALVC. Propagation delay again are very low, 9, 4.3 and 3 respectively. Power dissipation, the static power dissipation is again very low, 0.0016, 0.0008 and 0.0008 for the 3 gates respectively. Operational clock rate, maximum clock rate is 90, 100 and 150 respectively. We have just looked at some parameters of TTL and CMOS gates. So, by comparing those numbers, you can just see which gate is better and which particular gate is required for your particular implementation. Now, let us start with today's topic, Boolean algebra. Boolean algebra, how is it different from the conventional algebra? Well, Boolean algebra has two values. In conventional algebra, there are many values. Boolean algebra, since it is dealing with binary values, so its value will be 0, and its value will be 1. So, let us start by looking at some definitions of Boolean algebra. In Boolean algebra, we use literal, we talk about complement and we talk about variables. What is a variable? You may use variable programming. It is a symbol which represents a value. So, in Boolean algebra, a variable is usually a capital letter, alphabet letter A, B, C, Z, N. So, it represents a value. What is the value? Either it is going to be a 0 or a 1 because it is, of course, representing a Boolean value or a binary value. What is complement? The inverse of a variable. So, if variable A has the value 0, A bar would have the value 1. So, the bar represents the inverse of the complement. What is literal? Literal could be the variable or its complement. So, let us say B bar is a literal. B is its complement. Again, it is considered to be a literal. So, these are the three definitions which we have to use again and again. Boolean algebra, why are we using it? Basically, why do we use mathematics? One benefit of mathematics is that all real world problems can be represented in the form of equations. What is the benefit? We can check or implement in the form of mathematical equations. Let me give you an example. You take an iron rod let us say 1 inch diameter. You start heating it on one end. And you need to determine that on the other end, let us say 1 foot long rod is there. On the other end, at what time it will be 50 degrees temperature. Now, this entire heating of an iron rod can be modeled using a mathematical equation. Now, if you have an equation then you can perform you do not need to perform an experiment. You do not need to get an actual rod and heat that. Equations say you have different values done or you can calculate a answer. In fact, if you have an equation then you can use different rods different metal rods and different diameter rods. So, mathematics solves the problem. You do not have to use practically anything. Similarly, Boolean algebra allows you to represent digital circuits in terms of expressions. What will be the benefit? One benefit is that let us suppose you have a very complex, complicated circuit having different gates. How will that circuit perform? What output will come? Well, one way is to practically implement that circuit. You will need more time to spend your money on it. So, the easiest way would be to represent that circuit through a Boolean expression a mathematical expression of different input combinations you get the answer. So, you can easily check if that particular circuit is going to work correctly or otherwise. You can use a Boolean expression to simplify circuits. Let us suppose you have a circuit composed of 20 gates it performs a certain function. Now, you could implement the same function using 5 gates. Now, if you implement a circuit having 5 gates that means the circuit is going to be faster less propagation delay the power requirement would be reduced so it would be better than the previous circuit which uses more gates. Now, how would you simplify a circuit? By looking at the circuits it is very difficult to find out the simplified version of that circuit. A better way would be to write an expression of that original circuit and then using the rules and laws of Boolean algebra simplify the expression. You would get a simplified version of the expression and if you directly implement the circuit using that expression you would have a simple circuit. So, let us look at Boolean operations. In the beginning we talked about two Boolean operations which are important and are used throughout Boolean logic. Boolean addition and Boolean multiplication. Boolean addition is performed by an OR gate. So, two input's OR gate is A plus B. A plus B is known as a sum term because we are adding some term. The multiplication is performed by an AND gate. So, if you have a two input AND gate you have a product term A dot B. A dot B is known as the product term because it is a multiplication result. Let us look at some other sum terms and product terms and let us look at what would be the output of this sum term if one of the variables let us first look at Boolean addition. Boolean addition gives us a sum term. A sum term is described as sum of literals. So, as you can see A plus B is a sum term which is in fact a sum of literals. Another term A plus B bar again it is a valid sum term it is again a sum of literals. The third expression A bar plus B bar plus C is again a sum term composed of three literals. Now, a sum term is equal to one if any literal is equal to one considering A plus B A one ho jai here B one ho jai the output is going to be a one. Similarly, A plus B bar ko agar dekh ha jai in me say koi bhi literal agar one ho jai output one ho jai. A sum term is equal to zero if all are equal to zero. So, considering A bar plus B bar plus C the output or the sum term is zero if A bar is zero B bar is zero and C is zero. Similarly, considering the term A plus B bar the output is zero or the sum term is zero if A is zero and B bar is zero. Let us now look at Boolean multiplication. Boolean multiplication results in a product term product term is described as a product of literals. So, you have three product terms as can be seen AB, AB bar and A bar B bar C. Now, a product term is equal to one if all literals are equal to one. So, AB would give you a one result if A is one and B is one. Similarly, AB bar would give you a one result if A is one and B bar is one. A product term is equal to zero if any one literal is equal to zero. So, considering the term the product term A bar B bar C if any one literal is zero the entire term would be zero. So, if A bar is zero or B bar is zero or C is zero or all three terms literals are zeros the product would be zero. We have just looked at Boolean addition and Boolean multiplication we have looked at the sum term and the product term. Now, when you write Boolean expressions and solve Boolean expressions they are solved on the basis of certain rules, laws and theorems. Normal algebra may be it is done according to certain rules and theorems. So, now let us look at some rules and theorems being applied in Boolean algebra. There are three laws being applied in Boolean algebra the associative law the distributed law and the commutative law. These three laws are used in normal algebra and they are used in the same way. Then there are certain rules I think there are about 12 rules again we are going to be looking at those rules. Then we have two theorems So, again we would be looking at the two theorems. So, now let us have a look at the the three laws the rules and the theorems applied in Boolean algebra. The basic laws of Boolean algebra are the same as ordinary algebra and hold true for any number of variables. So, let us start by looking at the commutative law commutative law for addition states that sum of literal and beliteral that is a or b is equal to b or a if you have three literals a b and c the commutative law states that a or b or c is equal to c or b or a. Implementation wise an or gate is used. So, if you have a two input or gate it does not matter if you connect the literal a to the first input or to the second input in both cases the output would be the same. Let us consider the commutative law for multiplication. The commutative law for multiplication states that product of literal a and b is equal to product of b and a. Again implementation wise an and gate is used because multiplication of course is done using and gates. So, according to the commutative law for multiplication it does not matter if you connect the literal a to the first input or the second input of the and gate. The result would always be the same. Again you could have multiple literals a b c and d. The law holds true for multiple literals. Let us now consider the associative law for addition. The associative law for addition states that sum of b and c literals that is b or c odd with a is the same as a or b odd with c. Implementation wise you require a combination of two or gates. So, the inputs are 3 a b and c. So, any of these 3 inputs can be connected to any of the 3 literals a b or c. The output would be the same. Let us consider the associative law for multiplication. The associative law for multiplication states that the product of b and c b and c ended with literal a is the same as the product of literals a b ended with c. Implementation wise you would use a combination of 2 and gates. So, altogether you would have 3 inputs. Any of the 3 literals a b and c can be connected to any 3 inputs. The output would be the same. You could have 4 literals, 5 literals the associative law for multiplication would hold true. The distributive law states that b or c ended with a is equal to a b odd with a c. The left side of the expression that is b or c ended with a can be implemented by using a combination of AND gate and OR gate. The inputs b and c are odd together. The output of the OR gate is connected to an AND gate and the result is b or c ended with a. The right side of the expression a b the product of a b odd with the product of a and c can be implemented using 3 gates 2 AND gates and an OR gate. So, the first AND gate is connected to inputs a and b the product is a b the second AND gate is connected to inputs a and c the product is a c the outputs of the 2 AND gates are connected to an OR gate the output of the OR gate is product a b odd with product a c let us have a look at 12 rules of Boolean algebra the first rule a or 0 is equal to a this can be proved by assigning values to the variable a if a is 0 what is the result the result is 0 if a is 1 what is the result the result is 1 so the output is the same is equal to a the second rule a OR 1 is equal to 1 now by assigning values to the variable a 0 or 1 the output remains a 1 the third rule a product 0 so as we have discussed before anything multiplied by 0 gives you a 0 so a product 0 is equal to 0 the fourth rule a product 1 is equal to a we have studied this before as well anything multiplied by 1 gives you the same thing rule number 5 a OR a is equal to a again this can be proved by applying assigning values to the variable a so if a is 0 so it is 0 or 0 which is equal to 0 if a is 1 then the rule becomes 1 OR 1 the result is 1 rule number 6 OR a bar is equal to 1 again let us apply the two values to the variable a if a is 0 then the equation becomes 0 OR 1 the result is 1 if a is 1 the equation becomes 1 OR 0 the result is again a 1 rule number 7 a product a is equal to 0 again let us check if a has the value 0 so the equation becomes 0 and 0 the result is 0 if a is equal to 1 it is 1 and 1 the result is 1 rule number 8 a and a bar the result is 0 applying the two values if a is equal to 0 0 and 1 the result is 0 similarly if a is equal to 1 the equation becomes 1 and 0 the result is again a 0 rule number 9 a double bar is equal to a double bar means complemented twice so if you have a single bar that means complemented once if you have a double bar you complement the complement again you have just cancelled out the two complements so if a is equal to 0 the answer would remain a 0 if a is equal to 1 the answer would remain a 1 rule number 10 a odd with the product term ab the result is a we can prove this a is common in both the terms if you take a as a common you are left with 1 or b now 1 or b if you apply a rule 2 is equal to 1 so you are left with a product 1 which is equal to a rule 11 a or a bar b which is equal to a or b this can again be proved a can be multiplied with sum of b plus 1 the result is 1 so you you multiply a with 1 it would remain an a so you multiply a with b or 1 you simplify the equation this gives you ab plus a and the previous term a bar b now re-adjusting the expression you have a plus ab plus ab bar b now considering b which is common in terms ab and ab bar b if you simplify this the equation turns out to be a plus b the last rule rule number 12 a or b ended with a or c the result is a odd with the product term b c this can be proved by rewriting the equation so a multiplied by a gives you a a ended with c gives you a c b ended with a gives you a b b ended with c gives you b c now considering the first 3 terms a a c and a b you have a common a so if you remove the common a the remaining portion is equal to 1 so the equation simplifies to a plus b c now let us consider de Morgan's theorems there are 2 theorems the first theorem states the complement of a product of variables to the sum of the complements of the variables in form of an equation a b whole bar is equal to a bar plus b bar de Morgan's theorem the first theorem holds true for any number of variables implementation wise ab whole bar is implemented using anand gate the right side of the equation a bar plus b bar is implemented again by anand gate but the alternate symbol of the anand gate is used let us consider the second theorem de Morgan's second theorem states that the complement of sum of variables is equal to the product of the complements of the variables writing the expression it is a or b whole bar equals to a bar entered with b bar implementation wise you have a nor gate so a nor b gives you a plus b or a or b complemented the right side of the equation a bar entered with b bar can be implemented using the same nor gate but represented in its alternate symbolic form again the second theorem of de Morgan's holds true for multiple variables de Morgan's theorem can be applied to expressions having any number of variables so let us have a look the first expression x and y and z whole bar is equal to x bar plus y bar plus z bar similarly or y or z whole bar is equal to x bar entered with y bar entered with z bar de Morgan's theorem can be applied to a combination of other variables let's have a look at the expression the expression is a or b c this is considered to be a single term entered with another term which is a c odd with b both these terms are complemented so you have an entire bar over the product of the two terms these two terms according to de Morgan's theorem can be written as a plus b c whole bar plus a c plus b whole bar the plus of course indicates the or operation this equation can be further simplified to a bar b c bar plus a c bar b bar this can further be simplified to a bar entered with b bar plus c bar plus b bar entered with sum of a bar and c bar this expression can be further rewritten as a bar odd with a bar b bar odd with b bar c bar final expression is a bar b bar plus a bar c bar plus b bar c bar we have looked at the laws, rules and theorems used in Boolean algebra you need to practice these laws, rules and theorems you need to know about them because we would be applying these rules and laws throughout this course now look at the application of these Boolean expressions to analyze a circuit let us consider an expression a plus b entered with c now what kind of information does this expression convey well a odd or b means it to input or gate is being used the output of this is connected to an AND gate and the other input of the AND gate is connected to the variable c so basically a or b entered with c gives you three variables so if you write out the function of this or gate and AND gate circuit how many columns are there how many possible input combinations how many possible well since there are three input variables therefore you have eight possible combinations what is the output of this circuit again you can analyze that by applying the appropriate inputs let us say you start with the combination 0 0 and 0 so a or a or b is 0 or 0 which is 0 AND it with c anything AND it with 0 gives you a 0 so the output is going to be 0 instead of you using all the eight terms you can find the output just by looking at the equation c is entered with the or of a and b a or b entered with c now if a or b is a 1 and c is a 1 then you would get a 1 output if either a or b is a 0 or c is a 0 you would get a 0 output right so if we know this then we can easily determine for which particular combination of 0s and 1s you would get a 1 output let us have a look at another circuit and let us analyze that circuit using the boolean expression let us consider the expression a and b or with c bar and the entire term entered with d now what type of or what kind of information does this expression convey basically there are 4 variables so the circuit represented by this expression is going to have 4 inputs and of course a single output now by looking at the expression again the variables a and b are entered together 2 input and gate is required the output of the AND gate is ord with c bar that means a and a not gate is required the input of this not gate is connected to c the output of the not gate is c bar so the output of the not gate and the output of the AND gate are connected to the input of an or gate so it is a 2 input or gate the output of the or gate would of course be a and b ord with c bar we have the 4th variable variable d it is entered with the term a b plus c bar so how do we AND d with the a b plus c bar term an AND gate a 2 input AND gate now for which particular combinations of 1s and 0s for the inputs a b c and d the output is going to be a 1 let us again have a look at the expression the output is going to be 1 if d is 1 and the term a b plus c bar is a 1 now how would the term a b be a 1 if a b is a 1 and the little c bar is a 1 how is the AND term a b going to be 1 if both a and b are 1s the little c bar how is that going to be 1 if c is 0 so when do you obtain a 1 output when a is 1 b is 1 c is 0 and d is 1 so for this particular combination of 1s and 0s you obtain a 1 output now this can be verified by drawing out a function table so what is the size of the function table how many input columns are required basically there are 4 variables a b c and d so there are 4 input columns a b c and d how many input combinations are possible since there are 4 variables so 16 different input combinations are possible the output of course is 1 represented by f now let us write all the outputs for each of these 16 combinations starting with the first combination of 0 0 0 the output is 0 why because d is 0 considering the second input 0 0 0 1 now in this particular case d is 1 and c is 0 so both the conditions are satisfied the answer is or the output is 1 similarly for the input condition 0 1 0 1 again d is 1 and c is 0 the conditions are satisfied the output is again going to be a 1 let us consider the input 1 0 0 1 again d is 1 c is 0 the conditions are satisfied the output is again going to be a 1 let us consider inputs the inputs 1 1 0 1 in this particular case d is 1 c is 0 a and b are both 1s so all 4 conditions are satisfied the output is going to be a 1 the last input combination 1 1 1 1 d is a 1 c is a 1 so this particular condition is not satisfied but a and b both are 1s therefore the output is again going to be a 1 we have looked at 2 examples of analyzing boolean circuits or logic circuits using boolean expressions basically we would be using this method to analyze different circuits throughout the course so it is important that you learn to use these expressions to analyze circuits I would suggest you just write any simple expression of 3 or 4 variables then try to decipher that expression you implement a circuit from that which gates are there then I would suggest that you look at the expression and see which particular combination of inputs you get a 1 output you can verify if you make it a function table so if you have 4 variables in your expression you will get 4 columns 16 combinations all 16 combinations are written you calculate the output using the circuit you verify it by applying those values in the expression so both the things will verify each other so it is important that you practice with the analysis of circuits using these boolean expressions now the other thing which we said we use the expression is to simplify circuits if a circuit is made you will not be able to see it by simplifying it so the best way is to write an expression representing that circuit then apply the rules of boolean algebra laws and theorems which we have discussed that would give you a simpler expression when you have that simple expression then of course you can always implement a circuit using that expression the expression which we are going to use and simplify in the example is a 3 variable expression it represents a large circuit we are going to be simplifying this expression using the rules laws studied so far we would get a simpler expression then we would implement that expression we would obtain a circuit which of course would be very simple as compared to the original circuit the functionality of both the circuits would of course be the same so let us have a look the expression is composed of 3 terms the term b the second term b or c and it with a and the third term b or c and it with b if you simplify this expression consisting of 3 terms you have a b odd with a b odd with a c odd with b b odd with b c now some of these terms simplify is common so you just remove one of the terms b b simplifies to b so you end up with an expression a b odd with a c odd with b odd with b c this expression can be further simplified b and the term b c you have a common b so you end up with b ended with 1 plus c 1 plus c is equal to 1 so you end up with the term b now you have the expression a b odd with a c odd with b now b is the common literal in the terms a b and b so if you simplify these 2 terms you end up with literal b so the expression is b c let us have a look at the circuits represented by the original expression having the 3 terms and the simplified expression having the 2 terms b odd ac the original circuit is made up of 3 AND gates and 2 OR gates the OR gate used at the input has 2 inputs and the OR gate used at the output has 3 inputs altogether you require 5 gates let us now look at the simplified expression and the circuit resulting out of that simplified expression only 2 gates are required an AND gate and an OR gate so you have removed 3 gates the circuit is simpler less power is required and perhaps it is faster than the original circuit we have just looked at an example of simplification of Boolean expressions to implement simpler circuits again this method would be used throughout the course you would be writing expressions you would be simplifying them to simplify and implement simpler circuits so I would advise that you practice you just write some expressions you implement a circuit which represents that expression and then you simplify the expression and then implement the circuit you have to confirm that both the circuits are working how do you do that of course you can do the function through the tables you can check the values in both the expressions now we have been writing expressions different types of expressions there are 2 standard ways of writing expressions one is the sum of products form sum form before we look at the 2 ways of writing these expressions let me describe what is the sum of product form you have already been writing expressions in this form basically you have product terms abc it is a product term de bar it is another product term so abc plus de bar it is sum of product terms you have sum them up this kind of expression basically how you implement basically you would be using a combination of and gates and an or gate product terms by using and gates in all the terms when you add you would use an and gate let us look at the product of some terms product of some terms at the name indicates is some terms which are ended together so let us consider the sum term a plus b it is the sum term because you are implementing an or operation between 2 variables you have another sum term c or d now if you end these 2 terms together that is a or b ended with c plus d you get the sum of rather the product of some terms implement kase keringe you would be using a combination of or gates and and gate you sum terms a implement keringe using or gates in saaron ka joh output hai that would be connected to a single and gate let us have a look at an example which shows the representation of different sum of product terms and product of sum terms but before we do that normally any expression can be easily converted into a sum of product term expression let us look at sum of products form first there are 3 expressions the first expression is a and b odd with a b c so you have 2 product terms and they are odd together the second example is you have 3 product terms product term a b c the second product term c d e and the third product term b bar c d bar now all these 3 product terms are implemented using 3 gates 3 and gates and each of these 3 and gates have 3 inputs sum of products is implemented by using a single or gate having 3 inputs the third example is a bar b the second product term in this example is a bar b c bar and the third term is a c again a combination of and gates and a single or gate is used you would require 2 and gates having 2 inputs and a single and gate having 3 inputs you would of course require a single 3 input or gate now let us look at examples of product of sums form again there are 3 expressions the first expression has 2 sum terms which are multiplied together so the first term is a bar odd with b the second term sum term is a odd with b bar odd with c now both these 2 terms are ended together to give you the second expression has 3 sum terms the first sum term has a bar plus b bar plus c bar the second sum term is c plus d bar plus e and the third sum term is b bar plus c plus d all these sum terms are ended together now how many or gates and and gates are required well there are 3 or terms 3 or gates are required what should be the inputs of these or gates basically in each sum term there are 3 variables so 3 input or gates are required now the final expression is product of these 3 sum terms so you need to have 3 input and gate to and all the 3 sum terms the last expression again has 3 sum terms the first sum term is a plus b the second sum term is a plus b bar plus c and the third sum term is a bar plus c again 3 or gates are required 2 or gates have 2 inputs and a single or gate has 3 inputs the outputs of all these 3 gates are ended together by using a single 3 input and gate the implementation of the first sum of product expression again 2 and gates are used and a single or gate is used the second diagram shows the implementation of the product of sum terms 3 or gates are used and the results or the outputs of these or gates are ended together to gives you the product of sum terms any general expression can be added to the standard sum of product form let us consider the expression a b plus b ended with the sum of c d plus e f so you have a product term c d plus e f this entire term is ended with b and this entire term is added with the product term a b now if you simplify this you end up with a b c d plus b e f the original expression is not in the standard s o p form the simplified version has been converted into the standard sum of product form let us consider the second example you have 2 sum terms ended together so the first sum term is a plus b the second sum term is b plus c plus d now to convert this expression into the standard sum of product form you simplify the expression so you end up with the terms a b odd with a c odd with a d plus b plus b c plus b d if you simplify this expression removing the common terms you end up with a c plus a d plus b which is the standard sum of products form the third expression is a plus b bar odd with c whole bar now this is not a standard sum of products form now this has to be simplified so you end up with a plus p double bar ended with c bar now what did we say when you have double bars over an expression basically both the bars are removed so in this particular case a plus b double bar is simplified to a plus b and c bar remains as it is now if you simplify the expression you end up with a c bar plus b c bar so you have converted the original general expression into sum of products form today we have looked at boolean algebra we have simplified boolean expressions using some tools which we have studied we have also looked at simplifying circuits using boolean expressions we will continue with these boolean expressions in the next lecture hope to see you again for the half is aslam aleykum