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In this presentation we will study sumo product form it is also written as shop form where assistants for some or stands for off and P stands for products let×39’s start with it, you can see a 1/2table in front of you in which there are three variables a B and C and you already know if you want to find out teetotal number of combinations the total number of combinations then it is equal to 2 to the power n where n is the number of variables where n is the number of variables and here the number of variables are 3 so definitely 2 over3 will give us 8, so they are in total 8possible combinations and I have already written these 8 combinations starting from 0 0 0 all the way to 1 1 of 1 F is the output of this truth table and thesis our function and I have already written the value of the function for this different combinations for the first combination when an is low B's lowland C is low the function is also lowland in the same way it is a low high lowland for the last 4 cases it is a high sole me tell you one thing in boolean algebra you can express this function f can be expressed in two ways the first way is to express it in soft form top form that is our sum of product form and the second way is to express it in pass form POS form and these two forms are very important and very easy at the same time and definitely some problems will be there in your exam, and they are very easy you can easily understand these things and solve their problem so let'start with the soft form in this presentation and the coming presentation will explain to you the pause form the soft form is written when the functions high I will write the soft form onlywhenthe function is high so let×39’s see when it is high you can see for this case the value of f is 1, and it is when an is low×39’s high and C's low in the same waists 1 for the last 4 cases so the function f is equal to let's see for the first case a complement and B is 1 so Will write B and C is 0, so I will writer complement as I have already told you we will write the swap form for the true values or for the high values and a slow, so I have to write it as complement whereas B is 1, so there is no problem I will write B as it is and the Cis also low, so I will write Complement function f is 1, so I will write F I will not take its complementing the same way I will see for this case is 1 B is a 0 so B complement sees a 0so C complement for this case is 1 B is0 and C is 1 and for the second last case I have an is 1 B is 1 C is 0 so a BC complement and for the final case all the 3 variables are high so a B C this Call as the sum of product from this Call as the shop form I am calling it sum of product because you can see Have product this looks like product actually it is not product this dot is the end operator but at first it lookalike it is product, so I have the products, and then I am taking this sumo this product this is the sum however this is the or operator this one is their operator and this dot is the...
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