What is POS form in Boolean algebra?

POS, or Product of Sums form, is a way of expressing Boolean functions where the function is represented as a product of sum terms, used when the output is 0.

How is POS form different from SOP form?

POS form is used when the output is 0, while SOP (Sum of Products) is used when the output is high or 1. In POS, 0 is written as the variable, and 1 as its complement, opposite to SOP.

How are max terms identified in a truth table?

Max terms are identified when the output is low or 0 in a truth table. Each combination resulting in 0 is used to form a max term in POS.

What are the steps to convert an SOP form to a POS form?

To convert SOP to POS, identify all cases where the output is 0, use the variables in their original state for 0, and complemented state for 1, then apply De Morgan's law to get the POS form.

What are canonical and minimal POS forms?

Canonical POS form includes all variables in each max term either in normal or complemented form. Minimal POS form reduces the number of variables or terms using Boolean algebra while maintaining the function.

Why do we use De Morgan's law in converting forms?

De Morgan's law helps in complementing the entire expressions effectively, transitioning between SOP and POS forms by changing expressions' structures while maintaining logical equivalence.

Can the functionality of Boolean function change with form conversion?

No, converting between SOP and POS does not change the Boolean function's logical output; it only alters the representation form.

How does the distributive law apply in reducing POS form?

The distributive law helps to factor out common expressions and reduce complex terms in POS form, simplifying the equation while retaining logical equivalence.

Product of Sums (Part 1) | POS Form

00:11:06

https://www.youtube.com/watch?v=nXsiLPJfDZ4

Resumen

TLDRThis presentation delves into the Product of Sums (POS) form in Boolean algebra, complementing your existing knowledge of Sum of Products (SOP) form. It uses a three-variable truth table, with variables A, B, and C, which forms 8 combinations of outputs. The focus is on transforming the Boolean function Y, already in SOP form, into its POS form by considering only the cases where the output Y is 0. The presenter explains how, in contrast to SOP form where 0 is written as a complement and 1 as itself, in POS, 0 remains as it is, and 1 is complemented. The presentation also discusses canonical and minimal POS forms, showcasing their derivation and differences. De Morgan's law is applied to convert between forms, and Boolean algebra is used to reduce the canonical POS to minimal POS form, demonstrating distribution laws to simplify expressions. Emphasis is placed on practical approaches for solving Boolean expressions in exam settings, including direct application of known conventions without recalculating detailed complemented forms.

Para llevar

📘 POS form is used when the output is 0.
🔄 Difference between POS and SOP forms is in variable interpretations.
📋 Max terms are derived from cases where output is low.
🧠 Use De Morgan's law to convert expressions.
✂ Canonical POS form to minimal POS form involves reduction using Boolean algebra.
📏 Minimal POS form has fewer variables and terms but retains functionality.
🧮 Practical tips include directly using established conventions for efficiency.
🚀 SOP to POS conversion doesn't change function outcome, only its representation.
🔍 The distributive law simplifies equations in conversion processes.
🔁 Direct transformation from truth tables aids POS form writing.

Cronología

00:00:00 - 00:05:00
In this presentation, the speaker introduces the Product of Sums (POS) form, an abbreviation for a type of Boolean expression. The audience is expected to be familiar with the SOP (Sum of Products) form, which inversely relates to the POS. Using a truth table with three variables (A, B, C), the presenter demonstrates converting a Boolean function output described in SOP form into POS form. The POS form is essential when the output is low (zero), unlike SOP used for high (one) outputs. The speaker explains how to write POS by considering cases where the output is low, and defining the function accordingly using max terms, which are larger than the min terms of SOP. They emphasize the functionality of the expression doesn't change, regardless of its form but explains the derivation from SOP to POS by complementing both forms and using De Morgan's Law.
00:05:00 - 00:11:06
The speaker explains De Morgan's Law application in deriving POS from SOP, discussing how complementing the expression can simplify the transformation process. Canonical or standard POS form is introduced, highlighting its construction where every max term includes all variables. The conversion to a reduced or minimal POS form using Boolean algebra is explained. This minimal form involves simplifying expressions, highlighting common elements and using distributive laws, until the function includes less-comprehensive max terms. This reduces complexity, while still representing the same Boolean logic. The presentation concludes with the prospect of further examples in subsequent presentations, outlining the advantage of understanding both canonical and minimal forms for academic assessments.

Mapa mental

Preguntas frecuentes

What is POS form in Boolean algebra?
POS, or Product of Sums form, is a way of expressing Boolean functions where the function is represented as a product of sum terms, used when the output is 0.
How is POS form different from SOP form?
POS form is used when the output is 0, while SOP (Sum of Products) is used when the output is high or 1. In POS, 0 is written as the variable, and 1 as its complement, opposite to SOP.
How are max terms identified in a truth table?
Max terms are identified when the output is low or 0 in a truth table. Each combination resulting in 0 is used to form a max term in POS.
What are the steps to convert an SOP form to a POS form?
To convert SOP to POS, identify all cases where the output is 0, use the variables in their original state for 0, and complemented state for 1, then apply De Morgan's law to get the POS form.
What are canonical and minimal POS forms?
Canonical POS form includes all variables in each max term either in normal or complemented form. Minimal POS form reduces the number of variables or terms using Boolean algebra while maintaining the function.
Why do we use De Morgan's law in converting forms?
De Morgan's law helps in complementing the entire expressions effectively, transitioning between SOP and POS forms by changing expressions' structures while maintaining logical equivalence.
Can the functionality of Boolean function change with form conversion?
No, converting between SOP and POS does not change the Boolean function's logical output; it only alters the representation form.
How does the distributive law apply in reducing POS form?
The distributive law helps to factor out common expressions and reduce complex terms in POS form, simplifying the equation while retaining logical equivalence.

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Subtítulos

Desplazamiento automático:

00:00:04
in this presentation I will teach you
00:00:07
product of some form which is an
00:00:09
abbreviation for POS form you have
00:00:12
already studied the saab form so it will
00:00:14
be easy for you to understand I have
00:00:16
taken the same truth table in which we
00:00:18
have three variables ABC
00:00:20
so the total number of combination will
00:00:22
be 2 to the power 3 which gives us 8 I
00:00:25
have already written all the eight
00:00:27
combinations and output corresponding to
00:00:30
them why is the boolean function that we
00:00:32
want to represent as pause form we have
00:00:35
already represented this Y has soft form
00:00:38
now we want to represent it in pass form
00:00:41
this we have done and this we want to do
00:00:44
in this presentation so let's start with
00:00:46
it pause form is used when the output is
00:00:49
0 or low so let me write this point down
00:00:52
pause form is used when the output when
00:01:01
the output is a low or a zero whereas in
00:01:06
case of soft form we used it when the
00:01:08
output is high so let's see from this
00:01:11
truth table for what cases the output is
00:01:13
low for the first two cases the output
00:01:16
is low and for this case the output is
00:01:19
low so there are three cases for which
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your output is a low so we have to
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consider these three cases only while
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writing the pause form and if the
00:01:29
variable is a and it is zero if the
00:01:31
variable is a and it is zero then we
00:01:34
will write it as a and if it is one then
00:01:37
we will write it as a compliment this is
00:01:40
just opposite to what we have studied in
00:01:42
the soft form in soft form we studied if
00:01:45
the variable is a and it is zero we
00:01:48
write it as a compliment and if it is
00:01:51
one we write it as a this one is for
00:01:54
soft and this one is for pause so you
00:01:58
have to remember this thing it will be
00:02:00
very helpful to you when you write your
00:02:02
exam so let's move to the pause form
00:02:04
let's write it the function is y and for
00:02:08
the first case you can see a is 0 B 0
00:02:10
and C is 0
00:02:12
so we will write it as a or B or C and
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for the second case you can see a 0 B 0
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C is 1 so A or B or C complement we have
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to write it as C complement because it
00:02:30
is 1 from here you can see and for the
00:02:33
last case we have a s 0 B as one C as a
00:02:39
1 now this we call as this we call as
00:02:43
this 3 we will call as the max term the
00:02:48
max term and in case of soft form we had
00:02:53
the min term and we just call them next
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term or min term because of their size
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if you write a or B or C and if you
00:03:02
write a and P and C you can see their
00:03:05
size this is looking maximum because of
00:03:08
its size and this is looking minimum
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because of its size that is why we call
00:03:11
it as Max term and this one we call as
00:03:13
the min term there is no special reason
00:03:15
for this the functionality of the
00:03:17
function y will never change if you want
00:03:20
to represent it in pass form or in soft
00:03:22
form it will yield the certain output
00:03:24
depending upon the input and this pause
00:03:27
form written here I have obtained from
00:03:29
the soft form itself so let's try to
00:03:31
prove it that we can have the pause form
00:03:34
from the soft form and I will consider
00:03:37
the output when it is alone so why is
00:03:41
now why complement because you can see
00:03:43
in soft form when 0 we write compliment
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and here function y is 0 so I will write
00:03:49
y complement it is 0 for 3 cases let's
00:03:53
try to find out the first case in which
00:03:55
a 0 B 0 and C is 0 so I will write a
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complement and B complement and C
00:04:03
complement or in this case I have a as 0
00:04:07
BS 0 C as 1 so a complement and B
00:04:11
complement and C and for the final case
00:04:14
when the output is 0 we have a as 0 B as
00:04:17
one C as 1 so a complement BC now I want
00:04:22
the left hand side as the output Y
00:04:25
not why compliment I want it as Y so
00:04:28
what I can do I can take the compliment
00:04:30
of both the sides the right-hand side
00:04:33
and the left-hand side so taking the
00:04:35
compliment compliment of both sides
00:04:42
let's see what we have Y compliment and
00:04:46
then again it's complement we have a
00:04:48
complement B complement C complement are
00:04:51
a compliment B complement see on a
00:04:55
complement BC and I will take the
00:04:58
complement of the whole right-hand side
00:05:01
we will use the de Morgan's law I hope
00:05:03
you remember it and when you take the
00:05:05
complement of something already
00:05:07
complimented like if you want to take
00:05:10
the complement of a complement you will
00:05:12
have itself in the same wave the
00:05:15
left-hand side will have Y that's what
00:05:17
we want and we will use demorgan's
00:05:19
theorum now if you have a or b and you
00:05:23
want to take its complement you will
00:05:25
have a complement the complement of the
00:05:27
variable this R will now change to end
00:05:30
and the B will now be written as B
00:05:33
complement we will use this Allah here
00:05:36
and we have a complement B complement C
00:05:40
complement and the complement of whole
00:05:42
this R will change to end and we have a
00:05:46
complement be complement C then its
00:05:49
complement this R will change to end a
00:05:53
complement BC and its complement again
00:05:56
we have to use the de Morgan's law for
00:05:58
this three min terms and we have a this
00:06:03
end will change to R be this end will
00:06:07
change to our C okay and in the same way
00:06:11
we have a or B or C complement and A or
00:06:16
B complement or C complement this is the
00:06:21
Y now you may see that this two are same
00:06:25
and instead of going to the de Morgan's
00:06:29
law every time if someone asks you to
00:06:31
write the pause form it is not required
00:06:33
to use the de Morgan's theorem every
00:06:35
time to convert the sob form to the pass
00:06:38
form
00:06:39
we can directly write down the pause
00:06:40
form by this convention you can use this
00:06:44
convention you use a when it is zero you
00:06:47
is a compliment when it is one you just
00:06:49
compliment it what we are doing here de
00:06:51
Morgan's law does what it takes the
00:06:53
compliment we are taking the compliment
00:06:55
of Y compliment instead of doing this we
00:06:58
just take the compliment initially you
00:07:00
just take the compliment of this thing
00:07:02
and you have this convention use this
00:07:04
convention and directly write the pause
00:07:06
form like this now this pause form
00:07:09
written here is the canonical pause form
00:07:14
canonical pause form we also call it
00:07:18
standard pause form and we write it
00:07:21
directly from the truth table like we
00:07:23
have already did and in canonical or
00:07:25
standard pause form each Max term will
00:07:28
have all the variables in normal or
00:07:30
complemented form here we have three max
00:07:33
terms and in each of this Max term we
00:07:36
have all the variables in normal or
00:07:37
complemented form in the first max term
00:07:39
you can see we have ABC there are three
00:07:42
variables a b c and in this we have ABC
00:07:44
in normal form in this we have a B
00:07:47
normal form C in complemented form and
00:07:49
in the last Max term we have a normal
00:07:52
form B and C in complemented form so
00:07:54
this is our canonical pause form you can
00:07:57
use the boolean algebra to reduce this
00:07:59
canonical pause form to the minimal pass
00:08:02
form I am going to do it quickly and in
00:08:04
the next presentation we will solve few
00:08:05
more examples on the pass form so the
00:08:09
function y is equal to a or b or c and A
00:08:15
or B or C compliment and a or B
00:08:20
compliment or C complement and we want
00:08:23
to reduce it by using the boolean
00:08:24
algebra this is a good practice for you
00:08:26
because you have already studied the
00:08:28
boolean algebra in the first two terms
00:08:30
you can see we have a or B as common and
00:08:34
from distributive law if you remember if
00:08:36
we have a are of B and a are C we write
00:08:41
it as a or BC so let's consider a or B
00:08:45
as X so we have X or C from here you can
00:08:50
see
00:08:51
and we have X or C complements so it can
00:08:53
be written as X or C and let me slide
00:08:56
this board C complement so y is equal to
00:09:01
X or C and C complement and from the
00:09:05
last mixed term we have a or B
00:09:08
complement or C complement now C and C
00:09:12
complement is definitely equal to 0 and
00:09:14
X are with 0 is equal to X and X is
00:09:18
equal to a or b so we have a or B and
00:09:23
here we have a or B complement or C
00:09:26
complement again you can see a is common
00:09:29
so from distributive law we have a or b
00:09:34
and I have to add B with B complement or
00:09:39
C complement so let's try to find out
00:09:42
the function y by the distributive log
00:09:45
in a all I have to open this bracket we
00:09:48
have B & B complement or B and C
00:09:54
complement B & B complement is
00:09:56
definitely equal to 0 so we have a or B
00:10:01
and C complement and again using the
00:10:04
distributive law we have a or b and a or
00:10:10
C complement so this is this is the
00:10:14
minimal pause form because in this you
00:10:20
can see the first max term this is max
00:10:22
term and in the first Max term C is not
00:10:25
there whereas in the last mix term or
00:10:28
the second next term we have only two
00:10:29
max turns here B is not there so all the
00:10:32
variables are not present whether a
00:10:33
normal or complemented form thus we call
00:10:35
it as the minimal pause form it is very
00:10:38
easy and in exam they will give you the
00:10:41
minimal pause form and ask you to have
00:10:44
your canonical pass form there are two
00:10:46
ways to do it the first way is to use
00:10:48
the truth table you can find the values
00:10:50
of your output depending upon this
00:10:52
expression and write the pause form like
00:10:55
this that I have explained you initially
00:10:57
or you can have another method that I
00:10:59
will explain you separately in one
00:11:01
presentation so this is all for this
00:11:04
presentation
00:11:04
in the next one

Etiquetas

POS form
Boolean algebra
canonical form
minimal form
De Morgan's law
max terms
truth table
distributive law
converting forms