Product of Sums (Part 1) | POS Form

00:11:06
https://www.youtube.com/watch?v=nXsiLPJfDZ4

الملخص

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.

الوجبات الجاهزة

  • 📘 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.

الجدول الزمني

  • 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.

الخريطة الذهنية

Mind Map

الأسئلة الشائعة

  • 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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الترجمات
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التمرير التلقائي:
  • 00:00:04
    in this presentation I will teach you
  • 00:00:07
    product of some form which is an
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    abbreviation for POS form you have
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    already studied the saab form so it will
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    be easy for you to understand I have
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    taken the same truth table in which we
  • 00:00:18
    have three variables ABC
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    so the total number of combination will
  • 00:00:22
    be 2 to the power 3 which gives us 8 I
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    have already written all the eight
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    combinations and output corresponding to
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    them why is the boolean function that we
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    want to represent as pause form we have
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    already represented this Y has soft form
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    now we want to represent it in pass form
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    this we have done and this we want to do
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    in this presentation so let's start with
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    it pause form is used when the output is
  • 00:00:49
    0 or low so let me write this point down
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    pause form is used when the output when
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    the output is a low or a zero whereas in
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    case of soft form we used it when the
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    output is high so let's see from this
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    truth table for what cases the output is
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    low for the first two cases the output
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    is low and for this case the output is
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    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
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    variable is a and it is zero if the
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    variable is a and it is zero then we
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    will write it as a and if it is one then
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    we will write it as a compliment this is
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    just opposite to what we have studied in
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    the soft form in soft form we studied if
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    the variable is a and it is zero we
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    write it as a compliment and if it is
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    one we write it as a this one is for
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    soft and this one is for pause so you
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    have to remember this thing it will be
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    very helpful to you when you write your
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    exam so let's move to the pause form
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    let's write it the function is y and for
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    the first case you can see a is 0 B 0
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    and C is 0
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    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
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    is 1 from here you can see and for the
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    last case we have a s 0 B as one C as a
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    1 now this we call as this we call as
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    this 3 we will call as the max term the
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    max term and in case of soft form we had
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    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
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    write a and P and C you can see their
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    size this is looking maximum because of
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    its size and this is looking minimum
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    because of its size that is why we call
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    it as Max term and this one we call as
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    the min term there is no special reason
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    for this the functionality of the
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    function y will never change if you want
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    to represent it in pass form or in soft
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    form it will yield the certain output
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    depending upon the input and this pause
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    form written here I have obtained from
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    the soft form itself so let's try to
  • 00:03:31
    prove it that we can have the pause form
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    from the soft form and I will consider
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    the output when it is alone so why is
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    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
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    BS 0 C as 1 so a complement and B
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    complement and C and for the final case
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    when the output is 0 we have a as 0 B as
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    one C as 1 so a complement BC now I want
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    the left hand side as the output Y
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    not why compliment I want it as Y so
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    what I can do I can take the compliment
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    of both the sides the right-hand side
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    and the left-hand side so taking the
  • 00:04:35
    compliment compliment of both sides
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    let's see what we have Y compliment and
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    then again it's complement we have a
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    complement B complement C complement are
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    a compliment B complement see on a
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    complement BC and I will take the
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    complement of the whole right-hand side
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    we will use the de Morgan's law I hope
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    you remember it and when you take the
  • 00:05:05
    complement of something already
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    complimented like if you want to take
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    the complement of a complement you will
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    have itself in the same wave the
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    left-hand side will have Y that's what
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    we want and we will use demorgan's
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    theorum now if you have a or b and you
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    want to take its complement you will
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    have a complement the complement of the
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    variable this R will now change to end
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    and the B will now be written as B
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    complement we will use this Allah here
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    and we have a complement B complement C
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    complement and the complement of whole
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    this R will change to end and we have a
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    complement be complement C then its
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    complement this R will change to end a
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    complement BC and its complement again
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    we have to use the de Morgan's law for
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    this three min terms and we have a this
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    end will change to R be this end will
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    change to our C okay and in the same way
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    we have a or B or C complement and A or
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    B complement or C complement this is the
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    Y now you may see that this two are same
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    and instead of going to the de Morgan's
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    law every time if someone asks you to
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    write the pause form it is not required
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    to use the de Morgan's theorem every
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    time to convert the sob form to the pass
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    form
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    we can directly write down the pause
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    form by this convention you can use this
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    convention you use a when it is zero you
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    is a compliment when it is one you just
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    compliment it what we are doing here de
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    Morgan's law does what it takes the
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    compliment we are taking the compliment
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    of Y compliment instead of doing this we
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    just take the compliment initially you
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    just take the compliment of this thing
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    and you have this convention use this
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    convention and directly write the pause
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    form like this now this pause form
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    written here is the canonical pause form
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    canonical pause form we also call it
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    standard pause form and we write it
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    directly from the truth table like we
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    have already did and in canonical or
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    standard pause form each Max term will
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    have all the variables in normal or
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    complemented form here we have three max
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    terms and in each of this Max term we
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    have all the variables in normal or
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    complemented form in the first max term
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    you can see we have ABC there are three
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    variables a b c and in this we have ABC
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    in normal form in this we have a B
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    normal form C in complemented form and
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    in the last Max term we have a normal
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    form B and C in complemented form so
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    this is our canonical pause form you can
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    use the boolean algebra to reduce this
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    canonical pause form to the minimal pass
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    form I am going to do it quickly and in
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    the next presentation we will solve few
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    more examples on the pass form so the
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    function y is equal to a or b or c and A
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    or B or C compliment and a or B
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    compliment or C complement and we want
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    to reduce it by using the boolean
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    algebra this is a good practice for you
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    because you have already studied the
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    boolean algebra in the first two terms
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    you can see we have a or B as common and
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    from distributive law if you remember if
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    we have a are of B and a are C we write
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    it as a or BC so let's consider a or B
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    as X so we have X or C from here you can
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    see
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    and we have X or C complements so it can
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    be written as X or C and let me slide
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    this board C complement so y is equal to
  • 00:09:01
    X or C and C complement and from the
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    last mixed term we have a or B
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    complement or C complement now C and C
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    complement is definitely equal to 0 and
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    X are with 0 is equal to X and X is
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    equal to a or b so we have a or B and
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    here we have a or B complement or C
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    complement again you can see a is common
  • 00:09:29
    so from distributive law we have a or b
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    and I have to add B with B complement or
  • 00:09:39
    C complement so let's try to find out
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    the function y by the distributive log
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    in a all I have to open this bracket we
  • 00:09:48
    have B & B complement or B and C
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    complement B & B complement is
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    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
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    C complement so this is this is the
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    minimal pause form because in this you
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    can see the first max term this is max
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    term and in the first Max term C is not
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    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
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    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
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    this that I have explained you initially
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    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
الوسوم
  • POS form
  • Boolean algebra
  • canonical form
  • minimal form
  • De Morgan's law
  • max terms
  • truth table
  • distributive law
  • converting forms