(a)In a certain application, four inputs A,B,C,D (both true...

(a)In a certain application, four inputs A,B,C,D (both true and complemented forms available) are fed to logic circuit, producing an output F which operates a relay. The relay turns on When F(ABCD)=1 for the following states of inputs(ABCD): '0000', '0010', '0100', '1100', '1110'. States '0101', 0110, 0111, 1000, 1001 and '1010' do not occur(don't care) and for the remaining states, the relay is off. Minimize F with the help of a K map and realize it using Logic gates? (b) X and Y are the two keys of a Locker. The Locker will be opened when the two keys are turn ON. Draw a circuit using of logic gates that resembles the concept and write the truth table?

Transcript

00:01 Hello students here for 9a equation.

00:04 It is a four input circuit and the relay turns on only in the following five given conditions and six states are given as don't care.

00:14 So let's first draw the karnaugh map to get the view of the circuit.

00:26 So this vertical one represents ab and the horizontal one represents cd and the inputs are 00, 01, 11 and 10.

00:40 Here also 00, 01, 11, 10.

00:45 Now let's mark the states in which the relay is on 00, 00 this is so here put 1 and 00, 10 that means this state 01, 00 here, 11, 00 this state and 11, 10, 11, 10 that means this state and the states don't care are marked by crosses the states 0, 1, 0, 1.

01:18 So this state 0, 1, 1, 0 this state 0, 1, 1, 1, 1, 0, 0, 1, 0, 0 and 1, 0, 0, 1, 1, 0, 1, 0, 1, 0.

01:40 So these states are don't care states and the remaining states let's fill it with 0.

01:47 Now let's circle the greatest pairs or quads that is available.

01:53 Here one pair will be available that is this and this by rolling method and this is one pair and here a term is there.

02:05 Now let's simplify the write the boolean expression boolean expression as sop equation.

02:23 So here it is y is equal to a bar abc this is a bar b bar c bar d bar a bar b bar c bar d bar plus this state is a bar b bar c d bar plus this states are written as a bar b c bar d bar plus a bar ab ab since it is 1.

03:00 So ab c bar d bar plus the single state this is written as a b c d bar.

03:14 Now from this term and this term let's take the common terms outside.

03:18 So we get the expression that is here a bar b bar c bar is taken means d a bar b bar d bar is taken means remaining we get c bar plus c and that will be equal to 1 plus if here we take b c bar d bar outside and remaining a bar plus a this will be equal to 1 plus a b c d bar this as remains as it is.

03:55 So the expression is a bar b bar d bar plus b c bar d bar plus a b c d bar.

04:07 This is the expression we have needed and this expression can be implemented using a combination of not gate and gate and or gate...

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