How do you implement a system that receives as input 3 variables A, B, C each variable consisting of k bits and a flag variable F consisting of 2 bits and outputs a k bit variable Y based on the following rules. A B C and Y are considered to be in 2's complement form with k bits. If F= 00 then the output Y will compare between the values absolute value A-C minus absolute value of B and absolute value of A minus absolute value of C minus B. Namely Y= 000...000 if absolute value of A minus C minus absolute value of B is less than absolute value of A minus the absolute value of C minus B and Y = 111 otherwise.
sir in 1st prbm we have a case" less than or equal" ...so we can take 1 for 3 value ...then we got minimised expression ..but u take 0 instead of 1, why so ..plz give me clarity on this
in this question they said the output is one when binary value of input is less than three "now see in binary table the binary value of 011 is equal to three so the condition is false and the output is 0 because they said the output is one when binary value of input is less than three
Design a circuit that accepts binary numbers between 1 and 10 and generates an output equal to the number of 0’s in the input. Implement the circuit using only NOR gate. Please solve this question and post. 🙏🙏
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Sir in a) 0 1 1will also give 1 as output is 1 for less than or equal to 3.
correct
Ohh❤❤❤❤❤❤❤❤fantastic ❤❤❤ thank you ,sir for your effort💕💕💕💕💕💕💕💕💕💕
Jeee oyeee dhiman🎉
next level question
Sir I guess in rule (a) circuit diagram there should be NOT gate in x' ,y' and z'.
Thank you sir
How is 011 equal to 0 011 means 2^0+2^1=3 which is included in the condition less than or equal to 3
Awesome brother
Sir in part c you take 000 is 1 in even condition.
Exactly,
Design a combinational circuit of your choice which is used for ALU(arithmatic logic unit).justify your answer.
At 3:11 can we simplify F = y' + x' ?
thank you dr dhiman
for the first part, 3 also gives output 1
Thanks man 👏🏼
How do you implement a system that receives as input 3 variables A, B, C each variable consisting of k bits and a flag variable F consisting of 2 bits and outputs a k bit variable Y based on the following rules. A B C and Y are considered to be in 2's complement form with k bits. If F= 00 then the output Y will compare between the values absolute value A-C minus absolute value of B and absolute value of A minus absolute value of C minus B. Namely Y= 000...000 if absolute value of A minus C minus absolute value of B is less than absolute value of A minus the absolute value of C minus B and Y = 111 otherwise.
Thxq sir
Very very thxq
sir in 1st prbm we have a case" less than or equal" ...so we can take 1 for 3 value ...then we got minimised expression ..but u take 0 instead of 1, why so ..plz give me clarity on this
yeah exactly
thank you very much sir.
amazing
Sir for question (a) the condition is less than or equals to 3 but you wrote 0 in the output of 3 isn't that a mistake?
i got stuck on the same question. i think there is a little mistake, right?
Yess, that's a mistake we have to take 0 as even
@@abhishekpoudel1902 thx bro, have a good day
can we add buffer gate in option b
Thanks man
You're welcome!
i dont get it sir, when we put three zero's in 3 inputs in AND gates followed by OR gate the final output should zero right? the truth table is F= 1??
oh i get it we should put, NOT gate in every input , that makes sense in BAR'S. thanks!!!
Hi Sir! part b k logic gates wala portion naii kiya
Input is 3 and output is 1 and conditions is when more than 1 input variables are logic 0
Super sir
thank you so much
The output for binary value Three is also 1
Tell me about a site that has questions about logical design
Design a combinational circuit using a ROM. The circuit accepts a three‐bit number and outputs a binary number equal to the square of the input number
Please refer to chapter 7 videos
Sir u haven't considered 3 in the first case....... "Less than or equal to 3"
in this question they said the output is one when binary value of input is less than three "now see in binary table the binary value of 011 is equal to three so the condition is false and the output is 0
because they said the output is one when binary value of input is less than three
Implement A’C + ABC + BC’D + AB’C using decoder.plz help sir
Please watch the later videos in this chapter. such as 4.28, 4.29, 4.30
hello sir, I'm wondering why the gates for x'y' and x'z' are just AND gate. Shouldn't there be an inverter?
He used an a simpler way to represent the logic diagram. He just took the inverses as the starting points since the function doesn't have X, Y or Z.
In part 1 there is a condition output is 1 if it is less than or equal to 3 so you did a mistake
Design a circuit with four inputs and one output where the output is 1 if
the input is divisible by 3 or 7.
why 3 o's give 1 ? omgggg so confusing i'm lost
It means it's a complement of the other
Design a circuit that accepts binary numbers between 1 and 10 and generates an output
equal to the number of 0’s in the input. Implement the circuit using only NOR gate.
Please solve this question and post. 🙏🙏
Wrong re-read the homework less or equal to 3 the output is one . 3=3 therefore the output for 3 is one
Thank you Sir
Thank you sirji