mock test digital logic design

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  • เผยแพร่เมื่อ 22 ธ.ค. 2024

ความคิดเห็น • 9

  • @dn2358
    @dn2358 4 ปีที่แล้ว +3

    Thanks for uploading 👍

  • @narennaren436
    @narennaren436 2 ปีที่แล้ว +2

    6:40 Only one 4 bit adder is sufficient ... bit[1:0] is added with 2'b00, so no adder is required, for the rest of bits, only one 4-bit adder is sufficient!!

  • @Golukumar-vj6kk
    @Golukumar-vj6kk ปีที่แล้ว +1

    Nice explanation maaam ❤❤

  • @sohamdas7775
    @sohamdas7775 3 ปีที่แล้ว +1

    Great job brother

  • @RakeshPrabhakara
    @RakeshPrabhakara ปีที่แล้ว +1

    Simple answer is : We must analyze based on worst case scenario and worst case would be when A is 4'b1111 and the product would be 75 (1001011) which is 7-bits wide. two 4-bit adders can accommodate max of 8 bits which would make sense for outputting 1001011 including carry-forward. Even if A was 5-bits input, two 4-bit Adders would suffice since largest 5-bit number: 31 (5'11111) multiplied by 5 would give 155 or 8'b10011011. Seeing this, from an optimization point of view, we are hardly utilizing two 4-bit adders fully which is wastage of resource and at the same time, an inevitable option given the specification.

  • @mrigendrasingh4072
    @mrigendrasingh4072 3 ปีที่แล้ว +1

    Wouldn't u need AND gates (or multipliers) too if u want to multiply 101 and A first then add them...........in the first question. But, extra gates or devices are not allowed here.

  • @amitmishra7003
    @amitmishra7003 3 ปีที่แล้ว

    Four 4 bit adder is required in cascade

  • @saileshsaileshreddy3347
    @saileshsaileshreddy3347 2 ปีที่แล้ว

    Implement all Morris Manu even cc and seq circuits nothing more is there

  • @Golukumar-vj6kk
    @Golukumar-vj6kk ปีที่แล้ว +1

    In ring counter if you starts with 0001 then next three states will be 1000,0100,0010.
    Correct me if am wrong.
    If I am right then your written order for ring counter is mistakenly wrong.