- #1
mathmari
Gold Member
MHB
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Hey! :giggle:
a) Suppose a binary pick-up adder (carry-select) of $32$-bits, comprising $4$ sub-sections adders spreading carry of range $8$ bits. Show the values obtained internally in the circuit of this adder to perform the addition of the numbers $01011001101100111110110011001101$ and $10010110011111000001101100110110$. Specifically, show the input and output values on each $1$-bit full adder circuit, as well as the input and output values on all multiplexers in the circuit. If each complete $1$-bit adder circuit and each multiplexer circuit cause a delay of $2T$ in the execution of the operation, where $T$ is the delay time of an elementary gate, find the calculation time of the retained output.
b) Repeat the above if the $32$-bit adder includes $6$ sub-sections adders spreading carry, ranging in order from less to more important sections $2$, $4$, $4$, $6$, $6$, and $10$ bits.
The idea of that method is :
The adder is divided into two halves of equal length. And for both halves, the addition is started at the same time. In the left (more significant) half we don't know whether the carry input of the rightmost full adder is a $1$ or a $0$. Therefore, we do the addition of the left half twice at the same time, once with a $0$ at the carry input and once with a $1$. If the right half with its addition is done, we know the incoming carry of the left half. So we know which of the results is the right one, which we then select (select). The other (wrong) result is simply discarded.
Is that correct?
So we split $01011001101100111110110011001101$ into $a_1=0101100110110011$ and $a_2=1110110011001101$, right?
And we split also $10010110011111000001101100110110$ into $b_1=1001011001111100$ and $b_2=0001101100110110$, right?
Then we add $a_1+b_1$ wih carry $0$ and $a_1+b_1$ wih carry $1$.
We also calculate the addition $a_2+b_2$.
As for the addition $a_1+b_1$ wih carry $0$ : $$0101100110110011+1001011001111100=1111000000101111$$
As for the addition $a_1+b_1$ wih carry $1$ : $$0101100110110011+1001011001111100=1111000000110000$$
As for the addition $a_2+b_2$ :
$$1110110011001101+0001101100110110=0000100000000011\ \ \text{ with carry } 1$$ So we select the second case of $a_1+b_1$.Is everything correct? How could we continue? :unsure:
a) Suppose a binary pick-up adder (carry-select) of $32$-bits, comprising $4$ sub-sections adders spreading carry of range $8$ bits. Show the values obtained internally in the circuit of this adder to perform the addition of the numbers $01011001101100111110110011001101$ and $10010110011111000001101100110110$. Specifically, show the input and output values on each $1$-bit full adder circuit, as well as the input and output values on all multiplexers in the circuit. If each complete $1$-bit adder circuit and each multiplexer circuit cause a delay of $2T$ in the execution of the operation, where $T$ is the delay time of an elementary gate, find the calculation time of the retained output.
b) Repeat the above if the $32$-bit adder includes $6$ sub-sections adders spreading carry, ranging in order from less to more important sections $2$, $4$, $4$, $6$, $6$, and $10$ bits.
The idea of that method is :
The adder is divided into two halves of equal length. And for both halves, the addition is started at the same time. In the left (more significant) half we don't know whether the carry input of the rightmost full adder is a $1$ or a $0$. Therefore, we do the addition of the left half twice at the same time, once with a $0$ at the carry input and once with a $1$. If the right half with its addition is done, we know the incoming carry of the left half. So we know which of the results is the right one, which we then select (select). The other (wrong) result is simply discarded.
Is that correct?
So we split $01011001101100111110110011001101$ into $a_1=0101100110110011$ and $a_2=1110110011001101$, right?
And we split also $10010110011111000001101100110110$ into $b_1=1001011001111100$ and $b_2=0001101100110110$, right?
Then we add $a_1+b_1$ wih carry $0$ and $a_1+b_1$ wih carry $1$.
We also calculate the addition $a_2+b_2$.
As for the addition $a_1+b_1$ wih carry $0$ : $$0101100110110011+1001011001111100=1111000000101111$$
As for the addition $a_1+b_1$ wih carry $1$ : $$0101100110110011+1001011001111100=1111000000110000$$
As for the addition $a_2+b_2$ :
$$1110110011001101+0001101100110110=0000100000000011\ \ \text{ with carry } 1$$ So we select the second case of $a_1+b_1$.Is everything correct? How could we continue? :unsure: