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Start building the 4-bit adder subtractor circuit from the half adder until you reach the final circuit results.

1. Build the 1-bit half adder circuit and convert it into symbol (block).

2. Build the 1-bit full adder circuit and convert it into symbol (block).

3. Build the 4-bit adder circuit and convert it into symbol (block).

4. Build the 4-bit adder subtractor and convert it into symbol (block).

5. Build the final circuit 

1. Build the 1-bit half adder circuit and convert it into symbol (block).  

Write a detailed explanation about the full adders.

Insert your screenshot and the explanation here

I want to sum 2 binary digits (A,B) and the sum of 2 binary numbers needs (at most) 2 output digits;

so from truth table:

A

B

Sum

Carry

0

0

0

0

0

1

1

0

1

0

1

0

1

1

0

1

Sum:    S = (A.B’) + (A’.B) = A⨁ B

Carry:  C = A.B 

1. Build the 1-bit full adder circuit and convert it into symbol (block).

Write a detailed explanation about the full adders.

Insert your screenshot and the explanation here

I want to sum 3 binary digits (X  ,Y , Cin) and the sum of 3 binary numbers needs (at most) 2 output digits (S , Cout).

X

Y

Cin

Sum

Carry

0

0

0

0

0

0

0

1

1

0

0

1

0

1

0

0

1

1

0

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1

0

0

1

0

1

0

1

0

1

1

1

0

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1

1

Sum = (X’ .Y’ .Cin) + (X’ .Y .Cin’) + (X .Y’ .Cin’) + (X .Y .Cin)

= X’ .((Y’ .Cin) + (Y .Cin’))  +  X .((Y’ .Cin’) + (Y .Cin))

= X’ .(Y⨁ Cin)  +  X .(Y⨁ Cin)’

= X ⨁ Y ⨁ Cin

Carry = (X’ .Y .Cin) + (X .Y’ .Cin) + (X .Y .Cin’) + (X .Y .Cin)

= Cin .(X⨁ Y)  + X .Y

since: Sum [from Full Adder (F.A.)] = X ⨁ Y ⨁ Cin

     = Sum1 [from Half Adder (H.A.)]   ⨁  Cin

Carry [from Full Adder (F.A.)] = X .Y +  Cin .(X⨁ Y)

  = Carry1 [from Half Adder (H.A.)] +  Cin .(X⨁ Y)

I can use 1 Half Adder to get [Sum1 from H.A. = X⨁ Y] and [Carry1 from H.A. = X.Y]

then I can use anther 1 Half Adder with (first Sum form H.A.  &  Cin) to get:

===>  Sum =[(Sum1 from H.A.) ⨁ Cin] = X ⨁ Y ⨁ Cin   <=== that’s the final Sum

and Carry2 [from second Half Adder (H.A.)] = Cin .(X⨁ Y)

If I passed the two Carry outputs (Carry1 & Carry2) through OR gate; I will get the final Carry: 

===>   final Carry = Carry1 + Carry2 =  X .Y + Cin .(X⨁ Y)   <=== that’s the final Carry

So, we can use 2 H.A. & OR gate to construct Full Adder. 

1. Build the 4-bit adder circuit and convert it into symbol (block).  

Write a detailed explanation about how it works.

Insert your screenshot and the explanation here

since Half Adder Circuit (H.A.) can sum 2 digits

since Full Adder Circuit (F.A.) can sum 3 digits at once

It’s generally used Full Adder (F.A.) Circuits to sum 2 digits (1-digit from input x & 1-digit from input y) and add to them the carry (in the third input digit) from previous sum operation.

Ofcourse first 1-bit adder doesn’t need 3 input digits (because of there is no Carry);

So I used H.A. Circuit in first 1-bit Adder.

1. Build the 4-bit adder subtractor and convert it into symbol (block).  

Write a detailed explanation about how it works.

Insert your screenshot and the explanation here

It’s preferred to subtract binary numbers (Z= X - Y) using the 2’s complement method; Which says that we find the 1’s complement to the -ve number (Y) and then add 1 to it;

After that we can use the new input of Y to sum it normally with X using a 4-bit Adder Circuit.

⇒ The complement of every digit of Y = Y’    ⇒ Y’ = {y0’ , y1’ , y2’ , y3’}

so every digit of Y enters on NOT gate to become Y’

⇒ Then add 1 (as a carry) to the Sum of first 1-bit Full Adder(F.A.)

∴The output Z = X - Y [but X must be >= Y]

1. Build the final circuit 

• Include screenshots with 4 different random examples and explain each one:

• 2 addition

• 2 subtraction

• Write detailed explanation about how it works and how you did it.

Insert your screenshots and the explanation here

HOW I DID THIS CIRCUIT:

The Circuit works in 2 options. When I keep [selector = 0] the Circuit is a 4-bit Adder  and when I keep [selector = 1] the circuit is a 4-bit Subtractor.

∵The difference between 4-bit adder & 4-bit subtractor is in

1)the input Y (which is converted to Y’ in a 4-bit Subtractor)

2)& the third input digit in the first bit Full adder.

∴For input Y problem:

we need to find the equation that converts Y to Y’ (the 1’s complement) when the [selector = 1]. (i’ll use y0 digit to explain on)

selector

y0

f(y0)

0

0

0

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1

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0

f(y0) = selector ⨁ y0

⇒ ∴the problem of Y is solved when we connect every digit of Y with selector using X-OR gate.

∴For the problem of the third input digit in the first 1-bit Full adder:

we need to keep [the third input digit = 0] when the wanted operation is sum (selector=0) and keep [third input digit = 1] when the operation is subtraction (selector=1)

⇒ ∴ The third input digit in first bit Full adder = selector

Screenshot 1

When the input X=(0101) :{x3=0  , x2=1  , x1=0  , x0=1}

& when the input Y=(0111) : {y3=0  , y2=1  , y1=1  , y0=1}

∵ selector = 0 ∴ The operation is summing (Z = X + Y)

∴ Z0 = sum of three input digits (x0 , y0 , Cin) 

Z0= x0 + y0 + selector = 1 + 1 + 0    = 0  [& 1-carried to next bit Full Adder]

Z1= x1 + y1 + Cin = 0 + 1 + 1(carry)= 0 [& 1-carried to next bit Full Adder]

Z2= x2 + y2 + Cin = 1 + 1 + 1(carry)= 1 [& 1-carried to next bit Full Adder]

Z3= x3 + y3 + Cin = 0 + 0 + 1(carry)= 1 [& no carry]

Z4= last carry = 0

 ∴  Z = 01100

Screenshot 2

When the input X=(1111)     :{x3= 1   ,x2= 1   ,x1= 1   ,x0= 1}

& when the input Y=(1111) : {y3= 1   ,y2= 1   ,y1= 1   ,y0= 1}

∵ selector = 0 ∴ The operation is summing (Z = X + Y)

∴ Z0 = sum of three input digits (x0 , y0 , Cin) 

Z0= x0 + y0 + Cin = 1 + 1 + selector = 0  [& 1-carried to next bit Full Adder]

Z1= x1 + y1 + Cin = 1 + 1 + 1(carry) = 1 [& 1-carried to next bit Full Adder]

Z2= x2 + y2 + Cin = 1 + 1 + 1(carry) = 1 [& 1-carried to next bit Full Adder]

Z3= x3 + y3 + Cin = 1 + 1 + 1(carry) = 1 [& 1-carried to next bit Full Adder]

Z4= last carry = 1

 ∴  Z = 11110

Screenshot 3

When the input X=(1111)     :{x3= 1   ,x2= 1   ,x1= 1   ,x0= 1}

& when the input Y=(0101) : {y3= 0   ,y2= 1   ,y1= 0   ,y0= 1}

∵ selector = 1 ∴ The operation is subtract (Z = X - Y)

the circuit Subtracts using 2’s complement method:

∴ 1’s complement of Y = Y’ = 1010:  {y3’= 1   ,y2’= 0   ,y1’= 1   ,y0’= 0}

∴ Z0 = sum of three input digits (x0 , y0’ , Cin) 

Z0= x0 + y0’ + Cin = 1 + 0 + selector = 0  [& 1-carried to next bit Full Adder]

Z1= x1 + y1’ + Cin = 1 + 1 + 1(carry) = 1 [& 1-carried to next bit Full Adder]

Z2= x2 + y2’ + Cin = 1 + 0 + 1(carry) = 0 [& 1-carried to next bit Full Adder]

Z3= x3 + y3’ + Cin = 1 + 1 + 1(carry) = 1 [& 1-carried to next bit Full Adder]

Z4= last carry = 1 (note: we ignore last bit (Z4) in subtract operation & it have to = 1)

when [Z4 = 1] that means  [X>=Y] 

 ∴  Z = 1010

Screenshot 4

When the input X=(0111)     :{x3= 0   ,x2= 1   ,x1= 1   ,x0= 1}

& when the input Y=(0111) : {y3= 0   ,y2= 1   ,y1= 1   ,y0= 1}

∵ selector = 1 ∴ The operation is subtract (Z = X - Y)

the circuit Subtracts using 2’s complement method:

∴ 1’s complement of Y = Y’ = 1000:  {y3’= 1   ,y2’= 0   ,y1’= 0   ,y0’= 0}

∴ Z0 = sum of three input digits (x0 , y0’ , Cin) 

Z0= x0 + y0’ + Cin = 1 + 0 + 1(selector)= 0  [& 1-carried to next bit Full Adder]

Z1= x1 + y1’ + Cin = 1 + 0 + 1(carry)     = 0 [& 1-carried to next bit Full Adder]

Z2= x2 + y2’ + Cin = 1 + 0 + 1(carry)     = 0 [& 1-carried to next bit Full Adder]

Z3= x3 + y3’ + Cin = 0 + 1 + 1(carry)     = 0 [& 1-carried to next bit Full Adder]

Z4= last carry = 1 (note: we ignore last bit (Z4) in subtract operation & it have to = 1)

when [Z4 = 1] that means  [X>=Y] 

 ∴  Z = 0000