Answer :
Sure, let's go through each part of the quiz step-by-step:
### Part A - Meaning of Register Transfer Statements
#### a) [tex]\( \text{PX1}: R2 \leftarrow R1 + \overline{R3} + 1 \)[/tex]
This statement represents the addition of [tex]\( R1 \)[/tex] to the two's complement of [tex]\( R3 \)[/tex]. In essence, taking the one's complement ([tex]\(\overline{R3}\)[/tex]) and adding 1 to it, converts [tex]\( R3 \)[/tex] to its negative form, effectively making this a subtraction operation. Thus;
[tex]\[ R2 \leftarrow R1 + (\overline{R3} + 1) \][/tex]
Which simplifies to:
[tex]\[ R2 \leftarrow R1 - R3 \][/tex]
Therefore, the meaning is:
[tex]\[ R2 \leftarrow R1 - R3 \][/tex]
#### b) [tex]\( R3 \leftarrow M[AR] \)[/tex]
This statement means that the contents of the memory location whose address is held in the Address Register (AR) are being loaded into [tex]\( R3 \)[/tex]. Essentially, we are reading from memory into [tex]\( R3 \)[/tex].
Thus, the meaning is:
[tex]\[ R3 \leftarrow \text{Memory[Address in AR]} \][/tex]
### Part B
#### 1) Store the contents of register 3 in memory location 240
To store the contents of [tex]\( R3 \)[/tex] in memory location 240, we simply place the value in [tex]\( R3 \)[/tex] at this specific memory address. If [tex]\( R3 \)[/tex] contains the value 55, then memory location 240 will contain 55 afterwards.
So, memory[240] = 55.
#### 2) Perform a logical shift right operation on binary number 1100
A logical shift right operation shifts all bits of the binary number to the right by one place. So, for the binary number 1100:
[tex]\[ 1100 \quad(in\; binary) \Rightarrow 0110 \quad(after\; logical\; shift\; right) \][/tex]
The resulting binary number is 0110, which is 6 in decimal.
Therefore, the result of the logical shift right operation is 6.
#### 3) Perform a circular shift left operation on binary number 1101
A circular (or rotate) shift left operation moves the leftmost bit to the rightmost position. For the binary number 1101:
[tex]\[ 1101 \quad(in\; binary) \Rightarrow 1011 \quad(after\; circular\; shift\; left) \][/tex]
The resulting binary number is 1011, which is 11 in decimal.
Therefore, the result of the circular shift left operation is 11.
#### 4) If [tex]\( R1 = 1100 \)[/tex] and [tex]\( R2 = 1010 \)[/tex], perform [tex]\( R3 \leftarrow R1 \: \mathbf{OR} \: R2 \)[/tex]
To perform the bitwise OR operation between [tex]\( R1 \)[/tex] and [tex]\( R2 \)[/tex]:
[tex]\[ R1 = 1100 \][/tex]
[tex]\[ R2 = 1010 \][/tex]
[tex]\[ R3 = R1 \: \mathbf{OR} \: R2 \][/tex]
Bitwise OR operation on each corresponding bit:
[tex]\[ 1100 \, (R1)\][/tex]
[tex]\[ 1010 \, (R2)\][/tex]
[tex]\[ \mathbf{1110} \, (R3)\][/tex]
The resulting binary number is 1110, which is 14 in decimal.
Therefore, the contents of register [tex]\( R3 \)[/tex] after performing [tex]\( R1 \: \mathbf{OR} \: R2 \)[/tex] is 14.
### Part A - Meaning of Register Transfer Statements
#### a) [tex]\( \text{PX1}: R2 \leftarrow R1 + \overline{R3} + 1 \)[/tex]
This statement represents the addition of [tex]\( R1 \)[/tex] to the two's complement of [tex]\( R3 \)[/tex]. In essence, taking the one's complement ([tex]\(\overline{R3}\)[/tex]) and adding 1 to it, converts [tex]\( R3 \)[/tex] to its negative form, effectively making this a subtraction operation. Thus;
[tex]\[ R2 \leftarrow R1 + (\overline{R3} + 1) \][/tex]
Which simplifies to:
[tex]\[ R2 \leftarrow R1 - R3 \][/tex]
Therefore, the meaning is:
[tex]\[ R2 \leftarrow R1 - R3 \][/tex]
#### b) [tex]\( R3 \leftarrow M[AR] \)[/tex]
This statement means that the contents of the memory location whose address is held in the Address Register (AR) are being loaded into [tex]\( R3 \)[/tex]. Essentially, we are reading from memory into [tex]\( R3 \)[/tex].
Thus, the meaning is:
[tex]\[ R3 \leftarrow \text{Memory[Address in AR]} \][/tex]
### Part B
#### 1) Store the contents of register 3 in memory location 240
To store the contents of [tex]\( R3 \)[/tex] in memory location 240, we simply place the value in [tex]\( R3 \)[/tex] at this specific memory address. If [tex]\( R3 \)[/tex] contains the value 55, then memory location 240 will contain 55 afterwards.
So, memory[240] = 55.
#### 2) Perform a logical shift right operation on binary number 1100
A logical shift right operation shifts all bits of the binary number to the right by one place. So, for the binary number 1100:
[tex]\[ 1100 \quad(in\; binary) \Rightarrow 0110 \quad(after\; logical\; shift\; right) \][/tex]
The resulting binary number is 0110, which is 6 in decimal.
Therefore, the result of the logical shift right operation is 6.
#### 3) Perform a circular shift left operation on binary number 1101
A circular (or rotate) shift left operation moves the leftmost bit to the rightmost position. For the binary number 1101:
[tex]\[ 1101 \quad(in\; binary) \Rightarrow 1011 \quad(after\; circular\; shift\; left) \][/tex]
The resulting binary number is 1011, which is 11 in decimal.
Therefore, the result of the circular shift left operation is 11.
#### 4) If [tex]\( R1 = 1100 \)[/tex] and [tex]\( R2 = 1010 \)[/tex], perform [tex]\( R3 \leftarrow R1 \: \mathbf{OR} \: R2 \)[/tex]
To perform the bitwise OR operation between [tex]\( R1 \)[/tex] and [tex]\( R2 \)[/tex]:
[tex]\[ R1 = 1100 \][/tex]
[tex]\[ R2 = 1010 \][/tex]
[tex]\[ R3 = R1 \: \mathbf{OR} \: R2 \][/tex]
Bitwise OR operation on each corresponding bit:
[tex]\[ 1100 \, (R1)\][/tex]
[tex]\[ 1010 \, (R2)\][/tex]
[tex]\[ \mathbf{1110} \, (R3)\][/tex]
The resulting binary number is 1110, which is 14 in decimal.
Therefore, the contents of register [tex]\( R3 \)[/tex] after performing [tex]\( R1 \: \mathbf{OR} \: R2 \)[/tex] is 14.
