Fetch and Decode




Initially, the program counter PC is loaded with the address of the first instruction in the program. The sequence counter SC is cleared to 0, providing a decoded timing signal To. After each clock pulse, SC is incremented by one, so that the timing signals go through a sequence T0, T1, T2, and so on. The rnicrooperations for the fetch and decode phases can be specified by the following register transfer statements.

T0: AR <- PC

T,: IR <-M[AR], PC <- PC + 1

T2: D0, ••• , D7 <-Decode IR(12-14), AR <---IR(0-11), 1 <---IR(lS)

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Since only AR is connected to the address inputs of memory, it is necessary to transfer the address from PC to AR during the clock transition associated with timing signal T0• The instruction read from memory is then placed in the instruction register IR with the clock transition associated with timing

Fetch and Decode

 

 

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signal T1• At the same time, PC is incremented by one to prepare it for the address of the next instruction in the program. At time T2, the operation code in IR is decoded, the indirect bit is transferred to flip-flop I, and the address part of the instruction is transferred to AR . Note that SC is incremented after each clock pulse to produce the sequence To, T1, and T2•

Figure 5-8 shows how the first two register transfer statements are implemented in the bus system. To provide the data path for the transfer of PC to AR we must apply timing signal T0 to achieve the following connection:

1. Place the content of PC onto the bus by making the bus selection inputs 525150 equal to 010. 2. Transfer the content of the bus to AR by enabling the LD input of AR .

The next clock transition initiates the transfer from PC to AR since T0 = 1. In a-d'CY" ,tJ' .it is necessary to use timing signal T1 to provide the following connections in the bus system.

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1. Enable the read input of memory.

2. Place the content of memory onto the bus by making 525150 = Ill.

3. Transfer the content of the bus to IR by enabling the LD input of _ JR.

4. Increment PC by enabling the INR input of PC.

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The next clock transition initiates the read and increment operations since T, = 1.

Figure 5-8 duplicates a portion of the bus system and shows how T0 and T, are connected to the control inputs of the registers, the memory, and the bus selection inputs. Multiple input OR gates are included in the diagram because there are other control functions that will initiate similar operations.

 

 

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Frequently Asked Questions

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Ans: A program residing in the memory unit of the computer consists of a sequence of instructions. The program is executed in the computer by going through a cycle for each instruction. Each instruction cycle in turn is subdivided into a sequence of subcycles or phases. In the basic computer each instruction cycle consists of the following phases: view more..
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Ans: The last three waveforms in Fig. 5-7 show how SC is cleared when D3T4 = I. Output D3 from the operation decoder becomes active at the end of timing signal T2• When timing signal T4 becomes active, the output of the AND gate that implements the control function D3T4 becomes active. This signal is applied to the CLR input of SC. view more..
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Ans: The timing for all registers in the basic computer is controlled by a master clock generator. The clock pulses are applied to all flip-flops and registers in the system, including the flip-flops and registers in the control unit. The clock pulses do not change the state of a register unless the register is enabled by view more..
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Ans: Initially, the program counter PC is loaded with the address of the first instruction in the program. The sequence counter SC is cleared to 0, providing a decoded timing signal To. After each clock pulse, SC is incremented by one, so that the timing signals go through a sequence T0, T1, T2, and so on. The rnicrooperations for the fetch and decode phases can be specified by the following register transfer statements. view more..
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Ans: The timing signal that is active after the decoding is T3• During time T,, the control unit determines the type of instruction that was just read from memory. The flowchart of Fig. 5-9 presents an initial configuration for the instruction cycle and shows how the control determines the instruction type after the decoding. The three possible instruction types available in the basic computer are specified in Fig. 5-5. view more..
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Ans: Register-reference instructions are recognized by the control when 07 = 1 and I = 0. These instructions use bits 0 through 11 of the instruction code to specify one of 12 instructions. These 12 bits are available in IR(0-11). They were also transferred to AR during time T2• view more..
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Ans: In order to specify the rnicrooperations needed for the execution of each instruction, it is necessary that the function that they are intended to perform be defined precisely. Looking back to Table 5-2, where the instructions are listed, we find that some instructions have an ambiguous description. This is because the explanation of an instruction in words is usually lengthy, and not enough space is available in the table for such a lengthy explanation. We will now show that the function of the memory-reference instructions can be defined precisely by means of register transfer notation. view more..
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Ans: This is an instruction that performs the AND logic operation on pairs of bits in AC and the memory word specified by the effective address. The result of view more..
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Ans: This instruction adds the content of the memory word specified by the effective address to the value of AC. The sum is transferred into AC and the output carry C,., is transferred to the E (extended accumulator) flip-flop. The rnicrooperations needed to execute this instruction are view more..
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Ans: This instruction transfers the memory word specified by the effective address to AC. The rnicrooperations needed to execute this instruction are view more..
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Ans: This instruction stores the content of AC into the memory word specified by the effective address. Since the output of AC is applied to the bus and the data input of memory is connected to the bus, we can execute this instruction with one microoperation: view more..
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Ans: This instruction is useful for branching to a portion of the program called a subroutine or procedure. When executed, the BSA instruction stores the address of the next instruction in sequence (which is available in PC) into a memory location specified by the effective address. The effective address plus one is then transferred to PC to serve as the address of the first instruction in the subroutine. This operation was specified in Table 5-4 with the following register transfer: view more..
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Ans: The BSA instruction performs the function usually referred to as a subroutine call. The indirect BUN instruction at the end of the subroutine performs the function referred to as a subroutine return. In most commercial computers, the return address associated with a subroutine is stored in either a processor view more..
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Ans: This instruction increments the word specified by the effective address, and if the incremented value is equal to 0, PC is incremented by 1. The programmer usually stores a negative number (in 2's complement) in the memory word. As this negative number is repeatedly incremented by one, it eventually reaches the value of zero. At that time PC is incremented by one in order to skip the next instruction in the program. view more..
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Ans: A computer can serve no useful purpose unless it communicates with the external environment. Instructions and data stored in memory must come from some input device. Computational results must be transmitted to the user through some output device. Commercial computers include many types of view more..
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Ans: The input register INPR consists of eight bits and holds an alphanumeric input information. The 1-bit input flag FGI is a control flip-flop. The flag bit is view more..
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Ans: The final flowchart of the instruction cycle, including the interrupt cycle for the basic computer, is shown in Fig. 5-15. The interrupt flip-flop R may be set at any time during the indirect or execute phases. Control returns to timing signal T0 after SC is cleared to 0. If R = 1, the computer goes through an interrupt cycle. If R = 0, the computer goes through an instruction cycle. view more..
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Ans: 1. A memory unit with 4096 words of 16 bits each 2. Nine registers: AR, PC, DR, AC, IR, TR, OUTR, INPR, and SC 3. Seven flip-flops: I, S, E, R, lEN, FGI, and FGO 4. Two decoders: a 3 x 8 operation decoder and a 4 x 16 timing decoder view more..




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