ADD to AC




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

DIT,: DR <--- M[AR]

D1T5: AC <--- AC + DR, E <--- C,.,, SC <--- 0

The same two timing signals, T, and T5, are used again but with operation decoder D1 instead of D0, which was used for the AND instruction. After the instruction is fetched from memory and decoded, only one output of the operation decoder will be active, and that output determines the sequence of rnicrooperations that the control follows during the execution of a memory-reference instruction.



Frequently Asked Questions

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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: 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: 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: 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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Ans: he block diagram of the control logic gates is shown in Fig. 5-6. The inputs to this circuit come from the two decoders, the I flip-flop, and bits 0 through 11 of IR. The other inputs to the control logic are: AC bits 0 through 15 to check if AC = 0 and to detect the sign bit in AC( view more..
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Ans: The registers of the computer connected to a common bus system are shown in Fig. 5-4. The control inputs of the registers are LD (load), INR (increment), and CLR (clear). Suppose that we want to derive the gate structure associated with the control inputs of AR. We scan Table 5-6 to find all the statements that change the content of AR view more..
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Ans: The control gates for the seven flip-flops can be determined in a similar manner. For example, Table 5-6 shows that lEN may change as a result of the two instructions ION and !OF. view more..
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Ans: The 16-bit common bus shown in Fig. 5-4 is controlled by the selection inputs S2, S1, and 50• The decimal number shown with each bus input specifies the equivalent binary number that must be applied to the selection inputs in order to select the corresponding register. Table 5-7 specifies the binary numbers for S2S1S0 that select each register. Each binary number is associated with a Boolean variable x1 through x7, corresponding to the gate structure that must be active in order to select the register or memory for the bus. view more..




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