Design of Basic Computer




The basic computer consists of the following hardware components:

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

5. A 16-bit common bus

6. Control logic gates

7. Adder and logic circuit connected to the input of AC

The memory unit is a standard component that can be obtained readily from a commercial source. The registers are of the type shown in

 

 

 

Design of Basic Computer

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

are similar to integrated circuit type 74163. The flip-flops can be either of the D or JK type, as described in Sec. 1-6. The two decoders are standard components similar to the ones presented in Sec. 2-2. The common bus system can be constructed with sixteen 8 x 1 multiplexers in a configuration similar to the one shown in Fig. 4-3. We are now going to show how to design the control logic gates. The next section deals with the design of the adder and logic circuit associated with AC.



Frequently Asked Questions

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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: 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: 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: 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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