Activities Manual For Programmable Logic Controllers
5th Edition
ISBN: 9781259682476
Author: Frank Petruzella
Publisher: Mcgraw-hill Higher Education (us)
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Question
Chapter 9, Problem 7RQ
Program Plan Intro
Programmable Logic Controller (PLC):
- PLC is a specialized computer used in industrial segments to control machines and process.
- PLC is program that consists of a set of instructions which resembles the controlling functions needed to perform specific tasks.
- The function of PLC is similar to that of a relay and hence, in an industrial segment a relay is greatly replaced by a suitable PLC.
- When compared to general PC, a PLC is available in small and tiny sizes.
- The basic architecture of a PLC consists of an input-output interface module and a small Central Processing Unit (CPU) that runs by control
programming language.
Explanation of Solution
b.
Benefits of programming:
- It is beneficial to program an immediate input or immediate output instruction near the beginning of a program.
- It is because the I/O data files will have the latest status of input and output at the beginning of the program...
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When the INT instruction executes, what is the first task carried out by the CPU?
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Chapter 9 Solutions
Activities Manual For Programmable Logic Controllers
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Similar questions
- The INT instruction causes the CPU to go through four stages when it is used by a program.arrow_forwardi. Explain briefly, using a flow graphic and an Intel microprocessor, how the combination of two binary integers occurs. ii. Create a pseudocode and code in any of the basic assembly languages for the above flow chart. iii. For the reduction procedure, repeat steps i. and ii. above.arrow_forwardAssembly Language True or False: a. Given the way OR( ... ) works, the most likely outcome is a false result (that is, the value 0), not a true result (that is, the value 1). True or False? b. When the bits of register CL are changed by an HLA programmer, no other register will find its bits changed as a result. True or False?arrow_forward
- Explain in detail 1. What does a time complexity analysis usually count?A. The number of arithmetic and other operations required for the program to runB. The number of seconds required for the program to runC. The number of megabytes required for the program to runD. The number of seconds plus the number of megabytesarrow_forwardInstructions: 1. Construct a simple assembly language program that would satisfy the following conditions: The program should accept a single-digit input, from 0 to 9. Compare the single-digit input to 5. Prompt the user whether the input is below, equal, or above the value 5. Utilize compare and jump instructions. The program should contain at least one (1) procedure that is called and executed. Add comments in the important sections/code segments of your program. Note: It is highly recommended to use the Emu8086 Emulator in writing the program. If the emulator is currently not available, you may utilize any online assembler/compiler/emulator, as long as your program satisfies the six (6) conditions stated above. The instruction set of online assemblers/compilers/emulators may vary. 2. Explain the structure of your program using 4-7 sentences. 3. Include a screenshot of the output of your program. 4. Include in your answer sheet all the assembly language code of your program.arrow_forwardMUL & DIV instruction in Assembly Language Table 1 EAX EBX ECX EDX 10H 20H 2H 0H Refer to Table 1 above for the following tasks.Write your own code in Visual Studio 2019 to execute the instruction, then fill in the blanks with the correct value of the registers. Also, note the changes. **Note: Always start with the initial value of the registers given in Table 1 above for each question (a-f) This means you have to execute only one instruction per debugging session e.g. execute MUL BX and then debug and get the values of the registers and then stop. Then start again with MUL CX. Answer all questions in 32bit format XXXX XXXX e.g. 0000 1000 with only one white space in between a. MUL BX EAX EBX ECX EDX Answer Answer Answer Answer b. MUL CX EAX EBX ECX EDX c. MUL AX EAX EBX ECX EDX d. DIV BX EAX EBX ECX EDX e. DIV CX EAX EBX ECX EDX f. DIV AX…arrow_forward
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