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By Kilian C.T.
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Additional info for Modern Control Technology: Components and Systems
Sample text
To understand this, again examine the (V2 – V1) quantity, paying attention to the signs. In this case, [–2 µV – (–3 µV)] = +1 µV, which is positive. 3(c) is more straightforward. In this case, (V2 – V1) = (–2 µV – 3 µV) = –5 µV, which is clearly negative. Now consider the case where only a single input is required. There are two possibilities: The output will be either in phase or out of phase with the input. 4(a). 4(b). 4 − Single-input, open-loop amplifiers.
Before the program can be executed, it must first be loaded sequentially into memory. The op-code for the first instruction is loaded at the first address location, the op-code for the second instruction is loaded next in line, and so on. 4(b) shows a section of memory with a short program loaded in. The program listing includes the address, op-code, mnemonic, and a brief explanation. (A mnemonic is an English abbreviation of an instruction. 4(b) directs the CPU to get 1 byte of data from input port 01, add 1 to it, and send the result to output port 02.
The inputs are both negative, yet the output is positive. To understand this, again examine the (V2 – V1) quantity, paying attention to the signs. In this case, [–2 µV – (–3 µV)] = +1 µV, which is positive. 3(c) is more straightforward. In this case, (V2 – V1) = (–2 µV – 3 µV) = –5 µV, which is clearly negative. Now consider the case where only a single input is required. There are two possibilities: The output will be either in phase or out of phase with the input. 4(a). 4(b). 4 − Single-input, open-loop amplifiers.