birt gs1 128

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So far in this book you have seen the assembler calculator in operation calculating constant-value arguments for instructions such as

movlw (1 << GIE) | (1 << T0IE)

This instruction loads w register with a byte, destined for the INTCON register, which has the GIE (7) and T0IF (5) bits set. In this case, the assembler calculator is used to change bit numbers to actual values to be loaded into a byte. The trick in this statement is knowing that shifting one by the bit number converts the bit number into a constant value that will set the bit when loaded into a register. This is useful and avoids having to gure out what value is used for speci c bits being set. In the preceding example, if this trick had not been used, I would have to remember (or generate on a calculator) that bit 7 being set is the same as adding 128 and that bit 5 being set is the same as adding 32. The result of these two values is 160 decimal or 0x0A0. Using the assembler calculator, I didn t have to worry about any of this. To reset speci c bits, the same trick can be used, but the bits have to be reset, which is done by a bitwise inversion of the bits and then ANDing the result with the current value. XORing the set bit value with 0x0FF accomplishes the bitwise inversion. For example, to clear bits 4, 2, and 1 in the w register, the following instruction could be used:

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andlw 0x0FF ^ ((1 << 4) | (1 << 2) | (1 << 1))

If you were to do this manually, you would have to follow these steps:

1 Calculate the values for bits 4, 2, and 1 being set: (1 << 4) = 16 (1 << 2) = 4 (1 << 1) = 2

(1 << 4) | (1 << 2) | (1 << 1) = 16 | 4 |2 = 22 = 0x016 2 Calculate the inverse (XOR with 0x0FF): 0x0FF ^ 0x016 = 0x0E9 3 Put the value into the andlw instruction: andlw 0x0E9

30. B. Practice is essential to smooth delivery style. 5. References 28. Osborne and Osborne, Public Speaking (New York: Houghton Mif in, 1992). 34. S. Brydon and M. Scott, Between One and Many (CA: May eld, 1994). 10. Interview: C. S. Blanche, Professor of Public Speaking.

If you go through the manual process, you can see that there are more than seven opportunities for you to calculate constant values incorrectly or copy down the wrong value. Avoiding these manual calculations with their inherent opportunities for error is

9 9.1 Flow of cold water could be manipulated directly by the flow controller, while hot water is the product of the outputs of the flow and temperature controllers. 9.2 At 120 lb/min feed rate, oil flow must be infinite; at 80 lb/min, oil flow is reduced to 133 lb/min. At 160 F inlet, oil flow must be 964 lb/min; at 240 F inlet, it is reduced to.150 lb/min. 9.3 Process gain is O.O41 F/(lb/ min) at 100 lb/min feed, and 0.28a F/ (lb/min) at 80 Ib/min, a ratio of almost 7. Process and valve gain products are 16.8k and 40.3k for the same conditions, a ratio of only 2.4. 9.4 & = [llyc(T, - T&]/(l/Ak + 1/2C,). H e a t t r a n s f e r a p p e a r s t o b e linear with flow, but actually [I contains constant t,erms including metal conductivity and condensing film coefficient; in addition, the liquid film coefficient actually varies with the 0.8 power of flow. 9.5 Steam temperature in a drum boiler is controlled principally with spray. The once-through boiler has one less controlled variable (liquid level), which frees a manipulated variable (feedwater flow) for temperature control. 9.6 Coefficient k, = 7.716 X IOeG ft/rpm2, and kz = 0.01 ft/gpm2; N = 3,120 rpm. HHP at 3,120 rpm is 0.632 HP; at 3,600 rpm it is 0.948 HP.

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