Encrypting your link and protect the link from viruses, malware, thief, etc! Made your link safe to visit.

 What is the advantage to use constant array for Lookup Table? Why FIXED-LENGTH lookup table often used for MCU programming?

ANSWER

  • Generally, lookup table is used to read data and as a result, we almost write to a lookup table and this is the mistake we do. So in order to help to catch the unintended writes to lookup table, one should always declare array as constant.The const(used for constant array) keyword is to store the table in the flash memory, not in the RAM. This is because RAM is scarce in microcontrollers. This is the advantage that we have using constant array.
  • In MCU programming, we have less space when compared to other programmings and for using dynamic lookup table, we need much space. So we often use Fixed length look up table for MCU programming.

Hope this helps.

ST

Search This Blog

Labels

Report Abuse

Question: A 250-V, 4-pole, wave-wound d.c. series motor has 782 conductors on itsarmature. It has armature and series field resistance of 0.75 ohm. The motor takesa current of 40 A. Estimate its speed and gross torque developed if it has a flux per pole of 25 mWb Answer: Step 1 Mechanical Engineering homework question answer, step 1, image 1 Mechanical Engineering homework question answer, step 1, image 2 Step 2 Mechanical Engineering homework question answer, step 2, image 1 Step 3 Mechanical Engineering homework question answer, step 3, image 1

Question: A 250-V, 4-pole, wave-wound d.c. series motor has 782 conductors on itsarmature. It has armature and series field resistance of 0.75 ohm. The motor takesa current of 40 A. Estimate its speed and gross torque developed if it has a flux per pole of 25 mWb Answer: Step 1 Step 2 Step 3

QUESTION 6 (a) The bar shown in Figure Q2(a) is subjected to tensile load of 150 Kn. If the stress in the middle portions is limited to 160 N/mm², determine the diameter of the middle portion. Find also the length of the middle portion if the total elongation of the bar is to be 0.25 mm. E E = 2.0 + 105N/mm². (12 marks) 150 KN 10 cm DIA 10 cm DIA 150 KN 45 cm Figure Q6(a) (b) A brass bar, having cross-section area of 900 mm², is subjected to axial forces as shown in Figure Q2(b), in which AB = 0.6 m, BC = 0.8 m, and CD = 1.0 m. Find the total elongation of the bar. E = 1.0 + 105N/mm2 . (8 marks) Page 4 of 5 B D 40 KN 70 KN 20 KN 10 KN Figure Q6(b) (TOTAL = 20 MARKS)

  Question: Show transcribed image text Answer:

Contributors