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Chapter 04

Developing a Utility Library for the Propeller PCB

Page 9

Questions

Q14 Will the following code produce the same effect as previous code?
Suppose if the Propeller is rotating at a constant speed of RPM = 1200
It will make 20 number of turns in a second (1200/60 = 20)
and to move one degree it will take 137.888889 micro seconds
Assume that time taken to execute the LOOPING instructions is negligible.
What pattern the following code displays/renders:

float D = 137.888889;

unsigned int onSequence = 0x5555; // 0101 0101 0101 0101 Even LEDs are ON

int main()
{
	int i;
	while(1)
	{
		for(i=0; i< 90; i++) // First quadrant 
		{	switch_LEDs(onSequence, RED); 
			myDelay(D);
		}

		for(i=0; i< 90; i++) // Second quadrant  
		{	switch_LEDs(onSequence, BLUE); 
			myDelay(D);
		}

		for(i=0; i< 90; i++) // Third quadrant 
		{	switch_LEDs(onSequence, GREEN); 
			myDelay(D);
		}

		for(i=0; i< 90; i++) // Forth quadrant 
		{	switch_LEDs(onSequence, WHITE); 
			myDelay(D);
		}

	} // while(1)
} // main()

Imagine: What do you see ? Is it same as previous ? Scroll below to check your answer
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Answer: No change pattern remains the same as previous solution
To move one degree it will take 137.888889 micro seconds
To move one degree it will take D amount of time
To move one 90 degree it will take D x 90 amount of time
Since onSequence = 0x5555; // 0101 0101 0101 0101 Even LEDs are ON in RED for first 90° and
for next 90° the same LEDs are ON in BLUE color,
for next 90° the same LEDs are ON in GREEN color,
for next 90° the same LEDs are ON in WHITE color,

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