PwmOut
Example 1
The on-board LEDs can be dimmed via 4 of the 6 PWM found in LPC1768
#include "mbed.h"
PwmOut led1(LED1);
PwmOut led2(LED2);
PwmOut led3(LED3);
PwmOut led4(LED4);
int main()
{
while (1) {
for (float f = 0.0; f < 1.0; f += 0.001) {
led1 = f;
led2 = f*f;
led3 = (f+1.0)/2.0;
led4 = 1.0 - f;
wait(0.01);
}
}
}
Example 2
The changes in light intensity is a bit awkward. Let's smooth it using sinusoidal function
It seems that M_PI is not defined in the standard math.h header.
We need to ensure that the value (duty cycle) to PwmOut is positive. As sinusoidal function gives value between -1 to 1, we just need to add 1.0 to it.
#include "mbed.h"
#include <math.h>
#ifndef M_PI
#define M_PI 3.1415
#endif
PwmOut led1(LED1);
PwmOut led2(LED2);
PwmOut led3(LED3);
PwmOut led4(LED4);
int main()
{
float f = 0.0;
while (1) {
// Add 1.0 to ensure all values are positive
led1 = sin( (f ) * M_PI / 180.0) + 1.0;
led2 = sin( (f + 90.0) * M_PI / 180.0) + 1.0;
led3 = sin( (f + 180.0) * M_PI / 180.0) + 1.0;
led4 = sin( (f + 270.0) * M_PI / 180.0) + 1.0;
f += 1.0;
wait(0.01);
}
}
Information
Memory usage of the code above:
| Flash (kB) | RAM (kB) | |
|---|---|---|
| Code | 0.8 | 0.0 |
| Libraries | 15 | 0.3 |
| Total | 15 | 0.3 |
Let's commented out the wait(0.01) and time the program by letting it to display 1000 sets of value.
#include "mbed.h"
#include <math.h>
#ifndef M_PI
#define M_PI 3.1415
#endif
int main()
{
Timer t;
int count = 0;
printf("Timer start!\n");
t.start();
PwmOut led1(LED1);
PwmOut led2(LED2);
PwmOut led3(LED3);
PwmOut led4(LED4);
float f = 0.0;
while (count < 1000) {
// Add 1.0 to ensure all values are positive
led1 = sin( (f ) * M_PI / 180.0) + 1.0;
led2 = sin( (f + 90.0) * M_PI / 180.0) + 1.0;
led3 = sin( (f + 180.0) * M_PI / 180.0) + 1.0;
led4 = sin( (f + 270.0) * M_PI / 180.0) + 1.0;
f += 1.0;
count++;
//wait(0.01);
}
t.stop();
printf("The time taken was %f seconds\n", t.read());
}
Information
Memory usage of the code above:
| Flash (kB) | RAM (kB) | |
|---|---|---|
| Code | 0.8 | 0.0 |
| Libraries | 18 | 0.3 |
| Total | 18 | 0.3 |
The output of the program:
Timer start! The time taken was 0.140636 seconds
Now let's precompute the values to be display and store the results in an array. Let choose 256 records for each LED.
#include "mbed.h"
#include <math.h>
#ifndef M_PI
#define M_PI 3.1415
#endif
#define STP 256 /* Step */
#define LED 4 /* Number of LEDs that use PWM */
int main()
{
Timer t;
int count = 0;
printf("Timer start!\n");
t.start();
PwmOut Leds[] = {(LED1), (LED2), (LED3), (LED4)};
float y[STP][LED]; // Store all values
float res = 360 / STP; // Resolution (degree per step)
float x = 0.0; // Current x
float ph = 360 / LED; // Phase different btw two successive LEDs
for (int i = 0; i < STP; i++) {
for (int j = 0; j < LED; j++) {
y[i][j] = sin( (x + ph * j) * M_PI / 180.0) + 1.0;
}
x += res; // increment x by each resolution
}
while (count < 1000) {
for (int i = 0; i < STP; i++) {
for (int j = 0; j < LED; j++) {
Leds[j] = y[i][j];
}
count++;
//wait(0.01);
}
}
t.stop();
printf("The time taken was %f seconds\n", t.read());
}
Information
Memory usage of the code above:
| Flash (kB) | RAM (kB) | |
|---|---|---|
| Code | 0.8 | 0.0 |
| Libraries | 18 | 0.3 |
| Total | 19 | 0.3 |
The output of the program:
Timer start! The time taken was 0.045611 seconds
There is a slight increase in memory usage to store the value but there is a great improvement in speed.
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