Does fmodf() cause a hardfault in stm32? - c

Question

I am trying to create a modulated waveform out of 2 sine waves.
To do this I need the modulo(fmodf) to know what amplitude a sine with a specific frequency(lo_frequency) has at that time(t). But I get a hardfault when the following line is executed:
j = fmodf(2 * PI * lo_frequency * t, 2 * PI);
Do you have an idea why this gives me a hardfault ?
Edit 1:
I exchanged fmodf with my_fmodf:
float my_fmodf(float x, float y){
if(y == 0){
return 0;
}
float n = x / y;
return x - n * y;
}
But still the hardfault occurs, and when I debug it it doesn't even jump into this function(my_fmodf).
Heres the whole function in which this error occurs:
int* create_wave(int* message){
/* Mixes the message signal at 10kHz and the carrier at 40kHz.
* When a bit of the message is 0 the amplitude is lowered to 10%.
* When a bit of the message is 1 the amplitude is 100%.
* The output of the STM32 can't be negative, thats why the wave swings between
* 0 and 256 (8bit precision for faster DAC)
*/
static int rf_frequency = 10000;
static int lo_frequency = 40000;
static int sample_rate = 100000;
int output[sample_rate];
int index, mix;
float j, t;
for(int i = 0; i <= sample_rate; i++){
t = i * 0.00000001f; // i * 10^-8
j = my_fmodf(2 * PI * lo_frequency * t, 2 * PI);
if (j < 0){
j += (float) 2 * PI;
}
index = floor((16.0f / (lo_frequency/rf_frequency * 0.0001f)) * t);
if (index < 16) {
if (!message[index]) {
mix = 115 + sin1(j) * 0.1f;
} else {
mix = sin1(j);
}
} else {
break;
}
output[i] = mix;
}
return output;
}
Edit 2:
I fixed the warning: function returns address of local variable [-Wreturn-local-addr] the way "chux - Reinstate Monica" suggested.
int* create_wave(int* message){
static uint16_t rf_frequency = 10000;
static uint32_t lo_frequency = 40000;
static uint32_t sample_rate = 100000;
int *output = malloc(sizeof *output * sample_rate);
uint8_t index, mix;
float j, n, t;
for(int i = 0; i < sample_rate; i++){
t = i * 0.00000001f; // i * 10^-8
j = fmodf(2 * PI * lo_frequency * t, 2 * PI);
if (j < 0){
j += 2 * PI;
}
index = floor((16.0f / (lo_frequency/rf_frequency * 0.0001f)) * t);
if (index < 16) {
if (!message[index]) {
mix = (uint8_t) floor(115 + sin1(j) * 0.1f);
} else {
mix = sin1(j);
}
} else {
break;
}
output[i] = mix;
}
return output;
}
But now I get the hardfault on this line:
output[i] = mix;
EDIT 3:
Because the previous code contained a very large buffer array that did not fit into the 16KB SRAM of the STM32F303K8 I needed to change it.
Now I use a "ping-pong" buffer where I use the callback of the DMA for "first-half-transmitted" and "completly-transmitted":
void HAL_DAC_ConvHalfCpltCallbackCh1(DAC_HandleTypeDef * hdac){
HAL_GPIO_WritePin(GPIOB, GPIO_PIN_3, GPIO_PIN_SET);
for(uint16_t i = 0; i < 128; i++){
new_value = sin_table[(i * 8) % 256];
if (message[message_index] == 0x0){
dac_buf[i] = new_value * 0.1f + 115;
} else {
dac_buf[i] = new_value;
}
}
}
void HAL_DAC_ConvCpltCallbackCh1 (DAC_HandleTypeDef * hdac){
HAL_GPIO_WritePin(GPIOB, GPIO_PIN_3, GPIO_PIN_RESET);
for(uint16_t i = 128; i < 256; i++){
new_value = sin_table[(i * 8) % 256];
if (message[message_index] == 0x0){
dac_buf[i] = new_value * 0.1f + 115;
} else {
dac_buf[i] = new_value;
}
}
message_index++;
if (message_index >= 16) {
message_index = 0;
// HAL_DAC_Stop_DMA (&hdac1, DAC_CHANNEL_1);
}
}
And it works the way I wanted:
But the frequency of the created sine is too low.
I cap at around 20kHz but I'd need 40kHz.
I allready increased the clock by a factor of 8 so that one is maxed out:
.
I can still decrease the counter period (it is 50 at the moment), but when I do so the interrupt callback seems to take longer than the period to the next one.
At least it seems so as the output becomes very distorted when I do that.
I also tried to decrease the precision by taking only every 8th sine value but
I cant do this any more because then the output does not look like a sine wave anymore.
Any ideas how I could optimize the callback so that it takes less time ?
Any other ideas ?

Answer 1

Does fmodf() cause a hardfault in stm32?
It is other code problems causing the hard fault here.
Failing to compile with ample warnings
Best code tip: enable all warnings. #KamilCuk
Faster feedback than Stackoverflow.
I'd expect something like below on a well enabled compiler.
return output;
warning: function returns address of local variable [-Wreturn-local-addr]
Returning a local Object
Cannot return a local array. Allocate instead.
// int output[sample_rate];
int *output = malloc(sizeof *output * sample_rate);
return output;
Calling code will need to free() the pointer.
Out of range array access
static int sample_rate = 100000;
int output[sample_rate];
// for(int i = 0; i <= sample_rate; i++){
for(int i = 0; i < sample_rate; i++){
...
output[i] = mix;
}
Stack overflow?
static int sample_rate = 100000; int output[sample_rate]; is a large local variable. Maybe allocate or try something smaller?
Advanced: loss of precision
A good fmodf() does not lose precision. For a more precise answer consider double math for the intermediate results. An even better approach is more involved.
float my_fmodf(float x, float y){
if(y == 0){
return 0;
}
double n = 1.0 * x / y;
return (float) (x - n * y);
}
Can I not use any function within another ?
Yes. Code has other issues.

Answer 2

1 value every 10uS makes only 100kSPS whis is not too much for this macro. In my designs I generate > 5MSPS signals without any problems. Usually I have one buffer and DMA in circular mode. First I fill the buffer and start generation. When the half transmition DMA interrupt is trigerred I fill the first half of the buffer with fresh data. The the transmition complete interrupt is trigerred I fill the second half and this process repeats all over again.