How come similar sentence take different time to execute?

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How come similar sentence take different time to execute?

(11 posts) (3 voices)

Started 4 years ago by gulu2065
Latest reply from mbolivar

gulu2065
Member

I want to use the following sentence to generate several bits to configure my chip.
As planned, Pin 4 should output a 11001 bits synchronously with Pin 5, which is a CLK.
However, when I measure it on the scope, I find it take different time for each command and I get the five bits with different intervals.
Any suggestions?

void setup ()
{
pinMode(D4, OUTPUT);
pinMode(D5, OUTPUT);
}
void loop(){
myloop:
GPIOB_BASE->BSRR = BIT(6) | BIT(5);
GPIOB_BASE->BSRR = BIT(6+16) | BIT(5);
GPIOB_BASE->BSRR = BIT(6) | BIT(5);
GPIOB_BASE->BSRR = BIT(6+16) | BIT(5);
GPIOB_BASE->BSRR = BIT(6) | BIT(5+16);
GPIOB_BASE->BSRR = BIT(6+16) | BIT(5+16);
GPIOB_BASE->BSRR = BIT(6) | BIT(5+16);
GPIOB_BASE->BSRR = BIT(6+16) | BIT(5+16);
GPIOB_BASE->BSRR = BIT(6) | BIT(5);
GPIOB_BASE->BSRR = BIT(6+16) | BIT(5);
goto myloop;
}
Thanks a lot!

Posted 4 years ago #
gbulmer
Moderator
There are two interrupts, systick and USB running which may make the value jitter. You can disable them.

The delay between
```
GPIOB_BASE->BSRR = BIT(6+16) | BIT(5);
GPIOB_BASE->BSRR = BIT(6) | BIT(5);
```
will be different to
```
GPIOB_BASE->BSRR = BIT(6+16) | BIT(5);
goto myloop;
...
myloop:
GPIOB_BASE->BSRR = BIT(6) | BIT(5);
```
I can't think of much you can do about that.

SPI would be more accurate if you can get the clock speed correct.

(full disclosure: I am not a member of LeafLabs staff)
Posted 4 years ago #
gulu2065
Member

I find even this basic repeat do not produce a regular signal!(The same time delay between each bits) Why is that??
void setup ()
{
pinMode(D4, OUTPUT);
pinMode(D5, OUTPUT);
}
void loop(){
GPIOB_BASE->BSRR = BIT(6) | BIT(5);
GPIOB_BASE->BSRR = BIT(6+16) | BIT(5+16);
GPIOB_BASE->BSRR = BIT(6) | BIT(5);
GPIOB_BASE->BSRR = BIT(6+16) | BIT(5+16);
GPIOB_BASE->BSRR = BIT(6) | BIT(5);
GPIOB_BASE->BSRR = BIT(6+16) | BIT(5+16);
GPIOB_BASE->BSRR = BIT(6) | BIT(5);
GPIOB_BASE->BSRR = BIT(6+16) | BIT(5+16);
GPIOB_BASE->BSRR = BIT(6) | BIT(5);
GPIOB_BASE->BSRR = BIT(6+16) | BIT(5+16);
}

Posted 4 years ago #
gbulmer
Moderator
The overhead for calling loop() is quite significant compared to the time between two GPIOB_BASE->BSRR = ...

There is really a function called main() which is roughly:
```
int main(...) {
    ...
    setup();

    while (1) loop();
}
```
That while and call of loop() will be much slower than the goto, or:
```
while (1) {
    GPIOB_BASE->BSRR = BIT(6) | BIT(5);
    GPIOB_BASE->BSRR = BIT(6+16) | BIT(5+16);
    GPIOB_BASE->BSRR = BIT(6) | BIT(5);
    GPIOB_BASE->BSRR = BIT(6+16) | BIT(5+16);
    GPIOB_BASE->BSRR = BIT(6) | BIT(5);
    GPIOB_BASE->BSRR = BIT(6+16) | BIT(5+16);
    GPIOB_BASE->BSRR = BIT(6) | BIT(5);
    GPIOB_BASE->BSRR = BIT(6+16) | BIT(5+16);
    GPIOB_BASE->BSRR = BIT(6) | BIT(5);
    GPIOB_BASE->BSRR = BIT(6+16) | BIT(5+16);
}
```
You might adjust the timing between each GPIOB_BASE->BSRR = ... with __asm__("nop\n\t"); as mentioned in one of the other threads. That would allow you to get close to having the jump instructions for the goto or while match the gaps between the GPIOB_BASE->BSRR = ...

SPI is worth looking at.

Look at disabling the interrupts too:

See http://leaflabs.com/docs/lang/api/serialusb.html#lang-serialusb
I think it is USBSerial::end()

http://leaflabs.com/docs/libmaple/api/systick.html?highlight=systick
void systick_disable()
Posted 4 years ago #
gulu2065
Member

gbulmer,
what I mean is the difference between each GPIOB_BASE->BSRR = ... Not related to loop or while().
I mean,
GPIOB_BASE->BSRR = BIT(6) | BIT(5);
set both high and
GPIOB_BASE->BSRR = BIT(6+16) | BIT(5+16);
set both low
...
...
when repeat this, I hope I can see a regular signal with the same delay between the several bits. While the delay seems irregular and do not make sense.
Thanks

Posted 4 years ago #
gbulmer
Moderator
gulu2065 - I have done similar code. There are only two explanations I can think of:
1. other interrupts are casing the signal to jitter.
If you are observing this on a normal oscilloscope, the interrupts will be quite rare (about 1 ms) compared to the signal, so that might not be the cause.

2. The actual instruction sequence might not have enough registers to hold all of the constant values, so the sequence has an extra load register or two between them.
Test that hypothesis with the simpler sequence:
```
while (1) {
    GPIOB_BASE->BSRR = BIT(6) | BIT(5);
    GPIOB_BASE->BSRR = BIT(6+16);
    GPIOB_BASE->BSRR = BIT(6) | BIT(5);
    GPIOB_BASE->BSRR = BIT(6+16);
    GPIOB_BASE->BSRR = BIT(6) | BIT(5);
    GPIOB_BASE->BSRR = BIT(6+16);
    GPIOB_BASE->BSRR = BIT(6) | BIT(5);
    GPIOB_BASE->BSRR = BIT(6+16);
    GPIOB_BASE->BSRR = BIT(6) | BIT(5);
    GPIOB_BASE->BSRR = BIT(6+16) | BIT(5+16);
}
```
Are you checking on a dual trace oscilloscope?
If you are, you might be able to get the 5 cycles (bit 6) of that sequence on one trace, and the other signal (bit 5) on the other.
Then you might see the delay due to the loop on the display on bit 5 trace.
Posted 4 years ago #
gulu2065
Member

gbulmer,
I am a little confused by this
Then you might see the delay due to the loop on the display on bit 5 trace.
Did you want me to compare the two signal (bit(5)and bit (6))? Their produce are the same.

Do you have any other ideas to generate a 1.6 Mhz clock and at the same time generate several bits (less than 20) which is synchronously with the 1.6Mhz?
Thanks

Posted 4 years ago #
gbulmer
Moderator
gulu2065

Did you want me to compare the two signal (bit(5)and bit (6))?

Yes, on a dual trace oscilloscope.

Their produce are the same.

That sequence should toggle bit 6 of port B 5 times, and bit 5 once each time around the while loop.
If it doesn't, there is something going wrong.

To avoid confusion:
```
while (1) {
    GPIOB_BASE->BSRR = BIT(6) | BIT(5);
    GPIOB_BASE->BSRR = BIT(6+16);
    GPIOB_BASE->BSRR = BIT(6);
    GPIOB_BASE->BSRR = BIT(6+16);
    GPIOB_BASE->BSRR = BIT(6);
    GPIOB_BASE->BSRR = BIT(6+16);
    GPIOB_BASE->BSRR = BIT(6);
    GPIOB_BASE->BSRR = BIT(6+16);
    GPIOB_BASE->BSRR = BIT(6);
    GPIOB_BASE->BSRR = BIT(6+16) | BIT(5+16);
}
```
Do you have any other ideas to generate a 1.6 Mhz clock and at the same time generate several bits (less than 20) which is synchronously with the 1.6Mhz?

You could use:
- a similar code sequence, but with all the signal encoded so there is no loop, or
- SPI (running as a slave, with a timer to generate the correct baud rate), or
- a timer and DMA to write bit patterns to the port pins, or
- a table with the values for the port pin values:
```
uint32 pinValues[] = {
    BIT(6) | BIT(5);
    BIT(6+16);
    BIT(6) | BIT(5);
    BIT(6+16);
    BIT(6) | BIT(5);
    BIT(6+16);
    BIT(6) | BIT(5);
    BIT(6+16);
    BIT(6) | BIT(5);
    BIT(6+16) | BIT(5+16);
    ...
};
...
for (int i=0; i<sizeof(pinValues[])/sizeof(uint32); ++i) {
    GPIOB_BASE->BSRR = pinValues[i];
}
```
Edit: I made a mistake, the types should be uint32, and not uint16. I have corrected that in the code above.

(full disclosure: I am not a member of LeafLabs staff)
Posted 4 years ago #
mbolivar
LeafLabs

- a table with the values for the port pin values:

I have used this approach on Maple. With systick and USB disabled, it results in extremely regular timing.

However, if all you're doing is toggling a pin at a predefined rate, then PWM is fastest. See the forum posts referenced in the FAQ:

http://leaflabs.com/docs/faq.html#how-do-i-write-to-a-pin-at-high-speed

Posted 4 years ago #
gbulmer
Moderator

mbolivar - AFAIK, the program needs to generate a 1.6MHz clock signal, and read data coming in from a pin at that rate.

I think this may be exactly an SPI-style problem, but I also think an STM32F has just about enough raw horse-power to do it in software. It might migrate to SPI, or even SPI+DMA.

(full disclosure: I am not a member of LeafLabs staff)

Posted 4 years ago #
mbolivar
LeafLabs

Ah, good point.

Posted 4 years ago #

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