Example 1:
   ahb  freq. 48MHz
   apb1 freq. 24MHz i.e. a /2 prescaler ahb -> apb1
   apb2 freq. 48MHz i.e. same as ahb freq.

   tim5 is on apb1 which has a /2 prescaler => inp.freq. to timer is 2x
   apb1 freq.

   tim1 is on apb2 which has no prescaling => inp.freq. to timer is 1x
   apb2 freq.

   which means that both tim1 (apb2) and tim5 (apb1) has the same
   input clock frequency.

   tim5 has an input freq. 2*apb1 = 2*24MHz = 48MHz
   48 = 6*8, so with a prescaler of 6 will get you a timer input freq. of
   8MHz and an integer ts = 125ns

   (If you used a prescaler of 3, the inp. freq. would be 16MHz and the
   ts would be 62.5ns wich isn't integer, so no luck with that with
   theese routines wich has ns as the smallest resolution.)

   Now if timer inp. freq. = 8MHz, then by dividing by 8000 you get 1kHz
   so we use time_steps = 8000 to that end, and time_cnth will then
   have a timestep of 1ms giving you a variable with the time in ms.
   which can be handy.

   If a counters input freq. is f, then the first bit of the counter
   will change att half the freq., i.e. f/2, the second bit f/4, third
   f/8 etc.

   So by doing:

   int main(void) {
     // set sysclk = 48MHz, ahb = sysclk, apb1 = ahb/2, apb2 = ahb
     // set GPIO_CRH(GPIOE) to all CNF_OPP50
     time_init(5, 6, 8000, 125, 2);
     while (1) {
       time_get();
       // "<< 8" so we get the value in the high byte of gpioe
       GPIO_ODR(GPIOE) = time_cnth << 8;
   }

   you can measure that PE8 has a freq. of 500Hz, PE9 250Hz, PE10 125Hz etc.
   since time_cnth is incremented at 1kHz.

   Doning so I got PE8 freq. to 500.05 kHz wich is in the 100ppm region.

   If instead you do:

   while (1) {
    GPIO_ODR(GPIOE) = (TIM_CNT(time_timer) & 0x0ff) << 8;
   }

   You'll see that each bit has a period time (T):
     bit pin   T      f
      7  PE15 32us  ~31kHz
      6  PE14 16us  ~62kHz
      5  PE13  8us  125kHz
      4  PE12  4us  250kHz
      3  PE11  2us  500kHz
      2  PE10  1us  1MHz though irregular
      1  PE9  500ns 2MHz though irregular
      0  PE8  1.75us ~570kHz

   in bit 1 and 2 you see that the mcu isn't quick enough to read the
   counter and set the output bits. In bit 0 you might be fooled into
   thinking that the counter is at the wrong frequency.

   To verify that the sysclk is 48MHz, you can acess the MCO signal with:
   GPIO_CRH(GPIOA) |= GPIOCNF( CNF_APP50,  8); //  PA8  MCO
   and setting the MCO bits in RCC_CFGR to output sysclk, pllclk/2 or what
   you want to check.

   To check time_cntl, do:

   while (1) {
     time_get();
     GPIO_ODR(GPIOE) = (time_cntl & 0x0ff) << 8;
   }

   and see what pins have a correct freq. Above I set the shift to 2,
   so hopefully we will see a T of 1us (1MHz) in bit 0, but I get 9us,
   bit 1 I get 2us. In all bits you'll see jitter. Soo, time_cntl is
   useful to measure times greater than say 5us. To get better
   resolution you have to use hw triggers.