comparing frequency of two clocks

[arant]

Hi Everyone,

Does anyone have any idea of how to compare two reference clocks using
completely digital logic .The problem is something like this

There are two slow ref clocks (Khz) CLKA (rc clock) CLKB (crystal
clock) and a fast sys_clk (pll clock) Mhz range

If freq CLKA > freq CLKB then
op1 = 1;
elsif freq CLKA < freq CLKB op2 = 1
else
op3 = 1;
end if;

Eagerly awaiting some solutions

 

[Ralf Hildebrandt]

arant wrote:

Does anyone have any idea of how to compare two reference clocks using
completely digital logic .The problem is something like this

There are two slow ref clocks (Khz) CLKA (rc clock) CLKB (crystal
clock) and a fast sys_clk (pll clock) Mhz range

If freq CLKA > freq CLKB then
op1 = 1;
elsif freq CLKA < freq CLKB op2 = 1
else
op3 = 1;
end if;

Just an idea - you should think about it, before you use it:

Make two counters, one counting with CLKA and one with CLKB. Let them
count for a while (maybe until counter A has reached hundred or
whatever). Then check the counter values. If the values of the two
counters are equal, the clocks have almost the same frequency. If the
difference is one this may be because of jitter and you should assume,
that the frequencies are even then very similar. If the difference is 2
or greater the frequencies are different enough to decide which clock is
faster. As you count longer the answer becomes more reliable and the
difference of one is more likely because of jitter.


Ralf

 

[jacko]

arant wrote:

Hi Everyone,

Does anyone have any idea of how to compare two reference clocks using
completely digital logic .The problem is something like this

There are two slow ref clocks (Khz) CLKA (rc clock) CLKB (crystal
clock) and a fast sys_clk (pll clock) Mhz range

If freq CLKA > freq CLKB then
op1 = 1;
elsif freq CLKA < freq CLKB op2 = 1
else
op3 = 1;
end if;

frequency is 1/time, so some time has to pass to measure frequency.

if two counters are clocked by clk a and clk b then the difference sign
says which is faster,

to stop over or underflow, the sys clk can be xor gated with respective
clock to either clock both up, or both down depending on the sum being
above or below the centre.

the dynamic difference between the clocks could lead to the need for
infinite precision counters

cheers


> Eagerly awaiting some solutions

 

[Pascal Peyremorte]

arant a écrit :

Hi Everyone,

Does anyone have any idea of how to compare two reference clocks using
completely digital logic .The problem is something like this

There are two slow ref clocks (Khz) CLKA (rc clock) CLKB (crystal
clock) and a fast sys_clk (pll clock) Mhz range

If freq CLKA > freq CLKB then
op1 = 1;
elsif freq CLKA < freq CLKB op2 = 1
else
op3 = 1;
end if;

Eagerly awaiting some solutions

For the two slow clock do :
- start fast counter on positive edge,
- store the counter then clear it on next negative edge of the clock.
Set a flag signifying capture is done.

When both flag are set, reset them and compare the two captured values.

Pascal

 

[arant]

Hi Pascal ,

I think you are suggesting the use of a capcom unit .

One solution that I can think of is running two parallel counters on
the sys_clk for one period of the CLKA (count a) and CLKB (ount b) and
comparing the count values
Two main points :
-------------------------------
Clock Period measurement using a one shot pulse generator
on both the rising and falling edges of the slow clock

Start count on the rising edge pulse and stop and store count on the
falling edge pulse
of the slow clock

if count a = count b then
op3 = 1
elsif count a > count b then
op2 = 1 -- freq CLKB > freq CLKA
else -- freq CLKB < freq CLKA
op1 = 1
end if

By the way the requirement is for an SoC clock managment unit that
generates an interrupt/event to the CPU based on the clock comparison
operation.

Thanks

Pascal Peyremorte wrote:

arant a écrit :
Hi Everyone,

Does anyone have any idea of how to compare two reference clocks using
completely digital logic .The problem is something like this

There are two slow ref clocks (Khz) CLKA (rc clock) CLKB (crystal
clock) and a fast sys_clk (pll clock) Mhz range

If freq CLKA > freq CLKB then
op1 = 1;
elsif freq CLKA < freq CLKB op2 = 1
else
op3 = 1;
end if;

Eagerly awaiting some solutions

For the two slow clock do :
- start fast counter on positive edge,
- store the counter then clear it on next negative edge of the clock.
Set a flag signifying capture is done.

When both flag are set, reset them and compare the two captured values.

Pascal

 

[Thomas Stanka]

Hi,

arant schrieb:

I think you are suggesting the use of a capcom unit .

One solution that I can think of is running two parallel counters on
the sys_clk for one period of the CLKA (count a) and CLKB (ount b) and
comparing the count values
Two main points :
-------------------------------
Clock Period measurement using a one shot pulse generator
on both the rising and falling edges of the slow clock

Start count on the rising edge pulse and stop and store count on the
falling edge pulse
of the slow clock

You should think about your needs. Do you need a statement about the
length of a high pulse or a statement about the clock periode? For a
50% duty cycle there's no difference, else you might need to stop on
the next rising edge.

bye Thomas