Re: Delay a pulses freq. by x% (5v, 50%, variable frequency)



On Thu, 09 Oct 2008 17:49:26 -0500, John Fields
<jfields@xxxxxxxxxxxxxxxxxxxxx> wrote:

On Thu, 9 Oct 2008 14:08:06 -0700 (PDT), got2go@xxxxxxx wrote:

On Oct 9, 1:45 pm, John Fields <jfie...@xxxxxxxxxxxxxxxxxxxxx> wrote:
On Thu, 9 Oct 2008 11:56:17 -0700 (PDT), got...@xxxxxxx wrote:
Hello,

I'm looking for advise on how to delay a pulse that may vary in
frequency.

---
Delay it with respect to what?
---

Would like to delay its frequency by a specific percentage (ideally
adjustable delay percent).

---
You can the increase the pulse frequency, and you can lower it, but you
can't "delay it" without respect to a certain instant in time.

Are you perhaps looking for a way to vary the width of the pulse as its
frequency varies?
---

Are there any simple circuit schematics I could follow to do this ?

---
Maybe, if you can describe what you want, or your application,
accurately.

JF

Hi John,

Sorry for the lack of details here.

Delay it in respect to the original frequency.
Basically, decrease the frequency by a given percent.

Basically, if I have a pulse train at 500Hz, I would like to be able
to change that by, say, 10%, so the output would then be 450Hz.

And, as the frequency of that pulse train changes coming into the
circuit, it would always output 10% lower than the incoming frequency.

I hope that explains it a little better.

---
Yes, it does, but there's still no information on the frequency range of
the input signal, the range of frequency translation, and how accurate
you need the translated output to be.

I assume that the "50" in the subject line refers to the duty cycle of
the input and output; if so, how close does the output have to be to
50%?

BTW, what you want to do is called "frequency translation" or "frequency
shifting", not pulse delay. :-)

Anyway, for a 50% duty cycle in and a 10% translation down, input here's
one way to do it:

1. Use an up-counter to continuously determine the half-periods of the
incoming signal.

2. As each edge comes in, store the accumulated count in a register
(let's call it CNT1) and reset the counter so it'll start counting
again from zero.

3. Divide the contents of CNT1 by ten, add that number to CNT1 and store
the sum in CNT2.

4. Load a down-counter with the contents of CNT2 and have it count down
using the same clock as the up-counter. When it gets to zero have it
toggle a "D" type flip-flop and reload itself with the contents of
CNT2


That way CNT1 will refresh itself on its own terms, as will CNT2, the
down-counter will generate half cycles which will be 10% longer than the
input's and the flip-flop will automatically assemble a square wave
which will be 10% lower in frequency than the input signal.

---

Even simpler:


Vin>--+---------A _
| EXNOR Y--[R]--+--[Y = MX + B]--+--[VCO]--+-->Vout
+--[R]-+--B | | |
| | | |
[C] [C] | |
| | | |
GND>---------+-------------+----------------+---------+-->GND

JF
.



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