How to shift phase by 90 degrees between approx 10 Hz and 10

[Bernhard Kraemer]

Hello,

For an experiment, I need to build a little Lock-In Amplifier an my own
circuit. I do not have an external Lock-In and so this will be the cheaper
and better solution.

The circuit is nearly finished. What lacks is a circuit which should be
controlled by an external oscillator and produces a two output signals,
a TTL signal of the same frequency and another of 90 degrees
phase shifted. The frequency range should be about 10 Hz to 10 KHz (7
decades).

I thought about first producing a sine and a cosine signal which I am
going to convert to TTL afterwards.

In "The Art of Electronics" p 294, if found an interestin circuit using
the AD639. Unfortunately, our official provider Farnell doesn't offer the
IC saying it wouldn't be produced nomore.

There is another circuit on the next page using a lot of resitors and
capacities, but it only works for 2 decades instead of seven and
honestly, I would not even know how to calculate the necessary values

Finally, having tried another couple of ideas, I still don't figure out
how I can solve my problem. This is why I post on this forum. How could
I compose a circuit dephasing either two sinusoidal or digital signals
by 90 degrees (and if possible, without too much costful and placetaking
electronic circuitry)?

I hope someone could help me.

Yours,

Bernhard

 

[John Fields]

On 22 Apr 2005 12:48:25 GMT, Bernhard Kraemer
<bernhard.kramer@lien.uhp-nancy.fr> wrote:

Hello,

For an experiment, I need to build a little Lock-In Amplifier an my own
circuit. I do not have an external Lock-In and so this will be the cheaper
and better solution.

The circuit is nearly finished. What lacks is a circuit which should be
controlled by an external oscillator and produces a two output signals,
a TTL signal of the same frequency and another of 90 degrees
phase shifted. The frequency range should be about 10 Hz to 10 KHz (7
decades).

I thought about first producing a sine and a cosine signal which I am
going to convert to TTL afterwards.

In "The Art of Electronics" p 294, if found an interestin circuit using
the AD639. Unfortunately, our official provider Farnell doesn't offer the
IC saying it wouldn't be produced nomore.

There is another circuit on the next page using a lot of resitors and
capacities, but it only works for 2 decades instead of seven and
honestly, I would not even know how to calculate the necessary values

Finally, having tried another couple of ideas, I still don't figure out
how I can solve my problem. This is why I post on this forum. How could
I compose a circuit dephasing either two sinusoidal or digital signals
by 90 degrees (and if possible, without too much costful and placetaking
electronic circuitry)?

---


+----------------->I
+------------|------------+
| +-----+ | +-----+ |
+--|D Q|---+---|D Q|------->Q
| | | | |
4fOUT---------+--|> | +---|> _| |
| | | | | Q|O-+
| +-----+ | +-----+
| |
+------------+

The "D" type filp-flops are XX74s or 4013s, 4fOUT is 40hZ to 40kHz,
and I and Q ARE YOUR 10Hz to 10kHz outputs, in quadrature.

--
John Fields
Professional Circuit Designer

 

[Tam/WB2TT]

"Tim Shoppa" <shoppa@trailing-edge.com> wrote in message
news:1114175153.882654.83420@f14g2000cwb.googlegroups.com...

What lacks is a circuit which should be
controlled by an external oscillator and produces a two
output signals,
a TTL signal of the same
frequency and another of 90 degrees
phase shifted. The frequency range should be about 10 Hz to
10 KHz (7 decades).

If you allow the external clock to be 4 times the frequency of the
outputs, all you need is a divide-by-four clock generator.

************************************

You have to use the right kind of divider, though. Not two div-2 counters
in tandem.
Here is what you do:
1. Get 2 D flipflops.
2. Connect the two clock leads together, and to the input.
3. Connect the Q of the first stage to the D of the second.
4. Connect the Q* (QBAR) of the second to the D of the first.
5 The outputs are the Q of the first and second stages.

The Div-4 should simplify building the oscillator. In fact, you might want
to consider switching in different prescalers in addition to cover your
range.

Tam
********************

2-times works if you know that the input is a symmetrical square wave.

If you don't have the luxury of a 4x clock, you can make one witha PLL.
40Hz to 40kHz is possible with a single VCO and an edge-sensitive
phase detector.

10Hz to 10kHz is 3 decades, so maybe I misunderstand your 7 decades...
(8 octaves?)

Tim.

 

[Mac]

On Fri, 22 Apr 2005 11:56:54 -0500, John Fields wrote:

On 22 Apr 2005 12:48:25 GMT, Bernhard Kraemer
bernhard.kramer@lien.uhp-nancy.fr> wrote:


Hello,

For an experiment, I need to build a little Lock-In Amplifier an my own
circuit. I do not have an external Lock-In and so this will be the cheaper
and better solution.

The circuit is nearly finished. What lacks is a circuit which should be
controlled by an external oscillator and produces a two output signals,
a TTL signal of the same frequency and another of 90 degrees
phase shifted. The frequency range should be about 10 Hz to 10 KHz (7
decades).

I thought about first producing a sine and a cosine signal which I am
going to convert to TTL afterwards.

In "The Art of Electronics" p 294, if found an interestin circuit using
the AD639. Unfortunately, our official provider Farnell doesn't offer the
IC saying it wouldn't be produced nomore.

There is another circuit on the next page using a lot of resitors and
capacities, but it only works for 2 decades instead of seven and
honestly, I would not even know how to calculate the necessary values

Finally, having tried another couple of ideas, I still don't figure out
how I can solve my problem. This is why I post on this forum. How could
I compose a circuit dephasing either two sinusoidal or digital signals
by 90 degrees (and if possible, without too much costful and placetaking
electronic circuitry)?

---


+----------------->I
+------------|------------+
| +-----+ | +-----+ |
+--|D Q|---+---|D Q|------->Q
| | | | |
4fOUT---------+--|> | +---|> _| |
| | | | | Q|O-+
| +-----+ | +-----+
| |
+------------+

The "D" type filp-flops are XX74s or 4013s, 4fOUT is 40hZ to 40kHz,
and I and Q ARE YOUR 10Hz to 10kHz outputs, in quadrature.

Nice!

--Mac

 

[Bernhard Krämer]

Hello all,

Thank you very much for your suggestions and corrections. I already thought
about multiplying and then dividing the frequency to make use of a circuit
similar to yours.

But after having received your post, I finally figured out that my circuit
would give the same results and even better if I would use 4 x freq instead
of 1 x freq and multiplying it.

Sorry for the " 7 decades error "; I thought about another problem which has
nothing to do with this one while writing the post: another circuitry where
I have to worry about a frequency range from 10 mHz to 10 KHz. And this,
err ..., are only 6 decades, so I was still wrong. I am deeply ashamed
now
Next time, I will concentrate more while writing. Word.

Thank you a lot, I can hardly express how I appreciate your help!

Yours,

Bernhard




John Fields wrote:

On 22 Apr 2005 12:48:25 GMT, Bernhard Kraemer
bernhard.kramer@lien.uhp-nancy.fr> wrote:


Hello,

For an experiment, I need to build a little Lock-In Amplifier an my own
circuit. I do not have an external Lock-In and so this will be the cheaper
and better solution.

The circuit is nearly finished. What lacks is a circuit which should be
controlled by an external oscillator and produces a two output signals,
a TTL signal of the same frequency and another of 90 degrees
phase shifted. The frequency range should be about 10 Hz to 10 KHz (7
decades).

I thought about first producing a sine and a cosine signal which I am
going to convert to TTL afterwards.

In "The Art of Electronics" p 294, if found an interestin circuit using
the AD639. Unfortunately, our official provider Farnell doesn't offer the
IC saying it wouldn't be produced nomore.

There is another circuit on the next page using a lot of resitors and
capacities, but it only works for 2 decades instead of seven and
honestly, I would not even know how to calculate the necessary values

Finally, having tried another couple of ideas, I still don't figure out
how I can solve my problem. This is why I post on this forum. How could
I compose a circuit dephasing either two sinusoidal or digital signals
by 90 degrees (and if possible, without too much costful and placetaking
electronic circuitry)?

---


+----------------->I
+------------|------------+
| +-----+ | +-----+ |
+--|D Q|---+---|D Q|------->Q
| | | | |
4fOUT---------+--|> | +---|> _| |
| | | | | Q|O-+
| +-----+ | +-----+
| |
+------------+

The "D" type filp-flops are XX74s or 4013s, 4fOUT is 40hZ to 40kHz,
and I and Q ARE YOUR 10Hz to 10kHz outputs, in quadrature.

 

[Rich Grise]

On Sat, 23 Apr 2005 11:48:34 +0200, Bernhard Krämer top-posted:

Hello all,

Thank you very much for your suggestions and corrections. I already
thought about multiplying and then dividing the frequency to make use of a
circuit similar to yours.

But after having received your post, I finally figured out that my circuit
would give the same results and even better if I would use 4 x freq
instead of 1 x freq and multiplying it.

Sorry for the " 7 decades error "; I thought about another problem which
has nothing to do with this one while writing the post: another circuitry
where I have to worry about a frequency range from 10 mHz to 10 KHz. And
this, err ..., are only 6 decades, so I was still wrong. I am deeply
ashamed now

Well, you'd better still be, because that's still only three (3) decades.

A "decade" is "times ten". 10 Hz to 10 MHz is six (6) decades.
ten times ten times ten is three decades.
ten times ten times ten times ten times ten times ten is six decades.

Next time, I will concentrate more while writing. Word.

Excellent idea! I wish more would adopt that attitude.

Thank you a lot, I can hardly express how I appreciate your help!

Possibly interestingly, I think I know how you feel.

Cheers!
Rich

John Fields wrote:

On 22 Apr 2005 12:48:25 GMT, Bernhard Kraemer
bernhard.kramer@lien.uhp-nancy.fr> wrote:


Hello,

For an experiment, I need to build a little Lock-In Amplifier an my own
circuit. I do not have an external Lock-In and so this will be the
cheaper and better solution.

The circuit is nearly finished. What lacks is a circuit which should be
controlled by an external oscillator and produces a two output signals,
a TTL signal of the same frequency and another of 90 degrees phase
shifted. The frequency range should be about 10 Hz to 10 KHz (7
decades).

I thought about first producing a sine and a cosine signal which I am
going to convert to TTL afterwards.

In "The Art of Electronics" p 294, if found an interestin circuit using
the AD639. Unfortunately, our official provider Farnell doesn't offer
the IC saying it wouldn't be produced nomore.

There is another circuit on the next page using a lot of resitors and
capacities, but it only works for 2 decades instead of seven and
honestly, I would not even know how to calculate the necessary values


Finally, having tried another couple of ideas, I still don't figure out
how I can solve my problem. This is why I post on this forum. How could
I compose a circuit dephasing either two sinusoidal or digital signals
by 90 degrees (and if possible, without too much costful and
placetaking electronic circuitry)?

---


+----------------->I
+------------|------------+
| +-----+ | +-----+ |
+--|D Q|---+---|D Q|------->Q
| | | | |
4fOUT---------+--|> | +---|> _| |
| | | | | Q|O-+
| +-----+ | +-----+
| |
+------------+

The "D" type filp-flops are XX74s or 4013s, 4fOUT is 40hZ to 40kHz, and
I and Q ARE YOUR 10Hz to 10kHz outputs, in quadrature.

 

[Ted Edwards]

Mac wrote:

+----------------->I
+------------|------------+
| +-----+ | +-----+ |
+--|D Q|---+---|D Q|------->Q
| | | | |
4fOUT---------+--|> | +---|> _| |
| | | | | Q|O-+
| +-----+ | +-----+
| |
+------------+

The "D" type filp-flops are XX74s or 4013s, 4fOUT is 40hZ to 40kHz,
and I and Q ARE YOUR 10Hz to 10kHz outputs, in quadrature.


Nice!

It is. It is called a Johnson counter. A three bit version can be used
to generate thre-phase but needs a bit of logic to avoid it statring in
a different sequence.

Ted

 

[Ted Edwards]

Bernhard Krämer wrote:

Thank you very much for your suggestions and corrections. I already thought
about multiplying and then dividing the frequency to make use of a circuit
similar to yours.

But after having received your post, I finally figured out that my circuit
would give the same results and even better if I would use 4 x freq instead
of 1 x freq and multiplying it.

For more information on Johnson counters, see
< http://www.play-hookey.com/digital/johnson_counter.html >

The "illegal state" prevention circuit is not needed for a 2-bit one.

This will work from 0Hz to as high as counters will operate.

Ted

 

[John Fields]

On Sat, 23 Apr 2005 11:48:34 +0200, Bernhard Krämer
<Bernhard.Kramer@lien.uhp-nancy.fr> wrote:

Hello all,

Thank you very much for your suggestions and corrections. I already thought
about multiplying and then dividing the frequency to make use of a circuit
similar to yours.

But after having received your post, I finally figured out that my circuit
would give the same results and even better if I would use 4 x freq instead
of 1 x freq and multiplying it.

---
Bullshit.

Post your circuit.

--
John Fields
Professional Circuit Designer

 

[Mac]

On Sat, 23 Apr 2005 16:05:38 +0000, Rich Grise wrote:

On Sat, 23 Apr 2005 11:48:34 +0200, Bernhard Krämer top-posted:

Hello all,
[snip]
Sorry for the " 7 decades error "; I thought about another problem
which has nothing to do with this one while writing the post: another
circuitry where I have to worry about a frequency range from 10 mHz to
10 KHz. And this, err ..., are only 6 decades, so I was still wrong. I
am deeply ashamed now

Well, you'd better still be, because that's still only three (3) decades.

I think mHz is millihertz. Anyway, I choose to give the OP the benefit of
the doubt at this point. ;-)

[snip]

--Mac

 

[John Woodgate]

I read in sci.electronics.design that Bernhard Kraemer
<bernhard.kramer@lien.uhp-nancy.fr> wrote (in
<4268f298$0$79458$14726298@news.sunsite.dk&gt about 'How to shift phase
by 90 degrees between approx 10 Hz and 10 KHz ?', on Fri, 22 Apr 2005:

How could I compose a circuit dephasing either two sinusoidal or
digital signals by 90 degrees (and if possible, without too much
costful and placetaking electronic circuitry)?

Over seven decades of frequency, any analogue solution is big and
complicated. You use two all-pass filters, one shifting 45 degrees and
the other shifting 135 degrees. But for seven decades of frequency you
will need very high-order filters.

One solution is to modulate everything on to a carrier at a much higher
frequency and shift the phases at that frequency, but a modulator with a
seven-decade bandwidth may be just as difficult as the filters.

It may not be easier to do it digitally, either.
--
Regards, John Woodgate, OOO - Own Opinions Only.
There are two sides to every question, except
'What is a Moebius strip?'
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

 

[Mac]

On Fri, 22 Apr 2005 12:48:25 +0000, Bernhard Kraemer wrote:

Hello,

For an experiment, I need to build a little Lock-In Amplifier an my own
circuit. I do not have an external Lock-In and so this will be the cheaper
and better solution.

The circuit is nearly finished. What lacks is a circuit which should be
controlled by an external oscillator and produces a two output signals,
a TTL signal of the same frequency and another of 90 degrees
phase shifted. The frequency range should be about 10 Hz to 10 KHz (7
decades).

[snip]

I hope someone could help me.

Yours,

Bernhard

Just to be clear, does the external oscillator range from 10 Hz to 10 kHz?
Or is it a multiple of that or what?

As others have pointed out, that is not 7 decades, so please clarify.

I have several ideas for how to do this, but I need to know the answer to
this question first.

--Mac

 

[John Fields]

On Fri, 22 Apr 2005 14:32:53 -0400, "Tam/WB2TT"
<t-tammaru@c0mca$t.net> wrote:

"Tim Shoppa" <shoppa@trailing-edge.com> wrote in message
news:1114175153.882654.83420@f14g2000cwb.googlegroups.com...
What lacks is a circuit which should be
controlled by an external oscillator and produces a two
output signals,
a TTL signal of the same
frequency and another of 90 degrees
phase shifted. The frequency range should be about 10 Hz to
10 KHz (7 decades).

If you allow the external clock to be 4 times the frequency of the
outputs, all you need is a divide-by-four clock generator.

************************************
You have to use the right kind of divider, though. Not two div-2 counters
in tandem.
Here is what you do:
1. Get 2 D flipflops.
2. Connect the two clock leads together, and to the input.
3. Connect the Q of the first stage to the D of the second.
4. Connect the Q* (QBAR) of the second to the D of the first.
5 The outputs are the Q of the first and second stages.

The Div-4 should simplify building the oscillator. In fact, you might want
to consider switching in different prescalers in addition to cover your
range.

---
LOL, click on this:

news:lu9i619955pbdcp1rmukd27vdqvpaf64ss@4ax.com
--
John Fields
Professional Circuit Designer

 

[gwhite]

John Woodgate wrote:

One solution is to modulate everything on to a carrier at a much higher
frequency and shift the phases at that frequency,...

That will not work. There is no free lunch in implementing a hilbert
transform.

... but a modulator with a
seven-decade bandwidth may be just as difficult as the filters.

Why do you believe that?

 

[John Woodgate]

I read in sci.electronics.design that gwhite <gwhite@deadend.net> wrote
(in <426C5C43.78590373@deadend.net&gt about 'How to shift phase by 90
degrees between approx 10 Hz and 10 KHz ?', on Mon, 25 Apr 2005:

John Woodgate wrote:


One solution is to modulate everything on to a carrier at a much higher
frequency and shift the phases at that frequency,...

That will not work. There is no free lunch in implementing a hilbert
transform.

... but a modulator with a
seven-decade bandwidth may be just as difficult as the filters.

Why do you believe that?

It's irrelevant now anyway, because the 'seven decades' was a mistook.
--
Regards, John Woodgate, OOO - Own Opinions Only.
There are two sides to every question, except
'What is a Moebius strip?'
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

 

[bernhard.kramer@lien.uhp-]

John Fields a écrit :

On Sat, 23 Apr 2005 11:48:34 +0200, Bernhard Krämer
Bernhard.Kramer@lien.uhp-nancy.fr> wrote:


Hello all,

Thank you very much for your suggestions and corrections. I already thought
about multiplying and then dividing the frequency to make use of a circuit
similar to yours.

But after having received your post, I finally figured out that my circuit
would give the same results and even better if I would use 4 x freq instead
of 1 x freq and multiplying it.


---
Bullshit.

Post your circuit.

Hello John,

Here's my circuit as I planned it before I saw yours:


Schmitt Digital
Multiplier trigger Electronic
+-----+ +-----+ +-----+
sin(x) ---+--| | K(1 + cos(2x)) || K cos(2x) | +-+-| clk |
|--- Out1
| | X |-------------------||------------| | | |-----| |
+--| | || |-+-+ | |
|--- Out2
+-----+ +-----+ +-----+


The Digital Electronic would be the following:

1) A FlipFlop which would be divide the clk frequency. The result is Out1
2) Logic Table:

Clk Out1 || Out2
-----------++-----
0 0 0
0 1 1
1 0 1
1 1 0

You can verify yourself that Out1 would be phase shifted by 90° in
comparision
with Out2, and that the frequency of Out1 and Out2 equals the frequency
of sin(x).


Best,

Bernhard

 

[Winfield Hill]

bernhard.kramer@lien.uhp-nancy.fr wrote...

k(1+cos(2x)) Schmitt Digital
Multiplier | k cos(2x) trigger Electronic
+-----+ | | +-----+ +-----+
sin(x) ---+--| | | || | | +-+-| clk | |--- Out1
| | X |-----||----------| | | |-----| |
+--| | || |-+-+ | | |--- Out2
+-----+ +-----+ +-----+

The Digital Electronic would be the following:

1) A FlipFlop which would be divide the clk frequency. The result is Out1
2) Logic Table:

Clk Out1 || Out2
-----------++-----
0 0 0
0 1 1
1 0 1
1 1 0

You can verify yourself that Out1 would be phase shifted by 90° in
comparision with Out2, and that the frequency of Out1 and Out2 equals
the frequency of sin(x).

Be careful, Bernhard, with your sine-squared waveforms. It won't be
easy to get a reliable exact 50% duty-cycle 2f square-wave from your
Schmitt trigger. It may be back to the drawing board for you!


--
Thanks,
- Win

 

[The Phantom]

On Fri, 22 Apr 2005 14:41:00 +0100, John Woodgate <jmw@jmwa.demon.contraspam.yuk> wrote:

I read in sci.electronics.design that Bernhard Kraemer
bernhard.kramer@lien.uhp-nancy.fr> wrote (in
4268f298$0$79458$14726298@news.sunsite.dk&gt about 'How to shift phase
by 90 degrees between approx 10 Hz and 10 KHz ?', on Fri, 22 Apr 2005:

How could I compose a circuit dephasing either two sinusoidal or
digital signals by 90 degrees (and if possible, without too much
costful and placetaking electronic circuitry)?

Over seven decades of frequency,

How do you figure 7 decades?

any analogue solution is big and
complicated. You use two all-pass filters, one shifting 45 degrees and
the other shifting 135 degrees. But for seven decades of frequency you
will need very high-order filters.

One solution is to modulate everything on to a carrier at a much higher
frequency and shift the phases at that frequency, but a modulator with a
seven-decade bandwidth may be just as difficult as the filters.

It may not be easier to do it digitally, either.

 

[John Fields]

On Tue, 26 Apr 2005 12:18:36 +0200,
"bernhard.kramer@lien.uhp-nancy.fr"
<Bernhard.Kramer@lien.uhp-nancy.fr> wrote:

Hello John,

Here's my circuit as I planned it before I saw yours:


Schmitt Digital
Multiplier trigger Electronic
+-----+ +-----+ +-----+
sin(x) ---+--| | K(1 + cos(2x)) || K cos(2x) | +-+-| clk |
|--- Out1
| | X |-------------------||------------| | | |-----| |
+--| | || |-+-+ | |
|--- Out2
+-----+ +-----+ +-----+


The Digital Electronic would be the following:

1) A FlipFlop which would be divide the clk frequency. The result is Out1
2) Logic Table:

Clk Out1 || Out2
-----------++-----
0 0 0
0 1 1
1 0 1
1 1 0

You can verify yourself that Out1 would be phase shifted by 90° in
comparision
with Out2, and that the frequency of Out1 and Out2 equals the frequency
of sin(x).

---
Using Win's schematic for clarity:


k(1+cos(2x)) Schmitt Digital
Multiplier | k cos(2x) trigger Electronic
+-----+ | | +-----+ +-----+
sin(x) ---+--| | | || | | +-+-| clk | |--- Out1
| | X |-----||----------| | | |-----| |
+--| | || |-+-+ | | |--- Out2
+-----+ +-----+ +-----+

It seems you've opted to full-wave rectify sin(x), and then run the
rectified pulses through a capacitor to remove the DC component, then
square up the edges with a Schmitt trigger, (I'd use a comparator with
no hysteresis if I could get away with it) then get your quadrature
output by running the output of the Schmitt trigger through some
digital circuitry. As Win has stated, I believe you're going to run
into some problems with getting a 50% duty cycle square wave out of
the Schmitt trigger, but that may not be a problem depending on the
slop you can stand in your output waveform. Also, I'm interested in
seeing what your digital circuitry looks like so, if you don't mind
I'd appreciate it if you could post a schematic here or, if you don't
want to do ASCII, on alt.binaries.schematics.electronic.

Thanks.

--
John Fields
Professional Circuit Designer

 

[Fred Bartoli]

"John Fields" <jfields@austininstruments.com> a écrit dans le message de
news:1jv1719uuairppg4sdj45bm7p4v9pfks2c@4ax.com...

On Tue, 26 Apr 2005 12:18:36 +0200,
"bernhard.kramer@lien.uhp-nancy.fr"
Bernhard.Kramer@lien.uhp-nancy.fr> wrote:


Using Win's schematic for clarity:


k(1+cos(2x)) Schmitt Digital
Multiplier | k cos(2x) trigger Electronic
+-----+ | | +-----+ +-----+
sin(x) ---+--| | | || | | +-+-| clk | |--- Out1
| | X |-----||----------| | | |-----| |
+--| | || |-+-+ | | |--- Out2
+-----+ +-----+ +-----+

It seems you've opted to full-wave rectify sin(x), and then run the
rectified pulses through a capacitor to remove the DC component, then
square up the edges with a Schmitt trigger, (I'd use a comparator with
no hysteresis if I could get away with it) then get your quadrature
output by running the output of the Schmitt trigger through some
digital circuitry. As Win has stated, I believe you're going to run
into some problems with getting a 50% duty cycle square wave out of
the Schmitt trigger, but that may not be a problem depending on the
slop you can stand in your output waveform. Also, I'm interested in
seeing what your digital circuitry looks like so, if you don't mind
I'd appreciate it if you could post a schematic here or, if you don't
want to do ASCII, on alt.binaries.schematics.electronic.

You can have pretty good 50% duty cycle ratio with this:

VDD
|
|
V
k(1+cos(2x)) Schmitt Digital
Multiplier | k cos(2x) trigger Electronic
+-----+ | | +-----+ +-----+
sin(x) ---+--| | | || | | +-+-| clk | |--- Out1
| | X |-----||--+-------| | | |--+--| |
+--| | || | |-+-+ | | | |--- Out2
+-----+ | +-----+ | +-----+
| |
.-. .-.
| | | |
| | | |
'-' '-'
| || |
+-----||---------+
| || |
| /| |
| /-|-------'
'-----< |
\+|- VDD/2
\|

(created by AACircuit v1.28 beta 10/06/04 www.tech-chat.de)

The 50% duty cycle accuracy depends on the VDD to VDD/2 ratio accuracy.
However, it'll still be worse than a divided clock, unless you go to
considerable efforts.


--
Thanks,
Fred.