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

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

How do you figure 7 decades?

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.

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

I read in sci.electronics.design that Bernhard Kraemer wrote (in ) about 'How to shift phase by 90 degrees between approx 10 Hz and 10 KHz ?', on Fri, 22 Apr 2005:

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.

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

Nice!

--Mac

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:

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.

Excellent idea! I wish more would adopt that attitude. :-)

Possibly interestingly, I think I know how you feel. :-)

Cheers! Rich

For more information on Johnson counters, see <

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

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

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

--Mac

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

Why do you believe that?

I read in sci.electronics.design that gwhite wrote (in ) about 'How to shift phase by 90 degrees between approx 10 Hz and 10 KHz ?', on Mon, 25 Apr 2005:

It's irrelevant now anyway, because the 'seven decades' was a mistook.

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!

John Fields a écrit :

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:

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

--- 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