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

[Fred Bartoli]

On Tue, 26 Apr 2005 12:18:36 +0200,
"[email protected]-nancy.fr"


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.