On Mon, 16 Feb 2004 21:02:02 GMT, James Meyer
wrote:

On Mon, 16 Feb 2004 13:03:46 -0600, John Fields
posted this:


Starting with a perfect square wave at f1, bang the hell out of a diode
with it, and then bandpass it and the 3rd harmonic (f2) separately, then
mix them to get f1, f2, f1+f2, and f1-f2. Using a doubly balanced mixer
will get rid of f1 and f2, then notching out f1+f2 will leave f1-f2,
which will be 2f1, that non-existent second harmonic.

What purpose does the diode serve? You're already starting with a
"perfect" square wave.

---
Duhhh.... None, of course.

Thanks.
--
John Fields

In article , Paul Burridge wrote:
What's the maximum multiplication factor it's practical and sensible
to attempt to achieve in one single stage of multiplication? (Say from
a 7Mhz square wave source with 5nS rise/fall times.)

Not radio, but interesting nevertheless. The older Hewlett-Packard cesium
clocks, ie 5060/61/62 vintage multiplied a crystal oscillator up to 90 MHz in
several stages. This fed into a step-recovery diode that sits in a cavity, and
has 12.631... MHz applied to the SRD bias. The cavity selects the ***102nd***
harmonic ie 9180 MHz, and there are also sidebands at +/- 12.631.. MHz This is
then fed into a hi-Q cavity tuned to the upper sideband ie 9192.631... MHz
which is the desired cesium transition frequency.

Adjusting the whole thing was a bit fiddly, and there were also some
factory-set adjustments that you NEVER TOUCHED unless you had plenty of time
and a squillion dollars worth of test gear. This was all a 1960's design and
was a bit of a stretch. The newer (5071) clocks do things QUITE differently.

Steve Quigg

In article , Paul Burridge wrote:
What's the maximum multiplication factor it's practical and sensible
to attempt to achieve in one single stage of multiplication? (Say from
a 7Mhz square wave source with 5nS rise/fall times.)

Not radio, but interesting nevertheless. The older Hewlett-Packard cesium
clocks, ie 5060/61/62 vintage multiplied a crystal oscillator up to 90 MHz in
several stages. This fed into a step-recovery diode that sits in a cavity, and
has 12.631... MHz applied to the SRD bias. The cavity selects the ***102nd***
harmonic ie 9180 MHz, and there are also sidebands at +/- 12.631.. MHz This is
then fed into a hi-Q cavity tuned to the upper sideband ie 9192.631... MHz
which is the desired cesium transition frequency.

Adjusting the whole thing was a bit fiddly, and there were also some
factory-set adjustments that you NEVER TOUCHED unless you had plenty of time
and a squillion dollars worth of test gear. This was all a 1960's design and
was a bit of a stretch. The newer (5071) clocks do things QUITE differently.

Steve Quigg

Wadley loop recievers had to generate 33rd+ harmonic
Not quite OT but a great (old) idea
http://www.siliconchip.com.au/cms/A_30512/article.html

Wadley loop recievers had to generate 33rd+ harmonic
Not quite OT but a great (old) idea
http://www.siliconchip.com.au/cms/A_30512/article.html

I had a Yaesu FRG-7 receiver that used this lovely Wadley loop. If you
subscribe to the theory that every beep and bloop you hear as you tune
across the dial is a station, that is the receiver for you!

However, if you understand spurs and birdies, a different picture emerges.
Lots of noise, too!


"GPG" wrote in message
om...
Wadley loop recievers had to generate 33rd+ harmonic
Not quite OT but a great (old) idea
http://www.siliconchip.com.au/cms/A_30512/article.html

I had a Yaesu FRG-7 receiver that used this lovely Wadley loop. If you
subscribe to the theory that every beep and bloop you hear as you tune
across the dial is a station, that is the receiver for you!

However, if you understand spurs and birdies, a different picture emerges.
Lots of noise, too!


"GPG" wrote in message
om...
Wadley loop recievers had to generate 33rd+ harmonic
Not quite OT but a great (old) idea
http://www.siliconchip.com.au/cms/A_30512/article.html

Tell me how you will use that and I will tell you the answer.

"Paul Burridge" wrote in message
...
What's the maximum multiplication factor it's practical and sensible
to attempt to achieve in one single stage of multiplication? (Say from
a 7Mhz square wave source with 5nS rise/fall times.)
--

The BBC: Licensed at public expense to spread lies.

Tell me how you will use that and I will tell you the answer.

"Paul Burridge" wrote in message
...
What's the maximum multiplication factor it's practical and sensible
to attempt to achieve in one single stage of multiplication? (Say from
a 7Mhz square wave source with 5nS rise/fall times.)
--

The BBC: Licensed at public expense to spread lies.

On Mon, 16 Feb 2004 11:53:11 +0000,
said...
On Sun, 15 Feb 2004 16:46:32 -0700, Jim Thompson
wrote:

On Sun, 15 Feb 2004 23:48:47 +0000, Paul Burridge
wrote:

What's the maximum multiplication factor it's practical and sensible
to attempt to achieve in one single stage of multiplication? (Say from
a 7Mhz square wave source with 5nS rise/fall times.)

You ought to be able to answer that yourself... what's the spectral
roll-off of a square wave ??

I suppose it boils down to how much signal is left in the mush as the
harmonics get higher and higher. Knew I shoulda held on to that
spectrum analyser I used to have. :-(
I suppose that's the proper answer though: get the rise/fall times as
small and possible, measure the specral output and pick a suitable
harmonic with enough energy in it to set it 'comfortably' above the
noise floor?

Gee. I could have sworn Jim was hinting at the math approach.
Wouldn'tcha just love to predict that roll-off on paper and *then*
see it in real life? Starts with an "F", looks like a number,
sounds like a frog.

--
Best Regards,
Mike