On Fri, 12 Mar 2004 17:57:24 +0000, Paul Burridge
wrote:


Great. So without a spectrum analyser there's no way to tell? If I
examine the output of the multiplier, it's very messy. There's a
dominant 3rd harmonic alright (my frequency counter resolves it
without difficulty) but the scope trace reveals a number of 'ghost
traces' of different frequencies and amplitudes co-incident with the
dominant trace. All rather confusing. I suppose the only answer is to
build Reg's band pass filter and stick it between the inverter output
and the multiplier input? shrug

---
You may want to try something like this:


COUNTER SCOPE COUNTER
| | |
| | |
FIN--[50R]-+-[1N4148]---+----+-------+---FOUT
| |
[L] [C]
| |
GND----------------------+----+

The 50 ohm resistor is the internal impedance of a function generator,
and when I set it to output a square wave at 1.5VPP, I got 10.8kHz for
the fundamental of the tank. Then I tuned the function generator down
until I got a peak out of the tank, and here's what I found:

Fin Fout Vout fout/fin
kHz kHz VPP
-----|-----|------|---------
10.8 10.8 0.9 1.0
3.58 10.8 0.25 3.02 ~ 3
2.14 10.8 0.2 5.05 ~ 5

So with a square wave in there were no even harmonics and it was easy
to trap the 3rd and 5th harmonics with a tank.


Next, I tried it with a 3VPP sine wave in and got:

Fin Fout Vout fout/fin
kHz kHz VPP
-----|-----|------|---------
10.8 10.8 1.3 1.0
5.39 10.8 0.9 ~ 2.0
2.14 10.8 0.3 5.05 ~ 5

So it looks like the second and the fifth harmonics were there. There
were also some other responses farther down, but I just wanted to see
primarily whether the fifth had enough amplitude to work with, and
apparently it does, so I let the rest of it slide.

So, it looks like if you square up your oscillator's output to 50% duty
cycle you could get the 5th harmonic without too much of a problem. If
you can't, then clip the oscillator's output with a diode or make its
duty cycle less than or greater than 50%, and you ought to be able to
get the 5th that way.

--
John Fields