On Fri, 12 Mar 2004 16:08:15 +0000, John Woodgate
wrote:

I read in sci.electronics.design that Reg Edwards
wrote (in
et.com) about 'Extracting the 5th Harmonic', on Fri, 12 Mar 2004:
According to Fourier, at some mark-space ratios of a square wave certain
harmonics may be missing from the spectrum.


For a waveform like this (use Courier font):
_____
/ \ /
_____/ \____________/

with rise-time f, dwell time d, fall time r and period T, the harmonic
magnitudes are given by:

Cn = 2Aav{sinc(n[pi]f/T)}{sinc(n[pi][f+d]/T)}{sinc(n[pi][r-f]/T)},

where sinc(x)= {sin(x)}/x

There seems to be a number of opportunities for a harmonic to 'hide' in
a zero of that function.

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
--

The BBC: Licensed at public expense to spread lies.

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

On Fri, 12 Mar 2004 16:31:19 -0600, John Fields
wrote:

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.

Historical note: about 1960, a guy at HP was doing exactly this with
some new diodes, and he got way more higher harmonics than theory
predicts. To figure it out, they hooked up the just-invented HP185
sampling scope (which then used avalanche transistors to make its
sampling pulses) and discovered the diode reverse-recovery snap
phenom. Soon the scope itself was using this effect. They were
originally called Boff diodes, after the discoverer Frank Boff, but
the name didn't stick (wonder why?) and they became "snap diodes" and
later "step-recovery diodes". I think I may have the HP Journal
article around somewhere.

See page 31:

http://cp.literature.agilent.com/lit...5980-2090E.pdf


John

On Fri, 12 Mar 2004 15:02:30 -0800, John Larkin
wrote:

On Fri, 12 Mar 2004 16:31:19 -0600, John Fields
wrote:

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.

Historical note: about 1960, a guy at HP was doing exactly this with
some new diodes, and he got way more higher harmonics than theory
predicts. To figure it out, they hooked up the just-invented HP185
sampling scope (which then used avalanche transistors to make its
sampling pulses) and discovered the diode reverse-recovery snap
phenom. Soon the scope itself was using this effect. They were
originally called Boff diodes, after the discoverer Frank Boff, but
the name didn't stick (wonder why?) and they became "snap diodes" and
later "step-recovery diodes". I think I may have the HP Journal
article around somewhere.

See page 31:

http://cp.literature.agilent.com/lit...5980-2090E.pdf

---
:-)

--
John Fields

On Fri, 12 Mar 2004 15:02:30 -0800, John Larkin
wrote:

On Fri, 12 Mar 2004 16:31:19 -0600, John Fields
wrote:

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.

Historical note: about 1960, a guy at HP was doing exactly this with
some new diodes, and he got way more higher harmonics than theory
predicts. To figure it out, they hooked up the just-invented HP185
sampling scope (which then used avalanche transistors to make its
sampling pulses) and discovered the diode reverse-recovery snap
phenom. Soon the scope itself was using this effect. They were
originally called Boff diodes, after the discoverer Frank Boff, but
the name didn't stick (wonder why?) and they became "snap diodes" and
later "step-recovery diodes". I think I may have the HP Journal
article around somewhere.

See page 31:

http://cp.literature.agilent.com/lit...5980-2090E.pdf

---
:-)

--
John Fields

On Fri, 12 Mar 2004 16:31:19 -0600, John Fields
wrote:

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.

Historical note: about 1960, a guy at HP was doing exactly this with
some new diodes, and he got way more higher harmonics than theory
predicts. To figure it out, they hooked up the just-invented HP185
sampling scope (which then used avalanche transistors to make its
sampling pulses) and discovered the diode reverse-recovery snap
phenom. Soon the scope itself was using this effect. They were
originally called Boff diodes, after the discoverer Frank Boff, but
the name didn't stick (wonder why?) and they became "snap diodes" and
later "step-recovery diodes". I think I may have the HP Journal
article around somewhere.

See page 31:

http://cp.literature.agilent.com/lit...5980-2090E.pdf


John

Assuming you have edges quite a bit shorter than the period, it's easy
to see what pulse widths you want to avoid to maximize the 5th: don't
let 5*pi*width/period be an integer multiple of pi. So avoid
width/period = 1/5, 2/5, 3/5 or 4/5. 2/5 and 3/5 (40% and 60%) are
not all that far from 50%. There are lots of ways to get 50% (or
close to it). One is to slow the edges a bit, and put the result into
a Schmitt trigger with adjustable DC level; the DC level will then
adjust the period. You can servo the DC level with an integrator tied
to the output and referenced to (v(high)+v(low))/2, for high accuracy.
Maybe that's too complicated, though, if you want things small. Note
that by not causing dissipation of the fundamental or other harmonics,
a multiplier can be considerably more efficient than indicated by the
percentage available power in the selected harmonic. In other words,
don't put a filter on a logic output that shorts out the fundamental,
but rather one that looks like an open circuit to the fundamental,
etc.

Cheers,
Tom


Paul Burridge wrote in message . ..
On Fri, 12 Mar 2004 16:08:15 +0000, John Woodgate
wrote:

I read in sci.electronics.design that Reg Edwards
wrote (in
et.com) about 'Extracting the 5th Harmonic', on Fri, 12 Mar 2004:
According to Fourier, at some mark-space ratios of a square wave certain
harmonics may be missing from the spectrum.


For a waveform like this (use Courier font):
_____
/ \ /
_____/ \____________/

with rise-time f, dwell time d, fall time r and period T, the harmonic
magnitudes are given by:

Cn = 2Aav{sinc(n[pi]f/T)}{sinc(n[pi][f+d]/T)}{sinc(n[pi][r-f]/T)},

where sinc(x)= {sin(x)}/x

There seems to be a number of opportunities for a harmonic to 'hide' in
a zero of that function.

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

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

Assuming you have edges quite a bit shorter than the period, it's easy
to see what pulse widths you want to avoid to maximize the 5th: don't
let 5*pi*width/period be an integer multiple of pi. So avoid
width/period = 1/5, 2/5, 3/5 or 4/5. 2/5 and 3/5 (40% and 60%) are
not all that far from 50%. There are lots of ways to get 50% (or
close to it). One is to slow the edges a bit, and put the result into
a Schmitt trigger with adjustable DC level; the DC level will then
adjust the period. You can servo the DC level with an integrator tied
to the output and referenced to (v(high)+v(low))/2, for high accuracy.
Maybe that's too complicated, though, if you want things small. Note
that by not causing dissipation of the fundamental or other harmonics,
a multiplier can be considerably more efficient than indicated by the
percentage available power in the selected harmonic. In other words,
don't put a filter on a logic output that shorts out the fundamental,
but rather one that looks like an open circuit to the fundamental,
etc.

Cheers,
Tom


Paul Burridge wrote in message . ..
On Fri, 12 Mar 2004 16:08:15 +0000, John Woodgate
wrote:

I read in sci.electronics.design that Reg Edwards
wrote (in
et.com) about 'Extracting the 5th Harmonic', on Fri, 12 Mar 2004:
According to Fourier, at some mark-space ratios of a square wave certain
harmonics may be missing from the spectrum.


For a waveform like this (use Courier font):
_____
/ \ /
_____/ \____________/

with rise-time f, dwell time d, fall time r and period T, the harmonic
magnitudes are given by:

Cn = 2Aav{sinc(n[pi]f/T)}{sinc(n[pi][f+d]/T)}{sinc(n[pi][r-f]/T)},

where sinc(x)= {sin(x)}/x

There seems to be a number of opportunities for a harmonic to 'hide' in
a zero of that function.

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