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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 |
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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 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 |
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 |
The 5th harmonic should be only 14dB below the fundamental, although it will
drop fairly quickly as the sides of the input square wave deviate from vertical. Does the 3.44MHz have a 50% duty cycle? Are you filtering before amplifying (eg a high impedance 3 pole bandpass/highpass L-C filter with a gain of about 5 at 17.2MHz). Does the inverter supply a decent square wave under the load of the filter? If all else fails, could you reverse the process - generate 17.2MHz and divide it down to 3.44MHz? (Many) years ago I made a functional TV modulator for an Apple ][ PC by pulling out the 3rd harmonic of the 14.318MHz system clock. I know it was only 3rd harmonic, but it was at ~43MHz, so I would expect similar or better logic should be able to produce 17.2MHz for you. On Fri, 12 Mar 2004 13:56:10 +0000, Paul Burridge wrote: Hi all, Is there some black magic required to get higher order harmonics out of an oscillator? I'm only trying to get 17.2Mhz out of a 3.44Mhz source and am thus far failing spectacularly. I've tried everything I can think of so far to no avail. All I can get apart from the fundamental is a strong third harmonic on 10.32Mhz, regardless of what I tune for. I've tried passing the osc output through two successive inverter gates to sharpen it up, but still nothing beyond the third appears after tuned amplification for the fifth. I no longer have a spectrum analyser so can't check for the presence of a decent comb of harmonics at the input to the multiplier stage but can only assume the fifth is well down in the mush for some reason. I could change the inverters for schmitt triggers and gain a couple of nS but can't see that making enough difference. What about sticking a varactor in there somewhere? Would its non-linearity assist or are they only any good for even order harmonics? Any suggestions, please. I'm stumped! :( Tony (remove the "_" to reply by email) |
The 5th harmonic should be only 14dB below the fundamental, although it will
drop fairly quickly as the sides of the input square wave deviate from vertical. Does the 3.44MHz have a 50% duty cycle? Are you filtering before amplifying (eg a high impedance 3 pole bandpass/highpass L-C filter with a gain of about 5 at 17.2MHz). Does the inverter supply a decent square wave under the load of the filter? If all else fails, could you reverse the process - generate 17.2MHz and divide it down to 3.44MHz? (Many) years ago I made a functional TV modulator for an Apple ][ PC by pulling out the 3rd harmonic of the 14.318MHz system clock. I know it was only 3rd harmonic, but it was at ~43MHz, so I would expect similar or better logic should be able to produce 17.2MHz for you. On Fri, 12 Mar 2004 13:56:10 +0000, Paul Burridge wrote: Hi all, Is there some black magic required to get higher order harmonics out of an oscillator? I'm only trying to get 17.2Mhz out of a 3.44Mhz source and am thus far failing spectacularly. I've tried everything I can think of so far to no avail. All I can get apart from the fundamental is a strong third harmonic on 10.32Mhz, regardless of what I tune for. I've tried passing the osc output through two successive inverter gates to sharpen it up, but still nothing beyond the third appears after tuned amplification for the fifth. I no longer have a spectrum analyser so can't check for the presence of a decent comb of harmonics at the input to the multiplier stage but can only assume the fifth is well down in the mush for some reason. I could change the inverters for schmitt triggers and gain a couple of nS but can't see that making enough difference. What about sticking a varactor in there somewhere? Would its non-linearity assist or are they only any good for even order harmonics? Any suggestions, please. I'm stumped! :( Tony (remove the "_" to reply by email) |
On Fri, 12 Mar 2004 13:56:10 +0000, Paul Burridge
posted this: Hi all, Is there some black magic required to get higher order harmonics out of an oscillator? I'm only trying to get 17.2Mhz out of a 3.44Mhz source and am thus far failing spectacularly. I've tried everything I can think of so far to no avail. Is this a simulated circuit or a "real" one built with "real" components? I have at least one suggestion, but I need to know whether to send an LTspice netlist or a gif. Jim |
On Fri, 12 Mar 2004 13:56:10 +0000, Paul Burridge
posted this: Hi all, Is there some black magic required to get higher order harmonics out of an oscillator? I'm only trying to get 17.2Mhz out of a 3.44Mhz source and am thus far failing spectacularly. I've tried everything I can think of so far to no avail. Is this a simulated circuit or a "real" one built with "real" components? I have at least one suggestion, but I need to know whether to send an LTspice netlist or a gif. Jim |
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On Fri, 12 Mar 2004 20:41:49 +0000, Ian Bell wrote:
Reminds me of the old joke about the mathemetician, the physicist and the engineer. They were each shown into a room in the centre of which was £50 note / $100 bill (depending on which side of the pond you live). They were told they could walk half the distance to the money and stop. Then they could walk half the remaining ditance and so on until they got the money. The mathemetician worked out you would never reach the money so he didn't even try. The physicist, working to five decimal places was still there a week later. The engineer did three iterations, said 'That's close enough' and picked up the money. The moral is of course, horses for courses. Ian .......and I always believed John was an engineer, have some similar expressions which an instructor used the xmas holidays to derive JM ---- Jan-Martin, LA8AK, N-4623 Kristiansand http://home.online.no/~la8ak/ |
On Fri, 12 Mar 2004 20:41:49 +0000, Ian Bell wrote:
Reminds me of the old joke about the mathemetician, the physicist and the engineer. They were each shown into a room in the centre of which was £50 note / $100 bill (depending on which side of the pond you live). They were told they could walk half the distance to the money and stop. Then they could walk half the remaining ditance and so on until they got the money. The mathemetician worked out you would never reach the money so he didn't even try. The physicist, working to five decimal places was still there a week later. The engineer did three iterations, said 'That's close enough' and picked up the money. The moral is of course, horses for courses. Ian .......and I always believed John was an engineer, have some similar expressions which an instructor used the xmas holidays to derive JM ---- Jan-Martin, LA8AK, N-4623 Kristiansand http://home.online.no/~la8ak/ |
On Fri, 12 Mar 2004 16:55:51 +0000, Bob Stephens wrote:
On Fri, 12 Mar 2004 16:08:15 +0000, John Woodgate wrote: where sinc(x)= {sin(x)}/x I've never seen this terminology before. Is this standard math parlance or is it something of your own? Don't flame, I'm genuinely curious. Bob I see the sinc function all the time. I was introduced to it in school, in a signal processing class, and people at work use it fairly often. In my experience it seems that anyone who deals with signal processing or fft's is familiar with the sinc() function. And I've always heard it pronounced the same as the word "sink." --Mac |
On Fri, 12 Mar 2004 16:55:51 +0000, Bob Stephens wrote:
On Fri, 12 Mar 2004 16:08:15 +0000, John Woodgate wrote: where sinc(x)= {sin(x)}/x I've never seen this terminology before. Is this standard math parlance or is it something of your own? Don't flame, I'm genuinely curious. Bob I see the sinc function all the time. I was introduced to it in school, in a signal processing class, and people at work use it fairly often. In my experience it seems that anyone who deals with signal processing or fft's is familiar with the sinc() function. And I've always heard it pronounced the same as the word "sink." --Mac |
I read in sci.electronics.design that Tim Wescott
wrote (in . com) about 'Extracting the 5th Harmonic', on Fri, 12 Mar 2004: Yes, it's pronounced "sink", and it's quite common in signal processing. You define it as being the _limit_ of sin(x)/x as x - 0 because otherwise it's undefined at zero, and all the mathematicians in the crowd will curse at you for being yet another engineer who's treating math so casually. I don't fear the wrath of any mathematician. The limit is very firmly established as = 1 at a quite elementary level. Just consider the expansion of sin(x) = x - (x^3)/3! +..... Of course, it can be established more rigorously, but there is nothing wrong with the series expansion AFAIK. -- Regards, John Woodgate, OOO - Own Opinions Only. The good news is that nothing is compulsory. The bad news is that everything is prohibited. http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk |
I read in sci.electronics.design that Tim Wescott
wrote (in . com) about 'Extracting the 5th Harmonic', on Fri, 12 Mar 2004: Yes, it's pronounced "sink", and it's quite common in signal processing. You define it as being the _limit_ of sin(x)/x as x - 0 because otherwise it's undefined at zero, and all the mathematicians in the crowd will curse at you for being yet another engineer who's treating math so casually. I don't fear the wrath of any mathematician. The limit is very firmly established as = 1 at a quite elementary level. Just consider the expansion of sin(x) = x - (x^3)/3! +..... Of course, it can be established more rigorously, but there is nothing wrong with the series expansion AFAIK. -- Regards, John Woodgate, OOO - Own Opinions Only. The good news is that nothing is compulsory. The bad news is that everything is prohibited. http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk |
Fourier (Napoleonic era or earlier?) first used his analysis to study
conduction not of electric current but of of heat. That was long before the invention of the electric soldering iron. When the soldering iron (actually copper) arrived Fourier's analysis was already here to greet it. Then along came Oliver Heaviside who turned the World upside down by replacing jw with p. |
Fourier (Napoleonic era or earlier?) first used his analysis to study
conduction not of electric current but of of heat. That was long before the invention of the electric soldering iron. When the soldering iron (actually copper) arrived Fourier's analysis was already here to greet it. Then along came Oliver Heaviside who turned the World upside down by replacing jw with p. |
On Fri, 12 Mar 2004 15:32:23 +0000, Ian Bell wrote:
Paul Burridge wrote: Hi all, Is there some black magic required to get higher order harmonics out of an oscillator? I'm only trying to get 17.2Mhz out of a 3.44Mhz source and am thus far failing spectacularly. I've tried everything I can think of so far to no avail. All I can get apart from the fundamental is a strong third harmonic on 10.32Mhz, regardless of what I tune for. In RF circles, the 'normal' way to do this would be a simple Class C amplifier with a collector load tuned to the fifth harmonic. In calls C, conduction only occurs for a small fraction of a cycle which produces a correspondingly higher proportion of higher harmonics than a square wave. I've been waiting for someone to post this. I would only add "The drive level, and the bais point, will vary the amount of fifth (or whichever) you will see." It's as common as noses in RF, as Ian pointed out. Just look at the average two-way radio prior to frequency synthesisers. Crystal freqs were multiplied this way in transmitter chains and for receive injection, although use of fifth wasn't especially common because you normally had enough design control to use the more efficient *2, *3 or *4. |
On Fri, 12 Mar 2004 15:32:23 +0000, Ian Bell wrote:
Paul Burridge wrote: Hi all, Is there some black magic required to get higher order harmonics out of an oscillator? I'm only trying to get 17.2Mhz out of a 3.44Mhz source and am thus far failing spectacularly. I've tried everything I can think of so far to no avail. All I can get apart from the fundamental is a strong third harmonic on 10.32Mhz, regardless of what I tune for. In RF circles, the 'normal' way to do this would be a simple Class C amplifier with a collector load tuned to the fifth harmonic. In calls C, conduction only occurs for a small fraction of a cycle which produces a correspondingly higher proportion of higher harmonics than a square wave. I've been waiting for someone to post this. I would only add "The drive level, and the bais point, will vary the amount of fifth (or whichever) you will see." It's as common as noses in RF, as Ian pointed out. Just look at the average two-way radio prior to frequency synthesisers. Crystal freqs were multiplied this way in transmitter chains and for receive injection, although use of fifth wasn't especially common because you normally had enough design control to use the more efficient *2, *3 or *4. |
"Reg Edwards" wrote in message ... The Hyperbolic Cosine is pronounced Cosh. The Hyperbolic Sine is pronounced Shine. The Hyperbolic Tangent is pronounced Than with a soft Th. At least that's the way I've been doing it for the last 55 years. Have you ever noticed no one sits next to you at meetings? They don't seem to come up very often in conversation although they are just as fundamental in mathematics as are the trigonometrical functions. They crop up all over the place especially in transmission lines where they appear in complex form such as Tanh(A+jB). |
"Reg Edwards" wrote in message ... The Hyperbolic Cosine is pronounced Cosh. The Hyperbolic Sine is pronounced Shine. The Hyperbolic Tangent is pronounced Than with a soft Th. At least that's the way I've been doing it for the last 55 years. Have you ever noticed no one sits next to you at meetings? They don't seem to come up very often in conversation although they are just as fundamental in mathematics as are the trigonometrical functions. They crop up all over the place especially in transmission lines where they appear in complex form such as Tanh(A+jB). |
Firstly, thanks to everyone who's responded to this question. I've had plenty of valuable leads to follow up on, for which I am as ever very grateful. On Sat, 13 Mar 2004 00:16:28 GMT, James Meyer wrote: Is this a simulated circuit or a "real" one built with "real" components? It *is* actually a real one in this instance! Although I've simulated it as well, of course, but that hasn't provided any clues as to what might be causing the problem with the actual circuit. I have at least one suggestion, but I need to know whether to send an LTspice netlist or a gif. Send 'em both! -- The BBC: Licensed at public expense to spread lies. |
Firstly, thanks to everyone who's responded to this question. I've had plenty of valuable leads to follow up on, for which I am as ever very grateful. On Sat, 13 Mar 2004 00:16:28 GMT, James Meyer wrote: Is this a simulated circuit or a "real" one built with "real" components? It *is* actually a real one in this instance! Although I've simulated it as well, of course, but that hasn't provided any clues as to what might be causing the problem with the actual circuit. I have at least one suggestion, but I need to know whether to send an LTspice netlist or a gif. Send 'em both! -- The BBC: Licensed at public expense to spread lies. |
On Sat, 13 Mar 2004 10:00:52 +1000, Tony wrote:
The 5th harmonic should be only 14dB below the fundamental, although it will drop fairly quickly as the sides of the input square wave deviate from vertical. Does the 3.44MHz have a 50% duty cycle? Not quite, no. Why would that make any difference? I'd have thought any decent 'squarish wave' of the correct frequency with sharp rise/fall edges ought to do the trick? It's spewing out the 3rd quite nicely after all. How about I post a pic of the sig trace into the multiplier? I'll see if I can do that a bit later 2day... -- The BBC: Licensed at public expense to spread lies. |
On Sat, 13 Mar 2004 10:00:52 +1000, Tony wrote:
The 5th harmonic should be only 14dB below the fundamental, although it will drop fairly quickly as the sides of the input square wave deviate from vertical. Does the 3.44MHz have a 50% duty cycle? Not quite, no. Why would that make any difference? I'd have thought any decent 'squarish wave' of the correct frequency with sharp rise/fall edges ought to do the trick? It's spewing out the 3rd quite nicely after all. How about I post a pic of the sig trace into the multiplier? I'll see if I can do that a bit later 2day... -- The BBC: Licensed at public expense to spread lies. |
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 |
"Paul Burridge" wrote in message ... Hi all, I'm only trying to get 17.2Mhz out of a 3.44Mhz source and am thus far failing spectrally. I've tried everything I can think of so far to no avail. Any suggestions, please. I'm stumped! :( -- The BBC: blah de blah, yawn. Certainly. Next time you want to put your foot in your mouth don't remove it from the end of your leg. Then you won't be......... Otherwise 17MHz sits in the range of a 74HC4046 PLL. I shouldn't suggest such things lest you start asking other questions..... but. DNA |
"Paul Burridge" wrote in message ... Hi all, I'm only trying to get 17.2Mhz out of a 3.44Mhz source and am thus far failing spectrally. I've tried everything I can think of so far to no avail. Any suggestions, please. I'm stumped! :( -- The BBC: blah de blah, yawn. Certainly. Next time you want to put your foot in your mouth don't remove it from the end of your leg. Then you won't be......... Otherwise 17MHz sits in the range of a 74HC4046 PLL. I shouldn't suggest such things lest you start asking other questions..... but. DNA |
The Hyperbolic Cosine is pronounced Cosh.
The Hyperbolic Sine is pronounced Shine. The Hyperbolic Tangent is pronounced Than with a soft Th. At least that's the way I've been doing it for the last 55 years. ========================== Have you ever noticed no one sits next to you at meetings? ========================== I always thought it was due to B.O. But we live and learn! |
The Hyperbolic Cosine is pronounced Cosh.
The Hyperbolic Sine is pronounced Shine. The Hyperbolic Tangent is pronounced Than with a soft Th. At least that's the way I've been doing it for the last 55 years. ========================== Have you ever noticed no one sits next to you at meetings? ========================== I always thought it was due to B.O. But we live and learn! |
On Fri, 12 Mar 2004 15:32:23 +0000, Ian Bell wrote:
In RF circles, the 'normal' way to do this would be a simple Class C amplifier with a collector load tuned to the fifth harmonic. In calls C, conduction only occurs for a small fraction of a cycle which produces a correspondingly higher proportion of higher harmonics than a square wave. But if you want to filter the 5th, it's mighty handy not to have nuch 4th or 6th around. Is a Smith Chart how you map an "RF circle"? John |
On Fri, 12 Mar 2004 15:32:23 +0000, Ian Bell wrote:
In RF circles, the 'normal' way to do this would be a simple Class C amplifier with a collector load tuned to the fifth harmonic. In calls C, conduction only occurs for a small fraction of a cycle which produces a correspondingly higher proportion of higher harmonics than a square wave. But if you want to filter the 5th, it's mighty handy not to have nuch 4th or 6th around. Is a Smith Chart how you map an "RF circle"? John |
Paul Burridge wrote in message . ..
Hi all, Is there some black magic required to get higher order harmonics out of an oscillator? I'm only trying to get 17.2Mhz out of a 3.44Mhz source and am thus far failing spectacularly. I've tried everything I can think of so far to no avail. All I can get apart from the fundamental is a strong third harmonic on 10.32Mhz, regardless of what I tune for. I've tried passing the osc output through two successive inverter gates to sharpen it up, but still nothing beyond the third appears after tuned amplification for the fifth. I no longer have a spectrum analyser so can't check for the presence of a decent comb of harmonics at the input to the multiplier stage but can only assume the fifth is well down in the mush for some reason. Fifth harmonic frequency multipliers do exist, but it's usually much easier to double and triple your way to the final frequency if possible. (You just discovered this, I think!) The lack of even harmonics is typical of push-pull stages ... if you are messing around with CMOS gates, you might try using a TTL gate (which pulls low much stronger than it pulls high) or an open collector TTL gate, both with smmallish (100-200 ohm) pull-up resistors for doubling. Why not do a x3 followed by a x2 to get 17.2 MHz out of 2.866 MHz? Tim. |
Paul Burridge wrote in message . ..
Hi all, Is there some black magic required to get higher order harmonics out of an oscillator? I'm only trying to get 17.2Mhz out of a 3.44Mhz source and am thus far failing spectacularly. I've tried everything I can think of so far to no avail. All I can get apart from the fundamental is a strong third harmonic on 10.32Mhz, regardless of what I tune for. I've tried passing the osc output through two successive inverter gates to sharpen it up, but still nothing beyond the third appears after tuned amplification for the fifth. I no longer have a spectrum analyser so can't check for the presence of a decent comb of harmonics at the input to the multiplier stage but can only assume the fifth is well down in the mush for some reason. Fifth harmonic frequency multipliers do exist, but it's usually much easier to double and triple your way to the final frequency if possible. (You just discovered this, I think!) The lack of even harmonics is typical of push-pull stages ... if you are messing around with CMOS gates, you might try using a TTL gate (which pulls low much stronger than it pulls high) or an open collector TTL gate, both with smmallish (100-200 ohm) pull-up resistors for doubling. Why not do a x3 followed by a x2 to get 17.2 MHz out of 2.866 MHz? Tim. |
On Sat, 13 Mar 2004 11:50:13 +0000, Paul Burridge
wrote: On Sat, 13 Mar 2004 10:00:52 +1000, Tony wrote: The 5th harmonic should be only 14dB below the fundamental, although it will drop fairly quickly as the sides of the input square wave deviate from vertical. Does the 3.44MHz have a 50% duty cycle? Not quite, no. Why would that make any difference? As the duty cycle deviates from 50%, the even harmonics start to appear, so you need a better filter to keep them out. John |
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