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On Wed, 17 Mar 2004 08:58:18 -0600, John Fields
wrote: It's an FFT of what's coming out of Larkin's suggested two-stage bandpass filter. The first vertical marker (f1) goes through the first peak, 17.19MHz, which proves the fifth is in a 3.44MHz square wave. The second marker goes through 34.38, so so's the tenth. Curious that there should be a sizeable pass response at the tenth harmonic, isn't it? It doesn't appear to be *that* much down on the intended pass frequency although there appears to be no indexing for the y axis. Are you kidding? Count the high frequency cycles between the first leading edge and the same point on the second leading edge of the square wave and you'll find there are exactly five. Well, I admit I'm a bit of a greenhorn on these things, but to my eyes there appears to be some phase difference. I'll accept your word for it there isn't. What was the active device you used to generate them? --- Tektronics FG502. Here's the layout: [TEK FG502]-+-[BPF]-+-[TEK 2465A]---[HP5328A] | | | +-[HP54602B] | [TEK DC504] Many thanks. I hope it didn't involve you in too much setting-up time to investigate this and post your findings. -- The BBC: Licensed at public expense to spread lies. |
In article , Peter John Lawton
writes: The above will hold true at any fundamental frequency provided the rise and fall times are equal and each equal to 0.02 times the repetition period. Those numbers will change given faster or slower rise/fall times. All db calculated as 20 x Log (voltage). Width is determined at the baseline, not the 50% amplitude point. Len Anderson retired (from regular hours) electronic engineer person I wonder what happens to these numbers as the rise/fall time tends to zero? The harmonic content will increase...but also show dips depending on the percentage width relative to the period. I could present those (takes only minutes to run the program and transcribe the results) but that is academic only. The rise and fall times will NOT be zero due to the repetition frequency being high (repetition time short). Consider that a 3 MHz waveform has a period of 333 1/3 nSec and that Paul is using a TTL family inverter to make the square wave. Even with a Schmitt trigger inverter the t_r and t_f are going to be finite, possibly 15 nSec with a fast device (and some capacitive loading or semi-resonant whatever to mess with on- and off-times). 15 nSec is 4.5% of the repetition period, quite finite...more than I showed on the small table given previously. I'm sure someone out there wants to argue minutae on numbers but what is being discussed is a squarish waveform with a repetition frequency in the low HF range. Periods are valued in nanoSeconds and the on/off times of squaring devices are ALSO in nanoSeconds. There's just NOT going to be any sort of "zero" on/off times with practical logic devices used by hobbyists. What is not intuitive to me (and to others) is that harmonic energy of a rectangular waveform drops drastically by the 5th harmonic and is certainly lower than "obvious" numbers bandied about. But, also mentioned before by others is that shortening the rect- angular waveshape DOES increase the 5th harmonic, as evident by the approximate 12 db increase at 40 to 35 percent of the repetition period. An equivalent shortening happens in vacuum tube multipliers through biasing (self, fixed, or both) and that can be adjustable along with the drive level. It's not quite the same with bipolars since the overdrive effects are more saturation than in the self- bias conditions of tubes. It's close, though. From all indications of the Fourier series results, there's a definite reason why so few multipliers went beyond tripling. The amount of energy (relative to fundamental and taking into account the finite rise and fall times) of 4th and higher harmonics just isn't as much as intuition would have everyone believe! Len Anderson retired (from regular hours) electronic engineer person |
In article , Peter John Lawton
writes: The above will hold true at any fundamental frequency provided the rise and fall times are equal and each equal to 0.02 times the repetition period. Those numbers will change given faster or slower rise/fall times. All db calculated as 20 x Log (voltage). Width is determined at the baseline, not the 50% amplitude point. Len Anderson retired (from regular hours) electronic engineer person I wonder what happens to these numbers as the rise/fall time tends to zero? The harmonic content will increase...but also show dips depending on the percentage width relative to the period. I could present those (takes only minutes to run the program and transcribe the results) but that is academic only. The rise and fall times will NOT be zero due to the repetition frequency being high (repetition time short). Consider that a 3 MHz waveform has a period of 333 1/3 nSec and that Paul is using a TTL family inverter to make the square wave. Even with a Schmitt trigger inverter the t_r and t_f are going to be finite, possibly 15 nSec with a fast device (and some capacitive loading or semi-resonant whatever to mess with on- and off-times). 15 nSec is 4.5% of the repetition period, quite finite...more than I showed on the small table given previously. I'm sure someone out there wants to argue minutae on numbers but what is being discussed is a squarish waveform with a repetition frequency in the low HF range. Periods are valued in nanoSeconds and the on/off times of squaring devices are ALSO in nanoSeconds. There's just NOT going to be any sort of "zero" on/off times with practical logic devices used by hobbyists. What is not intuitive to me (and to others) is that harmonic energy of a rectangular waveform drops drastically by the 5th harmonic and is certainly lower than "obvious" numbers bandied about. But, also mentioned before by others is that shortening the rect- angular waveshape DOES increase the 5th harmonic, as evident by the approximate 12 db increase at 40 to 35 percent of the repetition period. An equivalent shortening happens in vacuum tube multipliers through biasing (self, fixed, or both) and that can be adjustable along with the drive level. It's not quite the same with bipolars since the overdrive effects are more saturation than in the self- bias conditions of tubes. It's close, though. From all indications of the Fourier series results, there's a definite reason why so few multipliers went beyond tripling. The amount of energy (relative to fundamental and taking into account the finite rise and fall times) of 4th and higher harmonics just isn't as much as intuition would have everyone believe! Len Anderson retired (from regular hours) electronic engineer person |
Paul Burridge wrote in message . ..
.... What leads you to believe I have enough 5th harmonic in *my* particular case? The trace on the web site you provided a link to. The fact that you're using HC logic (which has inherent rise and fall times rather faster than the square wave source I used in my experiment). Your avering that the duty cycle is very nearly 50%. That's not to say you aren't doing something to kill it, but it's NOT difficult to extract it. Note that the subject you put on this thread really nails it: all you need to do is extract (and possibly amplify, depending on the final power level you need) what's already there. Now go do it. But feeding the whole square wave to the amplifier stage is a BAD idea because you can inadventently change the duty cycle (as seen at that amplifier's output) to one where the fifth is nulled. If you only need a few milliwatts, you can get that from the square wave directly, if the source impedance is low enough, simply by using the proper filter. |
Paul Burridge wrote in message . ..
.... What leads you to believe I have enough 5th harmonic in *my* particular case? The trace on the web site you provided a link to. The fact that you're using HC logic (which has inherent rise and fall times rather faster than the square wave source I used in my experiment). Your avering that the duty cycle is very nearly 50%. That's not to say you aren't doing something to kill it, but it's NOT difficult to extract it. Note that the subject you put on this thread really nails it: all you need to do is extract (and possibly amplify, depending on the final power level you need) what's already there. Now go do it. But feeding the whole square wave to the amplifier stage is a BAD idea because you can inadventently change the duty cycle (as seen at that amplifier's output) to one where the fifth is nulled. If you only need a few milliwatts, you can get that from the square wave directly, if the source impedance is low enough, simply by using the proper filter. |
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Ignoring most of the previous posts in this thread (sorry - I deleted them) it
occurs to me that maybe the problem may be the relatively low impedance load on the buffer, whose finite output impedance therefore causes some waveform distortion and loss of 5th harmonic. So how about a hi-Z parallel tank coupling circuit; for 17.2MHz (say 4p7 || 18uH) into a load of 100-500 ohms (say a 1k pot + 100n to Gnd, to see the effect of load variations), then into another 74AC buffer? Tony (remove the "_" to reply by email) |
Ignoring most of the previous posts in this thread (sorry - I deleted them) it
occurs to me that maybe the problem may be the relatively low impedance load on the buffer, whose finite output impedance therefore causes some waveform distortion and loss of 5th harmonic. So how about a hi-Z parallel tank coupling circuit; for 17.2MHz (say 4p7 || 18uH) into a load of 100-500 ohms (say a 1k pot + 100n to Gnd, to see the effect of load variations), then into another 74AC buffer? Tony (remove the "_" to reply by email) |
On Thu, 18 Mar 2004 10:42:22 +1000, Tony wrote:
Ignoring most of the previous posts in this thread (sorry - I deleted them) it occurs to me that maybe the problem may be the relatively low impedance load on the buffer, whose finite output impedance therefore causes some waveform distortion and loss of 5th harmonic. So how about a hi-Z parallel tank coupling circuit; for 17.2MHz (say 4p7 || 18uH) into a load of 100-500 ohms (say a 1k pot + 100n to Gnd, to see the effect of load variations), then into another 74AC buffer? Tony (remove the "_" to reply by email) Actually the tank's output will be biased to half-rail, which won't necessarily bias the CMOS output buffer properly. May need to cap-couple to the buffer, with a feedback resistor so it will self-bias (another case of reading the post AFTER hitting "send"). Tony (remove the "_" to reply by email) |
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