On Thu, 13 Oct 2011 10:24:17 -0700, Jon Kirwan <jonk@infinitefactors.org> wrote:>On Thu, 13 Oct 2011 10:13:20 -0700 (PDT), George Herold ><gherold@teachspin.com> wrote: > >>On Oct 13, 12:59�pm, Jon Kirwan <j...@infinitefactors.org> wrote: >>> On Thu, 13 Oct 2011 09:39:21 -0700 (PDT), George Herold >>> >>> >>> >>> >>> >>> <gher...@teachspin.com> wrote: >>> >This is a continuation of the 50kHz VCO thread I started last week. �I >>> >tried the stepped sine wave idea as suggested by James A, and Phil >>> >H. >>> >The circuit clocks a MC14017 at 10x(F) to make a stepped sine wave at >>> >frequency (F). �The ten outputs from the 4017 are sent through >>> >appropriate resistors and into the summing junction of an opamp. >>> >Here�s a �scope shot of the stepped output overlaid with a sine >>> >wave. >>> >>> >http://imageshack.us/photo/my-images/560/tek0024.png/ >>> >>> >The resistor values were chosen to intersect the sine wave at each new >>> >phase. �(R(n) = 1/sin^2(n*18degrees)) >>> >>> >Approximate values, R0=open, R1=R9=105k, R2=R8=28.9k, R3=R7=15.3k, >>> >R4=R6=11k, R5=10k. �all 1% resistors. >>> >>> >Here�s the spectrum as recorded by an SRS770 spectrum analyzer. >>> >>> >http://imageshack.us/photo/my-images/839/stepsin.png/ >>> >>> >The 2nd harmonic is only down by 50dB. �I don�t understand why it�s so >>> >big. �Is there some way to do better than this? �The 9th and 11th >>> >harmonics are big and then the 19th and 21st. >>> >>> >Thanks George H. >>> >>> George, I don't remember the discussion. �Probably didn't >>> read it. �But if you are using a 4017 (decade counter), then >>> I'm guessing that you are enabling one resistor at a time >>> while disabling others (they tie to the summing junction from >>> each, moving output pin.) �This worries me a little, mostly >>> because of delay and the fact that you are turning off one >>> while turning on another, but don't control that very well. I >>> also don't know what you are doing to filter the steps. >>> >>> Anyway, I'd have wanted to consider, instead, a Gray-coded >>> (actually, the real inventor is Boudot, I think, but Bell >>> Labs was patenting everything in a flurry in the mid 1900's >>> and who could remember Boudot so long ago?) design where you >>> only change one of the outputs at a time. �Not two. >>> >>> Anyway, I'll let the big hitters who probably did read the >>> earlier thread tell you what is more likely. �Just something >>> that crossed my mind, is all. >>> >>> Jon- Hide quoted text - >>> >>> - Show quoted text - >> >>Hi Jon, Thanks for that. > >No problem. In the interim, I did a quick search for summing >junctions and Gray codes and came up with this link: > >http://www.wiseguysynth.com/larry/schematics/walsh/walsh.pdfI posted this 8 years ago... http://www.analog-innovations.com/SED/SineEqualsSumOfSquares.pdf I also have this book... "Sequency Theory, Foundations and Applications" Henning F. Harmuth Academic Press, 1977 ISBN: 0-12-014569-3 which covers Walsh Functions in gruesome detail ;-)> >I have NOT read it. But it looks about right to me as a zero >order approximation to what I was thinking about. I have >never considered doing what you are doing, but your writing >sprung two things immediately to mind. One is doing Fourier >analysis (which you should have already done, I imagine) and >the other is that you are possibly changing two outputs at >once and with that plus ripple carry stuff I get kind of >worried. All this would make me want to go to theory to >calculate my expectations and make sure they matched >experience in your testing. If I can't match them up, that >means I don't know enough and need to read more. > >>The 4017 is just too simple! > >Yeah. And you know what that means. Things should be as >simple as needed but no more so. > >>I'm hoping all >>the swithing transients can just be filtered away. I put a little >>tweaker pot on the smallest R5 resistor and was able to get everything >>down below 60dB, so I'm thinking this is just a resistor tolerance/ >>selection issue. >>I was just twisting different resistors together to get the >>approximate values, and didn't measure any of them. I'll try really >>nailing the values I want. >> >>Oh I'll add some multi-pole low pass on the back end of this.. but >>that will do nothing for the lower order harmonics. > >I completely understand the problem in trying to filter out >2nd harmonics from the 1st. So yeah, that's not really a >good answer. Which is why I didn't really suggest it, but >instead went to Gray codes and the like. > >Check out the link and see if it triggers anything. I will >read it a little later on, it's interesting to me regardless. >But it might apply from a cursory glance at it. > >Jon...Jim Thompson -- | James E.Thompson, CTO | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | Phoenix, Arizona 85048 Skype: Contacts Only | | | Voice:(480)460-2350 Fax: Available upon request | Brass Rat | | E-mail Icon at http://www.analog-innovations.com | 1962 | I love to cook with wine. Sometimes I even put it in the food.
Stepped sine wave
Started by ●October 13, 2011
Reply by ●October 13, 20112011-10-13
Reply by ●October 13, 20112011-10-13
On Oct 13, 5:13=A0pm, Phil Hobbs <pcdhSpamMeSensel...@electrooptical.net> wrote:> On 10/13/2011 12:39 PM, George Herold wrote: > > > > > > > This is a continuation of the 50kHz VCO thread I started last week. =A0=I> > tried the stepped sine wave idea as suggested by James A, and Phil > > H. > > The circuit clocks a MC14017 at 10x(F) to make a stepped sine wave at > > frequency (F). =A0The ten outputs from the 4017 are sent through > > appropriate resistors and into the summing junction of an opamp. > > Here=92s a =91scope shot of the stepped output overlaid with a sine > > wave. > > >http://imageshack.us/photo/my-images/560/tek0024.png/ > > > The resistor values were chosen to intersect the sine wave at each new > > phase. =A0(R(n) =3D 1/sin^2(n*18degrees)) > > > Approximate values, R0=3Dopen, R1=3DR9=3D105k, R2=3DR8=3D28.9k, R3=3DR7==3D15.3k,> > R4=3DR6=3D11k, R5=3D10k. =A0all 1% resistors. > > > Here=92s the spectrum as recorded by an SRS770 spectrum analyzer. > > >http://imageshack.us/photo/my-images/839/stepsin.png/ > > > The 2nd harmonic is only down by 50dB. =A0I don=92t understand why it==92s so> > big. =A0Is there some way to do better than this? =A0The 9th and 11th > > harmonics are big and then the 19th and 21st. > > > Thanks George H. > > 50 dB is only 0.3%, which isn't too bad. =A0 That might easily be due to > the output impedances of the 4017 drivers, or to the resistor > tolerances. =A0Does it get better or worse when you change VDD? =A0If so, > it's probably the output impedance. > > Cheers > > Phil Hobbs- Hide quoted text - > > - Show quoted text -Yeah, I was thinking about the output impedance. (I didn't looked at the outputs from the 4017). When I decreased the supply voltage the 2nd harmonic was roughly constant while everything else went down. I added a tweaker on the lowest resistance output and got everything below the 9th close to 60 dB down. Which is almost beer time, except it's only at 1kHz. George H.
Reply by ●October 13, 20112011-10-13
On Thu, 13 Oct 2011 16:08:49 -0700, Jim Thompson <To-Email-Use-The-Envelope-Icon@On-My-Web-Site.com> wrote:>On Thu, 13 Oct 2011 10:24:17 -0700, Jon Kirwan ><jonk@infinitefactors.org> wrote: > >>On Thu, 13 Oct 2011 10:13:20 -0700 (PDT), George Herold >><gherold@teachspin.com> wrote: >> >>>On Oct 13, 12:59�pm, Jon Kirwan <j...@infinitefactors.org> wrote: >>>> On Thu, 13 Oct 2011 09:39:21 -0700 (PDT), George Herold >>>> >>>> <gher...@teachspin.com> wrote: >>>> >This is a continuation of the 50kHz VCO thread I started last week. �I >>>> >tried the stepped sine wave idea as suggested by James A, and Phil >>>> >H. >>>> >The circuit clocks a MC14017 at 10x(F) to make a stepped sine wave at >>>> >frequency (F). �The ten outputs from the 4017 are sent through >>>> >appropriate resistors and into the summing junction of an opamp. >>>> >Here�s a �scope shot of the stepped output overlaid with a sine >>>> >wave. >>>> >>>> >http://imageshack.us/photo/my-images/560/tek0024.png/ >>>> >>>> >The resistor values were chosen to intersect the sine wave at each new >>>> >phase. �(R(n) = 1/sin^2(n*18degrees)) >>>> >>>> >Approximate values, R0=open, R1=R9=105k, R2=R8=28.9k, R3=R7=15.3k, >>>> >R4=R6=11k, R5=10k. �all 1% resistors. >>>> >>>> >Here�s the spectrum as recorded by an SRS770 spectrum analyzer. >>>> >>>> >http://imageshack.us/photo/my-images/839/stepsin.png/ >>>> >>>> >The 2nd harmonic is only down by 50dB. �I don�t understand why it�s so >>>> >big. �Is there some way to do better than this? �The 9th and 11th >>>> >harmonics are big and then the 19th and 21st. >>>> >>>> >Thanks George H. >>>> >>>> George, I don't remember the discussion. �Probably didn't >>>> read it. �But if you are using a 4017 (decade counter), then >>>> I'm guessing that you are enabling one resistor at a time >>>> while disabling others (they tie to the summing junction from >>>> each, moving output pin.) �This worries me a little, mostly >>>> because of delay and the fact that you are turning off one >>>> while turning on another, but don't control that very well. I >>>> also don't know what you are doing to filter the steps. >>>> >>>> Anyway, I'd have wanted to consider, instead, a Gray-coded >>>> (actually, the real inventor is Boudot, I think, but Bell >>>> Labs was patenting everything in a flurry in the mid 1900's >>>> and who could remember Boudot so long ago?) design where you >>>> only change one of the outputs at a time. �Not two. >>>> >>>> Anyway, I'll let the big hitters who probably did read the >>>> earlier thread tell you what is more likely. �Just something >>>> that crossed my mind, is all. >>>> >>>> Jon- Hide quoted text - >>>> >>>> - Show quoted text - >>> >>>Hi Jon, Thanks for that. >> >>No problem. In the interim, I did a quick search for summing >>junctions and Gray codes and came up with this link: >> >>http://www.wiseguysynth.com/larry/schematics/walsh/walsh.pdf > >I posted this 8 years ago... > >http://www.analog-innovations.com/SED/SineEqualsSumOfSquares.pdf > >I also have this book... > >"Sequency Theory, Foundations and Applications" >Henning F. Harmuth >Academic Press, 1977 >ISBN: 0-12-014569-3 > >which covers Walsh Functions in gruesome detail ;-)I picke up Walsh's original paper (redone, actually, in Latex and error corrected as it had a few in the original article) from the web, today. The paper is "A Closed Set of Normal Orthogonal Functions." I will be reading it more thoroughly over the next couple of days. Also, already listed the above link which is a nice, short overview with two examples in it. Finally, there are a bevy of books (some of them nearly 1300 pages in length) on the subject regarding making and building synthesizers. I'll be ordering a few before the end of the week. A whole world has opened up on this subject for me and I can bring over Laplace and Fourier. Actually, it is almost easy for me to see how to apply this with almost any starting wave shape, not just sine/cosine or square wave. Which probably isn't terribly practical, but interesting all the same. (Must be some mathematician out there has already explored the use of triangle, sawtooth, and pretty much any arbitrary basic shape.) I will read your PDF, as well. :) Jon
Reply by ●October 13, 20112011-10-13
On Thu, 13 Oct 2011 09:39:21 -0700, George Herold wrote:> This is a continuation of the 50kHz VCO thread I started last week. I > tried the stepped sine wave idea as suggested by James A, and Phil H. > The circuit clocks a MC14017 at 10x(F) to make a stepped sine wave at > frequency (F). The ten outputs from the 4017 are sent through > appropriate resistors and into the summing junction of an opamp. Here’s > a ‘scope shot of the stepped output overlaid with a sine wave. > > http://imageshack.us/photo/my-images/560/tek0024.png/ > > The resistor values were chosen to intersect the sine wave at each new > phase. (R(n) = 1/sin^2(n*18degrees)) > > Approximate values, R0=open, R1=R9=105k, R2=R8=28.9k, R3=R7=15.3k, > R4=R6=11k, R5=10k. all 1% resistors. > > > Here’s the spectrum as recorded by an SRS770 spectrum analyzer. > > http://imageshack.us/photo/my-images/839/stepsin.png/ > > The 2nd harmonic is only down by 50dB. I don’t understand why it’s so > big. Is there some way to do better than this? The 9th and 11th > harmonics are big and then the 19th and 21st.So -- how easily can you filter that out? IIRC you needed a 3:1 frequency span -- can you stand having nine filters or so, and switching between them (I suppose if you're frequency modulating the answer is "NO!"). -- Tim Wescott Control system and signal processing consulting www.wescottdesign.com
Reply by ●October 13, 20112011-10-13
On 10/13/2011 07:30 PM, George Herold wrote:> On Oct 13, 5:13 pm, Phil Hobbs > <pcdhSpamMeSensel...@electrooptical.net> wrote: >> On 10/13/2011 12:39 PM, George Herold wrote: >> >> >> >> >> >>> This is a continuation of the 50kHz VCO thread I started last week. I >>> tried the stepped sine wave idea as suggested by James A, and Phil >>> H. >>> The circuit clocks a MC14017 at 10x(F) to make a stepped sine wave at >>> frequency (F). The ten outputs from the 4017 are sent through >>> appropriate resistors and into the summing junction of an opamp. >>> Here�s a �scope shot of the stepped output overlaid with a sine >>> wave. >> >>> http://imageshack.us/photo/my-images/560/tek0024.png/ >> >>> The resistor values were chosen to intersect the sine wave at each new >>> phase. (R(n) = 1/sin^2(n*18degrees)) >> >>> Approximate values, R0=open, R1=R9=105k, R2=R8=28.9k, R3=R7=15.3k, >>> R4=R6=11k, R5=10k. all 1% resistors. >> >>> Here�s the spectrum as recorded by an SRS770 spectrum analyzer. >> >>> http://imageshack.us/photo/my-images/839/stepsin.png/ >> >>> The 2nd harmonic is only down by 50dB. I don�t understand why it�s so >>> big. Is there some way to do better than this? The 9th and 11th >>> harmonics are big and then the 19th and 21st. >> >>> Thanks George H. >> >> 50 dB is only 0.3%, which isn't too bad. That might easily be due to >> the output impedances of the 4017 drivers, or to the resistor >> tolerances. Does it get better or worse when you change VDD? If so, >> it's probably the output impedance. >> >> Cheers >> >> Phil Hobbs- Hide quoted text - >> >> - Show quoted text - > > Yeah, I was thinking about the output impedance. (I didn't looked at > the outputs from the 4017). When I decreased the supply voltage the > 2nd harmonic was roughly constant while everything else went down. I > added a tweaker on the lowest resistance output and got everything > below the 9th close to 60 dB down. Which is almost beer time, except > it's only at 1kHz. > > George H.Well, they come in HC too. ;) Cheers Phil Hobbs -- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 845-480-2058 hobbs at electrooptical dot net http://electrooptical.net
Reply by ●October 13, 20112011-10-13
On Oct 13, 6:30=A0pm, George Herold <gher...@teachspin.com> wrote:> On Oct 13, 5:13=A0pm, Phil Hobbs wrote: > > On 10/13/2011 12:39 PM, George Herold wrote: > > > > This is a continuation of the 50kHz VCO thread I started last week. ==A0I> > > tried the stepped sine wave idea as suggested by James A, and Phil > > > H. > > > The circuit clocks a MC14017 at 10x(F) to make a stepped sine wave at > > > frequency (F). =A0The ten outputs from the 4017 are sent through > > > appropriate resistors and into the summing junction of an opamp. > > > Here=92s a =91scope shot of the stepped output overlaid with a sine > > > wave. > > > >http://imageshack.us/photo/my-images/560/tek0024.png/ > > > > The resistor values were chosen to intersect the sine wave at each ne=w> > > phase. =A0(R(n) =3D 1/sin^2(n*18degrees)) > > > > Approximate values, R0=3Dopen, R1=3DR9=3D105k, R2=3DR8=3D28.9k, R3=3D=R7=3D15.3k,> > > R4=3DR6=3D11k, R5=3D10k. =A0all 1% resistors. > > > > Here=92s the spectrum as recorded by an SRS770 spectrum analyzer. > > > >http://imageshack.us/photo/my-images/839/stepsin.png/ > > > > The 2nd harmonic is only down by 50dB. =A0I don=92t understand why it==92s so> > > big. =A0Is there some way to do better than this? =A0The 9th and 11th > > > harmonics are big and then the 19th and 21st. > > > > Thanks George H. > > > 50 dB is only 0.3%, which isn't too bad. =A0 That might easily be due t=o> > the output impedances of the 4017 drivers, or to the resistor > > tolerances. =A0Does it get better or worse when you change VDD? =A0If s=o,> > it's probably the output impedance. > > > Yeah, I was thinking about the output impedance. (I didn't looked at > the outputs from the 4017). =A0When I decreased the supply voltage the > 2nd harmonic was roughly constant while everything else went down. =A0I > added a tweaker on the lowest resistance output and got everything > below the 9th close to 60 dB down. =A0 Which is almost beer time, except > it's only at 1kHz.The outputs are loaded. The effect of this overloading is to squish the upper part of the waveform, distorting it. It looks like you've tweaked values pretty optimally to compensate, but that's part- dependent. How about multiplying all those resistances by 10? These are the relative step-sizes you should be seeing (calculated from your resistor values, in 'scope divisions, scaled and offset to match the 'scope screen shot you posted): (view in fixed font) calc'd actual desired 0.2 0.2 0.2 1.05 1 0.93 3.15 2.9 2.82 5.47 5.2 5.17 7.22 7.05 7.07 7.8 7.8 7.8 (sequence repeats, falling) Looks pretty good, but FWIW steps 1-9 and 3-9 look slightly mis- matched. -- Cheers, James Arthur
Reply by ●October 13, 20112011-10-13
On Oct 13, 8:32=A0pm, Phil Hobbs wrote:> On 10/13/2011 07:30 PM, George Herold wrote: > > > > On Oct 13, 5:13 pm, Phil Hobbs =A0wrote: > >> On 10/13/2011 12:39 PM, George Herold wrote: > > >>> This is a continuation of the 50kHz VCO thread I started last week. ==A0I> >>> tried the stepped sine wave idea as suggested by James A, and Phil > >>> H. > >>> The circuit clocks a MC14017 at 10x(F) to make a stepped sine wave at > >>> frequency (F). =A0The ten outputs from the 4017 are sent through > >>> appropriate resistors and into the summing junction of an opamp. > >>> Here=92s a =91scope shot of the stepped output overlaid with a sine > >>> wave. > > >>>http://imageshack.us/photo/my-images/560/tek0024.png/ > > >>> The resistor values were chosen to intersect the sine wave at each ne=w> >>> phase. =A0(R(n) =3D 1/sin^2(n*18degrees)) > > >>> Approximate values, R0=3Dopen, R1=3DR9=3D105k, R2=3DR8=3D28.9k, R3=3D=R7=3D15.3k,> >>> R4=3DR6=3D11k, R5=3D10k. =A0all 1% resistors. > > >>> Here=92s the spectrum as recorded by an SRS770 spectrum analyzer. > > >>>http://imageshack.us/photo/my-images/839/stepsin.png/ > > >>> The 2nd harmonic is only down by 50dB. =A0I don=92t understand why it==92s so> >>> big. =A0Is there some way to do better than this? =A0The 9th and 11th > >>> harmonics are big and then the 19th and 21st. > > > >> 50 dB is only 0.3%, which isn't too bad. =A0 That might easily be due =to> >> the output impedances of the 4017 drivers, or to the resistor > >> tolerances. =A0Does it get better or worse when you change VDD? =A0If =so,> >> it's probably the output impedance. > > > > > Yeah, I was thinking about the output impedance. (I didn't looked at > > the outputs from the 4017). =A0When I decreased the supply voltage the > > 2nd harmonic was roughly constant while everything else went down. =A0I > > added a tweaker on the lowest resistance output and got everything > > below the 9th close to 60 dB down. =A0 Which is almost beer time, excep=t> > it's only at 1kHz. > > > Well, they come in HC too. ;)Cool. I didn't think they made those in 'HC, but DigiKey says you're right. -- Cheers, James Arthur
Reply by ●October 13, 20112011-10-13
On 10/13/2011 09:57 PM, dagmargoodboat@yahoo.com wrote:> On Oct 13, 8:32 pm, Phil Hobbs wrote: >> On 10/13/2011 07:30 PM, George Herold wrote: >> >> >>> On Oct 13, 5:13 pm, Phil Hobbs wrote: >>>> On 10/13/2011 12:39 PM, George Herold wrote: >> >>>>> This is a continuation of the 50kHz VCO thread I started last week. I >>>>> tried the stepped sine wave idea as suggested by James A, and Phil >>>>> H. >>>>> The circuit clocks a MC14017 at 10x(F) to make a stepped sine wave at >>>>> frequency (F). The ten outputs from the 4017 are sent through >>>>> appropriate resistors and into the summing junction of an opamp. >>>>> Here�s a �scope shot of the stepped output overlaid with a sine >>>>> wave. >> >>>>> http://imageshack.us/photo/my-images/560/tek0024.png/ >> >>>>> The resistor values were chosen to intersect the sine wave at each new >>>>> phase. (R(n) = 1/sin^2(n*18degrees)) >> >>>>> Approximate values, R0=open, R1=R9=105k, R2=R8=28.9k, R3=R7=15.3k, >>>>> R4=R6=11k, R5=10k. all 1% resistors. >> >>>>> Here�s the spectrum as recorded by an SRS770 spectrum analyzer. >> >>>>> http://imageshack.us/photo/my-images/839/stepsin.png/ >> >>>>> The 2nd harmonic is only down by 50dB. I don�t understand why it�s so >>>>> big. Is there some way to do better than this? The 9th and 11th >>>>> harmonics are big and then the 19th and 21st. >> >> >>>> 50 dB is only 0.3%, which isn't too bad. That might easily be due to >>>> the output impedances of the 4017 drivers, or to the resistor >>>> tolerances. Does it get better or worse when you change VDD? If so, >>>> it's probably the output impedance. >> >> >> >>> Yeah, I was thinking about the output impedance. (I didn't looked at >>> the outputs from the 4017). When I decreased the supply voltage the >>> 2nd harmonic was roughly constant while everything else went down. I >>> added a tweaker on the lowest resistance output and got everything >>> below the 9th close to 60 dB down. Which is almost beer time, except >>> it's only at 1kHz. >> >> >> Well, they come in HC too. ;) > > Cool. I didn't think they made those in 'HC, but DigiKey says you're > right. >It's the physicist's version of the 555. ;) Cheers Phil Hobbs -- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 845-480-2058 hobbs at electrooptical dot net http://electrooptical.net
Reply by ●October 13, 20112011-10-13
On Oct 13, 6:30=A0pm, George Herold <gher...@teachspin.com> wrote:> On Oct 13, 5:13=A0pm, Phil Hobbs wrote: > > On 10/13/2011 12:39 PM, George Herold wrote: > > > > This is a continuation of the 50kHz VCO thread I started last week. ==A0I> > > tried the stepped sine wave idea as suggested by James A, and Phil > > > H. > > > The circuit clocks a MC14017 at 10x(F) to make a stepped sine wave at > > > frequency (F). =A0The ten outputs from the 4017 are sent through > > > appropriate resistors and into the summing junction of an opamp. > > > Here=92s a =91scope shot of the stepped output overlaid with a sine > > > wave. > > > >http://imageshack.us/photo/my-images/560/tek0024.png/ > > > > The resistor values were chosen to intersect the sine wave at each ne=w> > > phase. =A0(R(n) =3D 1/sin^2(n*18degrees)) > > > > Approximate values, R0=3Dopen, R1=3DR9=3D105k, R2=3DR8=3D28.9k, R3=3D=R7=3D15.3k,> > > R4=3DR6=3D11k, R5=3D10k. =A0all 1% resistors. > > > > Here=92s the spectrum as recorded by an SRS770 spectrum analyzer. > > > >http://imageshack.us/photo/my-images/839/stepsin.png/ > > > > The 2nd harmonic is only down by 50dB. =A0I don=92t understand why it==92s so> > > big. =A0Is there some way to do better than this? =A0The 9th and 11th > > > harmonics are big and then the 19th and 21st. > > > > 50 dB is only 0.3%, which isn't too bad. =A0 That might easily be due t=o> > the output impedances of the 4017 drivers, or to the resistor > > tolerances. =A0Does it get better or worse when you change VDD? =A0If s=o,> > it's probably the output impedance. > > > Yeah, I was thinking about the output impedance. (I didn't looked at > the outputs from the 4017). =A0When I decreased the supply voltage the > 2nd harmonic was roughly constant while everything else went down. =A0I > added a tweaker on the lowest resistance output and got everything > below the 9th close to 60 dB down. =A0 Which is almost beer time, except > it's only at 1kHz.I posted (and Google lost) a long post, the gist of which was: (view in fixed font) desired, G.H. desired, scaled to George's values theoretical 'scope actual (volts) (volts) (div) (div) 0.000 0.000 0.2 0.2 0.093 0.079 0.93 1.0 0.323 0.287 2.82 2.8 0.577 0.545 5.17 5.2 0.768 0.753 7.07 7.05 0.832 0.832 7.8 7.8 (1rst column is the expected outputs based on your resistor values) So, the outputs are loaded, but you've tweaked the resistors from the reported values to compensate. That's device-dependent. Might wanna up the resistances or switch to 'HC. The actual waveform looks pretty good. Some of the steps look a little mis-matched, e.g. 2-8 and 3-7 -- Cheers, James Arthur
Reply by ●October 13, 20112011-10-13
On Oct 13, 8:40=A0pm, Tim <t...@seemywebsite.please> wrote:> On Thu, 13 Oct 2011 09:39:21 -0700, George Herold wrote: > > This is a continuation of the 50kHz VCO thread I started last week. =A0=I> > tried the stepped sine wave idea as suggested by James A, and Phil H. > > The circuit clocks a MC14017 at 10x(F) to make a stepped sine wave at > > frequency (F). =A0The ten outputs from the 4017 are sent through > > appropriate resistors and into the summing junction of an opamp. Here==92s> > a =91scope shot of the stepped output overlaid with a sine wave. > > >http://imageshack.us/photo/my-images/560/tek0024.png/ > > > The resistor values were chosen to intersect the sine wave at each new > > phase. =A0(R(n) =3D 1/sin^2(n*18degrees)) > > > Approximate values, R0=3Dopen, R1=3DR9=3D105k, R2=3DR8=3D28.9k, R3=3DR7==3D15.3k,> > R4=3DR6=3D11k, R5=3D10k. =A0all 1% resistors. > > > Here=92s the spectrum as recorded by an SRS770 spectrum analyzer. > > >http://imageshack.us/photo/my-images/839/stepsin.png/ > > > The 2nd harmonic is only down by 50dB. =A0I don=92t understand why it==92s so> > big. =A0Is there some way to do better than this? =A0The 9th and 11th > > harmonics are big and then the 19th and 21st. > > So -- how easily can you filter that out? =A0IIRC you needed a 3:1 > frequency span -- can you stand having nine filters or so, and switching > between them (I suppose if you're frequency modulating the answer is > "NO!"). > > -- > Tim Wescott > Control system and signal processing consultingwww.wescottdesign.com- Hid=e quoted text -> > - Show quoted text -Oh, the single low pass has to start before the 9th harmonic of the lowest frequency (~30kHz*) so I'm thinking around 200 kHz. The resistors have to take care of the lower harmonics. (which is why the 2nd is so disturbing) 9 or 10, 0.1% resistors are not 'out of the question' only ~$2 + the cost of placing them. Though 1% would be nicer. George H. *the specs are still a bit fluid, which is a good thing.
