Tim, You might like to include in your paper a discussion on complex sampling, and explain why the rate can be half that of real sampling yet still comply with the Nyquist criterion. Jeff

# What Nyquist Didn't Say

Started by ●September 29, 2006

Reply by ●September 30, 20062006-09-30

Reply by ●September 30, 20062006-09-30

Mike Monett wrote:> "Helmut Sennewald" <helmutsennewald@t-online.de> wrote: > > > "Mike Monett" <No@email.adr> schrieb im Newsbeitrag > > news:Xns984E39E43288BNoemailadr@208.49.80.251... > > >> miso@sushi.com wrote: > > >>> Note that the Bessel filter will ring at higher orders. I don't > >>> have my copy of Zverev handy, but I think the Bessel rings at > >>> 4th order and higher. > > >> Are you sure about that? Here is a 9th order 1 MHz Bessel for > >> LTspice. (Save as a CKT file) > > >> * UTS Mike Monett > >> * Converted From Micro Cap Source file to LTspice > >> * > >> C1 0 1 248.3PF > >> C2 0 2 1200.0PF > >> C3 0 3 2007.3PF > >> C4 0 4 2749.9PF > >> C5 0 Vout 7209.4PF > >> L1 1 2 1.8UH > >> L2 2 3 4.1UH > >> L3 3 4 5.9UH > >> L4 4 Vout 8.6UH > >> R1 Vin 1 50 > >> R5 0 Vout 50 > >> V1 Vin 0 DC 0 PULSE (0 1 0 0 0 2.5e-006 5e-006) > >> .TRAN 1e-008 10u 0 1n UIC > >> .PRINT TRAN V(VOUT) V(VIN) > >> .PLOT TRAN V(VOUT) V(VIN) > >> .PROBE > >> .END > >> ;$SpiceType=SPICE3 > > >> It doesn't ring. > > >> Regards, > > >> Mike Monett > > > Hello Mike, > > > It seems you have overlooked that LTspice assumes a certain rise > > and fall time if you specify t_rise=0 or t_fall=0. > > > V1 Vin 0 DC 0 PULSE (0 1 0 0 0 2.5e-006 5e-006) > > > LTspice assumes in this case trise = fall = 10%*Twidth = 250ns > > > This is the reason why you see no overshoot. > > > Now we use fast edges. > > > V1 Vin 0 DC 0 PULSE (0 1 0 1n 1n 2.5e-006 5e-006) > > > You will see about 1.6x% overshoot. By the way, I am not sure how > > precise the choosen component values are to get the ideal Bessel > > filter response. Bessel filters have indeed overshoot in the step > > response. > > > Best regards, > > Helmut > > Hi Helmut, > > Thanks for taking a look at this, and for pointing out that LTspice > changes the rise and fall times if you do not specify the values. I > am still learning how LTspice works, and these unexpected variations > from other SPICE programs can be major pitfalls. > > In this case, the amount of overshoot is very small, and I'm not > sure if the 1.6% is not caused by the way the component values are > rounded. > > I use this filter program for general filter design. The component > values were calculated for the theoretical values and the inductors > were rounded to 2 significant digits. This reflects the difficulty > in obtaining precision inductors, particularly at VHF and UHF. > > The capacitors were rounded to tenths of a pf. This is because the > program is used mostly for work at VHF and UHF, where the caps are > much smaller. It may well be that using the correct theoretical > values for the components may reduce or eliminate the small amount > of overshoot you measured. > > However, there is still the practical matter of obtaining inductors > and capacitors with the needed precision, and any practical filter > will have imperfect components. I have found the Bessel and > Equiripple to be much more tolerant of errors in component values > than other filter types, and still give good performance. > > If we are concerned about such small values of overshoot, we need to > repeat this using the correct theoretical values for the components. > Even so, an overshoot of 1.6% is not in the same category as the > overshoot from Butterworth or other non-linear phase filters, which > may have ten percent or more, depending on the sharpness of the > cutoff. In practise, the Bessel and Equiripple are considered linear > phase filters, and have negligible overshoot. > > For example, a Butterworth or other nonlinear group delay filter may > cause excessive timing errors when used to filter digital data. This > can be minimized by changing to a Bessel or Equiripple filter. > > So practically speaking, the Bessel and Equiripple distinguish > themselves from other filters by the constant group delay through > the filter bandpass, and the low or non-existant overshoot. > > Regards, > > Mike Monett > > Antiviral, Antibacterial Silver Solution: > http://silversol.freewebpage.org/index.htm > SPICE Analysis of Crystal Oscillators: > http://silversol.freewebpage.org/spice/xtal/clapp.htm > Noise-Rejecting Wideband Sampler: > http://www3.sympatico.ca/add.automation/sampler/intro.htmYou really need to look at theory versus what spice predicts. The input is an ideal step. As you do the convolution, the negative region of the impulse response subtracts from the result, making the signal drop in value, then the postive regions make the result increase in value, hence ringing.. The Gaussian impulse response is always positive, hence the convolution output can't decrease as time increases. Linear phase alone is not enough to stop ringing. The discrete time situation is easier to understand, especially if you consider a finite impulse response filter. The response of a FIR filter to a step input is the the sum of the tap coefficients from one to N. That is, the first output is the first tap. The second output is the sum of the first two taps, etc. If no tap is negative, then the output always rises, hence no ringing. No circuits were simulated in writing this post. ;-)

Reply by ●September 30, 20062006-09-30

On 30 Sep 2006 17:33:25 -0700, miso@sushi.com wrote: [snip]> >No circuits were simulated in writing this post. ;-)Bwahahahaha! ROTFLMAO! ...Jim Thompson -- | James E.Thompson, P.E. | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | Phoenix, Arizona Voice:(480)460-2350 | | | E-mail Address at Website Fax:(480)460-2142 | Brass Rat | | http://www.analog-innovations.com | 1962 | I love to cook with wine. Sometimes I even put it in the food.

Reply by ●September 30, 20062006-09-30

"Ban" <bansuri@masterweb.it> wrote:> > Bessel filters do have a small overshoot, which decreases at higher > orders. 2nd .43% > 4th .84% > 6th .64% > 8th .34% > 10th .06% > note the *very low* value for higher orders. In fact the frequency > response gets hardly better then. > The Gauss filter has indeed zero ringing, but a much larger transition > band.Ban, Thanks for the clarification. In most references, the Bessel is considered to have low, negligible, or no overshoot, especially when compared to Butterworth, Chebyshev and other types of filters. Your numbers confirm this. In practise, it is difficult to obtain the exact component values needed for the theoretical performance. Not only are the values non-standard, but it may be difficult to get the inductor "Q" values used in most calculations. So we can assume there will be some deviation from the theoretical performance, and the overshoot will probably increase slightly. However, it is still low enough to be difficult to measure, and the terms "low", "neglible" or "no overshoot" are quite descriptive. Regards, Mike Monett Antiviral, Antibacterial Silver Solution: http://silversol.freewebpage.org/index.htm SPICE Analysis of Crystal Oscillators: http://silversol.freewebpage.org/spice/xtal/clapp.htm Noise-Rejecting Wideband Sampler: http://www3.sympatico.ca/add.automation/sampler/intro.htm

Reply by ●September 30, 20062006-09-30

miso@sushi.com wrote: > You really need to look at theory versus what spice predicts. The > input is an ideal step. As you do the convolution, the negative > region of the impulse response subtracts from the result, making > the signal drop in value, then the postive regions make the result > increase in value, hence ringing.. The Gaussian impulse response > is always positive, hence the convolution output can't decrease as > time increases. > Linear phase alone is not enough to stop ringing. > The discrete time situation is easier to understand, especially if > you consider a finite impulse response filter. The response of a > FIR filter to a step input is the the sum of the tap coefficients > from one to N. > That is, the first output is the first tap. The second output is > the sum of the first two taps, etc. If no tap is negative, then > the output always rises, hence no ringing. Please see Ban's post and my reply. Most references describe the Bessel as having low, negligible, or no overshoot. Ban's numbers show a theoretical overshoot of less than 1%, decreasing with higher order. In any practical LCR filter, the Bessel will have such low overshoot as to be difficult or impossible to measure. So it doesn't make sense to split hairs between <1% and zero overshoot. DSP is a quite different story, and you can realize filters that are impossible in real life. But it would be difficult to implement DSP at higher frequencies. Regards, Mike Monett Antiviral, Antibacterial Silver Solution: http://silversol.freewebpage.org/index.htm SPICE Analysis of Crystal Oscillators: http://silversol.freewebpage.org/spice/xtal/clapp.htm Noise-Rejecting Wideband Sampler: http://www3.sympatico.ca/add.automation/sampler/intro.htm

Reply by ●September 30, 20062006-09-30

Jim Thompson wrote:> On 30 Sep 2006 17:33:25 -0700, miso@sushi.com wrote: > > [snip] > > > >No circuits were simulated in writing this post. ;-) > > Bwahahahaha! ROTFLMAO! > > ...Jim ThompsonWell a few neurons were tortured trying to recall all that theory. No wait, make that extraordinarily rendered.> -- > | James E.Thompson, P.E. | mens | > | Analog Innovations, Inc. | et | > | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | > | Phoenix, Arizona Voice:(480)460-2350 | | > | E-mail Address at Website Fax:(480)460-2142 | Brass Rat | > | http://www.analog-innovations.com | 1962 | > > I love to cook with wine. Sometimes I even put it in the food.

Reply by ●October 1, 20062006-10-01

miso@sushi.com wrote:> > You really need to look at theory versus what spice predicts. The > input is an ideal step. As you do the convolution, the negative > region of the impulse response subtracts from the result, making the > signal drop in value, then the postive regions make the result > increase in value, hence ringing.. The Gaussian impulse response is > always positive, hence the convolution output can't decrease as time > increases. > > Linear phase alone is not enough to stop ringing. > > The discrete time situation is easier to understand, especially if you > consider a finite impulse response filter. The response of a FIR > filter to a step input is the the sum of the tap coefficients from > one to N. That is, the first output is the first tap. The second > output is the sum of the first two taps, etc. If no tap is negative, > then the output always rises, hence no ringing. > > No circuits were simulated in writing this post. ;-)There are differences between analog and digital filters. Digital means an approximation of the desired characteristic in the passband, but the poles and zeros are modified to compensate for the modulation effects. aliasing is always happening, but it can be reduced below the noise floor with the analog input filter. Linear phase has a very undesirable side effect, it rings *before and after* the step, supposed to be more audible. And Audio is a very forgiving m�tier, because the ears themselves function as reconstruction filters, suppressing all those high frequency artifacts.. If you want to display the signal with a digital scope or ECG, you better start with 5 to 10 times the upper frequency rolloff, look at the specs there, nobody even considers Nyquist adequate, exept programmers having no idea of reality. -- ciao Ban Apricale, Italy

Reply by ●October 1, 20062006-10-01

"Tim Wescott" <tim@seemywebsite.com> wrote in message news:UaOdnVhcfetF6IDYnZ2dnUVZ_oudnZ2d@web-ster.com...> I've seen a lot of posts over the last year or so that > indicate a lack of understanding of the implications of > the Nyquist theory, and just where the Nyquist rate fits > into the design of sampled systems.Nyquist has a lot to say about how far you *can* go and little to say about how far you *should* go. Those of us who are forced to push the Nyquist limits may have a better appreciation of this. For example, compare the frequency response of a 5-pole IIR Butterworth low-pass filter designed to roll off at 1 KHz. with one designed to roll off at 10 KHz., both filters having been designed around a sampling frequency of 50 KHz. Things start to get nasty above Fs/10 and the harder you push Nyquist, the worse it gets. Nyquist says you can put a sharp +/- 500 KHz analog filter around a 21.4 MHz IF and sample it at 2.14 MHz. What Nyquist doesn't say is what that does to your signal-to-noise ratio. TANSTAAFL! Nyquist is often used by slick, hand-waving charlatans to over-sell their capabilities without having a clue about the real trade-offs involved. (And, I'll bet Nyquist didn't say *that*, either.)

Reply by ●October 1, 20062006-10-01

Ban wrote:>=20 > And Audio is a very forgiving m=E9tier, because the ears themselves fun=ction=20> as reconstruction filters, suppressing all those high frequency artifac=ts..=20 And audio is a very unforgiving matter, because the distortions of the=20 audio amp are quickly growing towards the high frequencies. You won't=20 hear the high frq artifacts as they are, but you will very well hear the = result of the nonlinear distortion of those. Vladimir Vassilevsky DSP and Mixed Signal Design Consultant http://www.abvolt.com

Reply by ●October 2, 20062006-10-02

["Followup-To:" header set to sci.electronics.design.] On Fri, 29 Sep 2006 13:55:32 -0700, John Larkin <jjlarkin@highNOTlandTHIStechnologyPART.com> wrote in Msg. <2t1rh2d56g4gngogguro04er47ckvrcv0t@4ax.com>> Don't you go knocking whiteboards. A big board with a nice fresh set > of markers will multiply my IQ by about 1.3 or so.But only with the smelly markers. The water-based stuff doesn't do it for me. robert