"Robert Gush" <robert@suesound.co.za> wrote in message news:a0f35ea5.0312100614.18bae255@posting.google.com...> stocksb@ieee.org (Benjamin M. Stocks) wrote in messagenews:<132e56ad.0312090748.4c829fe@posting.google.com>...> > "Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message > > > Is that the case? Do you know that it's a linear system? Yourdescriptions> > > have really been too cryptic for people to respond very well. > > > > I apologize for the cryptic nature (I didn't think I was being that > > cryptic but maybe I was) but this is a work related question so I need > > to be careful about how much I describe, I'm sure you can understand. > > It is a linear system except for a device at the input that prevents > > any DC and any positive voltages from entering. > > > > > You have not said if you want the frequency response of the system inthe> > > general case or if you want the frequency response of the system tothe> > > input signal. > > > > To the specific input signal. I want to know the system's response at > > a specified frequency. > > > a) Generate a frequency sweep. > b) Lowpass filter the output. > c) Plot the measured amplitude against the input frequencies. > voila! you have the frequency response. > > Unless we're all missing something...Robert, I *think* he was saying that the input to the linear system isn't under direct control. That it is fixed in its nonsinusoidal form but that frequency and amplitude are under his control. So, if the output is lowpassed just above 2x the frequency of the sinusoid of the sweep (which is input pre-rectifier), then the output amplitude will be a measure of frequency response at the fundamental only. Sort of like applying "describing functions" in control systems analysis. How'd you propose to sweep the LP filter to do this? Fred

# Rectified FFT

Started by ●December 5, 2003

Reply by ●December 10, 20032003-12-10

Reply by ●December 10, 20032003-12-10

Fred Marshall wrote: ...> > > Robert, > > I *think* he was saying that the input to the linear system isn't under > direct control. That it is fixed in its nonsinusoidal form but that > frequency and amplitude are under his control. So, if the output is > lowpassed just above 2x the frequency of the sinusoid of the sweep (which is > input pre-rectifier), then the output amplitude will be a measure of > frequency response at the fundamental only. Sort of like applying > "describing functions" in control systems analysis. How'd you propose to > sweep the LP filter to do this? > > FredIt doesn't have to be low pass. Band pass will do. Think "lock-in amplifier." Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������

Reply by ●December 10, 20032003-12-10

In re-reading Ben's response, I see that the full-wave rectified sinusoid is under your control. What's not clear to me is if you have *chosen* to make it a full-wave rectified sinusoid or if you have no control over it. The approach I described earlier will work if you have no control over it. However, if you *do* have control over the waveform at that point, then a classical approach with a sinusoidal input may be preferred because it avoids having to do the Fourier Transform. After having done that, you can determine the output from any full-wave rectified sinusoid as before. Hey Ben, we need to understand if we're even close here. It's really not likely that any of what we're talking about can be all that proprietary. Anyway, pictures sometimes help. Here are a couple: This is what I understand the system setup to look like: +-----+ +-----------+ +-----------+ | | s1 | Full Wave | s2 | Linear | s3 | cos +----->| Rectifier |----->| System/ |-----> | | f0/2 | -1 | f0 | Filter | f0 +-----+ +-----------+ 2f0 +-----------+ 2f0 3f0 3f0 4f0 4f0 5f0 5f0 . . This shows that a sinusoid (actually a cosine with positive peak at time=0) going into a full wave rectifier which outputs an average negative voltage as you've specified (thus the "-1" shown in the box). In general, the output will have sinusoidal components at f0 and all of its harmonics. This goes into the linear system. In general, the output s3 will have all the same sinusoidal components - but at differenct amplitudes and phases. I understand that you can control the amplitude and frequency of cos=s1 where I've labeled the frequency to be f0/2 and have not shown the amplitude on the diagram. A question is, can you also control directly the form of s2 or not? The approach I tried to suggest earlier worked like this: In general, s1=A*cos(wt) where A is the peak amplitude and w is the radian frequency and t is time. Using the notation from the figure: s1=A*cos(2*pi*f0/2*t) from this, the general form for s2 is: s2=A*[b0 + b1*cos(2*pi*f0/2*t) + phi1) + b2*cos(4*pi*f0/2*t + phi2) + b3*cos(6*pi*f0/2*t + phi3) + .....] where b0, b1, b2, ..... are the Fourier Series coefficients for the output of a general nonlinear such as the full wave rectifier. To be very specific, let's use the actual Fourier Series coefficients for a perfect full-wave rectifier: s2 = -A*[(2/pi)*(1 + (2/3)*cos(2*pi*f0/2*t) - (2/15)*cos(4*pi*f0/2*t) + (2/35)*cos(6*pi*f0/2*t) - .....] s2 = -A*[(2/pi)*(1 + (2/3)*cos(pi*f0*t) - (2/15)*cos(2*pi*f0*t) + (2/35)*cos(3*pi*f0*t) - .....] So, there is - a dc term with value -A*2/pi, - a fundamental sinusoid with amplitude -A*4/3*pi, - a second harmonic with amplitude +4*A/15*pi, - etc. These are the Fourier Series Coefficients scaled by the input amplitude "A". (I conveniently chose a cosine as input so that the entire Fourier Series would be made up of cosines with no phase terms). With this: s3= -A*[(2/pi)*(c0*1 + c1*(2/3)*cos(pi*f0*t + phi1) - c2*(2/15)*cos(2*pi*f0*t + phi2) + c3*(2/35)*cos(3*pi*f0*t + phi3) - .....] where c0, c1, c2 etc. are the amplitude responses of the linear system at dc, 1*f0, 2*f0, etc. and where phi1 is the phase shift of the linear system at 1*f0, phi2 at 2*f0, etc. The first method I suggested would have you calculate the Fourier Transform of the output at f0, 2*f0, etc. The Fourier Transform gives you the phase values directly - or at least to an arbitrary phase reference if you haven't registered the output relative to the input phase. The Fourier Transform magnitudes allow you to calculate the magnitude response of the linear system by dividing them by the input amplitudes in s2 which are the Foureir Series Coefficients scaled by the input amplitude "A" given above. So, if you compute the Fourier Transform of s3 at f0, let's say you get magnitude 0.5 and phase pi/4. Let's also say that the temporal epoch that you transformed was aligned to start and end on peaks of the input sinusoid - so it looks like a cosine input for phase alignment purposes. Let's also say that the input peak amplitude A=1.0. The input amplitude to the linear system is 4/3*pi and the output is 0.5. So, the linear system amplitude response at f0 is 4/6*pi and the phase is pi/4 - assuming that the rectifier introduces no phase shift or delay. This calculation is repeated at each frequency 2*f0, 3*f0, etc. giving you the "frequency response" for a sinusoidal input at f0/2. For each different value of f0, you get a list of amplitude and phase for each of the harmonic frequencies. Then, the process is repeated for each value of f0 of interest and you end up with a table of input frequency vs. amplitude and phase for each harmonic frequency number. Now, this assumes that you only had control on s1 and not on s2. If you do have control over s2, then the process can be different - although I'm not sure if it's better. It does avoid doing the Fourier Transform by doing a direct measurement. You input a sinusoid as s2 for measurement purposes and measure magnitude and phase of s3 and stick those results in a table for each frequency of interest - being careful to include measurements at all of the harmonic frequencies - actually in this case only the even harmonics and not even the fundamental because of the form of the Fourier Series. Then, you use those results the same as if you had calculated the Fourier Transform of the output and use the same process as above to create a table of input frequency vs. amplitude and phase for each harmonic frequency number. I do hope that you realize that there isn't one "frequency response" except at a chosen harmonic - for example at f0 you can talk about a frequency response as you vary f0. However, there are also outputs at 2*f0, 3*f0, etc. that have energy in them which you may or may not be interested in. If you are interested in them, then you might view the situation as multiple frequency responses which apply to the "harmonic numbers". That is, you could plot the frequency response for f0 and another for 2*f0 and another for 3*f0. These could be superimposed on a plot against f0 as long as it's clear that the scale for the response at 2*f0 is really compressed, etc...... If the linear system is a perfect lowpass filter at 240Hz, then you might end up with a plot that looks something like this: amplitude ^ (not to scale) | | | f0---------------------------------------+ | | | | | | 2*f0------------------+ | | | | | | | | | | | | | | | 3*f0--------+ | | | | | | | | | | | | | | | +--------------|----|----|------------------|---------+ 0 100 120 240 f0----> The way this plot would be read is this: Say the input sinusoid frequency is 50Hz Then the rectifier output fundamental is 100Hz. This plot says that the fundamental will have some appreciable amplitude that can be read off the chart. It says that the 2nd harmonic will have some appreciable amplitude that can be read off the chart. It says that the 3rd harmonic will *not* have appreciable amplitude. Below f0=80Hz, the 3rd harmonic will show up. 80Hz=240/3 Above f0=120Hz, the 2nd harmonic will drop out. 120Hz=240/2 Above f0=240Hz, the fundamental will drop out. Phase can be plotted similarly. So, either this matches your situation and helps or it doesn't. Does it??? Fred

Reply by ●December 11, 20032003-12-11

"Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message news:<KLqdnfto6IKG-kqiRVn-iw@centurytel.net>...> I *think* he was saying that the input to the linear system isn't under > direct control. That it is fixed in its nonsinusoidal form but that > frequency and amplitude are under his control. So, if the output is > lowpassed just above 2x the frequency of the sinusoid of the sweep (which is > input pre-rectifier), then the output amplitude will be a measure of > frequency response at the fundamental only. Sort of like applying > "describing functions" in control systems analysis. How'd you propose to > sweep the LP filter to do this? >Hi Fred, I am sure the OP indicated that he can generate any stimulus frequency. He did indicate that there may be interference present, but Jerry's suggestion of using a BP filter that is matched to this stimulus frequency may be adequate. Of course if the input signal is close to the interference then there will be a problem. I wonder if the OP were to stimulate it with a unit impulse - in the direction that the rectifier would pass - and then do the FFT on the result. Because of the large influence an interfering tone can have on this method it might be prudent (if the interfence is changing) to do a number of these 'frequency response checks' and average the results Regards Robert