Bode plot driving me nuts - Page 1
Forum section: electronics / beginners

author: Guille
author_rank: Regular Contributor
timestamp: April 13, 2023, 10:23:41 pm
content: I have to replicate a High Pass filter for some old speakers, however this filter is encased in resin and I can't destroy it to look at the component markings. I do know however the schematic configuration (not the values though). It is a T-Network like the one pictured below. I thought about measuring the response of the filter with a signal generator and an osci, then fitting this curve to the theoretical response curve of such a filter topology, thus finaly figuring out the component values.However I cannot solve this simple Kirchoff problem, I would highly appreciate it if you can spot the mistake, I've been at this for 3 nights nowI swear I'm suppoused to have a degree on this.

author: Benta
author_rank: Super Contributor
timestamp: April 13, 2023, 10:33:29 pm
content: As it is, your i3 is zero and i1=i2.How about adding a load (= getting i3 flowing)?BTW, AC voltages are normally symbolized with u, DC voltages with U.

author: Guille
author_rank: Regular Contributor
timestamp: April 13, 2023, 10:41:46 pm
content: Makes sense... However, the load is a pair of tweeters which are electromechanical devices that are not easily translatable to electrical models. Could I just take my measurements with an arbitrary resistor then calculate the Bode plot for such a circuit and fit the curve? Does this make sense if I'm then going to use that filter with a speaker? I guess it would be the same right?

author: Benta
author_rank: Super Contributor
timestamp: April 13, 2023, 10:45:50 pm
content: Nominally, speakers are characterized as 4 or 8 ohms, for tweeters also 16 ohms.I'd suggest you just use an 8 ohm resistance as load, no reason to go all complex there.

author: Dundarave
author_rank: Regular Contributor
timestamp: April 14, 2023, 12:13:16 am
content: Unless it’s a very critical application, you might consider Googling “passive high pass filter speaker circuit”, and do a little empirical futzing around with the typical component values you find there.Unless you’re a purist, sometimes it’s easier to simply assume that the original circuit was just a bog-standard passive filter, and see how it sounds.

author: bdunham7
author_rank: Super Contributor
timestamp: April 14, 2023, 12:53:11 am
content: I don't know that a specific filter response will result in only one correct answer for the component values, or if it does in theory, how a small error might translate into a larger component value error.I'd suggest a more direct approach. Assuming you're well equipped and have a decent LCR meter, you can probably measure the two capacitors in series easily enough. Then you can use a low-frequency stimulus and measure the capacitance or reactance of the inductor in series with each individual capacitor and this will tell you whether they are equal or not. If so, now you know the values. If not, you can make some more measurements.What is the impedance of the driver and what is the specified crossover frequency?

author: mawyatt
author_rank: Super Contributor
timestamp: April 14, 2023, 02:00:31 am
content: Quote from: Guille on April 13, 2023, 10:23:41 pmJust a quick look, as mentioned by Benta, think your problem with your analysis is that without an impedance at Vout, I3 is zero, and I1 = I2. Thus Vout is just the voltage across L since no current is flowing thru C2.Edit: You can use this to evaluate the components. Apply a test voltage at Vin and measure the open circuit voltage Vout and the input current, Vout is the voltage across L. Then short Vout and measure the input current and the output current. With these values and the test frequency you can evaluate the three components. You can do this at a few frequencies for verification.Best,

author: MrAl
author_rank: Super Contributor
timestamp: April 14, 2023, 09:18:34 am
content: Quote from: Guille on April 13, 2023, 10:23:41 pmHello there,A degree on what?There is no way you can calculate this without a load of some type, at the very least, a resistive load like 2 Ohms, 4 Ohms, 8 Ohms, 16 Ohms, 32 Ohms, and in some cases 64 Ohms. At least use 8 Ohms to get a feel for how this works. Heck, even use 10 Ohms just to see how it works.The reason for this is the transfer function changes enormously with a load vs without a load of any kind. The output is not an ideal voltage source, it changes with load, and it changes a lot.To illustrate this in detail, here is the transfer function with no load (and you can see how C2 factors out of the calculation as others have suggested):Vout/Vin=(s^2*C1*L1)/(s^2*C1*L1 1)and here is the transfer function with a resistive load Ro:Vout/Vin=(Ro*s^3*C1*C2*L1)/(Ro*s^3*C1*C2*L1 s^2*C2*L1 s^2*C1*L1 Ro*s*C2 1)See any difference?Now with s=jw the first becomes:(w^2*C1*L1)/(w^2*C1*L1-1)and the second becomes:(j*Ro*w^3*C1*C2*L1)/(j*Ro*w^3*C1*C2*L1 w^2*C2*L1 w^2*C1*L1-j*Ro*w*C2-1)See any difference?The first has phase shift of zero and the second has a non zero phase shift.With Ro=1,C1=1,C2=1,L1=1, the first becomes:w^2/(w^2-1)and the second becomes:(j*w^3)/(j*w^3 2*w^2-j*w-1)where the first above has phase shift:ph=0and the second:ph=atan2((2*w^5-w^3)/(w^6 2*w^4-3*w^2 1),(w^6-w^4)/(w^6 2*w^4-3*w^2 1))and the first has amplitude:w^2/abs(w^2-1)and the second:w^3/sqrt((w-w^3)^2 (1-2*w^2)^2)You can see there is a huge difference between the load and no load versions. It would be impossible to curve fit the filter without the load.What you could do is try to find a filter online that is used for the same purpose. Either that or maybe look up what your load impedance is.You also have to keep in mind that the filter may be designed with the enclosure in mind. That would then have to include some speaker box calculations or some trial and error to get it right with the actual real life load and real life speaker box.I do like the suggestions of looking at it as a 2 port network and going from there. You can even short the output for another reading, and short the input for another reading also. You may be able to determine the values that way, which absolutely works in theory. The only problem is that real life inductors have some strange characteristics when driven at unusual power levels. You may have to try a number of different frequencies and amplitudes to get a good look at the profile of this thing.For more information on this, you can look up 2 port networks.You'll have to make some decent measurements too.Just to note, the output current with the output shorted to ground is:Iout(s)=(s^2*Vin*C1*L1)/(s^2*C2*L1 s^2*C1*L1 1)and with this we can see that C2 is still in effect.

author: Terry Bites
author_rank: Super Contributor
timestamp: April 14, 2023, 03:14:07 pm
content: This (3rd order) filter has been designed for a specific source and load impedance.The filter will not work as expected without a load.As pointed out its impossible to analyse the network without it.Once you add a load you can analyse the two sections separately.You can assume that the source impedance is very low, try Rload/100.

author: Guille
author_rank: Regular Contributor
timestamp: April 14, 2023, 11:12:49 pm
content: Hi, thanks for your answer it definitely sounds like a quick and out of the box method to find the value of this components, sadly though my LCR meter is basicaly chinesium:http://68.168.132.244/LC200A_EN_manual.pdf.I don't think I can "use a low-frequency stimulus and measure the capacitance or reactance of the inductor in series with each individual capacitor" with this crap.QuoteThe drivers are 8Ω (2x16Ω in paralell) and the cutoff frequency is stated as 6.5 KHz.

author: Guille
author_rank: Regular Contributor
timestamp: April 14, 2023, 11:39:37 pm
content: QuoteBSc in physics, MSc in electro-optics, not that I feel too proud of those right now lol.QuoteIt's a good advice, however this is one of 4 speakers, if one is out of phase due to it using a different filter cancellations will affect sound quality and level.QuoteThe problem is I all I have at my disposal is a cheapo LCRhttp://68.168.132.244/LC200A_EN_manual.pdf, an analog osci and a cheapo signal generator.Do you think calculating the theoretical response with a known input and output will work? As long as I then curve fit the components testing with those same input and output resistors of course.QuoteThank you for the response, I will derive it myself later, at least to take back some of my selfsteem

author: bdunham7
author_rank: Super Contributor
timestamp: April 15, 2023, 12:29:59 am
content: Quote from: Guille on April 14, 2023, 11:12:49 pmIn fact, perhaps you can, it may be low enough as-is. Just try and measure the caps first and then do the second step to see if they are equal. Being able to select the measuring frequency would be helpful, but perhaps not necessary. If you pursue your plan of measuring the response, you'll need to carefully measure phase response somehow. You'll want to know whether it is Butterworth (most likely), Bessel or something else. Given your known parameters, I'm going to guess that your values are 2µF, 6µF and 0.15mH respectively. But don't take my word for it, take some measurements. If you measure both in series as 1.5µF, I might be right.What is an "osci"? A scope?

author: MrAl
author_rank: Super Contributor
timestamp: April 15, 2023, 04:33:04 am
content: Quote from: Guille on April 14, 2023, 11:39:37 pmHello again,Isnt the theoretical response approach all we have at the moment? That is, try to match the theory to the stimulus and measurements.In theory if you have 3 components you need at least 3 equations. I did forget to mention the ESR of the inductor though, which is often significant. That moves it up to 4 components, ignoring the ESR of the two capacitors for now.One approach may be to try to solve for the 4 components directly by writing some equations and trying to solve for them mathematically.One equation may be the transfer function, but another would be with the output shorted and driving the input, then another with the input shorted and driving the output, etc. It may be possible to solve for the internal components that way. Maybe with the output actually open you may be able to get an equation with just C1 and L1 (and the ESR of L1) which means that equation already has C2 eliminated. I'm going to look at this too.So basically you write the equations and then try to solve them simultaneously.Testing is always a good idea in fact usually mandatory when we go from theory to real life practice. To start though you could use a simulator. Once you think you have the right component values and connections then enter that info into the simulator and see if you can get the responses to match up with the measurements.The second phase of testing would be with the actual components. Once you get a set of actual real life components, you would of course connect them up and test to see if the responses match up with the original network. That would be a test of phase and amplitude, and frequency response.There is also the time domain view. Networks that are frequency sensitive also have time domain response which depends on the exponentials where the frequency response does not show this information unless you sweep over a wide range of frequencies. You may get away with just sweeping the frequency like that or also look at the unit step response. The unit step response would show how the network responds to quick changes in input amplitude.Take a look at hybrid parameters also that may help here. That's a way of characterizing networks like this.For practice, open a simulator and enter the network and some dummy values for the components, then try to come up with a way to solve for those components mathematically, and since you already know the values you can check to see that you calculated them all correctly from any stimulus you apply and response measurements you make. You can alter the network but only in ways that you can do in real life, such as shorting the output or keeping the output open or using a set value load for the output, etc.If you do not yet have a simulator you should really get one by downloading at least the LT Spice version which is quite good and is free. It's made by Linear Tech and many people use this one. The learning curve is not very steep.There is also math software such as Maxima which is also free. That can be an enormous help to solve the simultaneous equations.Just to note, you do have one advantage to start with that you dont usually have, and that is that you already know the internal topology of the network. That's a lucky break really.

author: geggi1
author_rank: Frequent Contributor
timestamp: April 15, 2023, 09:04:39 am
content: If you have the equipment to measure resonance just use the equation for calculation resonance and do some measurements.The you use the method of solving equations with 3 unknowns to get the component data.Measurement 1.Find resonance with signal on A side.Measurement 2.Find resonance with signal on B side.Measurement 3.Find resonance with capacitors in parallel.Then you just use this to solve the equations.

author: MathWizard
author_rank: Super Contributor
timestamp: April 21, 2023, 07:58:55 pm
content: I would use node voltages, and with the output current being zero, i.e. vout=V of L, then the 2nd node equation relates the open circuit voltages0=-jwC1*V1 - (j/wL - jwC)Vout , w/ Iout=0