Note To Don Kelly

Don,
Would you please give a solution to the following equation
for the response function (transfer function) of a driver
in a closed-box.
G(s) = s^2 Tc^2 / s^2 Tc^2 + s Tc / Qt + 1
The first term on the right, s^2 Tc^2 is the numerator.
Values are:
fo (resonance) = 58.31 Hz
f = 228 Hz
Qt = 0.793
Thanks
ns
Reply to
northstar
Loading thread data ...
I'm not Don but,
divide both top and bottom by the constant in the numerator giving
s^2 ----------------------------------------- s^2 + (1 / TcQt)s + 1/Tc^2
compare this to the standard 2nd order form
s^2 + 2*zeta*wn*s + wn^2
since you of course know that
w = 2*pi*f
you can substitute in your value for fo into the last term and get back Tc
then plug in the rest of your parameters to get the answer.
regards
Reply to
dave y.
----------------------------
---------- Taking into account what Dave has indicated, then Tc=1/wo so what you have becomes
G(s)=[(s/wo)^2]/[(s/wo)^2 +s/woQt +1]
or G(jw) =[(w/wo)^2]/[((w/wo)^2 -1) -jw/woQt] (s=-jw for a sinusoidal signal)
At w = 2*pi*228 =1423 rad/s and wo =366.7 rad/s this will give G(jw) =15.29/(14.29-j4.931) Magnitude = 1.115 and phase angle 17.88 degrees
Magnitude rises from 0 at f=0 to 1.26 at resonance and then tails off to 1 at high frequencies -like a high pass filter with a hump. Phase shifts from 180 to 90 at resonance and tails off to 0 at high frequencies.
The transient response of G(s) as given would lead to a mighty bang on energization so I suspect that inductance has been ignored.
I hope that Dave is checking to be sure that I have it straight.
Reply to
Don Kelly
Would this not be magnitude = 1.140?
ns
Reply to
northstar
----------------------- Please note errors: Magnitude should be 1.0114 and phase 19 degrees. Also magnitude at resonance is 0.793 (not 1/0.793 as I indicated)
I had worked out the correct values but obviously it was too close to bedtime.
Here are results from a quickie computer program. Hump is less than what I thought and not at fo f Magnitude 30 .272 50 0.663 58.13 0.793 60 0.818 80 0.970 100 0.663 140 1.021 150 1.020 160 1.019 180 1.016 200 1.014 220 1.012 228 1.0114 240 1.010 300 1.007 500 1.002 1000 1.001
The values at fo and at 228 agree with what I calculated (as corrected) using a hand calculator.
Reply to
Don Kelly
Thanks. If your math is correct, then the equation is an approximation, and even then only valid for the piston-band. The correct value for G(jw) at 228 Hz is 1.063. BTW, the expression is equation 19 from Small's December 1972 J.A.E.S paper. Note 1.063 is the value required to achieve the correct result in his efficiency equation (J.A.E.S. June 1972, equation 11)
ns
Reply to
northstar
--------------------- The calculations have been repeated and check. They fit the equation and parameters given. I have no doubt that some approximations are involved in the transfer function (one of which, I pointed out)- it is too simple to be otherwise-so I suspect that Small's analysis was reduced to the stage where it was easily used and gave results which were close enough (A more complex model, involving some hard to evaluate parameters, while ideally better, may not have produced better results but would require more effort).
I would suggest that all of the equations in the paper include approximations and some simplifications. The tests for parameters include approximations even though the instrumentation may be very good. I have seen the same with electromagnetic magnetic devices where measurements can be carefully made because it is difficult to sort out the individual parameters perfectly-and- after a single use- the parameters have changed somewhat. 5%, even with the best instrumentation is very good in many such situations. There are many situations like that in engineering. I suspect that the audio world is the same. Compare a typical measured frequency response with the calculated response. (See Beranek fig.7.8) .
I do remember the use of a reference frequency for efficiency calculation gives a slightly different efficiency (PAE) number than the actual efficiency (acoustic output/electrical input at the same frequency). The PAE method simplifies calculations and, as the differences are small - good enough for practical purposes. Is the efficiency given by Small the actual or the PAE ( I vaguely recall that it was the latter). We have gone over this before and I don't actually want to open up that can of worms. That could be the source of the difference.
What might be more interesting is actual measurement, of G(jw) to compare with the calculated results.
By the way, what two quantities are being compared in G(s) ?
Reply to
Don Kelly
clipped but for main data
-----------
You gave "corrected" results as:
----------- You originally gave (before "correction")
Would you please give the math for the "corrected" results? I have checked further, and the efficiency equation is the "culprit". It is an approximation, as you pointed out in our original conversations.
I agree regarding variation in measurements, however one can get very close and obtain repeatability in my business with proper control of room temperature, break-in time, good instrumentation, and care taken in measurements. Thanks for your time and sharing of your expertise and experience.
ns
Reply to
northstar
Dave
I have been lax, but better late than never. Thanks for your input.
ns
Reply to
northstar
-------- G(s) for sinusoidal signals becomes G(jw). Subsitute jw for s and you will get G(jw) =(jw)^2(Tc*2)/[(jw^2)(Tc^2) (jwTc) +1] and (jw)^2 = -w^2 since j=root (-1) mathematically The denominator of G(s) is, as indicated by Dave of the form s^2 +s(wo/Qt +wo^2 which leads to wo =1/Tc Substituting G(jw) =[-(w/wo)^2]/ {[-(w/wo)^2 +1] +j(w/wo)(1/Qt)} adjusting the sign gives the equation G(jw) =[(w/wo)^2]/[((w/wo)^2 -1) -jw/woQt]
That is the math The arithmetic follows: Noting that w/wo =f/fo and using a hand calculator at 228 Hz w/wo =228/58.31 =3.91 so w/wo =3.91*2 =15.29 G(jw) =15.29/[(15.29-1) -j3.91/0.793] =15.29/[14.29-j4.93] =15.21/[15.12 @ -19.04 degrees] =[15.21/15.12] @ 19.04 degrees Magnitude 15.21/15,12 =1.0115 Phase angle 19.04 degrees The computer program that I used is in APL so I won't give the details (special characters used) but for a list of frequencies chosen, it calculates:
Num =(f/fo)^2 Denom= sqrt {[((f/fo)^2) -1]^2 +(f/fo)/Qc)^2} Magnitude = Num/Denom
I didn't bother with phase but observation indicates a shift from 180 degrees, through 90 degrees at resonance to 0 at infinity. (or -90 to 0 to +90 depending on ones point of view- all relative).
I have no problem with your ability to get excellent, repeatable results and excellent accuracy in doing so. You have demonstrated this to me. The problem, as I see it is in models. Small tried to get a simplified model using relatively easy to find parameters that give a good fit to reality, to a complex situation- accepting that the model is not exact. That is good engineering. A better model may be possible but it may take a Cray computer to deal with it and the results will not actually be better because of such things as room dynamics which are unknown. Possibly Bill Gates could afford such a detailed design but would his ears know the difference?
- Don Kelly snipped-for-privacy@shawcross.ca remove the X to answer ----------------------------
Reply to
Don Kelly
This should be:
G(jw) = [(w/wo)^2]/[((w/wo)^2 -1) + j[(w/wo)/Qt] = 1.0115
I use:
G(jw)=r^2/(1-r^2)+j(r/Qt)= 1.0115 (1)
phase = arc tan 2dr/(1-r^2)=21.15 deg. (2)
where r = w/wo d = damping ratio = 1/2Qt
Note the last term in eq.1 may be written as:
j(2dr)
Bill Gates has a golden ear and a golden pottie, note.
ns
Reply to
northstar
Don
Looking back, I believe your original translation was correct except for a typo or brain-fart on a sign.
My question was: "Would you please give a solution to the following equation for the response function (transfer function) of a driver in a closed-box. G(s) = s^2 Tc^2 / s^2 Tc^2 + s Tc / Qt + 1 The first term on the right, s^2 Tc^2 is the numerator. Values are: fo (resonance) = 58.31 Hz f = 228 Hz Qt = 0.793"
--------
Your reply: "Taking into account what Dave has indicated, then Tc=1/wo so what you have becomes:
G(s)=[(s/wo)^2]/[(s/wo)^2 +s/woQt +1] or G(jw) =[(w/wo)^2]/[((w/wo)^2 -1) -jw/woQt] (s=-jw for a sinusoidal signal)"
End of reply.
-------
Now if we change w/wo to r for ease of presentation, G(jw) = r^2/[(r^2-1)-j(r/Qt)] , then change the sign G(jw) = r^2/[(r^2-1)+j(r/Qt)] G(jw) = 3.910^2/(3.910^2-1)+j(3.910/0.793)=1.0115
Want to run this through your program and note the values given, and I'll see if they match my measurements and calculations? These frequencies would be adequate, if convenient: 20 30 40 58.13 70 100 150 228 400 700 1,000
ns
Reply to
northstar
----------------------------
-------- Actually, the sign is correct. Since j=root(-1) then j^2 =-1 and (jw)^2 =-w^2 so I get -(w/wo)^2/[-(w/wo)^2 +jw/wo +1] and changing sign top and bottom, gives G(jw)=w/wo)^2]/[((w/wo)^2 -1) -jw/woQt] which is what I used.
In your substitution the expression G(jw) = r^2/[(r^2-1)-j(r/Qt)] (1) is the same as I have The second expression needs a change of sign of both parts of the numerator You should have: G(jw) = r^2/[(1-r^2)+j(r/Qt)] (2) You didn't do this.
However, it doesn't matter in this case if you are only looking at magnitude as (1-r^2)^2 = (r^2-1)^2 so the magnitude of the denominator doesn't change as you are taking the root of the sum of squares.
In both cases the magnitude is correct and the first expression (1) gives the reactive part of the numerator as negative, while the second gives an answer flipped 180 degrees. Your second expression simply changes the sign of the reactive part. I did get the magnitude of G(jw) as 1.0115 at 228 Hz. I hope that I did not make a typing boob when writing it down the first time.
F Hz Magnitude
Th APL program essentially takes the vector of frequencies and evaluates r=f/fo =w/wo then evaluates NUM =r^2 and DENOM = root[(NUM-1)^2 +(r^2)] and then makes a 2 column list of all frequencies and Magnitudes The program takes 3 lines but could be reduced to 1 or 2 lines.
Computations carried out to 64 bit precision
Reply to
Don Kelly
With dB = 20 log magnitude, then using your magnitudes gives results that match the near-field frequency response almost exactly. Neat. Thanks for the advice. BTW what is this APL program you speak of?
ns
Reply to
northstar
----------------------------
I'm glad that I could help.
APL is a programming language developed in the early 60's. It is a bit esoteric and uses special characters and the learning curve is a lot steeper than BASIC but the flexibility is great. It still exists but new versions are bloody expensive. Mine is an old DOS version. It is very good for vectors and matrices. I could quickly write a little program to solve this problem without worrying about some of the overhead that most languages have (i.e. C or C++) and allows getting on with the solution.
Don
Reply to
Don Kelly
I recall using APL, (A Programming Language), on a "dumb" terminal connected to our main frame. It was mid 1960's IIIRC, and I don't even recall what main frame it used. We soon switched to Fortran as the general engineering language. I remember both APL and Fortran running under DOS. You needed a key-cap overlay for APL.
I've been interested in matrix solutions of circuits for many years. The simple A,B,C,D, matrix for four-terminal networks is quite powerful. I have never found a program, which can run on a standard PC, to add, subtract, multiply and divide these matrices with complex elements.
I was never directly involved in IT, but used computers a lot in helping to set up military communication satellite systems for the USA, UK Min. of Defence, and NATO. Most of the computers used were special purpose devices for tracking/command of the satellite network.
I'm happy to hear that APL is alive and well!
Reply to
VWWall
----------------------------
-------- The older APL's, of which mine is (hey I'm retired) don't handle complex numbers per se (and a version which is essentially the same as mine is available on the net). but it is easy to set up programs (either write your own or use one of workspaces supplied where someone has done this. For a matrix use a combined matrix made of submatrices real part -imag part imag. real Now invert or , multiply by vector real imag
One way is to represent an element by a 2 vector and a vector by a 2,n array and a matrix by a 2,n,n array (which has to be torn apart as above for multiplication or inversion but not for addition. You can also check comp.lang.apl and get advice there. Every now and then, there is a summary of the available APL's (free and commercial) given by G Sirlin
However, Iverson, the inventor of APL has produced a language "J" which used standard keyboard characters and does allow complex numbers. Check it out on the net. As with APL there is a steep learning curve and the language used is taken from grammar. Look it up on the net. --
Don Kelly snipped-for-privacy@shawcross.ca remove the X to answer
Reply to
Don Kelly
Don:
I have Luis Pipe's "Matrix Methods for Engineering" , but I've seen no programs that compute the "A,B,C,D" matrix with complex elements. Perhaps specialized programs like "Spice", take their place.
Thanks for the reference. I haven't had need of that sort of thing recently, but it would be interesting to have a look.
I find that even in retirement, the time required increases to meet the time available. The time available isn't much more than when I was working full time. :-)
Reply to
VWWall
----------------------------
I had to write my own routines. I used the routines for power transmission line solutions. However, J may handle complex matrices directly. Possibly Maple can also do so. The way that I (and others) did it was to represent an ABCD matrix by a two layer array real and imag layers. (order 2,n,n) a single number is also represented by a 2 element vector and a complex vector will have a dimension 2,n Addition of two matrices is simply adding the real parts and adding the imaginary parts. Multiplication requires catenation to get a (2n),(2n) matrix and changing a vector to a 2n vector of real and imag Result is a 2n vector which can be changed back to the 2,n form. This is an analog of the non matric process (a+jb)(c+jd) =(ac-bd) +j(ad+bc) Inversion is a straightforward inversion of the (2n, 2n) matrix and stripping out the duplicate information. It all sounds complicated but it really is not a problem to define a number of vector matrix functions which are easy to use (and write) i.e. a CMUL b or CINV m It is not the most efficient way to do it in terms of storage and time etc but it sure as hell beats writing a multiloop Fortran or C process as the actual number crunching is built into APL's operators and let me get on with the actual solving of a problem rather than the bookkeeping and code debugging required by C.
Reply to
Don Kelly
Don Re former discussions, note that further work has cleared matters up to my satisfaction re our differences. Either of us could have been correct about (Bl)^2/Re, depending on how electromagnetic parameters are assesed. The key was to consider the matter from a force point of view, at least that is one way.
I appreciate the time and assistance you gave.
ns
Reply to
northstar

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