Dear Barrie,

I wish you good work for the next week.

I did hand calculations on the new circuit. This time they were quite more complicated.

These are the final results:

wo = ( 6 R C + 3 R C4 + R4 C4)/ [ (R C + R C4 + 5 R4 C4) R^2 C^2];

A = (R C Ta^2 - Ta Tb Tc + R4 C4 Tc ) / ( C R Ta^2);

where: Ta = R C + R C4 + 5 R4 C4; Tb = 5 R C + 4 R C4 + 6 R4 C4;

Tc = 6 R C + 3 R C4 + R4 C4.

Certainly it wasn't necessary to confirm that MNA implemented in spice is good, and that the results of your AC analysys were right, anyway the formulas gave these results:

fo = 3.272960203526416 MHz; A = 20.969008264462810

soon

S.

'

Simone:

.

Such sweet concord is so comforting.

.

So, how would you like to take on the 

project of classifying all RC networks

up to, say, N = 10?  By that time, the

max gain should be close to cost(pi).

.

Then I think this pure research might

deserve be written up as a short note

to Electronics Letters.

.

Barrie

Dear Barrie,

if I'll figure out how to do, I will be honored to have the chance to submit a note with you.

S.

dear Barrie and Simone:

I believe that the discrepancy between the critical gain value obtained by AC and time-domain analysis is large but perhaps not completely surprising, even considering the tight accuracy settings, since the circuit is at the edge of instability (complex conjugate poles with a real part close to zero or even positive..). This is a challenge for ordinary differential equation solvers, since not only signals, but also the components of numerical errors close to the oscillating frequency are amplified (or at least not damped) from sample to sample and are summed together in the steady-state waveform, so that the final error can be remarkably larger than what expected considering the error in the evaluation of the single step and reducing the time step does not improve the situation. I cannot describe it better or more rigorously, not being an expert in numerical methods. Considering this, AC analysis is more trustworthy (at least in this - unfortunately not so common - case where a true linear circuit is considered...) as confirmed by comparison with analytical results provided by Simone.

By the way I have also performed the same simulations, both AC and time-domain:

from AC analysis I got:

fo= 3.2729602 MHz for  A0= 20.9690082645

(and this is a further unnecessary confirm,...)

from time-domain simulations (same accuracy settings, probably different algorithm):

fo= 3.2711808 MHz for  A0= 20.9691

this result is perhaps closer to what obtained in AC for A0, but the discrepancy in the frequency is still large.

Dear Barrie,

I found the solution.

I don't know if it deserves a note on EL, since it is quite simple.

I assumed that the resistor is composed of identical R-C sections.

I calculated the Transmission matrix (ABCD)  of each section.

It is straightforward to obtain:

T1 = [ 1/(1 + sRC)  R;  sC  1];       (Matlab notation used)

Furthermore it also easy to show that the desired open loop gain is equal to

L(s) =  (-Gain) *  A;

i.e. L1(s) = -Gain * / (1 + sRC);

Thus a resistor composed of 10 R-C sections,  will have T10 = T1^10.

In this manner I found

L10(s) =-Gain * [ (sRC)^15 +10*(sRC)^14 + 35*(sRC)^13 + 62*(sRC)^12 + 79*(sRC)^11 + 92*(sRC)^10 + 78*(sRC)^9 + 49*(sRC)^8 + 47*(sRC)^7 + 34*(sRC)^6 + 7*(sRC)^5 + 8*(sRC)^4 + 9*(sRC)^3 + 1) ] / (sRC)^10;

By nulling the imaginary or real part (odd/even power) of these polinomial we can find fo and thus the maximum Gain.

soon

S.

.

Simone:

.

Terrific! I still think your work deserves to be documented

if not in EL then possibly in one of the ISCAS journals. 

From your set of results, can you quickly derive the AMAX

and fo values? I don't know how to do that.

.

Barrie

Dear Simone,

it seems to me there is something wrong in the expression of the ABCD matrix T1 reported above. I believe it should be:

T1 = [ (1 + sRC)  R;  sC  1]; 

(A is V1/V2 for I2=0, i.e. the reciprocal of the voltage transfer). I have also checked considering T1 as the cascade connection of a series resistor

T_R = [ 1  R;  0 1]; 

and of a shunt capacitor

T_C = [ 1 0; sC 1];

and T_R * T_C =T1 as above

Moreover, for the same reason, the loop gain for N sections should be expressed as

L_N(s) = -Gain * 1/A_N,

being A_N the "A" term of the ABCD Matrix T1^N.

Do you agree with me? If not, what is the mistake in my derivation?

By the way, I suspected there was something wrong from the expression of L10, where the degree of the numerator is larger than the degree of the denominator (non-causal behaviour)

Dear Paolo,

you are right.

I copied A from my previous calculation, where I solved many voltage dividers.

The first was Vr / VA, but  Vr = V2.  I didn't check well (and it was very late).

In minutes I'll give the new results.

thanks

S.

The new coefficients (these should be right):

0   0     0     0       0       0       0       0     0   1 1    A-T1

0   0     0     0       0       0       0       0     1   3 1    A-T2

0   0     0     0       0       0       0       1     5   6 1    A-T3

0   0     0     0       0       0       1       7   15 10 1    A-T4

0   0     0     0       0       1       9     28   35 15 1    A-T5

0   0     0     0       1     11     45     84   70 21 1   A-T6

0   0     0     1     13     66   165   210 126 28 1   A-T7

0   0     1   15     91   286   495   462 210 36 1   A-T8

0   1   17 120   455 1001 1287   924 330 45 1   A-T9

1 19 153 680 1820 3003 3003 1716 495 55 1   A-T10

and so on....

I'll go further and write all more clearly ASAP.

S.

The coeffs in the previous post, was related to a cascade of R-C sections, where in each section, R=Rtot/N and C = Ctot/N.

What if we replace R and C, with Rtot/N and Ctot/N ?

You should just multiply each of the previous coeff for (1/N^2)^(n), where n is the order of the coeffs, and N the corresponding number of sections (T9, N=9).

In the attached file you can find the "normalized" coeffs.

You can see how, increasing N, they converge to a value, which it is not so clear with the old ones.

This values should be the respective coefficient of the Taylor expansion of cosh(sqrt(s*Rtot*Ctot)).

S.

P.S. to better explain, I also attached the A term of the Transmission matrix T10.

'

Simone:

.

Unfortunately, your table is unreadable - even the

"expanded" image has the same problem.

.

Barrie

I found a huge number of reference on URC (Uniform RC).

In the interesting linked paper, Antinone and Brown, thirty years ago, experimented a cascade of RCR network to modeling a distributed RC line.

Anyway, here attached you can find my coeffs for the polynomial of the A term.

Furthermore I evaluated all 15 polynomial at wo=1/RC and compared the results with the correct value ( cosh(sqrt( 1i ) ). The attached pdf is more explanatory.

S.

http://ieeexplore.ieee.org/stamp/stamp.jsp?tp=&arnumber=1051922

.

Simone:

.

I'm a simpleton. I just need like to see a (simple) table of

AMAX and the frequency of oscillation as a function of N. 

'

The IEEE JSSC paper was not helpful. The sophistication

of modern IC technologies, and a vast array of simulation

tools takes care of such sorts of "transmission" problems.

.

My mission started out (simply) to show that the concept

of an op amp as having infinite bandwidth while exhibiting

significant open-loop gain is of extremely dubious value.

'

Barrie

dear Barrie and Simone:

I hope what follows could be useful. It is obtained by a direct numerical implementation of the ABCD matrix approach proposed by Simone (not passing through the analytical expressions of the transfer functions) for R=1ohm, C=1uF. Each section i has Ri=R/N and Ci=C/N (all equal). A very robust (altough not extremely numerical efficient... in present day PCs this is definitively not an issue) dicothomic search method is used to find the f0 critical frequency. I have considered the number of sections up to 30 since the convergence to cosh(pi) is rather slow... the 9-digits of A0max should be exact.

In the attached files, the plots of the critical gain A0max and of the 180° phase shift frequency f0 can be found. Moreover, the Matlab script rc_distr.m used for the computations is also attached.

N=3  f0=3508635.605552  A0max=29.000000

N=4  f0=3043624.660807  A0max=18.387755

N=5  f0=2941073.843647  A0max=15.436418

N=6  f0=2914597.143087  A0max=14.117882

N=7  f0=2912024.019437  A0max=13.395731

N=8  f0=2918120.753722  A0max=12.951433

N=9  f0=2927372.983889  A0max=12.656427

N=10  f0=2937543.204745  A0max=12.449622

N=11  f0=2947673.618592  A0max=12.298601

N=12  f0=2957355.202580  A0max=12.184731

N=13  f0=2966428.458208  A0max=12.096629

N=14  f0=2974850.823902  A0max=12.026996

N=15  f0=2982634.550757  A0max=11.970965

N=16  f0=2989816.474962  A0max=11.925186

N=17  f0=2996442.997165  A0max=11.887285

N=18  f0=3002562.598778  A0max=11.855541

N=19  f0=3008222.197978  A0max=11.828682

N=20  f0=3013465.494415  A0max=11.805751

N=21  f0=3018332.348337  A0max=11.786012

N=22  f0=3022858.691351  A0max=11.768899

N=23  f0=3027076.700206  A0max=11.753962

N=24  f0=3031015.089362  A0max=11.740846

N=25  f0=3034699.445295  A0max=11.729267

N=26  f0=3038152.562328  A0max=11.718992

N=27  f0=3041394.759995  A0max=11.709832

N=28  f0=3044444.173080  A0max=11.701631

N=29  f0=3047317.011532  A0max=11.694259

N=30  f0=3050027.790585  A0max=11.687608

Here you can find the table with the first (least) frequency and gain that make the circuit to oscillate, as a function of the number of section.

Since the polinomials have up to 15 order, many other roots are present, but the reported one is the least.

The first column is the order, the second is  wo*R*C , the third Amax.

soon

S.

It seems I'm six minutes, late.

It took me more time to work with symbolic solution :-)

i.e.:

A15 = (s^15*C^15*R^15)/191751059232884086668491363525390625+(29*s^14*C^14*R^14)/852226929923929274082183837890625+(14*s^13*C^13*R^13)/140284268300235271453857421875+(13*s^12*C^12*R^12)/74818276426792144775390625+(598*s^11*C^11*R^11)/2992731057071685791015625+(3542*s^10*C^10*R^10)/22168378200531005859375+(134596*s^9*C^9*R^9)/1477891880035400390625+(81719*s^8*C^8*R^8)/2189469451904296875+

(7106*s^7*C^7*R^7)/648731689453125+(58786*s^6*C^6*R^6)/25949267578125+(184756*s^5*C^5*R^5)/576650390625+(8398*s^4*C^4*R^4)/284765625+(6188*s^3*C^3*R^3)/3796875+(476*s^2*C^2*R^2)/10125+(8*s*C*R)/15+1;

however not handmade.

soon

S.

P.S. Apple service took my MacBook, so for the we I'll be offline.

.

Paolo:

.

Your table is exactly what I was hoping for.

Many thanks for it and the extremely well-

documented MATLAB script -- something

that is SO important for later reference.

.

These are very interesting solutions, even

though their down-to-earth practical value

is, of course. likely to be limited.

.

I am  left wondering if there is an intuitive

explanation for that unexpected minimum

in the frequency at around N = 6 /7.  

.

By the way, as a practical matter, the final

R and C section ought to be a half-section.

,

Such experimental adventures are not just

entertaining but also valuable exercises in

the pursuit of deeper understanding. I feel

these are all-too rarely pursued in today's

world of circuit theory by young designers

who are obliged to rush to dead-lines, and

therefore to only a superficial appreciation

of their field of endeavour.

.

As a further note in this thread:  Its header

should not read "....the critical value of A in

ideal transimped[a]nce amplifier" but rather

"....the maximum permissible value of gain

in idealized (infinite bandwidth) operational

amplifier"

.

Barrie

dear Barrie,

I have modified the Matlab script so that the last RC section is a half-section (being N the number of sections and C the total capacitance, Ci=2*C/(2*N-1) for i=1...N-1 and C_N=C/(2*N-1) for the last section, similarly for R). The modified script is in the attachment. The convergence of A0max to cosh(pi) and the minimum in f0 can be observed as before, even though the minimum is now for N=8,9. I just report the values of f0 and A0max up to 10, where the values for N=4 match with the results you have obtained from AC simulations.

I do not have any intuitive explanation for the minimum in f0 (at least so far). I have just observed that it interestingly does not depend on the values of R and C.

By the way, it is not completely clear to me the reason why the whole last section should be with half-valued elements. Is it related to the open-circuit termination?

I fully agree with your comments about designers and researchers rushing to dead-lines and probably research is just what can be done between one dead-line and the next one, and often this time interval tends to zero...

N=3  f0=4101329.007544  A0max=37.250000

N=4  f0=3272960.203526  A0max=20.969008

N=5  f0=3067382.667517  A0max=16.714421

N=6  f0=2997286.482791  A0max=14.904838

N=7  f0=2971406.351085  A0max=13.939846

N=8  f0=2963261.941398  A0max=13.354938

N=9  f0=2963041.937718  A0max=12.969932

N=10  f0=2966534.406279  A0max=12.701430

Not forgetting the sound shift from the Doppler effect as the deadline screams past..

'Paolo:

'

When I first posed this teaser - several years ago,

to make the point that the most appropriate model

of the typical op amps of the time, using dominant

pole compensation, is that of a simple integrator,

not the notion of an infinite-bandwidth amplifier

that was sometimes listed as the most basic form,

- I built RC lines for simulation containing various

total numbers of sections, up to 100,000 made by

a cut-and-paste sequence of a primitive of just ten

sections. (It didn't take long). The convergence to

11.592 became evident before I realized that this

was cosh(pi), by asking the question: What is the

attenuation of an ideal (totally distributed) RC line

when its phase lag is 180 degrees?  Of course, in

this approach, the frequency at which this occurs

must depend on total R and C.  But that fact had

no bearing on the critical gain of an amplifier that

magically never rolled off in its HF response. (Of

course, the very idea defies physics!)

.

Barrie