Hello again,
Carl:
I hate to get into feedback if we dont have to
We could look at that too though i guess, might be interesting. I guess we are going to assume that the circuit is going to be used as a filter where the output is across the cap and resistor.
Ratch:
Ok i checked my calculations. I used basic calculus to develop most of the formulas i posted previously (except for the well known one) but of course there is always the chance that i made an error in the math because even though it is basic there are plenty of calculations to do and get them all right. I couldnt find an error anywhere however so i did a basic maximum amplitude test using the formula for the output voltage (across cap and resistor) which isnt too hard to compute:
Vampl=R/sqrt((R-w^2*C*L*R)^2+w^2*L^2)
Using my quoted center frequency for F3 of 0.351404599520409 the maximum amplitude calculates out to be:
VamplMax=4.5003516037041
Now since i want to test this to find out if it is a true max, i look at one point slightly to the left by decrementing F3 by a tiny amount and another point slightly to the right by incrementing F3 by a tiny amount, and if the amplitude is greater than the max calculated above as VamplMax then F3 fails. If the amplitude for both these points is less than VamplMax however, then F3 passes and that means anything greater than F3 (within some small error limit) will produce less output and anything less than F3 will also produce less output. Ok, so first F3 to the left:
F_Left=F3-0.000001=0.351403599520409
and then F3 to the right:
F_Right=F3+0.000001=0.351405599520409
and note that the increment we are using is less than the difference between your quoted value for F3 and my quoted value, but just to make sure we are not missing some resonant point anyway we will define one more frequency:
F3_Test=0.35146
and that is your quoted value for F3, so we will test all three of these frequencies in the amplitude equation and see what we get.
Using all three frequencies we get three different values for w:
w_Left=2*pi*F_Left=2.20793393339665
w_Right=2*pi*F_Right=2.20794649976727
w_Test=2*pi*F_Test=2.20828830806134
and this gives us three different amplitudes but i'll repeat the max amplitude here first for quick reference:
VamplMax=4.5003516037041
Vampl_Left=4.50035160230076
Vampl_Right=4.50035160230075
Vampl_Test=4.50034729587394
Now an inspection of the max amplitude VamplMax compared to the left sample and the right sample show that both samples are less than the max so it is confirmed that VamplMax calculated above is the max amplitude so that means that F3 really does equal 0.3514045995 to 10 decimal places. Just to make sure we didnt miss anything, we also checked that fourth point and note that Vampl_Test is also smaller than VamplMax so that means the frequency of 0.35146 can not be the frequency F3 that causes a maximum output amplitude.
I think you can verify this using a circuit simulator, as long as you can zoom in on the response well enough and the output accuracy is set high enough to provide this good resolution and the built in solver method is a good one.
One last note...
It's easier to compare those amplitudes when the values are shown in code:
Code:
VamplMax= 4.5003516037041
Vampl_Left= 4.50035160230076
Vampl_Right= 4.50035160230075
Vampl_Test= 4.50034729587394