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note

With \$C_4=47\:\mu\text{F}\$ and \$f_0\approx 650\:\text{Hz}\$ the impedance is about \$5\:\Omega\$. (In quadrature.) If significantly lower frequencies are sought, then \$C_4\$ should be made proportionately larger.

notes

With \$C_4=47\:\mu\text{F}\$ and \$f_0\approx 650\:\text{Hz}\$ the impedance is about \$5\:\Omega\$. (In quadrature.) If significantly lower frequencies are sought, then \$C_4\$ should be made proportionately larger.

Also, for \$R\approx 10\cdot R_{_\text{LOAD}}\$ and \$R\approx 10\cdot R_{_\text{SOURCE}}\$, which is close to the above case, then set \$\gamma=1.9\$. And for \$R\gg R_{_\text{LOAD}}\$ and \$R\gg R_{_\text{SOURCE}}\$ set \$\gamma=\sqrt{3}\$.

Then it appears there's a fairly simple equation: \$f_0\approx\frac1{2\pi\,\gamma\,R\,C}\$.

note

With \$C_4=47\:\mu\text{F}\$ and \$f_0\approx 650\:\text{Hz}\$ the impedance is about \$5\:\Omega\$. (In quadrature.) If significantly lower frequencies are sought, then \$C_4\$ should be made proportionately larger.

(I won't go into how I got that, here.)

(1/(2*pi*C*sqrt(rload*rsource+3*R*(R+Rload+Rsource)))).subs({R:27e3,C:4.7e-9,Rsource:2.1e3,Rload:2000}).n()
674.120958877630

(1/(2*pi*C*sqrt(rload*rsource+3*R*(R+Rload+Rsource)))).subs({R:27e3,C:4.7e-9,Rsource:2.1e3,Rload:200*(10+2.2)}).n()
669.289215790583

(I won't go into the details of how I got that. But it's just finding the transfer function. Here, it's numerator is an odd power of s, so I set the real part of the denominator equal to zero and solved. An even power in the numerator, had that been the case, would have meant setting the imaginary part of the denominator to zero before solving.)

(1/(2*pi*C*sqrt(Rload*Rsource+3*R*(R+Rload+Rsource)))).subs({R:27e3,C:4.7e-9,Rsource:2.1e3,Rload:2000}).n()
674.120958877630

(1/(2*pi*C*sqrt(Rload*Rsource+3*R*(R+Rload+Rsource)))).subs({R:27e3,C:4.7e-9,Rsource:2.1e3,Rload:200*(10+2.2)}).n()
669.289215790583