I am trying to find the transfer function for the following second order model, but running into difficulty as I can't manipulate it into the standard form required for me to extract time constants.
The circuit is as follows:
So far I have worked from this equation for the impedance:
$${\frac{\left( {\it R1}+{\frac {1}{s{\it C1}}} \right) \left( {\it R2}+{
\frac {1}{s{\it C2}}} \right)}{{\it R1}+{\frac {1}{s{\it C1}}}+{\it R2}+{\frac {1}{s{\it C2}}}}}+R3$$
and factored out the frequency independent gain:
$$k={\frac {{\it R1}\,{\it R2}}{{\it R1}+{\it R2}}}+{\it R3}$$
to arrive at somewhere near what I think is standard form, but missing an \$\omega_{o}\$ term. I wanted to keep the factor of \$s^2\$ in the denominator unity.
I have this:
$$k.{\frac{{s}^{2}+{\frac { \left( {\it R1}\,{\it C1}+{\it R2}\,{\it C2}+{\it R3}
\,{\it C2}+{\it R3}\,{\it C1} \right) s}{ \left( {\it R1}\,{\it R2}+{
\it R3}\,{\it R1}+{\it R3}\,{\it R2} \right) {\it C2}\,{\it C1}}}+{
\frac {1}{ \left( {\it R1}\,{\it R2}+{\it R3}\,{\it R1}+{\it R3}\,{
\it R2} \right) {\it C2}\,{\it C1}}}}{{s}^{2}+{\frac { \left( {\it C2}+{\it C1} \right) s}{ \left( {\it R1}+
{\it R2} \right) {\it C2}\,{\it C1}}}}}$$
and I'm stuck. I have spent a day manipulating it in various ways, not taking the factor k out, but can't get it into a recognisable form. Intuitively I see it is a mixture of high-pass and band pass.
I wonder if part of my problem is that I am assuming an input current of \$I_{b}\$ for the calculation of transfer impedance \$Z={\frac {V_{b}} {I_{b}} }\$, when the circuit must be considered open circuit for transfer function calculation? Any help appreciated.
Best Answer
First,the proper term is driving point impedance since it is the ratio of the voltage across and current through a one-port.
Now, since the impedance goes to infinity as the frequency goes to zero, the factor you've pulled out is the high frequency asymptotic impedance but I don't think it's useful in this case to do that.
I think a transparent form for the driving point impedance is:
\$Z_{eq} = R_3 + \dfrac{1}{s(C_1 + C_2)}\dfrac{s^2(R_1R_2C_1C_2) + s(R_1C_1 + R_2C_2) + 1}{s(R_1 + R_2)C_1||C_2 + 1} \$
There are clearly two poles; one at \$s = 0 \$ and one at \$s =-\dfrac{1}{(R_1 + R_2)C_1||C_2}\$
The numerator is 2nd order so there are two zeros. You can factor the numerator to find the zeros (of the numerator) at \$s = -\dfrac{1}{R_1C_1} \$ and \$s = -\dfrac{1}{R_2C_2} \$
As \$s \rightarrow \infty \$, the 2nd term approaches
\$\dfrac{R_1R_2}{R_1 + R_2} = R_1||R_2 \$
The two parallel RC networks have an equivalent impedance given by:
\$Z = (R_1 + \dfrac{1}{sC_1})||(R_2 + \dfrac{1}{sC_2}) = \dfrac{R_1R_2 + \frac{R_1}{sC_2} + \frac{R_2}{sC_1} + \frac{1}{s^2C_1C_2}}{R_1 + R_2 + \frac{1}{sC_1} + \frac{1}{sC_2}}\$
\$ = \dfrac{s^2R_1R_2C_1C_2 + s(R_1C_1 + R_2C_2) + 1}{s^2(R_1 + R_2)C_1C_2 +s(C_1 + C_2)}\$
\$ = \dfrac{1}{s(C_1+C_2)}\dfrac{s^2R_1R_2C_1C_2 + s(R_1C_1 + R_2C_2) + 1}{s(R_1 + R_2)C_1||C_2 +1} \$