About dynamic equations and the transfer function for the circuit

Attempt Follow my setup $$v_1=u_2$$

$$i_1=C\frac{du_2}{dt},v_2=L\frac{di_2}{dt},i_2=\frac{u_2-v2}{R_2},i=\frac{u1-u_2}{R_1}$$

According to KCL: $$\i=i_1+i_2\$$

Let $$\s=\frac{d}{dt}\$$

Get $$\R_2(u_1-u_2)=Csu_2+\frac{u_2-Lsi_2}{R_2}\$$

My puzzle I doubt that the KCL I Used is wrong, and in my final results,I can not get the transfer function .

• why my latex doesn’t work? – jackson Mar 11 at 14:38
• Fixed it for you. Please fix the last equation as it was malformed originally and I don't know what was the intention. – Eugene Sh. Mar 11 at 14:41
• Equations on the middle of phrases use "\" before each "$". For separate equations use two consecutives "$". – Dirceu Rodrigues Jr Mar 11 at 14:42
• @EugeneSh. Thanks a lot~ – jackson Mar 11 at 14:43
• @DirceuRodriguesJr thank you~ – jackson Mar 11 at 14:43

The easiest way to determine the transfer function of this resonating network is to use the fast analytical circuits techniques or FACTs. The principle is simple: determine the various time constants of this circuit without writing a single line of algebra: just observe the drawing when the energy-storing elements are set in their high- or low-frequency states.

There are two energy-storing elements so this is a second-order network. The denominator obeys $$\D(s)=1+sb_1+s^2b_2\$$. There is a gain $$\H_0\$$ for $$\s=0\$$ obtained when the capacitor is open-circuited and the inductor replaced by a short: $$\H_0=\frac{R_2}{R_1+R_2}\$$. Then, reduce the excitation to 0 V (replace the input source by a short circuit) and "look" at the resistance offered by the capacitor ($$\L_2\$$ is short circuited) and the inductor ($$\C_1\$$ is open circuited) connecting terminals. You have two time constants $$\\tau_1=(R_1||R_2)C_1\$$ and $$\\tau_2=\frac{L_2}{R_1+R_2}\$$ and you can form $$\b_1=\tau_1+\tau_2\$$. The second-order time constant is obtained by considering $$\C_1\$$ short circuited while "looking" into $$\L_2\$$'s terminals: $$\\tau_{12}=\frac{L_2}{R_2}\$$ leading to $$\b_2=\tau_1\tau_{12}\$$.

The zero is obtained when you look at the impedance configuration which could lead to a transformed short circuit on the output, nulling the response: $$\L_2\$$ and $$\R_2\$$ in series lead to a zero defined as: $$\\omega_z=\frac{R_2}{L_2}\$$.

This is it, we can now assemble the pieces and rework the expression to fit the second-order polynomial form: $$\H(s)=H_0\frac{1+\frac{s}{\omega_z}}{1+\frac{s}{Q\omega_0}+(\frac{s}{\omega_0})^2}\$$ by identifying the terms with $$\b_1\$$ and $$\b_2\$$ in $$\D(s)\$$.

The below Mathcad sheet details these results and compare the response between the brute-force expression (high-entropy expression, no insight) and the low-entropy result in which all poles, zero and gain appear in a clear and well-ordered form. This is what FACTs bring you: the swift determination of passive/active network transfer functions by considering the physical time constants.

For future reference

Using the basic impedance divider Z1/(Z1+Z2) for all parts, the old way is tedious.

After you do it the hard way, a visual simulation will end up being more useful.

Every part has RLC components including wires, so beware what you leave out like the ESR for the Cap.

After you understand the math, the interactive visual SIM tools are more useful.

where $$\H(s)=\dfrac{Z_c//(R_2+Z_l)}{R_1+Z_c//(R_2+Z_l)}\$$ , where $$\Z_c=\dfrac{1}{sC} ~,~ Z_l=sL\$$ then
$$\H(s)=\frac { \frac {R_2}{R_1+R_2} + \frac{sL}{R_1}} {1+sR_2C+s^2LC} \$$ and so on...

You know intuitively, at DC, the gain is the R2,R1 ratio and at infinite f the gain is 0 unless you have ESR in the cap, which always exists.

So after you understand Mr. Kint's excellent solution, get into simulations.

The time constants,τ for the denominator roots indicate the 1st order τ1 = R2C and 2nd order, τ2 = √(LC) while the numerator indicates the rising s τ3=L/R1 effects.

Then add parasitic RLC values to simulate via's traces, SRF on caps with ESL etc. Simulate, 500A DC busbars for parasitic high impedance SMPS resonance with micro-ohm busbars, class E filters and more.

If you can imagine it, you can simulate it. Well, assuming that all the initial condtions are equal to $$\0\$$ we can write:

$$\mathcal{H}\left(\text{s}\right)=\frac{\frac{1}{\frac{1}{\left(\frac{1}{\text{s}\text{C}}\right)}+\frac{1}{\text{R}_2+\text{s}\text{L}}}}{\text{R}_1+\frac{1}{\frac{1}{\left(\frac{1}{\text{s}\text{C}}\right)}+\frac{1}{\text{R}_2+\text{s}\text{L}}}}=\frac{\text{R}_2+\text{Ls}}{\text{CL}\text{R}_1\text{s}^2+\text{C}\text{R}_1\text{R}_2\text{s}+\text{Ls}+\text{R}_1+\text{R}_2}\tag1$$