Magnitude of transfer functions of two filters

Question

I need to find the magnitude of transfer function of the follwing two filters. I tried to do the calculation but I get two functions that seems to be incorrect. Here is what I found

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1)

^{simulate this circuit – Schematic created using CircuitLab}

$$G(f)=V_{out}/V_{in}=\frac{2 \pi f L R_p}{\sqrt{\left(R R_p-4 \pi ^2 c f^2 L R R_p\right)^2+4 \pi ^2 f^2 L^2 (R+R_p)^2}}$$

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2)

^{simulate this circuit}

$$G(f)=V_{out}/V_{in}=\frac{R_p \sqrt{4 \pi ^2 c^2 f^2 R_s^2+\left(1-4 \pi ^2 c f^2 L\right)^2}}{\sqrt{4 \pi ^2 c^2 f^2 (R (R_p+R_s)+R_p R_s)^2+(R+R_p)^2 \left(1-4 \pi ^2 c f^2 L\right)^2}}$$

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Are those the correct transfer functions of the two filters?

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Edit : For circuit 2)

I found total impedance $$Z=R+\frac{1}{\frac{1}{\text{Rp}}+\frac{1}{-\frac{j}{2 \pi c f}+2 j \pi f L+\text{Rs}}}$$

I wrote the ratio

$$1/Z \cdot \frac{1}{\frac{1}{\text{Rp}}+\frac{1}{-\frac{j}{2 \pi c f}+2 j \pi f L+\text{Rs}}}=\frac{(Rp (-1 + 2 c f \pi (2 f L \pi - j Rs)))}{( Rp (-1 + 2 c f \pi (2 f L \pi - j Rs)) + R (-1 + 2 c f \pi (2 f L \pi - j (Rp + Rs))))}$$

Taking magnitude

$$\frac{\text{Rp} \sqrt{4 \pi ^2 c^2 f^2 \text{Rs}^2+\left(4 \pi ^2 c f^2 L-1\right)^2}}{\sqrt{\left(R \left(4 \pi ^2 c f^2 L-1\right)+\text{Rp} \left(4 \pi ^2 c f^2 L-1\right)\right)^2+(2 \pi c f R (-\text{Rp}-\text{Rs})-2 \pi c f \text{Rp} \text{Rs})^2}}$$

Which should be the same as above.

Nevertheless this does not seem right since, for resonance frequency the ratio is zero, while it should not be zero if \$R_s\neq 0\$.

Is there a simpler way to get the correct result?

Answer 1

Take the first example and make life a little easier by simplifying it. For instance, it can be shown that R and Rp can be combined into a single series resistor being fed from a new voltage source that is reduced by the potential divider ratio. You then get a new circuit that looks like this: -

Clearly the value R in the above is the parallel combination of the two resistors in the original question. Solving: -

G(s) = \$\dfrac{\dfrac{\dfrac{sL}{sC}}{sL+\dfrac{1}{sC}}}{R+\dfrac{\dfrac{sL}{sC}}{sL+\dfrac{1}{sC}}}\$ = \$\dfrac{L/C}{L/C+sLR+R/sC}\$

Multiply through by sC to get \$\dfrac{sL}{sL+s^2LCR+ R}\$

And finally G(s) = \$\dfrac{s\dfrac{1}{CR}}{s^2+s\dfrac{1}{CR}+\dfrac{1}{LC}}\$

To get back to the original TF you need to reduce the above TF by the voltage potential divider formed by the original R and Rp.

Answer 2

The transfer function of the first circuit can be determined in a few lines just by looking at the circuit. This what Fast Analytical Circuits Techniques (FACTs) bring you. Here we go:

Observe the circuit for \$s=0\$: open caps and short inductors--> \$H_0=0\$

Reduce \$V_{in}\$ to zero and look at the resistance driving \$C_1\$ while \$L_2\$ is in its dc state (a short circuit). Then look at the resistance driving \$L_2\$ while \$C_1\$ is in its dc state (an open circuit). You have the two first time constants:

\$\tau_1=0*C_1\$, \$\tau_2=L_2/(R_1||R_p)\$

When I say "drive", I mean the resistance you "see" in the terminals of \$C_1\$ or \$L_2\$ when you temporarily remove them from the circuit.

Then, look at the resistance driving \$C_1\$ while \$L_2\$ is in its high-frequency state (an open circuit). You have \$\tau_{21}=C_1(R_p||R_1)\$.

\$D(s)=1+s(\tau_1+\tau_2)+s^2\tau_2\tau_{21}=1+sL_2/(R_1||R_p)+s^2C_1L_2\$

\$D(s)=1+s/(\omega_0Q)+(s/\omega_0)^2\$ with \$\omega_0=1/\sqrt{L_2C_1}\$ and \$Q=R_1||R_p\sqrt{C_1/L_2}\$

For \$N(s)\$, consider three gains when \$C_1\$ is in its high-frequency state, \$L_2\$ is in its high-frequency state and finally, when both are in their high-frequency state:

\$H_1=0\$, \$H_2=R_p/(R_p+R_1)\$ and \$H_{12}=0\$

\$N(s)=H_0+s(H_1\tau_1+H_2\tau_2)+s^2H_{12}\tau_2\tau_{21}=sH_2\tau_2=sL_2/R_1\$

\$N(s)=s/\omega_z\$ with \$\omega_z=R_1/L_2\$

The complete transfer function in a low-entropy form is thus:

\$H(s)=\frac{s/\omega_z}{1 + s/(\omega_0Q) + (s/\omega_0)^2}\$

I did not write a single line of algebra, just inspecting the network. Look at the FACTs in an APEC seminar from 2016:

http://cbasso.pagesperso-orange.fr/Downloads/PPTs/Chris%20Basso%20APEC%20seminar%202016.pdf

Answer 3

For the second circuit, the principle remains the same:

\$s=0\$, \$H_0=R_p/(R_p+R_1)\$

\$\tau_1=(R_s+R_1||R_p)C_1\$, \$\tau_2=L_2/R_{inf} = 0\$ \$\tau_{12}=L_2/(R_s+R_1||R_p)\$

\$D(s)=1+s(\tau_1+\tau_2)+s²\tau_1\tau_{12} = 1 + s/\omega_0Q + (s/\omega_0)^2\$

with \$\omega_0=1/\sqrt{L_2C_1}\$ and \$Q=[1/(R_s+R_p||R_1)]\sqrt{L_2/C_1}\$

\$H_1=\frac{R_s||R_p}{R_1 + Rs||Rp}\$ \$H_2=R_p/(R_p+R_1)\$ \$H_{12}=R_p/(R_p+R_1)\$ \$N(s)=H_0 + s(H_1\tau_1+H_2\tau_2)+s^2H_{12}\tau_1\tau_{12}\$

\$N(s)=1+sR_sC_1+s^2L_2C_1\$

\$N(s)=1+s/\omega_0Q_n + (s/\omega_0)^2\$ with \$\omega_0\$ same as above and \$Q_n=(1/R_s)\sqrt{L_2/C_1}\$

\$H(s)=H_0\frac{1+s/\omega_0Q_n+(s/\omega_0)^2}{1 + s/\omega_0Q + (s/\omega_0)^2}\$

Again, thank you FACTs! : )