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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


1)

schematic

simulate this circuit – Schematic created using CircuitLab

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


2)

schematic

simulate this circuit

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


Are those the correct transfer functions of the two filters?


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?

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  • \$\begingroup\$ it is best you show your work. and easier for me to make Bode plots \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Feb 12 '17 at 2:40
  • \$\begingroup\$ Just enter the circuit in a simulator. It will create the bode plot for you. \$\endgroup\$ – mkeith Feb 12 '17 at 2:55
  • \$\begingroup\$ Sören - in principle, the transfer function can be found by applying the voltage divider rule. That is all. This should not be a problem for you- \$\endgroup\$ – LvW Feb 12 '17 at 8:18
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    \$\begingroup\$ I don't think you should get square roots in your result... \$\endgroup\$ – Douwe66 Feb 12 '17 at 8:36
  • \$\begingroup\$ Strictly speaking, these do not look like "transfer functions", but look like the magnitude of the transfer functions. If they are correct, they are good for bode plots for example but you need to go back one step if you really need the transfer functions. \$\endgroup\$ – rioraxe Feb 12 '17 at 21:17
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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: -

enter image description here

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.

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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

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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! : )

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