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Consider a simple LC network:

schematic

simulate this circuit – Schematic created using CircuitLab

Is there an easy way to compute its transfer function in the Laplace domain? I would consider as "transfer function" the ratio \$ H(s) = V_{out} (s) / V_{in} (s) \$.

I am looking for a hint to proceed (consider the branches? The voltages? The currents? KVL only or KCL only?) and I'm not asking others to do my homework.

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    \$\begingroup\$ The transfer function does not depend on L2 because you have no load connected - is this actually true? You should also model the inductor's series resistance because this prevents infinities occuring. \$\endgroup\$
    – Andy aka
    Jan 7 '15 at 15:54
  • \$\begingroup\$ This filter should be inserted in a circuit, so it will have a load. I did not draw it because it is not part of the filter, but yes, a generic load \$ R_L \$ is present. I prefer to keep the components ideal to show also its behaviour for \$ \omega = 0 \$ and \$ \omega \to \infty \$. \$\endgroup\$
    – BowPark
    Jan 7 '15 at 15:59
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    \$\begingroup\$ On both counts you are being naive - the load is an integral part of the filter and examining the L and C as ideal components will result in infinite input current at the resonant frequency of the circuit. At w=0 the transfer function is 1 and at w = infinite the TF is 0. \$\endgroup\$
    – Andy aka
    Jan 7 '15 at 16:02
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    \$\begingroup\$ "Easy" as in not learn how to do it from a textbook? There are some symbolic circuit solvers in the Laplace domain, e.g. qsapecng.sourceforge.net \$\endgroup\$
    – Fizz
    Jan 7 '15 at 16:03
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    \$\begingroup\$ The issue is that when you connect the load resistor to the above circuit, the transfer function itself will change \$\endgroup\$
    – nav
    Jan 7 '15 at 18:49
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A simple (mathematical!) way to compute a transfer function for a circuit is to find the voltage at the output using the impedances of the components.

For a simple \$L\$-\$C\$ circuit (i.e. if you remove \$L_2\$ from the above circuit), we can use voltage divider rule to find the voltage across \$C\$:

$$V_o = V_i \times \frac{Z_C}{Z_C + Z_L} $$

(Where \$Z_C\$ is the impedance of the capacitor (= \$1/j\omega C\$) and \$Z_L\$ is the impedance of the inductor (= \$j\omega L\$))

Which gives you the transfer function as

$$\frac{V_o}{V_i} = \frac{Z_C}{Z_C + Z_L} = \frac{1}{1-\omega^2LC} $$

If we add "\$L_2\$ + a series resistance \$R_L\$" parallel to \$C\$, then we need to consider the combined impedance of "\$C\$ parallel (\$L_2\$ series \$R_L\$)" in the place of \$Z_C\$ in the above equation. After doing all the math, it gives us something big for the final transfer function:

$$\frac{V_o}{V_i} = \frac{R_L}{R_L + j\omega(L_1+L_2) - \omega^2L_1R_LC - j \omega^3L_1C}$$


Now, if you want to see the transfer function without \$R_L\$, just set \$R_L = \infty\$ in the above equation.

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Is there an easy way to compute its transfer function

The transfer function of the circuit does not contain the final inductor because you have no load current being taken at Vout. You should also include a small series resistance like so: -

enter image description here

As you can see the transfer function (in laplace terms) is shown above and if you wanted to calculate real values and get Q and resonant frequency then here is the webpage.

If you want to involve load conditions then it becomes a lot more complex.

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Represent each component in their s-domain equivalents. Apply KCL at the node. What you'll end up with is one equation in two unknowns (namely Vout and Vx where Vx is the node). To resolve this problem realize that current through L2 is zero (since there is no load as others have mentioned) and therefore Vx is equal to Vout! Make the substitution to end up with one equation in one unknown. Rearrange to achieve the desired H(s) xfer function. Hope this helps!

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  • \$\begingroup\$ You should end up with H(s) = 1 / (1 + s^2*L1*C1). And as you can see L2 does not enter into the equation as expected. \$\endgroup\$ Jan 8 '15 at 1:42
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Simplest approach is to use \$sL\$ as the reactance of the inductor and \$1/sC\$ as the reactance of the capacitor. The TF, \$V_{\text{out}}(s)/V_{\text{in}}(s)\$, is then (output impedance)/(input impedance) = $$\frac{1/sC}{sL+1/sC} = \frac{1}{s^2LC + 1}$$ In the absence of a load impedance (hence no current through \$L_2\$), \$L_2\$ must be ignored. Note: I've used \$L=L_1\$, \$C=C_1\$. If you want to determine the frequency response of the circuit, just let \$s=j\omega\$.

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