Given is this circuit:
I know that:
$$G(j\omega)=\frac{\frac{1}{j\omega C}}{j\omega L+\frac{1}{j\omega C}+R}$$
But how do I calculate from this my cutoff frequency?
Electrical – Calculating cutoff frequency of RCL circuit
cutoff frequencylow pass
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Best Answer
In my opinion, it is faster to re-derive the transfer function of this second order circuit. You will see that using the FACTs described here, it is truly easy. You can obtain the transfer function by 1st determining the ratio \$H_0=\frac{V_{out}}{V_{in}}\$ for \$s=0\$: short the inductor and open the capacitor. You see from the below sketch that \$H_0\$=1. This is shown in the below sketch:
You then determine the time constants of this circuit when the excitation is reduced to 0 V (replace the input source by a short circuit) and "look" at resistance offered by the energy-storing connecting terminals when alternately placed into their dc or high-frequency state: Dc state for a cap. is an open circuit but a short circuit for an inductor and a short circuit for a cap. but an open circuit for an inductor is you place them into their HF state. As shown in the sketch, you determine the time constants for \$b_1\$ by inspection only: the resistance driving \$C_1\$ is \$R_1\$ then \$\tau_1=R_1C_1\$ and for \$\tau_2\$, there is an infinite resistance "seen" from \$L_2\$'s terminals when \$C_1\$s is in its dc state: \$\tau_2=\frac{L_2}{\infty}=0\$. The first term of the second-order polynominal form is thus \$b_1=\tau_1+\tau_2=R_1C_1\$. For the second term, you now look at the resistance offered by \$L_2\$'s terminals when \$C_1\$ is set in its HF state: you "see" \$R_1\$ leading to \$\tau_{12}=\frac{L_2}{R_1}\$. This is it, the second term \$B_2\$ is equal to \$b_2=\tau_1\tau_{12}=\frac{L_2}{C_1}\$. The denominator of this transfer function is the determined as \$D(s)=1+b_1s+b_2s^2=1+sR_1C_1+s^2L_2C_1\$. You can put it under the following canonical form \$D(s)=1+\frac{s}{\omega_0Q}+(\frac{s}{\omega_0})^2\$ in which your resonant frequency \$\omega_0=\frac{1}{\sqrt{L_2C_1}}\$ and \$Q=\frac{1}{\omega_0R_1C_1}\$. The complete transfer function is thus \$H(s)=H_0\frac{1}{1+\frac{s}{\omega_0Q}+(\frac{s}{\omega_0})^2}\$. The below Mathcad sheet shows the dynamic response plot:
For the -3-dB crossover point, use the canonical form to determine the magnitude when \$s=j\omega\$. The result is here:
You can see how the FACTs quickly got you to the results, without writing a single line of algebra with the associated risk of making mistakes. Just a few drawings that you can individually fix in case you have a deviation between the raw expression - \$H_{ref}\$ in the Mathcad sheet - and the low-entropy expression. A truly useful skill that I encourage students and engineers to acquire.