Bode plot parallel RLC circuit

Question

Bode plot

I want to find the natural frequency and bandwidth given this bode plot. I know that the natural frequency corresponds to the x value at the peak of the plot and the bandwidth is the difference of frequency at -3dB. However, I don't know how to extract that from the diagram. The answer is f0=39788 Hz which translates to wo=250000 rad/s and B=56500H z. How do I arrive at those values?

Answer 1

It is not possible to give an accurate answer to your question because the values read on a graph are not accurate enough. It should be necessary to have a numerical table instead of a graph.

The method to answer is very simple :

One read the frequency at the peak , $\text{f}_0\simeq 40$ kHz.

This is consistent with the expected value $39.788$ kHz. The deviation 0.212 kHz is due to the inaccuracy of reading on the graph.

Then compute $\omega_0=2\pi\text{f}_0$ .

The frequency limits for the bandwidth are at $24-3=21$ dB . See the figure below.

We read on the graph : $\quad\text{f}_1<\text{f} <\text{f}_2 \quad \begin{cases} \text{f}_1\simeq 21\text{ kHz}\\ \text{f}_2\simeq 77\text{ kHz}\end{cases}$

The frequency bandwidth is $\quad 77-21=56$ kHz

This is consistent with the expected value 56.5 kHz . The deviation 0.5 kHz is due to the inaccuracy of reading on the graph.