cutoff frequencyfilterhigh pass filtersimulation
I have made the following simulation with the circuit you can see below:
I'm asked to find the cut-off frequency for this filter. From the simulation, it looks like it \$f_c = 500 \text{Hz}\$ or somewhere around there.
However, if I use the formula for cut-off frequency I get something totally different.
\$f_c=\frac{R}{2\pi L}=\frac{500 \Omega}{2\pi \cdot 10 \text{mH}}=7960 \text{Hz}\$
Am I using the correct formula, or is my interpretation of the simulation wrong?
I hope someone can help me.
All three methods give the same answer.
The filter with the load resistor starts from -6dB, so the '-3dB point' is -9dB on that graph. Reading the first graph by eye looks consistent with just over 30Hz. Zoom into it for better accuracy.
In the second graph, for some reason you have taken the phase curve (the dotted line), which is the same height for 4 degrees shift on the right y axis and -3dB on the left y axis. You don't see the gain curve at this frequency because the y axis scale starts at -1dB (or maybe you've not plotted the gain curve). Replot your second graph with frequency going up to 100Hz, and gain axis from 0dB to -10dB. When you do, you'll see it agrees with the theoretical 31.8Hz figure you've got from calculation.
If you replace all components values in the circuit by 1 (\$1\;\Omega\$ and \$1\;F\$) and use the equation I derived here, then the below graphs show the dynamic response and I can extract the cutoff frequency to 53 mHz or 0.335 rad/s as you correctly found.
Best Answer
Your simulation looks more like 15 kHz (3 dB point) and the reason is because you haven't used enough resolution in your AC analysis and the graph is inaccurately interpolating between points that are too far apart: -
Use more resolution in your AC analysis like in this simulation I did: -
The 3 dB point is around 7900 Hz.
Also try using a logarithmic X axis.