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I have made the following simulation with the circuit you can see below:

enter image description here

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.

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    \$\begingroup\$ \$f_c\$ is the frequency at which the gain is \$-3db\$; why do you think \$f_c\$ is 500Hz? \$\endgroup\$
    – across
    Apr 2, 2020 at 9:07
  • \$\begingroup\$ Huh,I guess have misunderstood some things then. However, if \$f_c\$ occurs when gain is -3dB, then the simulation and calculation still doesn't match. \$\endgroup\$
    – Carl
    Apr 2, 2020 at 9:13
  • \$\begingroup\$ maybe the frequency is not f rather w? \$\endgroup\$
    – Mike
    Apr 2, 2020 at 9:15
  • \$\begingroup\$ Are you perhaps looking at the phase line (the dotted line) thinking it is the gain line? \$\endgroup\$
    – Justin
    Apr 2, 2020 at 17:41

2 Answers 2

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From the simulation, it looks like it fc=500Hz or somewhere around there.

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: -

enter image description here

Use more resolution in your AC analysis like in this simulation I did: -

enter image description here

The 3 dB point is around 7900 Hz.

Also try using a logarithmic X axis.

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    \$\begingroup\$ Very often it is more exact to use the phase shift of 45 deg. as an indication for 3dB cutoff. \$\endgroup\$
    – LvW
    Apr 2, 2020 at 11:06
  • \$\begingroup\$ For any 1st-order lowpass or highpass the phase shift at the 3dB frequency is ALWAYS 45 deg. Look into your phase response diagram. In many cases. this point is easier to find than the 3dB drop of the magnitude. \$\endgroup\$
    – LvW
    Apr 2, 2020 at 11:16
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Well, mathematically speaking we can write:

$$\mathcal{H}\left(\text{s}\right)=\frac{\text{sL}}{\text{sL}+\text{R}}\tag1$$

Using \$\text{s}=\text{j}\omega\$, we get:

$$\left|\underline{\mathcal{H}}\left(\text{j}\omega\right)\right|=\left|\frac{\text{j}\omega\text{L}}{\text{j}\omega\text{L}+\text{R}}\right|=\frac{\left|\text{j}\omega\text{L}\right|}{\left|\text{j}\omega\text{L}+\text{R}\right|}=\frac{\omega\text{L}}{\sqrt{\text{R}^2+\left(\omega\text{L}\right)^2}}\tag2$$

Now, we get:

  • $$\lim_{\omega\to0}\left|\underline{\mathcal{H}}\left(\text{j}\omega\right)\right|=0\tag3$$
  • $$\lim_{\omega\to\infty}\left|\underline{\mathcal{H}}\left(\text{j}\omega\right)\right|=1\tag4$$

Solving for the \$-3\space\text{dB}\$ point gives:

$$\frac{\omega\cdot10\cdot10^{-3}}{\sqrt{500^2+\left(\omega\cdot10\cdot10^{-3}\right)^2}}=\frac{1}{\sqrt{2}}\space\Longleftrightarrow\space$$ $$\omega=50000\space\text{rad/sec}\tag5$$

Which is the same as:

$$\frac{25000}{\pi}\space\text{Hz}\approx7957.75\space\text{Hz}\tag6$$

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  • \$\begingroup\$ Jan - just a formal correction: The last line must not be written as one single equation. Instead you should write down two separate equations. \$\endgroup\$
    – LvW
    Apr 2, 2020 at 11:05
  • \$\begingroup\$ @LvW I do not see what you mean, you can edit my answer is you want. \$\endgroup\$ Apr 2, 2020 at 11:08
  • \$\begingroup\$ Jan...it is pure math: On the left side of the equations is w....and on the right side we see the frequency f. That means w=f ???. It should read: 50000rad/sec are equivalent to f=7957.75 Hz (not equal) \$\endgroup\$
    – LvW
    Apr 2, 2020 at 13:01
  • \$\begingroup\$ @LvW I've studied mathematics and I see what you mean, but using rad/sec and Hz make a difference \$\endgroup\$ Apr 2, 2020 at 13:21
  • \$\begingroup\$ Yes - and therefore, I think, we must not equate the two ..(different units on both sides) \$\endgroup\$
    – LvW
    Apr 2, 2020 at 13:26

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