Hey guys,(adsbygoogle = window.adsbygoogle || []).push({});

I have a question about the Bode diagram of a second order high-pass filter (one with both a capacitance and an inductance). The Bode-diagram of the amplitude looks like this.

Here, the cutoff frequency is equal to about 10 kHz.

You see that the amplitude of the transfer function is greater than 1 at the cutoff frequency. But, since

[tex]P_u = |H(f)|^2 P_i[/tex]

where Pi is the power in and Pu is the power out,

then power is created since [tex]|H(f)|^2 > 1[/tex].

So if you apply a sinus with a frequency equal to the cutoff frequency to the filter, you create energy!

I know there must be a flaw in my reasoning but I cannot see where.

Thanks,

Yoran

**Physics Forums - The Fusion of Science and Community**

# Second order low-pass filter

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

Have something to add?

- Similar discussions for: Second order low-pass filter

Loading...

**Physics Forums - The Fusion of Science and Community**