- #1
yoran
- 118
- 0
Hey guys,
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.
http://dave.uta.edu/dillon/pspice/images/pst07g.jpg
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
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.
http://dave.uta.edu/dillon/pspice/images/pst07g.jpg
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
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