Second order low-pass filter

  1. 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.
    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.


    Last edited: Jan 15, 2009
  2. jcsd
  3. berkeman

    Staff: Mentor

    First, that's a highpass filter, not lowpass. Second, the gain in the passband is shown to be 0dB, which means that the amplitude out equals the amplitude in. No energy gain.
  4. Hi,

    Sorry about that, I picked the wrong image. Anyway, the same thing happens for a high-pass filter. If you look at f=10 kHz, then you can see that the gain is actually higher than 0 dB...
  5. berkeman

    Staff: Mentor

    Ah, I think I see what you're asking now. To calculate the power, you need to plot both voltage and current, especially around that resonance.... Can you intuit why there is no power gain at resonance now?
  6. Oh, I think I see. The output voltage is higher than the input voltage, but then the output current must be a smaller than the input current?
  7. berkeman

    Staff: Mentor

    I don't think that's quite it. Think about what the voltage and current are doing in a resonant LC circuit. When the voltage is peak, what is the current doing? When the current is peak, what is the voltage?
  8. Isn't it so that when the voltage is high, the current is low and vice-versa?
  9. berkeman

    Staff: Mentor

    If you mean in the time domain, then yes. Not in the frequency domain of the Bode Plot. You lose time-domain info in going to the Bode Plot, and that's what caused your initial confusion, I think.
  10. Ah wait I think I see it. I worked the equations out at the resonance frequency and I get the result that the output voltage is pure imaginary ([tex]V(f_r) = \frac{j\omega_r L}{R}V_{in}[/tex]) while the current is pure real ([tex]I(f_r) = \frac{V_{in}}{R}[/tex]). As you said, then the real power is equal to the product which is zero then? Because the power is purely imaginary?
  11. berkeman

    Staff: Mentor

    Not sure I follow that. Could you post the rest of your work, and what the circuit looks like?
  12. It's a second order high-pass filter. It looks like this.
    but without any concrete values for the components (it's an image I found on the Internet, not mine).

    I just realized I made a mistake. The current I gave at the resonance frequency is the current through the inductance, not the current through the output (when assuming that there is a load impedance parallel to the inductance).
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