
#1
Jan1509, 11:16 AM

P: 118

Hey guys,
I have a question about the Bode diagram of a second order highpass filter (one with both a capacitance and an inductance). The Bodediagram 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 



#2
Jan1509, 11:22 AM

Mentor
P: 39,708

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.




#3
Jan1509, 12:40 PM

P: 118

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



#4
Jan1509, 12:44 PM

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P: 39,708

Second order lowpass filter 



#5
Jan1509, 12:51 PM

P: 118

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?




#6
Jan1509, 01:03 PM

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P: 39,708





#7
Jan1509, 01:15 PM

P: 118

Isn't it so that when the voltage is high, the current is low and viceversa?




#8
Jan1509, 01:19 PM

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P: 39,708





#9
Jan1509, 01:34 PM

P: 118

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?




#10
Jan1509, 03:58 PM

Mentor
P: 39,708





#11
Jan1509, 04:25 PM

P: 118

It's a second order highpass 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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