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Homework Help: AC Filter Vout calculation.

  1. Jun 24, 2009 #1
    Does anyone know the formula for finding the voltage of Vout at a certain frequency in a AC filter circuit?

    I have a formula it looks like Source x (XL or XC) / Resistor + i (XL or XC)

    I can get it to work for some but not others. I have a feeling it needs to be rearranged depending on the situation.

    Also how can you tell if a circuit is High Pass, Low Pass, Bandpass, or Band Reject. I know what they are supposed to do, but I have difficulty identifying which is which. Does it have anything to do with a capacitor being completely charged and then discharging over time or the opposite with an inductor?

    I would appreciate any help.:tongue2:
  2. jcsd
  3. Jun 24, 2009 #2


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    Staff: Mentor

    In general, a filter is a frequency-dependent voltage divider. So in the simplest form, you have a 2-element filter, with a top impedance, a bottom impedance to ground. You drive the top element with your signal source, and tap your output signal out of the junction in the middle of the two elements.

    If the bottom element is a resistor, and the top element is a capacitor, what kind of filter is that?

    If the top element is an inductor, what kind of filter is that?

    If the bottom element is an inductor, and the top element is a capacitor, what kind of filter is that? How is it different from the answer for the CR filter above?

    A bandpass filter takes more than just two elements, because it has to combine both a LPF and a HPF together. You could cascade two such 2-element filters to make a BPF, or there are other ways to do it better. Can you say what a better BPF might look like? How are the LPF and HPF parts of that BPF implemented?
  4. Jun 24, 2009 #3


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    Staff: Mentor

  5. Jun 24, 2009 #4
    Hmm. I looked into the wikipedia page and read your post and am still having difficulties. I have a paper right next to me that tells me exactly what a filter is based on its components such as:
    Top: Inductor Vout: Capacitor = Low pass
    Top: Inductor Vout: Resistor = Low pass
    Top: Resistor Vout: Capacitor= Low pass
    Top: Resistor Vout: Inductor= High pass

    The problem is that I have no idea how these are found. I am thinking this way, for example:
    Top: Inductor Vout: Resistor = Low pass

    An Inductor when plugged into a source starts low and charges based on time, soo would it be low pass ,because it starts off low? Or am I approaching this the wrong way?

    Edit: I also looked at the filters section on this website under the AC tab.
  6. Jun 24, 2009 #5


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    Staff: Mentor

    No, think in terms of impedance versus frequency for the R, the L and the C elements.

    What is the Z(f) equation for each of these elements?

    Then think how the impedance divider is going to change over frequency as the elements' impedances change.

    So for example, the impedance of a capacitor is infinite at DC and zero at infinite frequency, right? So how does that affect the transfer function of a C-R filter?
  7. Jun 24, 2009 #6
    I'm sorry for being so dumb, but you have me completely lost.:frown: I don't know what this Z(f) formula you are speaking of is. The values I have to solve for are FC (Cutoff Frequency) = 1/ (2ΩRC) AV (Voltage Gain)= VOUT/ VIN AV(db) (decibels gain)= 20 Log AV

    I need to find the impedence of the capacitor or inductor and then I need to calculate the voltage across Vout at different frequencies. Finally I have to identify the type of filter.

    I have no problem with anything except for the voltage across Vout at different frequencies and identifying the filter.

    I'm sorry but what you tried to explain to me was way over my head.
  8. Jun 24, 2009 #7


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    Staff: Mentor

    Do these formulas look familiar for the impedances of these components?

    Inductor: [tex] Z(f) = 2 \pi j f L [/tex]

    Capacitor: [tex] Z(f) = \frac{1}{2 \pi j f C} [/tex]

    Resistor: [tex] Z(f) = R [/tex]

    Where j is the square root of -1.

    Now, use each of those impedance formulas in the impedance divider equation to get the output voltage in terms of the input voltage:

    [tex]Vout(f) = Vin(f) \frac{Zup(f)}{Zdown(f) + Zup(f)}[/tex]

    In general you will end up with a complex-valued equation for the transfer function, and in the end you plot the magnitude and phase of the complex transfer function. The Bode Plot of the AC filter generally just shows the magnitude, but can also show the phase.

    See this wikipedia.org article:


    And/or Chapter 1 in "The Art of Electronics" by Horowitz and Hill, starting with the "Impedance and Reactance" section.
  9. Jun 24, 2009 #8
    Yes they are familiar ,but I am used to them being called XL and XC instead, and my formulas are slightly different:

    XL=2πFL XC= 1÷(2πFC) and for the resistor in RLC type circuits I take "It" which is found by : Source/ZT times the resistor so.....

    It x resistor.

    The only one that was unfamiliar to me was the last one, and I'm not really sure how to read that. I was aware that Vout and Vin frequencies were the same.

    I know to find XL or XC at a certain frequency you use that frequency instead of the resonant frequency.
  10. Jun 24, 2009 #9
    I re-arranged that equation earlier and got it to calculate the correct voltage ,but the wrong phaser.

    Source x Vout / ( Component closest to source + i Vout)

    So using this equation With a Low pass filter with a 10 VAC source a 10mH inductor and Vout of 50 ohms

    XL at 400 HZ= 25.1 ohms

    using the equation to find the voltage at Vout:

    10 x 50 / (25.1 + i 50) = 8.93 with a phaser of -63.3

    The actual answer should be 8.93 with a phaser of -27

    So I'm doing something wrong with the phasers. I know I didn't enter XL into the equation with the phaser of 90 because it gives me a completely wrong voltage. Do you see what I'm doing wrong?
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