Simple Band-pass Filter - passband voltage drop

Discussion in 'Homework Help' started by Rtier, Mar 12, 2009.

  1. Rtier

    Thread Starter New Member

    Mar 12, 2009
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    0
    First, a quick thanks in advance. I am looking to have my process verified - I used the example numbers from the book to guide myself. I'm doing homework involving band-pass filters and I'm not confident I understand the voltage drop across a pass-band filter. I'm working with RC high pass and low pass filters in series, not sallen-key filters.

    The example in the book's section on band-pass filters has a high pass RC filter in series with a low pass RC filter. C-high = 1.5nF, R-high = 1kΩ; R-low = 40kΩ, C-low = 4pF.
    Low cut-off frequency = 106.1kHz
    High cut-off frequency = 994.72kHz.
    The book's analysis at ~500kHz shows a Vo ≈ 0.9Vi.

    In my analysis, I assumed Vi = 1V for simplicity, and obtained Vo = 0.874V.
    First, I figured out the reactance of both capacitors:
    Xc high = (2*pi*500000*1.5*10^(-9))^-1 = 212.2Ω
    Xc low = (2*pi*500000*4*10^(-12))^-1 = 79.6kΩ

    Next, since the diagram shows a high-pass filter as the first stage, I went through the Av for it first.
    Av high = (1+ (Xc/R)^2)^(-0.5)
    Av high = (1 + (212.2Ω/1000Ω)^2)^(-0.5) = 0.97822

    Then, the low pass filter.
    Av low = (1 + (R/Xc)^2)^(-0.5)
    Av low = (1 + (40kΩ/79.6kΩ)^2)^(-0.5) = 0.89348

    Multiplying these numbers together gives the Av for the entire bandpass filter, Av = 0.874Vin.

    Anyone care to comment? Reassurance would be very very nice at this point. Sorry for the MSPaint circuit drawing, I don't have access to MultiSim right now.
     
    Last edited: Mar 12, 2009
  2. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    The filters are so simple and their frequencies are so close together that they affect each other.

    The lowpass filter cuts high frequencies but it also cuts the frequency you want.

    The highpass filter cuts low frequencies but it also cuts the frequency you want.

    Your circuit is too simple to do what you want.
     
  3. Rtier

    Thread Starter New Member

    Mar 12, 2009
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    0
    I edited the original post to reflect the proper frequencies. I had mistakenly noted (through poor notation) that the low pass filter cut off frequencies above 106.1kHz and the high pass cut off frequencies below 994.72kHz. It works the other way around.
     
  4. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,281
    326
    You ask for comment, but I think it would be more fruitful if you asked a question.

    Are you questioning the apparent discrepancy between the book's analysis giving an Av of .9, versus your value of .874?

    Perhaps they are just rounding the result to one digit.

    However, your value of .874 is only correct if there is a unity gain buffer between the stages. That's the implicit assumption you make when you simply multiply the Av of stage 1 times the Av of stage 2.

    If the stages are cascaded without a buffer, the overall Av is .87205. You have to analyze the complete cascade of 4 components, all at once.

    This value is pretty close to .874, and the impedances of the two stages are scaled in order to prevent the second stage from loading the first stage very much, but there is a small loading effect, which accounts for the correct Av of .87205 compared to your value of .874.
     
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