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Homework Help: Chebyshev Passaband Filter

  1. Apr 9, 2012 #1
    1. The problem statement, all variables and given/known data
    Get the transfer function of a second order Chebyshev passband filter, with central frequency f0 = 1 [kHz], lower cutoff frequency fc=670 [Hz], 3dB ripple in the pass-band and 30dB of gain in the central frequency.

    2. Relevant equations

    Maximum allowed variation in passband transmission [itex]A_{max}=10log(1+ε^2)[/itex]
    Transfer function of Chebyshev filter [itex] T(s)=\frac{Kw_{p}^N}{ε2^{N-1}(s-p_{1})...(s-p_{N})}[/itex]
    Chebyshev filter poles [itex] p_{k}=-w_{p}sin(\frac{(2k-1)\pi}{2N})sinh(\frac{1}{N}sinh^{-1}(\frac{1}{ε}))+jw_{p}cos(\frac{(2k-1)\pi}{2N})cosh(\frac{1}{N}sinh^{-1}(\frac{1}{ε})), k=1,2,...,N[/itex]
    N is the order of the filter
    K is the gain
    (expressions taken from "Microelectronic Circuits", Sedra, 5th edition)
    3. The attempt at a solution


    My first attempt, to this problem was to calculate ε through [itex]A_{max}[/itex] and get the two poles(which are conjugated) through the expression given, considering [itex]w_{p}[/itex] as [itex]670\times2\pi[/itex]. I then had all that was needed to build the transfer function. The problem is that this is the transfer function for a low pass filter. I have no idea how to get the pass band filter at this point, and im also not sure if what i did is correct.
  2. jcsd
  3. Apr 9, 2012 #2

    rude man

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  4. Apr 10, 2012 #3
    I made a search in the web before coming here and also found that document, but it didnt helped :S .
  5. Apr 10, 2012 #4


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    Staff: Mentor

    Have you visited that site recently, rude man? :frown:
  6. Apr 10, 2012 #5


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    Staff: Mentor

    I recall from my uni days that there are transforms to convert the LP to a HP, and also to a BP. That's why design equations deal only with the LP. Searching, I found mention on wikipedia. http://en.wikipedia.org/wiki/Frequency_transformations#Bandform_transformation

    I think you are just looking for a second order system, of the form A.s / (s² + bs +c)
    so it won't have ripple as such, there's only the one peak in the response, it's bandwidth being measured between the pair of -3dB points.

    Beyond this, I'm of no help here, sorry.
  7. Apr 10, 2012 #6

    rude man

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    see below
    Last edited: Apr 10, 2012
  8. Apr 10, 2012 #7

    rude man

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    SORRY, N/O. I'll try to figure out how I got their site. definitely did, only yesterday. Stay tuned, it's worth it.

    OK, found the problem - the site is www.nuhertz.com. Scroll down to the "Filter Free" downlink.
  9. Apr 14, 2012 #8
    I finally found the answer and the mistake i was making :D. To design the filter i was considering initially a second order low pass Chebyshev filter, with the specified characteristics, and then i applied the frequency transformation to make it a band pass. However to make the transformation what i needed was a first order low pass Chebyshev filter that as a transfer function of the form:


    where S is the normalized frequency with respect to the central frequency:


    I then calculated the pole [itex]p_{1}[/itex] of the filter, using the expression i took from the "Microelectronic Circuits", which gave me [itex]p_{1}=-1[/itex]. Thus the transfer function for the low pass filter is:


    To make it a band pass filter i used a frequency transformation:

    [itex]S\rightarrow\frac{S^{2}+1}{2\xi S}[/itex]​

    After making the transformation i ended up with a band pass second order Chebyshev filter , with the desired characteristics, of the form:

    B is the bandwidth

    In order to have the gain of 30dB at the central frequency i multiplied the transfer function for a constant k and calculated what value it had to have at [itex]s=jw_{p}[/itex].Then i finally got the desired result:


    Thanks for all the replies!

    Note: Thanks for the link for that software rude man :) it will be useful
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