1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Hi,

    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

    User Avatar
    Homework Helper
    Gold Member

  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

    NascentOxygen

    User Avatar

    Staff: Mentor

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

    NascentOxygen

    User Avatar

    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

    User Avatar
    Homework Helper
    Gold Member

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

    rude man

    User Avatar
    Homework Helper
    Gold Member

    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:

    [itex]T(S)=\frac{1}{S-p_{1}}[/itex]​

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

    [itex]S=\frac{s}{w_{p}}[/itex]​

    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:

    [itex]T(S)=\frac{1}{S+1}[/itex]​

    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:

    [itex]T(S)=\frac{s}{\frac{s^{2}}{B}+s+\frac{w_{p}^{2}}{B}}[/itex]​
    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:

    [itex]T(S)=\frac{s}{\frac{s^{2}}{Bk}+\frac{s}{k}+\frac{w_{p}^{2}}{Bk}}[/itex]​

    Thanks for all the replies!

    Note: Thanks for the link for that software rude man :) it will be useful
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Chebyshev Passaband Filter
  1. Piezoelectric filter (Replies: 0)

  2. Bandpass Filter (Replies: 9)

  3. FIR Filter (Replies: 1)

Loading...