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Math Help - Digital IIR filters

  1. #1
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    Digital IIR filters

    I need to design a number of digital IIR filters based upon transfer functions given in the s-domain. I have applied the bilinear transform to give the transfer functions in the Z-domain, however I need to expand and simplify the equations to place them in the form below:

    Digital IIR filters-form.jpg

    Once the equations are in the above form I can get the IIR filter coefficients and implement them into my filters, however I'm not quite sure how to go about it. Would anybody like to take a crack for me?

    The three equations are as follows;

    1. Digital IIR filters-1.jpg

    2. Digital IIR filters-2.jpg

    3. Digital IIR filters-3.jpg

    An example is given below;

    Digital IIR filters-eg.jpg

    Any help would be very much appreciated!
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  2. #2
    Lord of certain Rings
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    Well it only involves expanding like this: (1 \pm z^{-1})^2 = 1 + z^{-2} \pm 2z^{-1}

    Then you have to group coefficients with a common degree. I will work out the second one as an example:

    Lets write x = \frac{\omega_4 ^2}{\omega_3}, y = \frac{\omega_4}{Q_4}, w = \omega_4 ^2 for ease of typing.

    Now,

    H(z) = \dfrac{\dfrac{2x(1 - z^{-1}) + w (1 + z^{-1})}{1 + z^{-1}}}{\dfrac{4(1 - z^{-1})^2 + 2y(1 - z^{-1})(1 + z^{-1}) + w (1 + z^{-1})^2}{(1 + z^{-1})^2}}


    H(z) = \dfrac{2x(1 - z^{-1})(1 + z^{-1}) + w (1 + z^{-1})^2}{4(1 - z^{-1})^2 + 2y(1 - z^{-1})(1 + z^{-1}) + w (1 + z^{-1})^2}


    H(z) = \dfrac{2x(1 - z^{-2})+ w (1 + z^{-1})^2}{4(1 - z^{-1})^2 + 2y(1 - z^{-1})(1 + z^{-1}) + w (1 + z^{-1})^2}

    Now use (1 \pm z^{-1})^2 = 1 + z^{-2} \pm 2z^{-1} to expand.


    H(z) = \dfrac{2x(1 - z^{-2})+ w(1 + z^{-2} + 2z^{-1})}{4(1 + z^{-2} - 2z^{-1}) + 2y(1 - z^{-2})+ w (1 + z^{-2} + 2z^{-1})}

    Group terms with common powers...

    H(z) = \dfrac{(w - 2x)z^{-2} + (2w)z^{-1} + (w + 2x)}{(4 - 2y + w) z^{-2} +(2w - 8)z^{-1} + (w + 2y +4)}

    Thus comparing it with your first image:

    b_2 = w - 2x, b_1 = 2w, b_0 = w + 2x, a_2 = w -2y + 4,a_1 = 2w - 8, a_0 = w + 2y +4

    So substitute back x = \frac{\omega_4 ^2}{\omega_3}, y = \frac{\omega_4}{Q_4}, w = \omega_4 ^2 and get the final answer...
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  3. #3
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    Hi,

    Thanks for that.

    Things are a little clearer now, however I'm still a little bit unsure about what you did.
    Is there any way you could break it down a little more?

    Sorry, maths really isn't a strong point of mine (which is why I'm here).

    Cheers!
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  4. #4
    Lord of certain Rings
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    Quote Originally Posted by RyanW View Post
    Hi,

    Thanks for that.

    Things are a little clearer now, however I'm still a little bit unsure about what you did.
    Is there any way you could break it down a little more?

    Sorry, maths really isn't a strong point of mine (which is why I'm here).

    Cheers!
    Well I think I am not sure what is it that you do not understand. Tell me the step that you dont understand and I will explain...
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  5. #5
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    To be honest, the only part which I really get is the substitution at the end.
    The rest I can sort of follow when its written in front of me, however I'm not confident I could come up with the same answer by myself, and I really need to be sure of my answers to make sure the filters are doing exactly what I want.
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  6. #6
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    Quote Originally Posted by RyanW View Post
    To be honest, the only part which I really get is the substitution at the end.
    The rest I can sort of follow when its written in front of me, however I'm not confident I could come up with the same answer by myself, and I really need to be sure of my answers to make sure the filters are doing exactly what I want.
    Unless this is course/home work get a computer algebra system (Maxima is opensource and free) and use the rational simplification tools on these expressions.

    RonL
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  7. #7
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    Quote Originally Posted by CaptainBlack View Post
    Unless this is course/home work get a computer algebra system (Maxima is opensource and free) and use the rational simplification tools on these expressions.

    RonL
    No its not course/home work, its work related, however obviously not something I do every day. I'll give Maxima a try and see how it goes.
    Thanks for the tip!
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  8. #8
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    Ok, I tried Maxima, and its just confused me further!
    When I try and simlify it seems to give huge answers which don't really resemble what I need. Any tips?
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  9. #9
    Grand Panjandrum
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    Quote Originally Posted by RyanW View Post
    Ok, I tried Maxima, and its just confused me further!
    When I try and simlify it seems to give huge answers which don't really resemble what I need. Any tips?

    The attachment is a Maxima session (you will need to divide top and bottom through by z^2 to finnish.

    (the radscan function is accessible using the simplify (r) button, the ordinary simplify will also do the job)

    RonL
    Attached Thumbnails Attached Thumbnails Digital IIR filters-gash.jpg  
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  10. #10
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    Thanks for your help so far, however, I'm still not getting the correct answer when I input the example.
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