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Math Help - DSP; Difference equation from a transfer function

  1. #1
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    DSP; Difference equation from a transfer function

    Hi,

    This is a question relating to DSP. I wasn't sure what section to put it in. (move it if you so wish)


    I have the transfer function:


    0.5 - 0.3 z-1
    ------------- = H(z)
    1 + 0.25 z -1


    I need to obtain a difference equation ( i.e in the form y(z)= )


    My problem is I cant find in text books, or on the web, a simple explanation for converting said transfer function to a difference equation. (please note this is a difference equation not a differential equation).


    I found the following link to be the most useful thing so far, (if you explore the site a bit). But I still don't understand how the process is done.
    [ http://ccrma.stanford.edu/~jos/fp3/Z_Transform_Difference_Equations.html ]


    Thanks for any help anyone can provide,


    Bob
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by 1bob1 View Post
    Hi,

    This is a question relating to DSP. I wasn't sure what section to put it in. (move it if you so wish)


    I have the transfer function:


    0.5 - 0.3 z-1
    ------------- = H(z)
    1 + 0.25 z -1


    I need to obtain a difference equation ( i.e in the form y(z)= )


    My problem is I cant find in text books, or on the web, a simple explanation for converting said transfer function to a difference equation. (please note this is a difference equation not a differential equation).


    I found the following link to be the most useful thing so far, (if you explore the site a bit). But I still don't understand how the process is done.
    [ http://ccrma.stanford.edu/~jos/fp3/Z_Transform_Difference_Equations.html ]


    Thanks for any help anyone can provide,


    Bob
    H(z)=\frac{0.5-0.3z^{-1}}{1+0.25 z^{-1}}

    Write:

    y(n)=H(z)x(n)

    or:

    (1+0.25 z^{-1})y(n)=(0.5-0.3z^{-1})x(n)

    or:

    y(n)+0.25y(n-1)=0.5x(n)-0.3x(n-1)

    rearranging:

    y(n)=0.5x(n)-0.3x(n-1)-0.25y(n-1)

    That is z^{-1} may be considered to be the unit delay operator.

    CB
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  3. #3
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    Thanks for that, makes things a bit clearer. I'l keep staring at it, it'l click soon.
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  4. #4
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    Christchurch
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    DSP: difference equation to Z-domain transfer function to bode plot

    I've implemented a filter in a DSP and now wish to get its bode plot. I need some help determining first the z-domain transfer function (I've done this once but would like a second opinion), and then I would like to get a bode plot from that transfer function - how do I do this?

    Thanks Dave


    The difference equation is:-

    y(n) = [ intError(n-1) + { Igain }{ x(n) } ] + [ { Pgain }{ x(n) } ]

    where

    intError(n) = [ intError(n-1) + { Igain }{ x(n) } ]
    Igain is a constant
    Pgain is a contstant



    Quote Originally Posted by CaptainBlack View Post
    H(z)=\frac{0.5-0.3z^{-1}}{1+0.25 z^{-1}}

    Write:

    y(n)=H(z)x(n)

    or:

    (1+0.25 z^{-1})y(n)=(0.5-0.3z^{-1})x(n)

    or:

    y(n)+0.25y(n-1)=0.5x(n)-0.3x(n-1)

    rearranging:

    y(n)=0.5x(n)-0.3x(n-1)-0.25y(n-1)

    That is z^{-1} may be considered to be the unit delay operator.

    CB
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