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Math Help - Sampling and the Z transform

  1. #1
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    Sampling and the Z transform

    Hello,

    I'm trying to follow along with an example in a book, but there are two steps in an example and I just can't see how you go from one to the next. The problem is this:

    The function f(t) = t is sampled every interval T. The Z transform of the function is then...

    The solution is provided and worked through:

    F(Z) = \sum\limits_{k=0}^\infty \frac{kT}{z^k}

    = T \left(\frac{1}{z}+\frac{2}{z^2}+\frac{3}{z^3}+... \right\)

    = \frac{T}{z} \left(1+2z^{-1}+3z^{-2}+4z^{-3}+... \right\)

    I understand up to this point. Now here is the bit I don't get:

    = -Tz\frac{d}{dz} \left(1+z^{-1}+z^{-2}+z^{-3}+... \right\)

    It looks like the derivative of 'something' has been taken, but I just can't make the connection... Can anyone shed any light on this please?

    The rest of the solution is below, and includes the reverse of this 'mystery' operation! (which I also don't understand... I'd love to know what's going on here.

    = -Tz\frac{d}{dz} \left(1-\frac{1}{z} \right\)^{-1} = \frac{T}{z} \left(1-\frac{1}{z} \right\)^{-2} = \frac{Tz}{(z-1)^2}
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  2. #2
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    Re: Sampling and the Z transform

    Quote Originally Posted by halfnormalled View Post

    = \frac{T}{z} \left(1+2z^{-1}+3z^{-2}+4z^{-3}+... \right\)

    I understand up to this point. Now here is the bit I don't get:

    = -Tz\frac{d}{dz} \left(1+z^{-1}+z^{-2}+z^{-3}+... \right\)
    Easiest way to understand this first little problem is to work backwards.

    Starting with

    = -Tz\frac{d}{dz} \left(1+z^{-1}+z^{-2}+z^{-3}+\cdots \right\)

    Taking the derivative we get = -Tz \left(0-z^{-2}-2z^{-3}-3z^{-4}+\cdots \right\)

    as we are aiming to get a z in the denominator then re-write this as

    = -T\frac{z^2}{z} \left(-z^{-2}-2z^{-3}-3z^{-4}+\cdots \right\)

    Now multiply the expression through by -z^2 giving

    = \frac{T}{z} \left(1+2z^{-1}+3z^{-2}+4z^{-3}+\cdots \right\)

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  3. #3
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    Re: Sampling and the Z transform

    Quote Originally Posted by halfnormalled View Post
    Hello,

    I'm trying to follow along with an example in a book, but there are two steps in an example and I just can't see how you go from one to the next. The problem is this:

    The function f(t) = t is sampled every interval T. The Z transform of the function is then...

    The solution is provided and worked through:

    F(Z) = \sum\limits_{k=0}^\infty \frac{kT}{z^k}

    = T \left(\frac{1}{z}+\frac{2}{z^2}+\frac{3}{z^3}+... \right\)

    = \frac{T}{z} \left(1+2z^{-1}+3z^{-2}+4z^{-3}+... \right\)

    I understand up to this point. Now here is the bit I don't get:

    = -Tz\frac{d}{dz} \left(1+z^{-1}+z^{-2}+z^{-3}+... \right\)

    It looks like the derivative of 'something' has been taken, but I just can't make the connection... Can anyone shed any light on this please?

    The rest of the solution is below, and includes the reverse of this 'mystery' operation! (which I also don't understand... I'd love to know what's going on here.

    = -Tz\frac{d}{dz} \left(1-\frac{1}{z} \right\)^{-1} = \frac{T}{z} \left(1-\frac{1}{z} \right\)^{-2} = \frac{Tz}{(z-1)^2}

    This is just one of the usual tricks used to sum the series:

    1+2z^{-1}+3z^{-2}+4z^{-3}+...

    Consider the series for:

    \frac{1}{1-\frac{1}{z}}=1+z^{-1}+z^{-2}+z^{-3}+...

    we can formally differentiate both sides to get:

    -\frac{1}{z^2}\times \frac{1}{(1-\frac{1}{z})^2}=-z^{-2}-2z^{-3}-3z^{-4}-4z^{-5}+...

    .........---........... =-z^{-2}(1+2z^{-1}+3z^{-2}+4z^{-3}+...)

    The author should not have done the summation in place but mentioned what the sum was and referend to an appendix or book of tables for a justification (also I don't think this is the best way of doing it).

    CB
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