# Sampling and the Z transform

• Aug 17th 2011, 02:16 PM
halfnormalled
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}$
• Aug 17th 2011, 02:46 PM
pickslides
Re: Sampling and the Z transform
Quote:

Originally Posted by halfnormalled

$= \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\)$

Do you follow?
• Aug 17th 2011, 09:00 PM
CaptainBlack
Re: Sampling and the Z transform
Quote:

Originally Posted by halfnormalled
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