Z-transform equation

Hi guys! I've got a pretty big pain in the ass - a course work in DSP. To
be honest with you I'm a network engineer and electronics are a bit far
away from my area of interests, but anyway I have to graduate. I read at
least 15 different explanations on the Z-transform and I'm still not sure
what I'm doing. Here's what I have:

1) A discrete series: x(0)=0, x(T)=-2.5, x(2T)=-3, x(4T)=1.25, x(5T)=-5,
x(nT)=0 where n>=6

2) As far as I understand this should be a unilateral series, right?

3) The next thing I saw was the general formula here:

http://content.answers.com/main/content/wp/en/math/e/d/4/ed4eb34024a58e8b0159dfbaa66c985e.png

4) So the question is: what s the X(z) equation? Damn, I can't get the
principle, though I'm sure it's pretty easy. Can you help me, anyone?
Thank you very much in advance!

Greetings, Stefan Stefanov

"thegrave" <thegrave2000@yahoo.com> writes:

> Hi guys! I've got a pretty big pain in the ass - a course work in DSP. To
> be honest with you I'm a network engineer and electronics are a bit far
> away from my area of interests, but anyway I have to graduate. I read at
> least 15 different explanations on the Z-transform and I'm still not sure
> what I'm doing. Here's what I have:
>
> 1) A discrete series: x(0)=0, x(T)=-2.5, x(2T)=-3, x(4T)=1.25, x(5T)=-5,
> x(nT)=0 where n>=6
>
> 2) As far as I understand this should be a unilateral series, right?
>
> 3) The next thing I saw was the general formula here:
>
> http://content.answers.com/main/content/wp/en/math/e/d/4/ed4eb34024a58e8b0159dfbaa66c985e.png
>
> 4) So the question is: what s the X(z) equation? Damn, I can't get the
> principle, though I'm sure it's pretty easy. Can you help me, anyone?
> Thank you very much in advance!
>
> Greetings, Stefan Stefanov

Stefan,

You have everything you need - just plug it in. You're only about 5 minutes
away from your solution (if that). (Yes, it's unilaterial in this case.)

Also, you do know that x(kT) is equivalent to x[k], don't you? ...
The difference between the two is that kT is a point in time, 
while k is an index into a sequence (x[k]).

So just plug in x(kT) for x[k] in the definition you have.
-- 
%  Randy Yates                  % "How's life on earth? 
%% Fuquay-Varina, NC            %  ... What is it worth?" 
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http://home.earthlink.net/~yatescr

"thegrave" <thegrave2000@yahoo.com> wrote in message 
news:4tednSu5rMlDJTPYnZ2dnUVZ_u2mnZ2d@giganews.com...
> Hi guys! I've got a pretty big pain in the ass - a course work in DSP. To
> be honest with you I'm a network engineer and electronics are a bit far
> away from my area of interests, but anyway I have to graduate. I read at
> least 15 different explanations on the Z-transform and I'm still not sure
> what I'm doing. Here's what I have:
>
> 1) A discrete series: x(0)=0, x(T)=-2.5, x(2T)=-3, x(4T)=1.25, x(5T)=-5,
> x(nT)=0 where n>=6
>
> 2) As far as I understand this should be a unilateral series, right?
>
> 3) The next thing I saw was the general formula here:
>
> http://content.answers.com/main/content/wp/en/math/e/d/4/ed4eb34024a58e8b0159dfbaa66c985e.png
>
> 4) So the question is: what s the X(z) equation? Damn, I can't get the
> principle, though I'm sure it's pretty easy. Can you help me, anyone?
> Thank you very much in advance!
>
> Greetings, Stefan Stefanov

Stefan,

Your problem is what I call a "language" problem - as in "math is mostly 
about language".....

You already have the equations:
                 inf
X(z) = Z{x[n]} = sum{x[n]*z^-n}
                 n=0

It may help to expand the "sum" for clarity:

inf
sum{x[n]*z^-n} = x[0]*z^-0 + x[1]*z^-1 + x[2]*z^-2 + ......
n=0

The precursors to the equations at the top say this:

X(z) is the Z transform of a sequence of samples x[.] and is computed by the 
sum expression.

It is normal to assert that x[0] means "the value of x at time=0"
But, you might also say: "the zeroeth indexed sample contains the value of x 
at time=0".
Then, by extension, x[1] means the 1th indexed sample contains the value of 
x at time 1*T where T is the sample interval.

Often, we normalize to T=1 which makes the normalized sample rate be 1Hz. 
I'm going to use this convention now:

Note in the sum expression that the x value at time=0 is multiplied by z^-0 
or 1.0.
Note in the sum expression that the x value at time=T=1 is multiplied by 
z^-T = z^-1  .... and so forth.
So, you might say that z^-p means a delay of p seconds or, unnormalized, pT 
seconds.

Accordingly, the sum provides a polynomial in z or 1/z=z^-1 if you will.  It 
tells us by inspection that x[0] is undelayed from t=0; x[1] is delayed by 
one unit delay; x[2] is delayed by two unit delays, etc.
z^-1 corresponds to a unit delay and z^-n corresponds to n unit delays 
(where n is an integer of course).

This is a very basic observation related to the z transform.

using your example:

x=[0 -2.5 -3 ?  1.25 -5 0 0 0 0 0 0 0 .......]
the "?" because you didn't specify x[3]

which can be written:
x=[0 -2.5 -3 ?  1.25 -5]

Here the sample time is normalized to "1 sample time" or perhaps you might 
consider that to be 1 second.  It depends on how you like to carry constant 
factors around.  Most of the time the transforms are expressed in terms of 
unit delays; that is, on a normalized time (and thus frequency) scale.

I hope this might help.

Fred 

Thank you very much guys, you've been very helpful! I understood what you
meant exactly but it took me time because now I'm very far away from maths
and needed some time to refresh my memory. Good luck and wish you all the
best!