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

# Z-transform equation

Started by ●January 17, 2007

Reply by ●January 17, 20072007-01-17

"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 StefanovStefan, 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?" %%% 919-577-9882 % 'Mission (A World Record)', %%%% <yates@ieee.org> % *A New World Record*, ELO http://home.earthlink.net/~yatescr

Reply by ●January 17, 20072007-01-17

"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 StefanovStefan, 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

Reply by ●January 19, 20072007-01-19