Rewriting x(k) So i can calc a Z Transform

[Evo8]

Homework Statement

Let x(k)=
0 k<0
0.6^k 0<=k<=21
0 k>21

Calculate X(z)

Homework Equations

Z transform

The Attempt at a Solution

Ok so i think calculating the Z transform of this won't be too difficult for me at all. I just have a hard time with this type of notation. So what i would like to some help on is how to rewrite the equation x(k).

I understand that x(k)=0.6^k and for K is between 0 and 21. Any values outside of that x(k) = zero... So how do I assemble X(k) in a form that I can do a z transform to it? It seems like it should be easy but I just don't see it. Something with a summation maybe and a limit of 0 -> 21?

Thanks for any comments guys!

 

[I like Serena]

Hi Evo8!

X(z) is the z-transform.

It is given by:
X(z)\overset{\mathrm{def}}{~=~}\mathcal{Z}\{x(k)\}\overset{\mathrm{def}}{~=~}\sum_{k=-\infty}^{\infty} x(k)z^{-k}

 

[Evo8]

Hi I like Serena.

I understand the z transform formula. What I don't understand is what my x(k) actually is to do the z transform too.

Thanks for your help!

 

[I like Serena]

I'm not sure I understand your question.

You're meant to substitute your definition of x(k) in the summation...
Then adjust the bounds, since it's only non-zero in part of its range.

 

[Evo8]

Yes but my problem is I don't really understand what my x(k) is in this case.

x(k)=0.6^k with the bounds 0->21 How do I apply this? I don't know why this is hard for me to see. I am sure its simple. I just don't fully understand the notation I guess.

Would it just be

X(z)=(0.6^k) (z^-1)?

 

[I like Serena]

There is a summation over k:
X(z)=... + x(-1)z^{-(-1)} + x(0)z^{-0} + x(1)z^{-1} + ...

And
..., \quad x(-1)=0, \quad x(0)=0.6^0, \quad x(1)=0.6^1, \quad ..., \quad x(21)=0.6^{21}, \quad x(22)=0, \quad ...

 

[Evo8]

Ok I think I understand. I think my issue was more understaning how the summation works. I am still on the edge of understanding. Either way this is what I have come up with after following a similar example from my book.

x(k)=\sum(0.6^k) 0->21

X(z)= \sum(0.6^k z^-k) 0->21

X(z)=\sum(0.6^k (1/z^k)) 0->21

X(z)=\sum((0.6/z)^k) 0->21

X(z)=1/(1-(0.6/z))

X(z)=z/0.4

Am I on the right track? I used the generalized geometric series to get this form. Well I tried to anyway.

Thanks again for your help!

 

[I like Serena]

Let's see...

Ok I think I understand. I think my issue was more understaning how the summation works. I am still on the edge of understanding. Either way this is what I have come up with after following a similar example from my book.

x(k)=\sum(0.6^k) 0->21

No, x(k) itself is not a summation, but a function with a value that depends on the value of k.

X(z)= \sum(0.6^k z^-k) 0->21

X(z)=\sum(0.6^k (1/z^k)) 0->21

X(z)=\sum((0.6/z)^k) 0->21

X(z)=1/(1-(0.6/z))

Looking good, except for that last step.
That is the sum of a geometric series summing to infinity.
However, you should only sum to k=21.

X(z)=z/0.4

Am I on the right track? I used the generalized geometric series to get this form. Well I tried to anyway.

Thanks again for your help!

Geometric series is the way to go.
Just get the right formula for it. ;)

 

[Evo8]

Ok so up until

\sum(\frac{0.6}{z})^{k}

Im still on the right track?

The formula for the geometric series that I have is

\sum z^{k}=\frac{z^{m}}{1-z},m≥0 and lzl<1

My m in this case would equal 0 because my limits are 0-21. so the z in the numerator becomes 1 and the z in the denominator is my z which equals (0.6/z). This was my reasoning anyway. Do I just need to have notation instructing to stop the sum at 21?

Like lzl≤21?

 

[I like Serena]

Ok so up until

\sum(\frac{0.6}{z})^{k}

Im still on the right track?

Yep!

The formula for the geometric series that I have is

\sum z^{k}=\frac{z^{m}}{1-z},m≥0 and lzl<1

My m in this case would equal 0 because my limits are 0-21. so the z in the numerator becomes 1 and the z in the denominator is my z which equals (0.6/z). This was my reasoning anyway. Do I just need to have notation instructing to stop the sum at 21?

Like lzl≤21?

I don't know that formula.
Where did you get it?

Here's the formula I mean:
http://en.wikipedia.org/wiki/Geometric_series#Formula

 

[Evo8]

I got that formula from my DSP book. Ill take a look at the one you have posted from wikipedia and see if i can understand how to apply it.

Thanks again.

 

[Evo8]

Ok so I looked back and I am unsure if I am using the geometric series correctly.

This is what I have for a definition in my book.

However I made a stupid mistake on my Algebra...

Up to this point I am ok.. X(z) = \sum(\frac{0.6}{z})^{k}

Now when I apply the geometric series I get X(z)=\frac{1}{1-(\frac{0.6}{z})}

Now when I simplify it should be X(z)=\frac{z}{z-0.6}

This looks ok to me i think. Especially since there is an example that is very similar in my book that I followed. The only problem is I don't have the limit of 21 defined anywhere. The example in the book is 0->∞ My problem above is 0->21... Where does this upper limit of 21 come into play?

Here is the example for your reference.

The X_{c}(z) in this example is what I believe to be very similar to my problem above.


In addition I have to find the region of convergence of my X(z). I am a bit confused on what the ROC actually is but I am still reading that section over in my book. I may have a few questions on this shortly. Not sure if its worth starting another thread for though.

Thanks,


Edit: If my X(z) is correct would my ROC simply be 21>lzl.0.6 ?

 

[I like Serena]

Sorry, but your X(z) is not correct yet.
As you already surmised the upper limit of 21 should go somewhere.
All the formulas that you have shown sum up to infinity.

To sum a series r^k up to some limit n, you need:
\sum_{k=0}^n r^k = {1 - r^{n+1} \over 1 - r}

Alternatively you can write your sum as:
\sum_{k=0}^n r^k = \sum_{k=0}^\infty r^k - \sum_{k=n+1}^\infty r^k



When you sum up to infinity, there is a chance that the sum does not exist, for instance because it tends to infinity.
The region of convergence is where the sum exists.
But let's discuss that when you have your X(z).

 

[Evo8]

Hmmm...ok. I think I understand what you mean.

Going by the first formula you posted I would get

\frac{1-(\frac{0.6}{z})^{22}}{1-(\frac{0.6}{z})}

If I understand this correctly I see now what you mean by the series I had first written was to ∞ not to 21...

However I need to do something else here to simplify I assume? Or is this the final form? I don't see anything else I can do algebraically except for writing out the entire series. I may be missing something though as my algebra skills are sub par in my opinion.

 

[I like Serena]

Nope. That's it! :)
As far as I'm concerned that's the final form.
There's no useful algebraic simplification.

As for the region of convergence.
For which values of z is this formula well defined?

 

[Evo8]

I think it would be 0≤z≤21 right?

Or would it be 0.6≤z≤21 ...? If I were to guess I feel like my first answer is correct. I am still a little confused on this though.

 

[I like Serena]

What happens if you substitute z=0?
Does it exist?

What if z=22?

Btw, note that z can be any complex number, not just a real number.

 

[Evo8]

Well when you substitute z=o it does not exist. Thats for sure. So good point there.

If you substitute z=22 it does indeed exist. So maybe my second answer is more correct? Or should it be 0.6≤z≤22?

Or am I still way off track here?

 

[I like Serena]

You're still a bit off track. ;)

What if z=0.6?
What if z=23?
What if z=0.1?
What if z=i?

 

[Evo8]

Well I tried all of those values and

z=0.6 comes up undefined of course so that's out
z=23 comes out at just over 1. 1.02679...
z=0.1 is some large number.
z=i comes up as a complex number with real and non real parts. I am not sure how to understand this.

I can see that the lower limit (i think) has to larger then 0.6 to keep the expression form being 0. So z>0.6 ... ?

As for an upper limit it seems like 23 and higher would work no problem but were looking for the limit of the convergence right so it would be ≤ 21?

So 0.6< z ≤21?

 

[I like Serena]

Actually, in this case convergence is not an issue.
Convergence is only an applicable issue if you have a series that sums to infinity.
(There's a special convergence criterium, which is mentioned in your book in that case.)
This series doesn't. It stops at k=21.

So the only criterium is that the expression is well defined.
The only thing that can go wrong in this case, is division by zero.
Which values of z would make you divide by zero?
Those are off.
All other values are good.

 

[Evo8]

Ok, After taking another look it looks like the only value that causes a problem is 0.6. Above or below are just fine. So I just need to notate that 0.6 is not to be used?

Also I just noticed I am asked to find "the radius of convergence" not the region but i feel like this is the same thing just a different term correct?

 

[I like Serena]

What about z=0?
And what about z=i?Radius of convergence?
That makes no sense in this case.

For a regular geometric series \sum r^k that is summed to infinity, the ROC is the unit disk in the complex plane given by |r| < 1.
Then you can define a radius of convergence, which is the radius of the disk.

But you don't have a disk, so it seems to me that the question is badly phrased.

 

[Evo8]

I think you are correct. It should most likely be Region of convergence. My book doesn't even speak of radius of convergence and I don't really remember seeing it during lectures either.

So we can scratch radius of convergence and believe it to be region of convergence.

Right z=0 would also pose a problem not sure why I didnt see that. z=i I am slightly confused by. Of course when i substitute this in I get a complex number on the output. Something along the lines of

0.74 - 0.44*i which i don't see being a problem either. Maybe I am missing something though. Since you mentioned it I feel like its something of interest...

 

[I like Serena]

Indeed, z=i is not a problem. :)
The reason I mentioned it, is that you said "above or below 0.6".
But i is neither above or below, it is just different.

So what do you think the ROC is?

 

[Evo8]

Hmm well i think we have come to the conclusion that the two values that will throw the eq off is z=0 and z=0.6.

So would the ROC be
z≠0
z≠0.6
All other values of z are good. I am just not sure how to properly express this with numbers and/or symbols..

 

[I like Serena]

I believe that what you wrote is just fine.

It is clear and unambiguous.
That is all that is required mathematically.
If you want you could write:
ROC=\mathbb{C} \backslash \{0, 0.6\}

 

[Evo8]

Sounds good. I think that notation is a little bit cleaner. I really appreciate your help!