# Signal Z-transform

Discussion in 'Homework Help' started by Peytonator, Nov 5, 2011.

1. ### Peytonator Thread Starter Active Member

Jun 30, 2008
105
3
Hi,

Please could you help me with this z-transform question? (See the attachment).

Z[ y(k+2) ] = z^2*Y(z) - z^2*y(0) - z*y(1)

and

Z[ u(k-1) ] = z^-1 * U(z) - 0 = z^-1 * z/(z-1)

Then just multiply Z[y(k+2)]*Z[u(k-1)]

Is that correct?

Many thanks

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2. ### steveb Senior Member

Jul 3, 2008
2,433
469
No, that's not a valid operation. Multiplication in the time domain transforms to convolution in the frequency domain.

You are on the right track by using the time shifting property, but you can consider the shifting of the unit step function directly because normally we assume that a u(k) is implied in the unilateral transform. So, use the time shifting property to get a function with u(k), and then use the given transform along with the time shifting property as you did originally (with a different value of delay).

3. ### Peytonator Thread Starter Active Member

Jun 30, 2008
105
3
Hi Steve,

Thanks for always helping out with these z-transform questions

Sadly on this one I'm still stuck... don't really understand what you mean. Could you explain a little more please?

4. ### steveb Senior Member

Jul 3, 2008
2,433
469

OK, first of all, do you understand what I mean about not being able to just multiply the individual transforms? It's important to understand the duality of multiplication and convolution. Multiplication in the time domain is convolution in the frequency domain, and vice versa.

So, if you are ok with that, solving can be done a few ways. First, you could transform your Y(z) back to the time domain and then generate the time dependent function y(k+2)u(k-1), then transform back. I recommend you do it that way first.

Then, you can try a more elegant approach after that.

5. ### Peytonator Thread Starter Active Member

Jun 30, 2008
105
3
Yes I understand not not multiplying individual transforms.

I think I get it after re-reading your previous post.

Z[ f(n+3)u(n) ] = Z[ y(n+2)u(n-1) ]

where F(z) = 1/z * Y(z)

right?

6. ### steveb Senior Member

Jul 3, 2008
2,433
469
Yes, that would be a good way to start.

Then, you want to work along the lines that you started originally. You want to write y(n+3)u(n) in terms of y(n) so that you can determine the transform of y(n+3) in terms of the Y(z). There are a few ways to do this, but since you already seemed to be moving in the correct direction, I can say one example starts as follows.

y(n+3)u(n)=y(n+3)u(n+3)+y(0)d(0)+y(1)d(1)+y(2)d(2)

where d(n) is the delta function.

EDIT: Sorry, it should be the following.
y(n+3)u(n)=y(n+3)u(n+3)-y(-1)d(-1)-y(2)d(-2)-y(-3)d(-3)

Last edited: Nov 7, 2011