# Signal Z-transform

#### Peytonator

Joined Jun 30, 2008
105
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 #### steveb

Joined Jul 3, 2008
2,436
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 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).

#### Peytonator

Joined Jun 30, 2008
105
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?

#### steveb

Joined Jul 3, 2008
2,436
... don't really understand what you mean. Could you explain a little more please?

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.

#### Peytonator

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

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

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

right?

#### steveb

Joined Jul 3, 2008
2,436
Yes I understand not not multiplying individual transforms.

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

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

right?
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)

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