Sampling and the Z transform

Hello,

I'm trying to follow along with an example in a book, but there are two steps in an example and I just can't see how you go from one to the next. The problem is this:

The function is sampled every interval . The Z transform of the function is then...

The solution is provided and worked through:

I understand up to this point. Now here is the bit I don't get:

It looks like the derivative of 'something' has been taken, but I just can't make the connection... Can anyone shed any light on this please?

The rest of the solution is below, and includes the reverse of this 'mystery' operation! (which I also don't understand... I'd love to know what's going on here.

Re: Sampling and the Z transform

Re: Sampling and the Z transform

Quote:

Originally Posted by

**halfnormalled** Hello,

I'm trying to follow along with an example in a book, but there are two steps in an example and I just can't see how you go from one to the next. The problem is this:

The function

is sampled every interval

. The Z transform of the function is then...

The solution is provided and worked through:

I understand up to this point. Now here is the bit I don't get:

It looks like the derivative of 'something' has been taken, but I just can't make the connection... Can anyone shed any light on this please?

The rest of the solution is below, and includes the reverse of this 'mystery' operation! (which I also don't understand... I'd love to know what's going on here.

This is just one of the usual tricks used to sum the series:

Consider the series for:

we can formally differentiate both sides to get:

.........---...........

The author should not have done the summation in place but mentioned what the sum was and referend to an appendix or book of tables for a justification (also I don't think this is the best way of doing it).

CB