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A geometric series is a sequence in which each term is derived from the previous term by multiplying it by a fixed, non-zero number known as the common ratio. In the context of Z-transform, you often encounter geometric series when dealing with sequences where subsequent terms decrease or increase exponentially. This concept is crucial when solving problems involving the Z-transform of certain sequences, such as in cases (a) and (b) from the exercise.
In these cases, the Z-transform results in an expression of a geometric series:

• For case (a): the series starts with the term \(z^{-2}\) and has a common ratio of \(z^{-1}\).
• The formula for evaluating such a geometric series is \(\frac{a}{1-r}\), where \(a\) is the first term and \(r\) is the common ratio.

This results in the expression \(\frac{z^{-2}}{1-z^{-1}}\). Understanding how to manipulate and reduce a geometric series is key in handling Z-transforms efficiently.

A linear sequence is where each term increases by a fixed amount from the previous term. One example of a linear sequence is case (c) from the original exercise, where the sequence is represented by \(f[k] = 3k\). Here, every step increases by the same amount, and the challenge is incorporating this linearity into the Z-transform.
To solve such cases:

• Recognize that the sequence grows linearly from zero, that is, the increment is a constant integer.
• The formula for transforming a linear sequence involves using the known series sum equations. Specifically, one can use the sum formula \(\sum_{k=0}^{\infty} k x^k = \frac{x}{(1-x)^2}\) to handle such cases.

The resulting transformation, particularly for case (c), requires adjusting the series using a multiplier, and would look like \(\frac{3z^{-1}}{(1-z^{-1})^2}\). Recognizing and applying sum formulas is crucial when dealing with linear sequences.

An exponential function involves sequences where each term is a constant power of a base. In the context of sequences and Z-transform, it typically refers to sequences that grow or decay exponentially. For instance, the sequence in case (d) from the original exercise is a perfect example, represented by \(f[k] = \mathrm{e}^{-k}\).
This exponential decay can be handled using Z-transform by employing knowledge of geometric series, as exponential functions often form geometric series:

• The challenge is to identify the exponential base and the decay rate.
• For example, the sequence \(\mathrm{e}^{-k}\) transforms based on the sum \(\frac{1}{1- \/e^{-1}z^{-1}}\).

By understanding how exponential decay translates into a Z-transform, one can effectively solve cases involving exponential functions, recognizing their behavior and applying the appropriate mathematical series formula.

A constant sequence is composed of repeated occurrences of the same number. It's one of the simplest forms of sequences and is represented by case (g) in the exercise with \(f[k] = 2\) for all \(k \geq 0\). The Z-transform for such sequences is straightforward but requires attention to how constants affect the formula:

• A constant sequence can be viewed as an infinite sum, so it utilizes the formula for an infinite series of constants, which is \(\frac{c}{1-z^{-1}}\), where \(c\) is the constant.
• Thus, for case (g), the Z-transform simplifies into an expression \(\frac{2}{1-z^{-1}}\).

This formula transformation highlights how a constant term, when transformed, becomes a geometric series with no actual change in value. Constant sequences are foundational and illustrate a simple yet crucial principle in handling Z-transforms efficiently.