Compute the Z-transform of a^{-n} and step function

In summary, to find the Z-transform of x[n] = (0.5)^{-n} * u[n-1], we can use the definition of Z-transform and rewrite the expression as x(z) = \sum_{n=1}^\infty (2z^{-1})^{n}. Then, we can apply the geometric series to simplify it further. However, since the series diverges, this indicates that the Z-transform does not exist for this expression.
  • #1
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Homework Statement


Compute the Z-transform of

[tex]x[n] = (0.5)^{-n} * u[n-1][/tex]

And find the ROC (Region of Convergence)

Homework Equations


Z-Transform for discrete time
[tex]x(z) = \sum_{n=0}^\infty x[n] * z^{-n}[/tex]

The Attempt at a Solution



To solve this, I used the definition of Z-transform

[tex]x(z) = \sum_{n=1}^\infty (0.5)^{-n} * z^{-n}[/tex]

Note that the summation starts from n = 1 due to the time shift to the right by 1.

Simplifying this, we get

[tex]x(z) = \sum_{n=1}^\infty (0.5z)^{-n}[/tex]

Here's where I'm confused as hell. Apparently, we can't apply the geometric series to simplify this further since the expression is powered to the negative n.

Furthermore, what's contradicting about this problem is that the series diverges. Does this indicate that the z-transform does not exist at all?

Any help will be appriciated.
 
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  • #2
Try a little rewrite:

[tex]x(z) = \sum_{n=1}^\infty (0.5^{-1}z^{-1})^{n}[/tex] or

[tex]x(z) = \sum_{n=1}^\infty (2z^{-1})^{n}[/tex] Now use your geometric series...
 
  • #3

Firstly, it is important to note that the Z-transform is only defined for signals that are absolutely summable, i.e. the sum of the absolute values of the signal must converge in order for the Z-transform to exist. In this case, since the series (0.5)^{-n} diverges, the Z-transform of x[n] does not exist.

However, we can still discuss the Region of Convergence (ROC) for this signal. The ROC is the set of values of z for which the Z-transform converges. In this case, since the series diverges, the ROC will be empty, i.e. there is no region of convergence for x[n].

Now, moving on to the step function u[n-1]. The Z-transform of a step function is given by:

U(z) = \sum_{n=0}^\infty u[n-1] * z^{-n}

= \sum_{n=1}^\infty z^{-n}

= \frac{1}{1-z^{-1}}, |z| > 1

Here, we can apply the geometric series since the expression is powered to the negative n. The ROC for the step function is |z| > 1, which means that the Z-transform exists for all values of z outside the unit circle in the z-plane.

In conclusion, the Z-transform of x[n] does not exist, while the Z-transform of u[n-1] exists for all values of z outside the unit circle.
 

Related to Compute the Z-transform of a^{-n} and step function

1. What is the Z-transform of a^-n?

The Z-transform of a^-n is given by:
X(z) = 1/(1-az^-1)
where a is a constant and z^-1 represents the delay operator.

2. How is the Z-transform of a^-n derived?

The Z-transform of a^-n can be derived by assuming a geometric series for the expression a^-n and using the definition of the Z-transform.

3. What is the significance of the step function in the Z-transform of a^-n?

The step function represents the initial condition of the system and is necessary for the complete representation of the Z-transform of a^-n.

4. Can the Z-transform of a^-n be used for solving difference equations?

Yes, the Z-transform of a^-n can be used to solve difference equations in the time domain and is useful in analyzing discrete-time systems.

5. How can the Z-transform of a^-n be applied in signal processing?

The Z-transform of a^-n is commonly used in signal processing for analyzing and filtering discrete-time signals and can also be used for designing digital filters.

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