Help with Z Transform Calculations

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In summary, the conversation discusses finding the z transform and region of convergence for different sequences, using the z transform to perform convolution, and finding the causal signal for a given z transform. The steps and equations for solving each problem are provided, as well as the final results and regions of convergence. The poster also requests for someone to check if their solutions are correct.
  • #1
DrunkEngineer
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Homework Statement


I. Find the z transform and ROC of each of the ff sequence
1.[tex] x(n) = 2\delta{n} + 3(\frac{1}{2})^{n}u[n] - (1/4)^{n}u(n)[/tex]


II. Use the Z transform to perform the convolution of the following sequence.
[tex]x[n] = 3^{n}u(-n)[/tex]
[tex]h[n] = (0.5)^{n}u(n)[/tex]


part III. Find the causal signal x(n) fo the following z transforms
[tex]X(Z) = \frac{ 1-z^{-1} }{(1+z^{-1})(1+\frac{1}{2}z^{-1})}[/tex]
pls go to number 3 because latex cannot be edited


Homework Equations


properties and table of z transform are found here
http://en.wikipedia.org/wiki/Z_transform



The Attempt at a Solution


I. Find the z transform and ROC of each of the ff sequence
1.[tex] x(n) = 2\delta{n} + 3(\frac{1}{2})^{n}u[n] - (1/4)^{n}u(n)[/tex]
[tex]X(z) = 2(1) + 3\frac{1}{ 1-\frac{1}{2}z^{-1} } - \frac{1}{1-\frac{1}{4}z^{-1} }[/tex]
ROC: All z, 1/2 <z and 1/4 < z
the total ROC is z > 1/2

II. Use the Z transform to perform the convolution of the following sequence
[tex]x[n] = 3^{n}u(-n)[/tex]
since [tex]x(-n)[/tex]'s z transform is [tex]X(z^{-1})[/tex]

[tex]X_1(Z) = \frac{1}{1-3z}[/tex]


[tex]h[n] = (0.5)^{n}u(n)[/tex]
[tex]X_2(Z) = \frac{1}{1-\frac{1}{2}z^-1}[/tex]
[tex]x[n]*h[n][/tex] is equivalent to [tex]X_1(Z)X_2(Z)[/tex]

[tex]X_1(Z)X_2(Z) =\frac{1}{(1-3z)(1-\frac{1}{2}z^-1)}[/tex]

using wolfram alpha to solve partial fractions
[tex]\frac{2z}{(1-3z)(2z-1)}[/tex]

[tex]\frac{2}{3z-1} - \frac{2}{2z-1}[/tex]

then simplify
[tex]\frac{ 2z^{-1}\frac{1}{3} }{ 1-\frac{1}{3}z^{-1} } - \frac{ 2z^{-1}\frac{1}{2} }{1-\frac{1}{2}z^{-1} }[/tex]
the region of convergence is 1/3 < z and 1/2 < z
then the total ROC is 1/2

the inverse z transform is : using the time shifting property z^-1X(z) = u(n-1)
[tex]\frac{2}{3}(\frac{1}{3})^{n}u(n-1) -(\frac{1}{2})^{n}u(n-1)[/tex]

part III. Find the causal signal x(n) fo the following z transforms
[tex]X(Z) = \frac{ 1-z^{-1} }{(1+z^{-1})(1+\frac{1}{2}z^{-1})}[/tex]
using wolfram alpha
[tex]\frac{2(z-1)z}{z+1}{2z+1}[/tex]
[tex]\frac{3}{2z+1}-\frac{4}{z+1}+1[/tex]

Making it into a z transform
[tex]\frac{\frac{1}{2}3z^{-1}}{1-(-\frac{1}{2})z^{-1}} -\frac{4z^{-1}}{(1-(-z^{-1})} +1[/tex]
the region of convergence is
-1/2 < z, -1 < z, and the entire plane of z
the total region of convergence is the entire plane of z

the inverse z transform is
[tex]x(n) = \frac{3}{2}(-\frac{1}{2})^{n}u(n-1) - 4(-1)^{n}u(n-1)[/tex]

can you check if this is correct?
i mean the ROC etc.
 
Last edited:
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  • #2
Anybody kno z transform?
bump for 1st page
 

1. What is a z-transform and why is it useful?

A z-transform is a mathematical technique used in signal processing and control systems to convert a discrete time-domain signal into a complex frequency-domain representation. It is useful because it allows us to analyze and manipulate signals using algebraic operations, making it easier to design and implement filters and controllers.

2. How is the z-transform related to the Fourier transform?

The z-transform is closely related to the Fourier transform, as it can be thought of as the discrete version of the continuous Fourier transform. Both transforms convert signals from the time-domain to the frequency-domain, but the z-transform is specifically designed for discrete signals, while the Fourier transform is used for continuous signals.

3. How do I calculate the z-transform of a signal?

To calculate the z-transform of a signal, you can use the formula: X(z) = ∑[x(n) * z^(-n)], where x(n) is the signal and z^(-n) is the z-transform variable raised to the power of -n. Alternatively, you can use a table of common z-transform pairs to find the transform of a specific signal.

4. What is the inverse z-transform and how is it used?

The inverse z-transform is the process of converting a z-transformed signal back to the time-domain. This is useful for implementing digital filters and controllers, as these systems operate in the time-domain. The inverse z-transform can be calculated using a formula or a table of common pairs, similar to the z-transform.

5. How can the z-transform be applied in real-world applications?

The z-transform has many practical applications, such as designing digital filters and controllers for systems like audio processing, image processing, and control systems. It is also used in communication systems for equalization and error correction. Additionally, the z-transform is a fundamental concept in digital signal processing and is used extensively in fields like telecommunications, biomedical engineering, and robotics.

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