H(z) = (z2 + 0.5z - 0.5)/(z2 + 1.5z + 0.5)

In summary, the conversation was about solving a difference equation and finding the partial fraction form of the equation. The solution involves multiplying the fractions by different values and combining them into one fraction. The expert confirms that this is the correct method for solving the problem.
  • #1
asdf12312
199
1

Homework Statement


img.png


Homework Equations


H(z) = Y(z)/X(z)

The Attempt at a Solution


I realized this wasn't in partial fraction form because the 1+z-1+0.5z-2 has non-real roots. I multiplied the 1st fraction part by z1 and the 2nd fraction by z2, then I combined them into one fraction and I think I am able to get a difference equation at the end, but is the way I am doing it right? Or is there an easier way to do this problem that I am missing.
 
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  • #2
asdf12312 said:
I multiplied the 1st fraction part by z1 and the 2nd fraction by z2, then I combined them into one fraction

Yes, that's how you do it.

H(z) = 1/(1+0.5z-1) - 1/(1+z-1+0.5z-2) = z/(z+0.5) - z2/(z2+z+0.5).

Combine the fractions.
 

Related to H(z) = (z2 + 0.5z - 0.5)/(z2 + 1.5z + 0.5)

1. What is the inverse z-transform problem?

The inverse z-transform problem is a mathematical problem that involves finding the original discrete-time signal from its z-transform. It is essentially the reverse process of finding the z-transform of a signal.

2. What are the applications of the inverse z-transform?

The inverse z-transform is commonly used in signal processing and control systems to analyze and design discrete-time systems. It is also used in image processing, digital filters, and time series analysis.

3. How is the inverse z-transform calculated?

The inverse z-transform can be calculated using various techniques such as partial fraction expansion, power series expansion, and residue method. The choice of method depends on the complexity of the z-transform and the desired accuracy of the solution.

4. What is the difference between the inverse z-transform and the inverse Laplace transform?

The inverse z-transform and the inverse Laplace transform are both methods of finding the original signal from its transform. However, the inverse z-transform is used for discrete-time signals while the inverse Laplace transform is used for continuous-time signals.

5. What are some common properties of the inverse z-transform?

Some common properties of the inverse z-transform include linearity, time-shifting, scaling, and convolution. These properties allow for the manipulation and analysis of signals in the z-domain to be translated back to the time-domain.

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