Solve Inverse Z-Transform for H(z)=1/(1-2/3*z-1)

In summary, the conversation is about finding h(n) for the given H(z) function using the residue method. The attempt at a solution involves using the formula for residue and simplifying incorrectly, leading to a wrong result. The correct approach is to simplify by (3z-2)/3 when calculating the residue.
  • #1
Drao92
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0

Homework Statement


H(z)=1/(1-2/3*z-1)
h(n)=?

Homework Equations


I tried to use the residue method to solve this problem but it doesn't give me the good result and i am not sure if i don't know how to use the method or this problem can't be solved using this method.
Also, its pretty obvious that the result is h(n)=(2/3)n because it looks like a geometric progression

The Attempt at a Solution


I used the formula of residue method.
h(n)=residue(H(z)*zn-1) when z=2/3
The result is 3*(2/3)n
Its this result wrong because when we calculate Z transform of (2/3)n which is (1-(2/3)n*z-n )/(1-(2/3)*z-1) we ignore (2/3)n because is 0 when n=infinity?
 
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  • #2
Drao92 said:

Homework Statement


H(z)=1/(1-2/3*z-1)
h(n)=?

Homework Equations


I tried to use the residue method to solve this problem but it doesn't give me the good result and i am not sure if i don't know how to use the method or this problem can't be solved using this method.
Also, its pretty obvious that the result is h(n)=(2/3)n because it looks like a geometric progression

The Attempt at a Solution


I used the formula of residue method.
h(n)=residue(H(z)*zn-1) when z=2/3
The result is 3*(2/3)n
Its this result wrong because when we calculate Z transform of (2/3)n which is (1-(2/3)n*z-n )/(1-(2/3)*z-1) we ignore (2/3)n because is 0 when n=infinity?
You have
$$z^{n-1} H(z) = \frac{z^{n-1}}{1-\frac{2}{3z}} = \frac{z^n}{z-\frac{2}{3}}.$$ How did you get ##h(n)=3(2/3)^n## from that?
 
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  • #3
I think i found what what I am doing wrong.
I made H(z)=3z/(3z-2) then i multiply by zn-1 and simplify by (3z-2) but i must simplify by (3z-2)/3 when calculating the residue... From now ill stick only to z-x form for denominator.
Eveyrthing is clear now, thanks!
 
Last edited:

Related to Solve Inverse Z-Transform for H(z)=1/(1-2/3*z-1)

1. What is the Inverse Z-Transform?

The Inverse Z-Transform is a mathematical operation that converts a function from the Z-domain to the time-domain. It is the reverse of the Z-Transform and is used to solve difference equations and analyze discrete-time systems.

2. How do I solve Inverse Z-Transforms?

To solve an Inverse Z-Transform, you can use various techniques such as partial fraction decomposition, power series expansion, or contour integration. The specific method used will depend on the complexity of the function and the desired form of the solution.

3. What is the formula for Inverse Z-Transform?

The general formula for Inverse Z-Transform is given by:

f(n) = (1/2πi) ∫γ H(z)zn-1dz

Where f(n) is the time-domain function, H(z) is the given Z-Transform function, and γ is a contour in the Z-plane that encloses all the poles of H(z).

4. How do I solve Inverse Z-Transform for a specific function?

To solve Inverse Z-Transform for a specific function, you can follow these steps:

1. Write the given function in the form of H(z).

2. Identify the poles of H(z) by setting the denominator equal to zero.

3. Determine the contour γ that encloses all the poles.

4. Apply the general formula for Inverse Z-Transform to solve for the time-domain function f(n).

5. Can I use a computer or calculator to solve Inverse Z-Transforms?

Yes, you can use a computer or calculator to solve Inverse Z-Transforms. Many mathematical software programs and graphing calculators have built-in functions for calculating Inverse Z-Transforms. However, it is important to have a good understanding of the underlying principles and techniques for solving Inverse Z-Transforms in order to interpret and verify the results obtained from a computer or calculator.

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