Can someone help me check if my inverse Z transform solutions are correct?

In summary, the conversation is about solving two equations using the inverse Z transform by partial fractions and long division methods. The person is struggling because the two methods do not match, even though the teacher says they should. They are unsure if they should use partial fractions in z or z^{-1} and are reminded that the degree of the numerator should be less than the degree of the denominator. They are also told that if neither way works, they can use long division to get a remainder and then use partial fractions on the remainder. However, in both problems, using partial fractions in z^{-1} should work. The person is then asked to show their work.
  • #1
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Homework Statement



i need to solve this 2 equations, got to find the inverse Z transform by partial fractions expension and long division method each, but in each one in none of the two methods match, and according to the teacher, they do match!...can anyone see if the do match?

Homework Equations




X(z)= ((z^2)-0.7z)/(z-1*z-0.6)

and

X(z)= (1+2z^-1)/(1-z^-1)^2


The Attempt at a Solution

 
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  • #2
Did you do partial fraction in z or [tex]z^{-1}[/tex] ?

Remember that to use partial fraction you need the degree of the numerator< degree of the denominaotr (in whatever variable you choose z or [tex]z^{-1}[/tex])

Sometimes you can't use z because the degree of the numerator>= degree of the denominator so you can switch to [tex]z^{-1}[/tex] or vice versa... so you divide or multiply the numerator and denominator by a power of z to switch from one to the other...

But sometimes neither way works... degree of the numerator >=degree of the denominator with both ways... then you'll have to do one step of long division to get a remainder... so

P(z)/Q(z) = A(z) + R(z)/Q(z)... and now you can use partial fraction on R(z)/Q(z)...

But in both your problems, partial fractions in [tex]z^{-1}[/tex] should be no problem.

Can you show your work?
 

FAQ: Can someone help me check if my inverse Z transform solutions are correct?

1. What is the purpose of an inverse Z transform solution?

An inverse Z transform solution is used to convert a discrete-time signal from the Z domain back to the time domain. This is useful in understanding the behavior and characteristics of a system in the time domain.

2. How is an inverse Z transform solution calculated?

An inverse Z transform solution is calculated using partial fraction decomposition, where the Z transform is broken down into simpler terms that can be converted back to the time domain using known Z transform pairs.

3. Are there any limitations to using an inverse Z transform solution?

Yes, there are some limitations to using an inverse Z transform solution. It may not be possible to find an inverse Z transform for every discrete-time signal, and there may be multiple possible solutions for a given signal.

4. Can an inverse Z transform solution be used for continuous-time signals?

No, an inverse Z transform solution is only applicable for discrete-time signals. For continuous-time signals, the Laplace transform is used instead.

5. Are there any real-world applications of inverse Z transform solutions?

Yes, inverse Z transform solutions are commonly used in fields such as signal processing, control systems, and digital communication, where discrete-time signals are encountered and need to be analyzed in the time domain.

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