Is it better to take the inverse z-transform with respect to z or z^-1?

[vvkannan]

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Hello all,

I have a question on z-transforms.The question states

"determine the casual signal x(n) having the z-transform
X(z) = 1/(1-2z^-1)(1 - z^-1)^2".


This is how i tried

If i solve this by rewriting as

X(z)/z = z^2 /(z-2)(z-1)^2 and using partial fractions

i get x(n) as 4(2^n) u(n) - 3u(n) - nu(n).

But if i try to find the value by writing the partial fractions for the given function as it is
(i.e) A/(1 - 2z^-1) + B/(1 - z^-1) + C /(1 - z^-1)^2
i get A,B,C as 4,-2,-1
and hence x(n) as 4(2^n) - 2u(n) - n u(n).As seen i get the coefficient as -2 instead of -3 for the second term.


So is it right on my part to take partial fractions as i have done in the second case and solve the given function in z-inverse as it is or it should be solved after converting to a function in z instead of z^-1?
Thank you

 

[Valkarie]

.The answer is that you should solve the given function by converting it to a function in z instead of z^-1. In this case, it would be X(z) = 1/(1-2z)(1 - z)^2. Then solve for the coefficients of the partial fractions. This will give you a correct x(n).

 

[Mene]

for your question and for sharing your approach to solving this problem. The inverse z-transform can be a tricky concept to grasp, but it is an important tool in signal processing and control systems analysis.

To answer your question, both approaches are valid. The inverse z-transform can be taken with respect to either z or z^-1, but it is important to be consistent throughout the problem. In your first approach, you used z^-1 throughout the problem, so it is correct to take the inverse z-transform with respect to z^-1. In your second approach, you used z throughout the problem, so it is correct to take the inverse z-transform with respect to z.

As for the discrepancy in the coefficients, it is likely due to a calculation error in one of the two approaches. It is always a good idea to double check your work and make sure all the steps are correct when solving problems like this.

I hope this helps clarify the inverse z-transform for you. Keep practicing and you will become more comfortable with this concept. Good luck with your studies!