# inverse z-transform

Discussion in 'Homework Help' started by vvkannan, Feb 28, 2009.

1. ### vvkannan Thread Starter Active Member

Aug 9, 2008
138
11
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.

if i solve this by rewriting as

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

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

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

i dont know where iam going wrong.
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?

Thank you

2. ### GirishC Active Member

Jan 23, 2009
58
0
Your technique for partial fraction is wrong. carefully look on how to solve repeated root by partial fraction method.

3. ### vvkannan Thread Starter Active Member

Aug 9, 2008
138
11
Thank you for responding.

do you mean the second method?

A and C are solved as usual.
For B heres how i did it

differentiating [1/(1 - 2z^-1)] with respect to z^-1 and subtituting
z^-1 = 1 and i get -2.
is this method wrong?
Or is it wrong to find the partial fraction without inverting the z^-1.
Thank you

4. ### GirishC Active Member

Jan 23, 2009
58
0
I find A = 4, C = -1

so I substitue Z = 0 and solve for equation

Z^2/(Z-2)(Z-1)^2 = A/(Z-2) + B/(Z-2) + c/(Z-1)^2

I get B = -3

so which is equal to your first answer

5. ### vvkannan Thread Starter Active Member

Aug 9, 2008
138
11
Yes GirishC i get A=4,B=-3 AND C=-1 when i convert the given function into a rational function in Z

But my question is why am i not getting this answer when i try to find inverse z-transform keeping the given function as it is.(i.e) without converting it into a function of Z but keeping it in z^-1 form.
Hope you understand what i mean

6. ### GirishC Active Member

Jan 23, 2009
58
0
Your denominator has lesser power(Z^-1) than numerator (Z^0).

Girish

7. ### vvkannan Thread Starter Active Member

Aug 9, 2008
138
11
Yes the numerator has greater power in Z which means when divided it will result in positive powers of Z.

positive powers of Z would result in an anticasual signal.

So to get the casual signal we need lesser powers of z in numerator,correct?

So on the other hand if it was asked to find an 'anticasual' signal can i do it the second way? that is keeping the powers of Z in numerator greater than denominator?
Thank you ,your last answer cleared me up a bit