Question

Find the inverse Z-Transform of -
f(z)=(z+2)/(z^2-2z+1) ,|z|>1

Answer 1

So i have
\(f(z)=\dfrac{1}{(z-1)}+\dfrac {3}{(z-1)^2} = \dfrac{z^{(-1)}}{(1-z^{(-1)} )}+\dfrac{3 z^{(-2)}}{(1-z^{(-1)} )^2} \)

Answer 2

I know the INV Z Transform of 1st term, but how about the 2nd term ??

Answer 3

or is there any other way to do this ?

Answer 4

convolution ?

Answer 5

What if you use Partial Fraction for \(\large \frac{X(z)}{z}\)

Answer 6

Sorry that is F(z) there..

Answer 7

tried that..no use..

Answer 8

Let me try this. :)

Answer 9

I don't know this seriously as In Signals, I have not reached here, but I see what I can do in this.. :)

Answer 10

what subject is this?

Answer 11

Z transform in Maths ...

Answer 12

:/

Answer 13

do you know pdes?

Answer 14

Partial DE's ...yes, but I don't think that should be used here...

Answer 15

it's because I'm in that course and I don't understand what this person is talking about. you can read my question here http://math.stackexchange.com/questions/922338/find-a-solution-of-u-xxu-yy-axbyc-for-given-real-constants-a-b-and?noredirect=1#comment1904098_922338

Answer 16

Let X(z) = F(z)/z

Answer 17

\[X(z) = \frac{F(z)}{z} = \frac{z+2}{z(z-1)^2} = \frac{2}{z} - \frac{2}{z-1} + \frac{3}{(z-1)^2}\]

Check if it is right?

Answer 18

Did you get the same Hardik when you tried this method?

Answer 19

ok, i see where you're going

Answer 20

2-2z/(z-1) +3z/(z-1)^2

Answer 21

Now multiply by z both the sides, you will get F(z) again..

Answer 22

Yes..

\[Z^{-1}(\frac{z}{(z-1)^2}) = n \cdot u[n]\]

Answer 23

\(2\delta [n] -2u[n]+3 n u[n]\)
?

Answer 24

Yep, according to me..

Answer 25

wolf answer is a lil bit different...
http://www.wolframalpha.com/input/?i=inverse+z+transform&f1=+(z%2B2)%2F(z%5E2-2z%2B1)&f=InverseZTransformCalculator.transformfunction_+(z%2B2)%2F(z%5E2-2z%2B1)&f2=z&f=InverseZTransformCalculator.variable1%5Cu005fz&f3=n&f=InverseZTransformCalculator.variable2%5Cu005fn

Answer 26

anyways, i'll ignore wolf and go with your method/answer .
thanks! :)

Answer 27

But if n is greater than 0, then \(\delta[n]\) will be or not?

Answer 28

its not unilateral Z-Transform,
n can be negative too

Answer 29

Okay..

Does Long Division Method apply here??

Answer 30

I think, I can help you till here only because my knowledge is very limited here, I am on Convolution now, then Fourier Series, Transform, then Laplace and then Z-transform will come.. :( I want to go in this sequence in which they have their origin, so still a long journey to cover.. :)

Answer 31

good luck!

Answer 32

Thanks.. :)

Answer 33

Till then clear your concept on this, I will be in comfort then.. :P

Answer 34

sure!

Answer 35

and what we can do is to write the Z inverse transform for 1/(1-z^-1) as you did
and then we square it like above ^