Question

Need help in finding inverse Z transform

Answer 1

\[\Large Z^{-1} [ \frac{1}{(z-\frac{1}{2})(z-\frac{1}{3})}]\]
Case 1 : \[\Large \frac{1}{3}<|z| < \frac{1}{2}\]
Case 2:
\[\Large \frac{1}{2}<|z|\]

Answer 2

I am at this step after going through partial fractions and stuff :
\[\Large Z^{-1}[6 (\frac{1}{z-\frac{1}{2}} - \frac{1}{z-\frac{1}{3}})]\]
I don't know how to proceed according to the given cases here.

Answer 3

@DLS

Answer 4

lol :|

Answer 5

ofcourse except me, if there is anyone.

Answer 6

@ikram002p is offline
@ganeshie8 is offline
@Astrophysics is offilne

Aaj ka din hi kharaab hai.

Answer 7

yeah I noticed. :| its very rare for all 3 of them to be simultaneously offline :/

Answer 8

Is z a matrix or a complex number

Answer 9

complex number

Answer 10

Hint :

\[\dfrac{1
}{(z-1/2) (z-1/3) z} = \dfrac{24}{2 z-1}-\dfrac{54}{3 z-1}+\dfrac{6}{z}\\~\\~\\\implies \dfrac{1
}{(z-1/2) (z-1/3)} = \dfrac{24z}{2 z-1}-\dfrac{54z}{3 z-1}+6\]

Answer 11

Can't relate :/ can't we proceed with the step I left it at ?

Answer 12

you may proceed if you know that inverse transform for \(\dfrac{1}{z-b}\)

Answer 13

what do you not understand in rsadhvika's reply ?

Answer 14

do you mean z/z-b?

Answer 15

the step where you have left have terms in form \(\dfrac{1}{z-b}\)
not \(\dfrac{z}{z-b}\)

Answer 16

thats the reason rsadhvika has massaged the partial fractions a bit to get the terms into familar form : \(\dfrac{z}{z-b}\)

Answer 17

ohh..okay..I get it why the form is changed.

Answer 18

now how do we proceed differently for both the cases ?

Answer 19

\[\dfrac{1
}{(z-1/2) (z-1/3) z} = \dfrac{24}{2 z-1}-\dfrac{54}{3 z-1}+\dfrac{6}{z}\\~\\~\\\implies \dfrac{1
}{(z-1/2) (z-1/3)} = \dfrac{24z}{2 z-1}-\dfrac{54z}{3 z-1}+6\\~\\~\\
=12\dfrac{z}{z-1/2} -18\dfrac{z}{z-1/3} + 6
\]

Answer 20

look up your inverse tranform table

Answer 21

yep, I'm at this point.

Answer 22

12 (1/2)^k-18(1/3)^k + ...

Answer 23

scared of taking inverse transform of 1 ?

Answer 24

can't recall :P

Answer 25

http://mathfaculty.fullerton.edu/MATHEWS/C2003/ztransform/ZTransformTable/ZTransformTable.9.1.gif

Answer 26

first row

Answer 27

yeah..its delta N..its the first time im using it though..quite surprising :/

Answer 28

have you used unit step function before ?

Answer 29

yes :O i am able to recall now

Answer 30

then we may get the answer in terms of unit step functions

Answer 31

but how will the ans vary in both the cases ?

Answer 32

look at 4th row in previous link

Answer 33

yes, what next?

Answer 34

I am at this step after going through partial fractions and stuff :
\[\Large Z^{-1}[6 (\frac{1}{z-\frac{1}{2}} - \frac{1}{z-\frac{1}{3}})]\]
I don't know how to proceed according to the given cases here.

Answer 35

use that

Answer 36

\[ Z^{-1}[6 (\frac{1}{z-\frac{1}{2}} - \frac{1}{z-\frac{1}{3}})]\\~\\

=6\left[(1/2)^{k-1}u(k-1) -(1/3)^{k-1}u(k-1) \right]
\]

Answer 37

for \(k\ge 1\) that simplifies to
\[6\left[(1/2)^{k-1}-(1/3)^{k-1}\right]\]

Answer 38

I'd rather proceed with rsadhvika's simplification :| this formula isn't in my textbook.

Answer 39

Okay, if you see both answers are equivalent

Answer 40

Yep..both answers would definitely be equivalent.
but can't we directly do the z transform after rsadhvika's step ? I am just confused with the case 1 and 2 things, I haven't studied complex no.s in much detail/ or have forgotten it :/