Question

Inverse Z- Transform (for rvc)

Answer 1

Find the inverse Z transform of

F(z)=1(z−3)(z−2)

if ROC is
i. |z| <2
ii. 2 < |z| < 3
iii. |z| > 3

Answer 2

thank you so much @inkyvoyd
http://prntscr.com/db9bnv :(

Answer 3

\[ \rm F(z) = \frac{ 1 }{ (z-3)(z-2) }\]

Answer 4

lol its okay @skullpatrol ^-^

Answer 5

who can help me with this???
Please explain me ROC and the steps to solve the probelm

Answer 6

@Callisto @ganeshie8 @hartnn @whpalmer4

Answer 7

@imqwerty com'ere please <3

Answer 8

ooooo or maybe @Whitemonsterbunny17 :D
She's smartt

Answer 9

okay ill return here after 3 hrs ^-^

Answer 10

First using partial fractions to write
\[
\frac{1}{(z-3) (z-2)}=\frac{1}{z-3}-\frac{1}{z-2}
\]

Answer 11

Then
\[
\mathcal{Z}_z^{-1}\left[\frac{1}{z-2}\right](n)=2^{n-1} \theta (n-1)
\]

Answer 12

and
then
\[
\mathcal{Z}_z^{-1}\left[\frac{1}{z-3}\right](n)=3^{n-1} \theta (n-1)
\]
Compine and you are done

Answer 13

\[
\theta(x)=0 \text { for } x <0\\
\theta(x)=1 \text { for } x \ge0\\
\]

Answer 14

First split \(F(z)\) into partial fractions:
\[F(z)=\frac{1}{(z-3)(z-2)}=\frac{a_1(z-3)+a_2(z-2)}{(z-3)(z-2)}=\frac{a_1}{z-2}+\frac{a_2}{z-3}\]For the numerators to match, you need to have \(a_1,a_2\) satisfy
\[(a_1+a_2)z-3a_1-2a_2=1\implies\begin{cases}a_1+a_2=0\\[1ex]-3a_1-2a_2=1\end{cases}\]Solving, you get \(a_1=-1\) and \(a_2=1\).

So,
\[F(z)=\frac{1}{z-3}-\frac{1}{z-2}\]Recall that for \(|x|<1\), you have
\[\sum_{n=0}^\infty x^n=\frac{1}{1-x}\]
You can rewrite \(F(z)\) as
\[\begin{align*}
F(z)&=\frac{1}{z-3}-\frac{1}{z-2}\\[1ex]
&=\frac{1}{z}\frac{1}{1-\frac{3}{z}}-\frac{1}{z}\frac{1}{1-\frac{2}{z}}\\[1ex]
&=\frac{1}{z}\sum_{n=0}^\infty \left(\frac{3}{z}\right)^n-\frac{1}{z}\sum_{n=0}^\infty\left(\frac{2}{z}\right)^n\\[1ex]
&=\sum_{n=0}^\infty\left(3^n-2^n\right)z^{-(n+1)}\\[1ex]
&=\sum_{n=1}^\infty\left(3^{n-1}-2^{n-1}\right)z^{-n}
\end{align*}\]which corresponds to the \(Z\) transform of the sequence
\[\begin{cases}a_0=0\\[1ex]a_n=3^{n-1}-2^{n-1}&\text{for }n\ge1\end{cases}\]but this only holds for the ROC \(|z|>3\) (the intersection of the ROC for both series). I'm tempted to say the inverse transform doesn't exist in the other two cases but I'm not sure...

Answer 15

Any idea if @rvc is supposed to use the bilateral transform?

Answer 16

@eliesaab
hey hi

Answer 17

@ganeshie8
please help

Answer 18

as HolsterEmission completed the partial fraction
after that he took a term common

im not able to understand which to take
confused

Answer 19

I'm not sure what you're confused about, but I'll take a guess at what it might be. (I assume it's the algebra above, but correct me if it's something else.)

The \(Z\) transform for a sequence \(x_n\), if it exists, is given by the series \(X(z)=\displaystyle\sum_{n=0}^\infty \frac{x_n}{z^n}\). The idea is to then rewrite \(X(z)\) as a series with decreasing powers of \(z\).

Consider a simpler example, say \(X(z)=\dfrac{1}{1-z}\). Right away, exploiting the fact about geometric series, we could write this as \(\displaystyle\sum_{n=0}^\infty z^n\), but this a series with increasing powers. To get something more appropriate, we can factor out a power of \(z\) in the denominator of \(X(z)\) and change the sign:
\[\frac{1}{1-z}=\frac{1}{z}\frac{1}{\frac{1}{z}-1}=-\frac{1}{z}\color{red}{\frac{1}{1-\frac{1}{z}}}\]The red term is what we want. This is a geometric series with ratio \(\dfrac{1}{z}\), so we can write this as
\[\frac{1}{1-z}=-\frac{1}{z}\color{red}{\sum_{n=0}^\infty\left(\frac{1}{z}\right)^n}=-\frac{1}{z}\sum_{n=0}^\infty\frac{1}{z^n}\]Next, distribute the factor of \(\dfrac{1}{z}\) and you have
\[\frac{1}{1-z}=-\sum_{n=0}^\infty \frac{1}{z^{n+1}}=-\left(\frac{1}{z}+\frac{1}{z^2}+\cdots\right)\]and shift the starting index of the series so that it's in terms of \(z^n\):
\[\frac{1}{1-z}=-\sum_{n=1}^\infty\frac{1}{z^n}=-\left(\frac{1}{z}+\frac{1}{z^2}+\cdots\right)\]This transform corresponds to a sequence that starts with \(0\) and each successive term is \(-1\), i.e.
\[\begin{cases}a_0=0\\[1ex]a_n=-1&\text{for }n\ge1\end{cases}\]Indeed,
\[X(z)=-\frac{0}{z^0}-\frac{1}{z^1}-\frac{1}{z^2}-\cdots=-\left(\frac{1}{z}+\frac{1}{z^2}+\cdots\right)\]

Answer 20

thanks @HolsterEmission
:)
thanks everyone