Question

Find the inverse Z transform of F(z).

Answer 1

\[\Large F(z) = \frac{2z^2 - z}{2z^2-2z+2}\]

Answer 2

@ganeshie8 @dan815

Answer 3

I tried to get partial fractions of \(\Large \frac{F(z)}{z}\) but didn't get anywhere..and can't get poles out of this either.

Answer 4

@Jemurray3

Answer 5

After long division I have :
\[\Large 1 + \frac{1}{2}z^{-1} - \frac{1}{2}z^{-2}\]
so comparing it with standard z transform
\[\Large f(0) + f(1) z^{-1} + f(2) z^{-2} + f(3)z^{-3}....\]
f(0) = 1
f(1) = 1/2
f(2) = -1/2
I can't bruteforce the function from this :/

Answer 6

Why can't you get the poles from this?

Answer 7

Wolfram gives \[\Huge (-1)^\frac{1}{3}\]

Answer 8

meanwhile..@ganeshie8 can you help with this ? :|
\[\Large \frac{3z^2+5}{z^4}\]
The pole = 0 of 4th order..but everything turns 0.

Answer 9

Are you using contour integrals from this, or are you trying to do it some other way?

Answer 10

for this *

Answer 11

Also, if you consider
\[z^2 - z + 1\]
then the quadratic formula gives the roots as
\[\frac{1}{2} \pm i\frac{\sqrt{3}}{2} = e^{\pm i\pi/3}\]
so it can be factored:

\[z^2-z+1 = (z-e^{i\pi/3})(z-e^{-i\pi/3})\]

So those are the poles.

Answer 12

so I put them one by one right..
\[\Large \frac{2z^2-2}{z-e^\frac{-i \pi}{3}}\]
Now putting z = e^ipi/3
and similarly for the other root.
I'm not quite sure how answer is cos(npi/3)..
and what about the 2nd one ?

Answer 13

@Jemurray3

Answer 14

Are you using the contour integral approach?

Answer 15

no no..:/ inversion of z transform using poles

Answer 16

Okay, so dividing by z and expanding in partial fractions:

\[ \frac{2z-1}{2(z-e^{i\pi/3})(z-e^{-i\pi/3})} =\frac{1}{2}\left( \frac{A}{z-e^{i\pi/3}} + \frac{B}{z-e^{-i\pi/3}}\right) \]
so
\[2z-1 = A(z-e^{-i\pi/3}) + B(z-e^{i\pi/3} )\]

and so when \(z=e^{i\pi/3}\),

\[ 2e^{i\pi/3}-1 = 2A\sin(\pi/3)\]
and when \(z=e^{-i\pi/3}\)
\[2e^{-i\pi/3}-1 = -2B\sin(\pi/3)\]

so

\[\frac{F(z)}{z} = \frac{1}{4\sin(\pi/3)} \left( \frac{2e^{i\pi/3}-1}{z-e^{i\pi/3}}-\frac{2e^{-i\pi/3}-1}{z-e^{-i\pi/3}}\right)\]

so the inverse transform becomes
\[f(n) = \frac{1}{4\sin(\pi/3)}\left((2e^{i\pi/3}-1)e^{in\pi/3}-(2e^{-i\pi/3}-1)e^{-in\pi/3}\right)\]

\[f(n) = \frac{1}{4\sin(\pi/3)}\left(4\sin([n+1]\pi/3) - 2\sin(n\pi/3)\right)\]
\[f(n) = \frac{1}{4\sin(\pi/3)}\left(4\sin(n\pi/3)\frac{1}{2} + 4\cos(n\pi/3)\sin(\pi/3) - 2\sin(n\pi/3)\right) \]
and so
\[f(n)= \cos(n\pi/3)\]