DLS
  • DLS
Find the inverse Z transform of F(z).
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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DLS
  • DLS
\[\Large F(z) = \frac{2z^2 - z}{2z^2-2z+2}\]
DLS
  • DLS
@ganeshie8 @dan815
DLS
  • DLS
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.

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DLS
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@Jemurray3
DLS
  • DLS
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 :/
anonymous
  • anonymous
Why can't you get the poles from this?
DLS
  • DLS
Wolfram gives \[\Huge (-1)^\frac{1}{3}\]
DLS
  • DLS
meanwhile..@ganeshie8 can you help with this ? :| \[\Large \frac{3z^2+5}{z^4}\] The pole = 0 of 4th order..but everything turns 0.
anonymous
  • anonymous
Are you using contour integrals from this, or are you trying to do it some other way?
anonymous
  • anonymous
for this *
anonymous
  • anonymous
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.
DLS
  • DLS
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 ?
DLS
  • DLS
@Jemurray3
anonymous
  • anonymous
Are you using the contour integral approach?
DLS
  • DLS
no no..:/ inversion of z transform using poles
anonymous
  • anonymous
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 \[f(n)= \cos(n\pi/3)\]

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