Z Trandform matlab input

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Christopher Guerrero
Christopher Guerrero am 2 Jun. 2017
Kommentiert: Star Strider am 2 Jun. 2017
I have the following equation:
x(n)= 3(0.25^n)u(n)-(0.75^(n-1))u(n-1)
hot would I input this to solve using matlab?
  1 Kommentar
Christopher Guerrero
Christopher Guerrero am 2 Jun. 2017
More specifically sorry for not putting this I am looking for how to put this in and get the ztransform answer out.

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Star Strider
Star Strider am 2 Jun. 2017
If you have the Symbolic Math Toolbox:
syms n t z
u(n) = heaviside(n);
x(n)= 3*(0.25^n)*u(n)-(0.75^(n-1))*u(n-1);
X = ztrans(x, n, z); % Z-Transform
[Xn,Xd] = numden(X);
Xd = expand(Xd);
H(z) = Xn / Xd; % Transfer Function
H(z) =
-(- 48*z^3 + 40*z^2 + 17*z - 3)/(32*z^3 - 32*z^2 + 6*z)
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Christopher Guerrero
Christopher Guerrero am 2 Jun. 2017
>> syms n t z >> u(n) = heaviside(n)
u(n) =
heaviside(n)
>> x(n)= 3*(0.26^n)*u(n)-(-0.75^(n-1))*u(n-1)
x(n) =
3*(13/50)^n*heaviside(n) + (3/4)^(n - 1)*heaviside(n - 1)
>> x = ztrans(x, n, z)
x(z) =
(1/((4*z)/3 - 1) + 1/2)/z + 3/((50*z)/13 - 1) + 3/2
>> pretty(x)
1
------- + 1/2
4 z
--- - 1
3 3
------------- + -------- + 3/2
z 50 z
---- - 1
13
I used this for just the z transform does this look correct?
Star Strider
Star Strider am 2 Jun. 2017
That looks like a continued fraction expansion of some sort. The approach I took created what appears to be a transfer function with distinct numerator and denominator polynomials.
I’ve not done a hand version of the conversion of a difference equation to a z-transform in decades. I’m comfortable with whatever the Symbolic Math Toolbox calculates!

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