Generally we all know what a Fibonacci series is. I had 1st seen it in my computer science class where I used the recursive relation to find nth term of Fibonacci series, when I was 1st taught recursive functions. Today my friend told me that he solved the recurrence using power series method to get the golden ratio's and the relation. So I also wanted to write about it and derive the same with z-transform which I learned in my DSP class, one of the few classes I read the books of. So here we are and lets get started.

So lets see the Fibonacci series first...

*1,1,2,3,5,8,13,.....*
If y(n) be the nth term of the Fibonacci sequence then the recurrence relation below clearly satisfies the relation below...

**y(n) = y(n-1) + y(n-2) ** --------------(1)

we also have 2 initial conditions:

**1) y(0) = y(-1) + y(-2) = 1**
**2) y(1) = y(0) + y(-1) = 1**
from these conditions we see that y(-1) = 0 and y(-2)=1 work this out and make sure u get this.

To take the z-transform of the eq. (1) we need to know about one-sided or unilateral z-transform. Its not very different from two-sided z-transform. Its just that in two-sided z-transform needs signal to be specified in the range (-∞,∞). Since input is applied at a finite time sequence say n_{0 }therefore we need signal to be specified in the range [n0,∞).

The **unilateral or one-sided z-transform** may be defined as:

and the time shifting property is also a bit different just a bit modified and can be intuitively proved.Which I am not going to do here.

**TIME DELAY PROPERTY:**

Taking z-transform by applying time delay property, we get..

**Y(z) = [z**^{-1} Y(z) + y(-1)] + [z^{-2}Y(z) + y(-2) + y(-1)z^{-1}]

or

**Y(z) = 1/(1 - z**^{-1} - z^{-2})

or

**Y(z) = z**^{2}/(z^{2 }– z^{1} -1)

or

**Y(z) = z**^{2}/(z - p_{1})(z - p_{2}) where p_{1}=(1 + √5)/2 and p_{2}=(1 – √5)/2

or

**Y(z) = 1/(1 - p**_{1} z^{-1})(1 - p_{2}
z^{-1})

using partial fraction, we get :

**Y(z) = A**_{1}/(1 - p_{1}
z^{-1}) + A_{2}/(1 - p_{2} z^{-1})

where A_{1} = 1/(1 – p_{2}p_{1}^{-1})
= p_{1}/(p_{1} – p_{2}) = p_{1} / √5

and A2 = 1/(1 – p_{1}p_{2}^{-1})
= p_{2}/(p_{2} – p_{1}) = -p_{2} / √5

Now we know that or can be easily verified that:

**UZ(a**^{n} u(n)) = 1/(1 – az^{-1}) where UZ(.) is unilateral z-transform

so we finally get,

**y(n) = [A**_{1}(p_{1})^{n}
+ A_{2}(p_{2})^{n }]u(n)

substituting A1 and A2 we get:

_{ }

** y(n) = (1/√5)[(p**_{1})^{n+1}
- (p_{2})^{n+1 }]u(n) ---------------(2)

** y(n) = (1/√5)[(p**_{1})^{n+1}
- ( **p**_{1_conjugate}_{})^{n+1 }]u(n)
where p

_{1}=(1 + √5)/2 and p

_{2} = p

_{1_conjugate}=(1 – √5)/2

equation (2) above is the formula which works as nth term of Fibonacci series.

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