Z Transform

ScottB

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Sep 7, 2013
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Alright, this is probably an easy question for most people but I'm studying Z-transform at the moment and I dont fully understand how the solution to the one question was obtained.

The question

Z-transform of: 1/4^k

Solution: 4z/4z-1

Im struggling to find the steps to the solution.. can anyone help me out?
 
According to Wikipedia, the (single-sided) z-transform of a sequence \(\displaystyle x(k)\) is

\(\displaystyle \displaystyle \sum_{k=0}^{\infty} x(k)z^{-k}\)

In your case

\(\displaystyle \displaystyle \sum_{k=0}^{\infty} 4^{-k}z^{-k} = \sum_{k=0}^{\infty}(4z)^{-k}\)

Which is a straight-forward geometric series that converges for \(\displaystyle |1/4z|<1\).
 
Thanks but I understand that you can sum any geometric sequence with a ratio less than one
the formula for it (by taylors series or the binomial expansion) is:

S = a/1-r

now in this case i would assum a = 4 and r = 1 ( the coefficient of k )
but in this case it wouldnt work because u would obtain 0 as ur denominator.

1/4^k is obviously a series of fractions and i think that is what is giving the odd soultion.

I understand the basics but I'm struggling to figure out the proccess to get to the answer of 4z/4z - 1 ..

unless my above formula for S is wrong
 
Thanks but I understand that you can sum any geometric sequence with a ratio less than one
the formula for it (by taylors series or the binomial expansion) is:
S = a/1-r
now in this case i would assum a = 4 and r = 1 ( the coefficient of k )
but in this case it wouldn't work because u would obtain 0 as ur denominator.

In this case for \(\displaystyle k = 0\) we have \(\displaystyle (4z)^{-0}=1\) so \(\displaystyle a = 1~\&~r=\frac{1}{4z}.\)

Thus \(\displaystyle \dfrac{1}{1-\frac{1}{4z}}=\dfrac{4z}{4z-1}~.\)
 
In this case for \(\displaystyle k = 0\) we have \(\displaystyle (4z)^{-0}=1\) so \(\displaystyle a = 1~\&~r=\frac{1}{4z}.\)

Thus \(\displaystyle \dfrac{1}{1-\frac{1}{4z}}=\dfrac{4z}{4z-1}~.\)

Thank you so much, I understand it now
 
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