Relationship between DFT and Z transform

We already learn about what is DFT and what is Z transform, So now here this article gives the information about the relationship between DFT and Z transform to know more details about DFT as well as Z transform.

The Z transform of sequence x(n) is,

           ∞

X(Z) = ∑  x(n) . Z ^–n

           n=-∞

We know that at Z = e ^–jω

            ∞

X(Z) = ∑  x(n) . e ^–jωn

           n = -∞

It means that X(Z) is evaluated on the unit circle. 

Now suppose X(Z) is sampled at N equally spaced point on the unit circle, then we have 

ω = 2πK / N

Now if X(Z) is evaluated at Z =  e ^–jωk/n then by putting equation we get; 

            ∞

X(Z) = ∑  x(n) . e^{-j2πKn / N}

           n = -∞

At Z = e^j2πK/N

In the equation, if x(n) is a causal sequence and has N number of the sample then we can write an equation

            N-1

X(K) = ∑   x(n) . e^{-j2πKn / N}

           n = 0

At Z = e^j2πKn/N

^{X(K) = X(Z)}  At Z = e^j2πKn/N

We already learn about what is DFT and what is Z transform, So now here this article gives the information about the relationship between DFT and Z transform to know more details about DFT as well as Z transform.

The Z transform of sequence x(n) is,

           ∞

X(Z) = ∑  x(n) . Z ^–n

           n=-∞

We know that at Z = e ^–jω

            ∞

X(Z) = ∑  x(n) . e ^–jωn

           n = -∞

It means that X(Z) is evaluated on the unit circle. 

Now suppose X(Z) is sampled at N equally spaced point on the unit circle, then we have 

ω = 2πK / N

Now if X(Z) is evaluated at Z =  e ^–jωk/n then by putting equation we get; 

            ∞

X(Z) = ∑  x(n) . e^{-j2πKn / N}

           n = -∞

At Z = e^j2πK/N

In the equation, if x(n) is a causal sequence and has N number of the sample then we can write an equation

            N-1

X(K) = ∑   x(n) . e^{-j2πKn / N}

           n = 0

At Z = e^j2πKn/N

^{X(K) = X(Z)}  At Z = e^j2πKn/N