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

^{–j}ω
∞

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

^{–j}ω
∞

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}
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}