- #1

- 7

- 0

## Homework Statement

As an example, if we find a DFT of x[n]={1,1,0,1}

the result will be X(m)={3,1,-1,1}

## Homework Equations

My Question is that as we know DFT holds symmetry property, why this answer does not void for that property?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Shaheers
- Start date

- #1

- 7

- 0

As an example, if we find a DFT of x[n]={1,1,0,1}

the result will be X(m)={3,1,-1,1}

My Question is that as we know DFT holds symmetry property, why this answer does not void for that property?

- #2

- 1,444

- 4

Let's take the definition from Wikipedia:

[tex]X_k=\sum_{n=0}^{N-1}x_n\,e^{-i2\pi kn/N}[/tex]

For real data you have then the symmetry:

[tex]X_k=\bar{X}_{N-k},\, (k=1,...,N).[/tex]

In your case [itex]N=4[/itex], and you have [itex]X_0=3,X_1,=1,X_2=-1,X_3=1,[/itex] all real.

From the symmetry property you should have [itex]X_4=X_0,X_3=X_1,X_2=X_2[/itex].

And you have it ([itex]X_4[/itex] can be thought of as defined by the symmetry property).

[tex]X_k=\sum_{n=0}^{N-1}x_n\,e^{-i2\pi kn/N}[/tex]

For real data you have then the symmetry:

[tex]X_k=\bar{X}_{N-k},\, (k=1,...,N).[/tex]

In your case [itex]N=4[/itex], and you have [itex]X_0=3,X_1,=1,X_2=-1,X_3=1,[/itex] all real.

From the symmetry property you should have [itex]X_4=X_0,X_3=X_1,X_2=X_2[/itex].

And you have it ([itex]X_4[/itex] can be thought of as defined by the symmetry property).

Last edited:

Share: