DFT Symmetry Property: Why Does the Answer Not Void This Property?

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The discussion addresses the symmetry property of the Discrete Fourier Transform (DFT) using the example of x[n]={1,1,0,1}, resulting in X(m)={3,1,-1,1}. It questions why the computed DFT does not seem to violate the symmetry property, which states that for real data, X_k should equal the complex conjugate of X_{N-k}. With N=4, the values X_0=3, X_1=1, X_2=-1, and X_3=1 are confirmed to satisfy the symmetry condition. The symmetry property holds true as X_4 can be interpreted through this relationship, reinforcing that the DFT maintains its expected characteristics. Thus, the DFT result aligns with the symmetry property for real-valued sequences.
Shaheers
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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?
 
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Let's take the definition from Wikipedia:
X_k=\sum_{n=0}^{N-1}x_n\,e^{-i2\pi kn/N}
For real data you have then the symmetry:
X_k=\bar{X}_{N-k},\, (k=1,...,N).
In your case N=4, and you have X_0=3,X_1,=1,X_2=-1,X_3=1, all real.

From the symmetry property you should have X_4=X_0,X_3=X_1,X_2=X_2.
And you have it (X_4 can be thought of as defined by the symmetry property).
 
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