# Discrete Fourier transform real signals

The discrete Fourier transform is defined as: $$S(k)=\sum_{n=0}^{N-1}s(n)e^{-j\frac{2\pi}{N}kn}\quad k=0,...,N-1$$ I read that real signals $s(n)$ are: $$S(l)=S(N-l)^*$$ where $S(N-l)^*$ is the conjugated complex number of $S(N-l)$.

I am trying to prove that, but I can't get it right. I can't accept it without a proof.^^

-
Your formula for $S(n)$ is incorrect since $n$ is an argument on the left side and a index of summation on the right. Please correct your formula before attempting any proofs. – Dilip Sarwate May 8 '12 at 1:17

## 1 Answer

Put $e^{2\pi j/N}=:\omega$ and note the rules $\omega^*=\omega^{-1}$, $\omega^N=1$. It follows that the formula for the discrete Fourier transform can be written as $$S(k):=\sum_{n=0}^{N-1} s(n)\omega^{-kn}=\sum_{n=0}^{N-1} s(n)(\omega^*)^{kn}\ .$$ Now take on both sides the conjugate. If the $s(n)$ are real you get $$S(k)^*=\sum_{n=0}^{N-1} s(n)\omega^{kn}=\sum_{n=0}^{N-1} s(n)\omega^{-(N-k)n}==S(N-k)\ .$$

-