\begin{claim}

If $x(n)$ is a real-valued signal of an odd function, then its fourier transform is odd and imaginary. In other words, 

\begin{nicebox}
$$x \in \left[ \Z \to \R \right] \mbox{ AND } x(n) = -x(-n) \mbox{ } \forall n$$
$$\Rightarrow$$
$$-X(-\omega) = X(\omega) \mbox{ AND } X \in \left[\R \to i\cdot\R \right]$$
\end{nicebox}

\end{claim}


To show that the frequency response is odd, we need to show $X(\omega) = -X(-\omega)$. To show that the frequency response is purely imaginary, we need to show that the frequency response is equal to the negative of the conjugate of the frequency response, namely $X(\omega) = -X^*(\omega)$.


\begin{proof}

First show that the frequency response is odd by noting some properties of the DTFT:
\begin{align*}
x(n) &\ftp X(\omega) \\
x(-n) &\ftp X(-\omega) \\
-x(-n) &\ftp -X(-\omega) \\
\end{align*}

Putting it all together,

\begin{align*}
x(n) &= -x(-n) \\
\ftrans{x(n)} &= \ftrans{-x(-n)} \\
X(\omega) &= -X(-\omega)
\end{align*}

Now we can use conjugate symmetry to show that the frequency response is imaginary. This means that if $x$ is real-valued, then $X(\omega) = X^*(\omega)$. 

\begin{align*}
X(\omega) &= -X(-\omega) \quad \mbox{since $X$ is odd}\\
X(\omega) &= X^*(-\omega) \quad \mbox{since $x$ is real}\\
X^*(-\omega) &= -X(-\omega) \\
X^*(\omega) &= -X(\omega) \\
\end{align*}

\end{proof}

Here is a more rigorous proof:

\begin{proof}

To show that the frequency response is odd:

\begin{align*}
X(\omega) &= \sum \limits_{n=-\infty}^{\infty}x(n)e^{-i\omega n} \\
X(\omega) &= \sum \limits_{m=-\infty}^{\infty}x(-m)e^{i\omega m} \quad \mbox{let }m=-n \\
X(\omega) &= -\sum \limits_{m=-\infty}^{\infty}x(m)e^{i\omega m} \quad \mbox{since $x$ is odd} \\
X(\omega) &= -X(-\omega) \\
\end{align*}

To show that the frequency response is purely imaginary:

\begin{align*}
X(\omega) &= \sum \limits_{n=-\infty}^{\infty}x(n)e^{-i\omega n} \\
X(\omega) &= \sum \limits_{m=-\infty}^{\infty}x(-m)e^{i\omega m} \quad \mbox{let }m=-n \\
X^*(\omega) &= \sum \limits_{m=-\infty}^{\infty}x^*(-m)e^{-i\omega m} \\
X^*(\omega) &= \sum \limits_{m=-\infty}^{\infty}x(-m)e^{-i\omega m} \quad \mbox{since $x$ is real-valued}\\
X^*(\omega) &= -\sum \limits_{m=-\infty}^{\infty}x(m)e^{-i\omega m} \quad \mbox{since $x$ is odd}\\
X^*(\omega) &= -X(\omega)\\
\end{align*}

Since the frequency response is equal to the negative of its conjugate, it is purely imaginary. 

\end{proof}


Now we make make the general claim that the fourier transform of a real-valued odd signal is equal to the product of $i$ and the imaginary part of the signal's frequency response:

\begin{align*}
\ftrans{\frac{x(n) - x(-n)}{2}} &= i\Imag{\frac{X(\omega)-X(-\omega)}{2}} \\
\ftrans{x_o(n)} &= i\Imag{\frac{1}{2}X(\omega)-\frac{1}{2}X(-\omega)} \\
\ftrans{x_o(n)} &= i\Imag{\frac{1}{2}X(\omega)+\frac{1}{2}X(\omega)} \quad \mbox{since } X(\omega) = -X(-\omega) \\
\ftrans{x_o(n)} &= i\Imag{X(\omega)} \\
\end{align*}


\begin{nicebox}
$$x \in \left[ \Z \to \R \right]$$
$$\Rightarrow$$
$$x_o(n) \ftp i\Imag{X(\omega)} $$
\end{nicebox}


Signals of Odd Functions