The Fourier Transform of this function

\begin{equation} f(n)=u(n)-u(n-m) \end{equation}

(where $u$ is the unity step)


\begin{equation} F(\omega)=\frac{\sin(\omega m/2)}{\sin(\omega/2)}e^{-i(m-1)/2} \end{equation}

The phase of $F$ is: \begin{equation} \phi=\angle{\frac{\sin(\omega m/2)}{\sin(\omega/2)}}+\angle{e^{-i(m-1)/2}} \end{equation}

My question is how to find the phase of: \begin{equation} \angle{\frac{\sin(\omega m/2)}{\sin(\omega/2)}} \end{equation}


As it is a real number, $0$ or $\pi$ (modulo $2k\pi$, $k\in\mathbb{Z}$), depending on its sign. Since $e^{(0+2k\pi)\imath}=1$, and $e^{(\pi+2k\pi)\imath}=-1$, every positive real number can be written as $r = |r|e^{(0+2k\pi)\imath}$, and every negative real number as $r = |r|e^{(\pi+2k\pi)\imath}$. And there is an indeterminacy for $0$, since any phase would suit.

Finally, it is conventional to choose, for each number, a "simple" phase, for instance that looks constant, or "relatively continuous". Yet to be clean, on each non-zero number the phase is defined (modulo $2k\pi$).

  • $\begingroup$ Thank you, but can you please elaborate how this is calculated? $\endgroup$ – Adam Dec 9 '15 at 17:40
  • $\begingroup$ I hope the details provided in the edited answer suit you fine. $\endgroup$ – Laurent Duval Dec 9 '15 at 18:58
  • $\begingroup$ Thanks you again your explanation is very helpful. $\endgroup$ – Adam Dec 9 '15 at 19:08

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