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Fourier transform of DC

Let $X[n]=1$ for all $n$ (this repeats with any desired integer period $N>1$). From the preceding discussion, we expect to find that

\begin{displaymath}
{\cal FT} \left \{ X[n] \right \} (k) =
\left \{
\begin{...
...}
N & {k=0} \\
0 & {k=1, \ldots, N-1}
\end{array} \right .
\end{displaymath}

We will often need to know the answer for non-integer values of $k$ however, and for this there is nothing better to do than to calculate the value directly:

\begin{displaymath}
{\cal FT}\left \{ X[n] \right \} (k) =
{V ^ {0}} X[0] +
{V ^ {1}} X[1] +
\cdots +
{V ^ {N-1}} X[N-1]
\end{displaymath}

where $V$ is, as before, the unit magnitude complex number with argument $-k\omega$. This is a geometric series; as long as $V \not= 1$ we get:

\begin{displaymath}
{\cal FT} \left \{ X[n] \right \} (k) =
{{
{V^N} - 1
} \over {
V - 1
}}
\end{displaymath}

We now symmetrize the top and bottom in the same way as we earlier did in Section 7.3. To do this let:

\begin{displaymath}
\xi = \cos(\pi k / N) - i \sin(\pi k / N)
\end{displaymath}

so that ${\xi^2} = V$. Then factoring appropriate powers of $\xi$ out of the numerator and denominator gives:

\begin{displaymath}
{\cal FT} \left \{ X[n] \right \} (k) =
{\xi^{N-1}}
{{
{\xi^N} - {\xi^{-N}}
} \over {
\xi - {\xi^{-1}}
}}
\end{displaymath}

It's easy now to simplify the numerator:

\begin{displaymath}
{\xi^N} - {\xi^{-N}} =
\left (\cos(\pi k) - i \sin(\pi k) ...
...eft (\cos(\pi k) + i \sin(\pi k) \right )
= - 2 i \sin(\pi k)
\end{displaymath}

and similarly for the denominator, giving:

\begin{displaymath}
{\cal FT} \left \{ X[n] \right \} (k) =
\left ( {
\parbo...
...
} \right )
{{
\sin(\pi k)
} \over {
\sin(\pi k / N)
}}
\end{displaymath}

Whether $V=1$ or not, we have

\begin{displaymath}
{\cal FT} \left \{ X[n] \right \} (k) =
\left ( {
\parbo...
...(\pi k (N-1)/N) - i \sin(\pi k (N-1)/N)
} \right )
{D_N}(k)
\end{displaymath}

where ${D_N}(k)$, known as the Dirichlet kernel, is defined as

\begin{displaymath}
{D_N}(k) =
\left \{
\begin{array}{ll}
N & {k= 0} \\
{...
...pi k / N)
}}
& {k\not=0,\; -N < k < N}
\end{array} \right .
\end{displaymath}

Figure 9.1 shows the Fourier transform of $X[n]=1$, with $N=100$. The transform repeats every 100 samples, with a peak at $k=0$, another at $k=100$, and so on. The figure endeavors to show both the magnitude and phase behavior using a 3-dimensional graph projected onto the page. The phase term

\begin{displaymath}
\cos(\pi k (N-1)/N) - i \sin(\pi k (N-1)/N)
\end{displaymath}

acts to twist the values of ${\cal FT} \left \{ X[n] \right \} (k)$ around the $k$ axis with a period of approximately two. The Dirichlet kernel ${D_N}(k)$, shown in Figure 9.2, controls the magnitude of ${\cal FT} \left \{ X[n] \right \} (k)$. It has a peak, two units wide, around $k=0$. This is surrounded by one-unit-wide sidelobes, alternating in sign and gradually decreasing in magnitude as $k$ increases or decreases away from zero. The phase term rotates by almost $\pi $ radians each time the Dirichlet kernel changes sign, so that the product of the two stays roughly in the same complex half-plane for $k>1$ (and in the opposite half-plane for $k < -1$). The phase rotates by almost $2\pi $ radians over the peak from $k=-1$ to $k=1$.

Figure 9.1: The Fourier transform of a signal consisting of all ones. Here N=100, and values are shown for $k$ ranging from -5 to 10. The result is complex-valued and shown as a projection, with the real axis pointing up the page and the imaginary axis pointing away from it.
\begin{figure}\psfig{file=figs/fig09.01.ps}\end{figure}

Figure 9.2: The Dirichlet kernel, for $N$ = 100.
\begin{figure}\psfig{file=figs/fig09.02.ps}\end{figure}


next up previous contents index
Next: Shifts and phase changes Up: Properties of Fourier transforms Previous: Properties of Fourier transforms   Contents   Index
Miller Puckette 2006-12-30