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

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

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:

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

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:

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

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

$$\xi = \cos(\pi k / N) - i \sin(\pi k / N)$$

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

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

It's easy now to simplify the numerator:

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

and similarly for the denominator, giving:

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

Whether $V=1$ or not, we have

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

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

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

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

$$\cos(\pi k (N-1)/N) - i \sin(\pi k (N-1)/N)$$

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.

Figure 9.2: The Dirichlet kernel, for $N$ = 100.