Fourier transform of a sinusoid

We can use the phase shift formula above to find the Fourier transform of any complex sinusoid ${Z^n}$ with frequency $\alpha$, simply by setting $X[n]=1$ in the formula and using the Fourier transform for DC:

$${\cal FT} \left \{ {Z^n} \right \} (k) = {\cal FT} \left \{ 1 \right \}(k + {{\alpha } \over {\omega}})$$

$$= \left [ \cos(\Phi(k) + i \sin(\Phi(k))\right ] {D_N}(k + {{\alpha } \over {\omega}})$$

where ${D_N}$ is the Dirichlet kernel and $\Phi$ is an ugly phase term:

$$\Phi(k) = \pi (k + {{\alpha } \over {\omega}}) (N-1)/N)$$

Figure 9.3: Fourier transforms of complex sinusoids, with $N$ = 100; the frequencies are (a) $2\omega$ ; (b) $1.5\omega$.

If the sinusoid's frequency $\alpha$ is an integer multiple of the fundamental frequency $\omega$, the Dirichlet kernel is shifted by an integral value. In this case the zero crossings of the Dirichlet kernel line up with integer values of $k$, so that only one partial is nonzero. This is pictured in Figure 9.3 part (a).

Figure 9.4: A complex sinusoid with frequency $\alpha =1.5\omega =3\pi /N$, forced to repeat every $N$ samples. ($N$ was arbitrarily set to 100; only the real part is shown.)

Part (b) shows the result when the frequency $\alpha$ falls halfway between two integers. The partials have amplitudes falling off roughly as $1/k$ in both directions, measured from the actual frequency $\alpha$. That the energy should be spread over many partials, when after all we started with a single sinusoid, might seem surprising at first. However, as shown in Figure 9.4, the signal repeats at a period $N$ which disagrees with the frequency of the sinusoid. As a result there is a discontinuity at the beginning of each period.