Aliasing of discrete Fourier modes ($e^{2\pi i(m+qN)n/N}=e^{2\pi imn/N}$)

For integers $m,q,n$, adding $qN$ to a frequency index changes the complex exponential by $e^{2\pi iqn}=1$. Thus different continuous Fourier modes become identical discrete Fourier modes on the sampling grid. This is aliasing, caused by sampling before any discrete Fourier transform is computed.

For $N=8$, compare the continuous functions $\cos (2\pi mt)$ and $\sin (2\pi mt)$ with $m=1$ and $m=9$, sampled at $t=n/8$, $n=0,\dots ,7$. With $a=\sqrt{2}/2$, both cosines give exactly the first row and both sines give exactly the second:The nine-cycle sine agrees in sign as well as magnitude with the one-cycle sine: the frequency indices differ by $8$, rather than by a sign reversal. Consequently, either cosine produces the same discrete Fourier transform, $X_1=X_7=4$ with all other coefficients zero; either sine produces $X_1=-4i$, $X_7=4i$, with all other coefficients zero.

An observed low frequency therefore does not establish that physical high frequencies are absent. To exclude such alternatives, the Nyquist–Shannon sampling theorem requires a specified class of bandlimited functions. For arbitrary pure sinusoids with frequencies at most $B$, choose sampling rate $f_s>2B$; equality needs care because $\sin (2\pi (f_s/2)t)$ vanishes at every $t=n/f_s$. The finite-energy L2 space version of the Nyquist–Shannon sampling theorem permits equality under its own hypotheses. In either case, exact continuous reconstruction generally requires the entire infinite sample sequence; eight observed samples alone do not supply it. Demanet derives the grid aliasing condition as a difference by an integer multiple of the sampling angular frequency.^[ref]