Now when we talk explicitly about the periodicity of the Fourier transform, we prefer to use the term Discrete Fourier Series, or DFS for short. So the DFS is just a DFT with the periodicity explicit. The DFS maps an N-periodic signal onto an N-periodic sequence of Fourier coefficients. And the inverse DFS maps an N-periodic sequence of Fourier coefficients onto an N-periodic signal. But, mathematically, the two are the same, it's just a conceptual difference. However, DFS helps understand, for instance, why we said that the natural shift for finite-length sequences is actually a periodic shift. Let's consider an N-periodic sequence x tilde of [n]. For periodic sequences, shifts are well-defined, we can take x tilde of [n- big M] and it's just a normal shift of an infinite two-sided sequence. If we take the DFS of this periodic sequence with a shift, we can work out very easily that the DFD coefficient for index k is equal to the DFD coefficient for index k of the original sequence without the shift. So x[k] is simply the DFS[ tilde x[n] ]. With a phase shift term, e to the minus j, 2 pi over N, Big M times k. Similarly, it's very easy to verify that the inverse Discrete Fourier Series of a vector of DFS coefficients, multiplied by a phase delay factor, is simply the periodic sequence delayed by M. So this term, e to the minus j, 2 pi over N, Mk is a delay factor in the frequency domain. So what happens for finite length signals? We saw that in this case a shift is not a well defined operation, and we need to embed the signal into an infinite length signal either by building a finite support or a periodic signal. Now we can always build x tilde of [n] as x[n modulus big N]. Now mathematically, the DFS of this periodic signal is equal to the DFT of the original signal. So the inverse DFT of the DFT of the signal, times this delay factor that we showed in the previous slide, is numerically equal to the inverse DFS of the same quantity. But we know that the inverse DFS of this will be a delayed periodic sequence. And if we go back to the definition of how we build this periodic sequence, this will be simply x[ (n- big M) modulus big N ]. What we have shown here is that if we think of the four-year representation of a finite-length signal, circular shifts are the natural extension of the shift operator to finite-length signals. Because the underlying four-year representation is the same as that we would use for a periodic signal built on the finite-length signal.