# Duality in the discrete-time Fourier series

Suppose $$g[n]$$ is periodic with fundamental period $$N$$ and $$f[k]$$ being its Fourier coefficients i.e. $$f[k] = \frac{1}{N}\sum_{n=}g[n]e^{-j\frac{2\pi}{N}nk}$$ In more convenient notation $$g[n]\leftrightarrow f[k]$$ It's easy to prove $$f[n]\leftrightarrow \frac{1}{N}g[-k]$$ This is called "Duality in the discrete-time Fourier series". My problem is applying duality in the case of $$x[n-n_0] \leftrightarrow a_ke^{-jk(\frac{2\pi}{N})n_0}$$ And $$\sum_{r=}x[r]y[n-r]\leftrightarrow Na_kb_k$$ I think for the first case duality should gives$$x[n]e^{jk(\frac{2\pi}{N})m} \leftrightarrow \frac{1}{N}a_{m-k} \tag{1}\label{*}$$ and the second gives $$x[n]y[n] \leftrightarrow \frac{1}{N^2} \sum_{l = }a_lb_{-k-l} \tag{2}\label{**}$$ Edit: Let $$g[n] = \sum_{r=}x[r]y[n-r]$$ and $$f[k] = Na_kb_k$$. So according to the mentioned property $$Nx[n]y[n] \leftrightarrow \frac{1}{N}\sum_{r=}a_rb_{-k-r}$$

Dividing by $$N$$ leads to $$\eqref{**}$$. Similarly if $$g[n] = x[n-n_0]$$ and $$f[k] = a_ke^{-jk(\frac{2\pi}{N})n_0}$$ I think then $$x[n]e^{-jn(\frac{2\pi}{N})n_0}\leftrightarrow a_{-k - n_0}$$ Let $$m = -n_0$$ and we have \eqref{*}.

Obviously these results are wrong but I don't know what's my mistake.

This is taken from Signals and systems by Alan V. Oppenheim:

• But we don't know how you came up with these results, so we can't know what your mistake is. – Matt L. Jun 22 '20 at 9:03
• @MattL. You're right, sorry for that. I will add the details. – S.H.W Jun 22 '20 at 9:05
• @MattL. Please see the edit. – S.H.W Jun 22 '20 at 9:32

Based on the references and the definitions given, here is a proof.

Given: $$x[n] \leftrightarrow a[k], \ and$$ $$y[n] \leftrightarrow b[k]$$

Fourier Series expansion of $$x[n]e^{j\frac{2\pi}{N}nm}$$ would be derived as follows: $$\frac{1}{N}\sum^{N-1}_{n=0}x[n]e^{j\frac{2\pi}{N}nm}e^{-j\frac{2\pi}{N}nk} = \frac{1}{N}\sum^{N-1}_{n=0}x[n]e^{-j\frac{2\pi}{N}(k-m)n} = a[k-m]$$ The factor $$\frac{1}{N}$$ gets consumed in the definition of Fourier Series coefficients. Therefore, the following relationship can be stated: $$x[n]e^{j\frac{2\pi}{N}nm} \leftrightarrow a[k-m]$$

Similarly, for the second case of multiplication in time-domain, you need to figure out that the inverse Fourier Series relation or the synthesis expression will be the following: $$x[n] = \sum^{N-1}_{k=0}a[k]e^{j\frac{2\pi}{N}kn}$$ I am not adding the proof of the above which can be easily derived by plugging the expression for $$a[k] = \frac{1}{N}\sum^{N-1}_{n=0}x[n]e^{-j\frac{2\pi}{N}kn}$$ on the RHS.

Then all we are required to do is proving that the inverse Fourier Transform of convolution of $$a[k]$$ and $$b[k]$$ would be product $$x[n]y[n]$$. Which is as follows: $$\mathcal{FS}^{-1}(a*b)[k] = \sum^{N-1}_{k=0}(a*b)[k]e^{j\frac{2\pi}{N}nk} = \sum^{N-1}_{k=0}( \sum^{N-1}_{l=0}a[l]b[k-l]) e^{j\frac{2\pi}{N}nk}$$ $$=>\mathcal{FS}^{-1}(a*b)[k] = (\sum^{N-1}_{l=0}a[l]e^{j\frac{2\pi}{N}ln})(\sum^{N-1}_{k=0}b[k-l]e^{j\frac{2\pi}{N}(k-l)n})$$ $$=> \mathcal{FS}^{-1}(a*b)[k] = \mathcal{FS}^{-1}a[k] \mathcal{FS}^{-1}b[k] = x[n]y[n]$$ I have replaced $$m = (k-l)$$ in the proof above to get $$y[n]$$.

Hence both the equations (5.69) and (5.71) are true and correctly given in the book.

I think you are trying to use equation (5.67) to prove equation (5.69) and (5.71). That is not correct, because (5.67) is stating Duality of Fourier Series expression between Continuous Time and Discrete Time. Equation (5.69) and equation (5.71) are duality of Shift-Property and Convolution Property. The second reason you cannot use Duality Expression (5.67) is because (5.67) states that if $$\mathcal{FS}g[n] = f[k]$$, then $$\mathcal{FS}f[n] = \frac{1}{N}g[−k]$$. That is it takes Fourier Series of the Fourier Series itself assuming $$f[n]$$ in time-domain. But in (5.69) and (5.71), the time-domain sequences remain $$x[n]$$ and $$y[n]$$ only, so (5.67) does not apply here. There are 3 Duality Properties stated in equation (5.67), (5.69) and (5.71). They are not the consequence of one another. They are a consequence of similarity in definition of Continuous time Fourier Series and Discrete Time Fourier Series.

• @S.H.W Why do you think that the DFT of $\sum x[m]y[n-m]$ is $Na_k b_k$? Where are you getting the factor $N$ from? – DSP Rookie Jun 22 '20 at 15:25
• @S.H.W DFS is just periodization of DFT. So, can you add a little bit of of detail on the factor $N$ in the convolution property? I am not sure that the DFS will have that scaling of $N$. It should just be $a_k b_k$. – DSP Rookie Jun 22 '20 at 15:34