Understanding Analytic Signals

Question

I know that a function $f_a\in \text{L}^2(\mathrm{R})$ is an analytic signal if its Fourier transform is 0 for negative frequencies, i.e.

$$ \hat{f_a}(\omega)=0$ if $\omega<0 $$

We can characterize an analytic function by its real part $f = \text{Re}[f_a]$, and the Fourier transform of this function is

$$ \hat{f_a}(\omega)=\left\{\begin{matrix} 2\hat{f}(\omega) & \text{if }\omega\geq0 \\ 0 & \text{if }\omega<0 \end{matrix}\right. $$

Now if we have a discrete signal $f[n]$ of size $N$, we can also compute $f_a[n]$ by setting the negative frequency components of its discrete Fourier transform to 0.

Now why is it that if we want $\text{Re}[f_a]=f$, then we must have:

$$ \hat{f_a}[k]=\left\{\begin{matrix} \hat{f}[k] && \text{if }k=0,N/2\\ 2\hat{f}[k] && \text{if }0<k<N/2\\ 0 && \text{if } N/2<k<N \end{matrix}\right. $$

Could that be explained?

Answer 1

Analytic signals to get a but more tricky in the discrete domain because of the periodicity in the time domain and the circular nature of the convolution. This article treats that fairly well http://www.andrewduncan.ws/air/

Answer 2

It's not clear what you're asking, but I think you want to know why the DFT of a signal has extra conditions for it to real.

For the continuous case: the 2 is there because we need a function to be symmetric to have a real Fourier transform.

In the discrete case, the DFT obey the symmtery:

$$ X_{N-k} = X_{-k} = X_k^* $$ i.e. conjugate symmetry (half the values are just complex conjugates of the other half). Thus $X_0$ and $X_{N/2}$ are real numbers whilst the remainder of the DFT is specified by $N/2-1$ complex numbers (hence the 2 in the second line).

Also, the answers to this question.

offer counterexamples to your first statement - that analytic functions have vanishing negative frequencies.