# Understanding Analytic Signals

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

Could that be explained?

• I thought a function was analytic on the line or plane if it can be approximated by a convergent series. Sep 4 '13 at 8:27
• @Tom Kealy: The OP confuses analytic functions and analytic signals. Sep 4 '13 at 10:26
• @chirlu I mentioned that in my answer below. Sep 4 '13 at 10:54
• Your characterization of an analytic signal in the continuous-frequency domain is incorrect. See Wikipedia's definition.
– Peter K.
Sep 4 '13 at 12:01

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/

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.