I am given an amplitude spectrum (no phase) and the envelope of a signal. What I know about the original signal is, that it can be thought of as wavepacket like structure (sinusoids under a bell curve) and the spectrum is narrow banded. To be able to further analyse the original signal I would like to recover the input signal from those two given functions.

My questions are: Can I recover the signal from its amplitude spectrum and envelope or would you advice me to do otherwise.

To clarify what is given:

The signal is thought of wavepacket structure, which is

$$x(t) \approx \sum_n\frac{1}{\sigma_n\sqrt{2\pi}}e^{-\dfrac{1}{2}\left(\dfrac{x-\mu_n}{\sigma_n}\right)^2} \cdot \sin(2\pi f_n x)$$

with $n \gt 1$, $\lvert\mu_k-\mu_l\rvert \lt \sigma_k, \sigma_l$ and if $\mu_k \gt \mu_l$ then $f_k \lt f_l$.

The amplitude spectrum is the absolute of the Fourier transform of the signal, which is

$$a(\omega) = \lvert\mathcal{F}\{x(t)\}\rvert$$

The envelope can be thought of as absolute value of the analytic signal, which is

$$e(t) = \lvert x(t) + j \cdot \mathcal{H}\{x(t)\}\rvert$$

I'm not sure if it is possible or even that beneficial to recover the signal and I am pleased for any answer or every hint.

  • $\begingroup$ Say the original signal is a sinusoid. Any phase shift of it will have the same $a(\omega)$ and $e(t)$, so you can't know which one of them is the original. Do you allow such ambiguity as long as the synthesized signal matches $a(\omega)$ and $e(t)$? $\endgroup$ Jun 5, 2015 at 7:06
  • $\begingroup$ @OlliNiemitalo As I said it is a wavepacket like structure meaning: $x(t) \approx \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2} \cdot sin(2\pi f \cdot x)$ This should be fairly unambigous given the envelope and frequency amplitude. $\endgroup$
    – wagnerpeer
    Jun 5, 2015 at 11:02
  • $\begingroup$ It will give at least an approximation of your three unknowns to find: 1) the highest peak in the spectrum, 2) the highest peak in the envelope, and 3) variance of the envelope. $\endgroup$ Jun 5, 2015 at 12:07
  • $\begingroup$ @OlliNiemitalo Sorry, I realized the given information is incomplete or at last the description of the signal was. Let's make it more complex to get to the point... $\endgroup$
    – wagnerpeer
    Jun 5, 2015 at 18:04
  • $\begingroup$ Your signal model is as general as it gets: The wave atoms span the whole signal space, so your restriction is meaningless. Also the answer is no. The magnitude spectrum Is invariant with respect to time inversion for instance, so any analytic signal with time inversion symmetry of the magnitude will not be uniquely defined by your constraints. Similar constructions can be made with frequency symmetry etc. $\endgroup$
    – Jazzmaniac
    Jun 7, 2015 at 10:09

2 Answers 2


One counterexample is two wave packets of identical spread and amplitude, well separated in time so that their envelopes do not overlap, and of high enough (differing) well-separated frequencies such that their analytic signals can be considered Gaussian-enveloped complex exponentials. The Hilbert envelope will thus consist of two identical well-separated Gaussians. The amplitude spectrum will consist of well-separated Gaussians of identical magnitude. There is not enough information to assign the frequencies to the wave packets.

Two wave packets

So in general you cannot resolve the original signal, not even up to an ambiguity of a constant phase shift over all frequencies.

With sufficiently large time spreads of the two wave packets their frequencies need not be far apart to still be well separated in the spectrum, satisfying your constraint for a narrow-banded signal.

  • $\begingroup$ And again, I have to admit that this would be easy but ambiguous. I have to refine the question. $\endgroup$
    – wagnerpeer
    Jun 7, 2015 at 11:26

You could try matching the theoretical Fourier transform of your equation to your data using an optimization framework such as

$\min_{\mu, \sigma, f}{ ||a_{observed}(w) - |\mathcal{F}(x(t))|||^2_2}$

Use some derivate-free optimization method if you don't want to get your hands dirty


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