Take the 2-minute tour ×
Signal Processing Stack Exchange is a question and answer site for practitioners of the art and science of signal, image and video processing. It's 100% free, no registration required.

I created a discrete signal sampled at 44.1kHz that is 330Hz for 1 second, 360Hz for 0.5s, and 390Hz for 0.25s. I did an FFT transformation of this, and got the magnitude graph below:

enter image description here

I don't understand how this magnitude information (combined with the phase information not shown) can reconstitute the original signal. I know it does -- I'm not asking for a proof. I'm hoping someone can supply the intuition for how that strange jagged FFT image ends up reconstructing the original signal.

share|improve this question
2  
That image cannot reconstruct the original signal. You need the phase information, too. Otherwise it could be a tone made up of 330Hz 360Hz and 390Hz all at the same time. –  endolith Dec 10 '12 at 19:23
add comment

2 Answers

up vote 4 down vote accepted

Each frequency burst is equivalent to a rectangular window on an infinite sinusoid. A rectangular window in the time domain is the same as circular convolution with a Sinc function in the frequency domain. Thus, you end up with a spike with side humps (the spike for the sinusoid, and the humps due to the convolution with the transform of the rectangular window) for each frequency.

Then, since the FFT operation is linear, the 3 sine bursts or, equivalently, the 3 spiky hump groups, just sum additively in both the time and the frequency time domain, respectively.

You are correct in that the phase information is important for reconstruction. The FFT phase results includes the information about the location of each (non-overlapping in your case) rectangular window for each sine burst. Twist (or untwist) each spiky hump group's phase in the frequency domain, and you will move the burst's location (the rectangular window) in the time domain.

share|improve this answer
add comment

As hotpaw2 already has stated, the time envelope information is included in the phase. Let's use the following example: A unit impulse has a flat magnitude frequency response and a zero phase. If you delay that unit impulse by, say, 1s, the magnitude response remains flat. However, the phase response has changed dramatically and is a linear with a very steep slope. The amount of delay can often be recovered from the phase response by calculating the group delay. In the case of the delayed impulse has a group delay of 1s.

The same thing applies to your three sine wave blips: the magnitude response mostly is given by the length and window shape of the blip and the delay is determined by the group delay in that frequency area.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.