So, this command must be equivalent to Eq. I can try to expand Eq. It is now in a form that I can recognise Matlab's fft expression in there but this is where I am stuck - how can I show that this last equation is indeed calculated by the command. I have a feeling that there might be a way to use the shift theorem to show that if you circularly shift the vectors through ifftshift and fftshift then you can cancel the extra phase factors, and the fft can be applied.
As a quick sanity check on Eq. Continuing with the OP's approach we will show how only in the case that N is an even number of samples, the following expression that was derived does match the operations noted with ifftshift and fftshift. We will also show the equivalent expression for the case that N is odd.
At the end we provide additional intuitive insight into what is occurring, which does not have anything to do with zero-phase. Below are further details going back to the more general question of the OP:. For a Discrete Fourier Transform DFT to match samples of the Continuous-Time Fourier Transform CTFT , the signal unless sampled and therefore periodic in frequency must have no spectral occupancy beyond the sampling rate, or will otherwise deviate due to the effect of the aliasing from those higher frequencies.
Further the CTFT is a continuous time integration involving the function x t , and any summation expressions on samples of x t are numerical approximations such as Forward Euler, Backward Euler, Tustin, etc whose results improve as the sampling rate is increased but to my understanding none can provide an exact match, so similarly an exact match to the underlying CTFT cannot be derived from the DFT, although can be closely approximated for sufficiently large sampling and no aliasing this last sentence must be confirmed.
This is not "assumed" but clearly defined. It is directly clear that the above two operations are exactly what is performed by ifftshift and fftshift respectively.
This is a matter of personal selection no different than deciding what range is desired for the x-axis on a plot since the DFT repeats over any range selected. For example an even sequence: [ 9 2 5 3 1 12] would map to [1. The time index must be defined if used any differently from the DFT definition. Therefore it is useful to summarize for reference the Fourier Transform time and frequency shift properties. Answer : ifftshift is required to make the sampled Gaussian symmetric modulo N in discrete time-domain.
That is why use of ifftshift makes phase response zero-phase. And, with this DFT definition you do not have to use ifftshift. Both the time and frequency domain signals are discrete.
This implies they are periodic in the other domain: when you discretize a signal in one domain you force it to be periodic in the other. The periodicity also means that there is really no start or end point. You can actually start summing at any point in time and you will get the same result.
At this point it becomes more a less a question of convention. For most time domain signals, the first one makes more sense and it's also the cleanest mathematically so that's one of choice at the moment. Note that the phase zero reference point of an FT is where the cosine or real component of all the exponential basis vectors is 1. So that does not meet the criteria for being the phase zero reference of an FT.
This works because all the DFT basis vectors are circular, thus any rotation of the input data just results in a shift to needed phase reference. Sign up to join this community. The best answers are voted up and rise to the top. Stack Overflow for Teams — Collaborate and share knowledge with a private group. Create a free Team What is Teams? Learn more. Why do we have to rearrange a vector and shift the zero point to the first index, in preparation for an FFT?
Ask Question. I will call this 'swapped' from now on. That is, time zero or frequency zero are at the centre of such an array. Thus, if you start with a naturally ordered spectrum say, a desired filter frequency response centred at DC then you must make it swapped before you use the ifft.
The function fftshift 'unswaps' an array, and the inverse function ifftshift swaps one. The fft also requires a similar idiom. The reason things are like this can be explained in two ways, I think.
Second, that the fft implements a particular DFT equation that is a sum up from zero no negative time or frequency and so the inputs must be shifted from a DC-centred array. Now to the real question. What is the effect if people do not do it right? If we look only at magnitude of the result, then the result is unchanged by any circular shift.
And since probably most people do look only at the magnitude of the result of an FFT, perhaps this is why the incorrect idiom is most common? But if we look at complex numbers, or phase, then there is a clear difference and the most common idiom is wrong. The results were incorrect at least, differd from my expectation until I adopted the correct idiom I have outlined above. I'll ask them.. You might want to be aware that matlab, in its early days, had a number of amateurishly implemented signal processing functions.
These things might have improved with the redesign of the signal processing toolbox of recent years, but don't haev too high hopes. Because few people are willing or able to make the effort to learn the basics.
Read the current newspapers, about the Gulf oils spill, and project to whatever subject you might be interested in. Rune Reply Start a New Thread. Just post one copy but include all group names, separated by commas, on the newsgroup line. For eg:. Given the title of the book you are studying, I assume you are working with images. Then the proper way is to call both fftshift and ifftshift before and after you call [i]fft. In the book "Computational Fourier Optics, A Matlab Tutorial" by David Voelz, it is written that a call to fftshift is needed before a call to fft or ifft , but in the MATLAB documentation of fftshift it's only written that this command rearranges the outputs of fft , fft2 , and fftn by moving the zero-frequency component to the center of the array.
Asked By: user Answered By: Tikoloche. Answered By: Adriaan. Answered By: Jun Ouyang. Answered By: NSredar.
0コメント