According to Wikipedia, for any function $x(t)$ with the fourier transform of $\hat x(t) = X(f)$, we can generate a discrete sample by "sampling" $x(t)$ at $T$-separated values to get a sequence $x[n] = Tx(nT), n\in \mathbb{Z}$ (it further says that the result holds as long as $x[n]$ is proportional to $x(nT)$). For the Discrete-time fourier transform of this sequence, we have $$X_{1/T}(f) = X_{2\pi}(2\pi f T) \triangleq \sum\limits_{n=- \infty}^{\infty}Tx(nT)e^{-i2\pi fTn}=\sum\limits_{k=-\infty}^{\infty}X(f-k/T)$$ where the penultimate term follows from replacing $x[n]$ in the definition of DTFT and the ultimate term follows from the application of the Poisson Summation Formula. I can intuitively appreciate this to mean that the DTFT represents a "periodic summation" of the continuous Fourier transform. But in regards to this, I have a couple of doubts:

  1. While the RHS of the equation depends on $x(t)$, the LHS depends on where exactly we sample, i.e. we would get a different sequence if we translate our sampling points to the right or left (or translate $x(t)$). How would then the equality hold for all sampling cases? Since the plot of the fourier "coefficient" magnitudes would remain unchanged as long as the frequency decomposition of $x(t)$ remained unchanged, how would then the plot of DTFT coefficient magnitudes always represent the periodic summation of the former?

  2. What about the zero case, that is, our sampling frequency leads us to get only zeros in our series despite $x(t)$ not being zero everywhere? I understand that the DTFT must then be zero, and this case is a trivial case not applicable in practical scenarios, but in the derivation of the above relation, did we use the fact that the $x[n]$ are non zero somewhere (I am not sure of the assumptions behind the derivation of this relation)?


1 Answer 1


I managed to solve the question at least to a semi-sure moderate degree, and am posting, assuming no other answers were coming forth (happy to correct/alter based on any valid comments).

As for 1, while the plot of the "coefficient" magnitudes (i.e. the magnitudes of the fourier transform values) would remain the same with translation in time of $x(t)$, but the exact coefficients will change (translation in time domain $\rightarrow$ phase shift in the frequency domain), and the poisson summation formula implies a relation of DTFT to the periodic summation of the actual fourier transform values (not their amplitudes). Therefore, translation does change the periodic summation since it is defined as summing all values spaced by $T$ starting from $x(0)$.

As for 2, the summation of the complex value of the fourier transform does sum to zero in such a case. The mathematical proof of this fact can be considered to be the poisson summation formula itself, since no assumptions on the function's form were made in proving it. The Poisson Summation formula related a periodic summation of the function to the periodic summation of the fourier transform, and the latter to the DTFT. Therefore, if the periodic summation of the function itself is zero (for a particular period), then so will the DTFT.

(Since I did understand many related things while trying to solve this, the intuition below seems quite large, for it also includes the other things I appreciated while simulating the specific case. Posting here with the hope that it might be useful for someone.)

Intuition for a specific case:

All of the above can be verified for atleast a specific case by simulations. Below is the simulation for $$f(t) = \cos(2\pi t)e^{-\pi t^2 / 9}$$ a conveniently chosen even function. As such, the fourier transform is entirely real (since the even part of the time domain function maps to a real part in the frequency domain). Moreover, the shapes of the peaks in the fourier transform amplitudes are gaussian (and hence symmetric about the points $1$ and $-1$, the "core" frequency. This allows us to appreciate the intuition well.

We will look at $g(t) = f(t - t_0)$ for varying shift times $t_0$, from $0$ to $0.25$ when $g(0)$ becomes $0$ and we can then get our all-zero sampling sequence. In each case, we will be sampling at $1 \text{ Hertz}$, its core frequency, because the we need this frequency (or its multiple) to get the all-zero sequence. (For the remaining parts, the colors will remain consistent for the specific $t_0$) enter image description here

Firstly, we just plot the fourier transforms for different $t_0$ to get the idea of how a shift in time domain maps to a phase shift in frequency domain (see wikipedia for the exact relation, and for better visualisation of this phenomenon). In our case, for $t_0 = 0$, we have a perfectly even function and hence the "phase" of our frequency (i.e. the imaginary components of the fourier transform values) are zero and the transform is entirely real. As we start shifting towards a mixed function (neither odd, nor even, in entirety, because at $t_0 > 0$ we have a non-symmetric envelope), we have increasing imaginary component and the real component both decreases in value and also "inverts" beyond a shift (the exact nature of this change is equivalent to a rotation of each complex vector associated with the frequency but at different speeds - again, see wikipedia). But at all times, the imaginary as well as the real (i.e. the fourier transform itself) is symmetric locally about the points $1$ and $-1$. enter image description hereenter image description hereenter image description here

Next is the DTFT. We first take a high sampling frequency, just to illustrate the point of the effect of a shift on DTFT. Then, we can take the $1 \text{ Hertz}$ frequency to illustrate the zero summing. I am not plotting the periodic sums, but they perfectly coincide anyway. The form of the DTFT illustrates the nature of what it does, namely, taking the fourier transform and adding it to a shifted version of itself (the periodic sum) for all shifts a multiple of a fixed number. The gaussian humps may coincide while adding if they are spaced close by, such that the tails of the DTFT humps are thicker (for lower sampling frequencies), or they may not, such that the humps are distinct (higher sampling frequencies. Here, the fourier transform can be recovered from our DTFT. See the Shannon-Nyquist Sampling Theorem intuition below). Below are the plots for $G_{1/T}$ with $t_0 = 0.15$ (to allow for non-zero imaginary components) for different $T$, the sampling period (time difference between two samples; the samples are then $Tg(nT)$ with the $T$ constant to make the math easier). Also note that the amplitudes of the DTFT do not directly represent the summated amplitudes of the fourier transform, and at low sampling frequencies, may give lower peaks. enter image description hereenter image description hereenter image description here

Finally, for our $T = 1$, I've plotted the DTFT for different $t_0$, and as we can appreciate from the fourier transforms above, as we approach a $t_0$ of $0.25$, the real component has undergone an inversion. Looking at the real component first, we know that since our original function was real, the real component is an even function. There are two humps in all the fourier transforms. The spacing between them is $2$ hertz, since both are centered at $1$ and $-1$. Now, for the $t_0=0.25$, each of these is further composed of two opposite sign humps, with the actual $1$ and $-1$ points being zero - and the two humps are arranged akin to an odd function in that small locality about the $\pm 1$ points, that is, on either side, the value is the negative of the value at the corresponding point on the other side of $\pm 1$. We can clearly appreciate why the DTFT with a spacing of $1$ must give us zero for the core frequencies since they are zero anyway; but for the other ones, taking the sum at a frequency just right to the $-1$ point will reach in its summation a point just right to the $1$ point (since the periodic sum is at a spacing of $1$), which, due to the nature of the humps as described above on the other side of the y-axis, is the exact negative of this value and it cancels out. Same for frequency values to the left of this point. In the imaginary part of the DTFT, we have a similar case with similarly spaced humps, and the imaginary component is an odd function for the real function we are using. Both the peaks are symmetric locally about $\pm 1$. Therefore, if we sample any point on one of the humps, with a spacing of $2$, we would hit, in our periodic sum, a point in the hump on the other side of the y-axis and because of the "local" symmetry and the fact that the humps are negatives of each other, it would be the exact negative of our value and cancel out. Overall, the DTFT cancels out. This is possible only with a spacing equal to the "core" frequency of $1$ in our case! enter image description hereenter image description hereenter image description here

For the Shannon-Nyquist Sampling Theorem intuition (since our function is close to a band-limited function), we can appreciate this easily by noticing that in order to fully recreate the original fourier transform from the DTFT, we need that the periodic sums include no sum at all, i.e. the humps do not coalesce or add when the periodic sum is taken and every value of the DTFT is just a fourier transform value shifted without addition. For this, we notice, that if the maximum frequency (the rightmost and the leftmost edges of the fourier transform humps) are less than the half of sampling frequency in our periodic sum, the smallest jump of values between two points on fourier transform which are added in the periodic sums would be the sampling frequency, and in even the smallest jump, we have traversed the entire region of the non-zero fourier transform values. So, no matter what point we choose among the non-zero values, the next value to be added will be zero in the periodic summation because the largest spacing between the non-zero values was smaller than the smallest jump in the periodic summation.


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