One of the forms of the Poisson summation formula is

$$ \sum_{n=-\infty}^{\infty} T\cdot x(nT)\ e^{-i 2\pi f T n}\; {=} \; \sum_{k=-\infty}^{\infty} X\left(f - k/T\right),$$

where $x(nT)$ are samples of a continuous function $x(t)$, and $X(f)$ the fourier transform of $x(t)$. The RHS is a periodization of $X(f)$.

What I couldn't understand is the following, we can define two functions $x_{1}(t)$ and $x_{2}(t)$ that are equal at the sample points, $x_{1}(nT)=x_{2}(nT)$. We have $$ \sum_{n=-\infty}^{\infty} T\cdot x_{1}(nT)\ e^{-i 2\pi f T n}\; {=} \; \sum_{k=-\infty}^{\infty} X_{1}\left(f - k/T\right),$$ and $$ \sum_{n=-\infty}^{\infty} T\cdot x_{2}(nT)\ e^{-i 2\pi f T n}\; {=} \; \sum_{k=-\infty}^{\infty} X_{2}\left(f - k/T\right),$$

where $X_{1}(f)$ and $X_{2}(f)$ are the Fourier transforms of $x_{1}(t)$ and $x_{2}(t)$ respectively.

But since $x_{1}(nT)=x_{2}(nT)$, we then also have $$ \sum_{n=-\infty}^{\infty} T\cdot x_{1}(nT)\ e^{-i 2\pi f T n}=\sum_{n=-\infty}^{\infty} T\cdot x_{2}(nT)\ e^{-i 2\pi f T n},$$

Which means $$\boxed{\sum_{k=-\infty}^{\infty} X_{1}\left(f - k/T\right){=} \; \sum_{k=-\infty}^{\infty} X_{2}\left(f - k/T\right)}$$

But we can choose the functions $x_{1}(t)$ and $x_{2}(t)$, even though they are equal at the sample points, in such a way that their transforms $X_{1}(f)$ and $X_{2}(f)$ are quite different from each other, and, therefore, so will their periodizations. And so why should this last equation be true in general?

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    $\begingroup$ $X_1(\cdot)$ and $X_2(\cdot)$ are different, but their periodic sums are not. $\endgroup$ Commented Feb 1, 2020 at 4:10

3 Answers 3


$x(t)$ is a continuous-time signal with Fourier transform $X(f)$. There is no restriction whatsoever on the bandwidth of $x(t)$. If the signal is sampled at intervals of $T$ seconds, then the $n$-th sample of $x(t)$ is $x_n = x(nT)$. The OP correctly states that $$ \sum_{n=-\infty}^{\infty} T\cdot x(nT)\ e^{-i 2\pi f nT} =\sum_{n=-\infty}^{\infty} T\cdot x_n\ e^{-i 2\pi f nT} = \sum_{k=-\infty}^{\infty} X\left(f - k/T\right).\tag{1}$$ Regardless of whether $X(f)$ is bandlimited or not, the sum on the right is a periodic function of the frequency $f$ with period $\frac 1T$. There is no requirement that $X(f)$ and $X\left(f-\frac 1T\right)$ have non-overlapping support. The OP then wonders: if there is another signal $y(t)$ with different Fourier transform $Y(f)$ but $y(t)$ and $x(t)$ are equal to one another at the sampling instants $nT$, that is, $y(nT) = x(nT) = x_n$ for all $n$ and so $$\sum_{n=-\infty}^{\infty} T\cdot x(nT)\ e^{-i 2\pi f nT} = \sum_{n=-\infty}^{\infty} T\cdot y(nT)\ e^{-i 2\pi f nT} = \sum_{n=-\infty}^{\infty} T\cdot x_n\ e^{-i 2\pi f nT} ,$$ then $(1)$ implies that $$\sum_{k=-\infty}^{\infty} X\left(f - k/T\right)=\sum_{k=-\infty}^{\infty} Y\left(f - k/T\right).\tag{2}$$ Why would such equality hold?

Well, there are infinitely many different signals that all have the same set of sample values $\{x_n\}$, not just $x(t)$ and $y(t)$. But among all these signals with the the same set of sample values $\{x_n\}$, there is only one signal $x_0(t)$ ---- most well-beloved, perhaps even adored, on dsp.SE ---- that not only has sample values $\{x_n\}$ but also Fourier transform $X_0(f)$ whose support is $\left(-\frac{1}{2T}, \frac{1}{2T}\right)$; that is, $X_0(f) =0$ for $|f| \geq \frac{1}{2T}$. Thus, in the sum $\sum_{k=-\infty}^{\infty} X_0\left(f - k/T\right)$, the term $X_0\left(f - k/T\right)$ occupies the frequency band $\left(-\frac{k-1}{2T}, \frac{k+1}{2T}\right)$ and doesn't overlap at all with any other term $X_0\left(f - k^\prime/T\right)$: they occupy disjoint frequency bands. Put another way, for all real numbers $f$, the sum $$\sum_{k=-\infty}^{\infty} X_0\left(f - k/T\right),\tag{3}$$ equals the sum $$ \sum_{k=-\infty}^{\infty} X\left(f - k/T\right)\tag{4}$$ but is different from the sum in $(4)$ in that for any real number $f$, no more than one of the terms $X_0\left(f - k/T\right)$ can be nonzero. In contrast, in the sum in $(4)$, for any choice of $f$, more than one term is typically nonzero. The special signal $x_0(t)$ is the only one of the myriad signals with sample values that can be reconstructed (e.g. by ideal low-pass filtering) from the sample values $\{x_n\}$. But what of the other signals with the same sample values? Well, their Fourier transforms are not restricted to have support $\left(-\frac{1}{2T}, \frac{1}{2T}\right)$; the support extends beyond that and might even be the entire frequency axis, and so for any given $f$, more than one term can be nonzero as stated earlier. Thus, all the other signals with sample values $\{x_n\}$ are effectively what is commonly called undersampled; the sampling rate is not high enough, and so their spectra alias into the band $\left(-\frac{1}{2T}, \frac{1}{2T}\right)$ and the result of this aliasing is exactly $X_0(f)$ in the band $\left(-\frac{1}{2T}, \frac{1}{2T}\right)$.

In summary, $(2)$ holds because the two different undersampled signals (with identical sample values) have the same spectrum $X_0(f)$ in the band $\left(-\frac{1}{2T}, \frac{1}{2T}\right)$ after aliasing is taken into account.

  • $\begingroup$ Would you mind giving a reference, or the name of the theorem, that establishes the equality of (3) and (4) (an intuitive outline as to why would be even better)? $\endgroup$
    – Hilbert
    Commented Feb 7, 2020 at 22:23

But we can choose the functions $x_1(t)$ and $x_2(t)$, even though they are equal at the sample points, in such a way that their transforms $X_1(f)$ and $X_2(f)$ are quite different from each other, and, therefore, so will their periodizations.

That's a wrong conclusion. If $x_1(nT)=x_2(nT)$ $\forall n$ then the discrete-time Fourier transforms of the respective sequences must be identical, and so must be their periodized spectra.

Note that $x_1(nT)=x_2(nT)$ and $x_1(t)\neq x_2(t)$ implies that at least one of the two functions is not sampled according to the sampling theorem, i.e., the function is not completely represented by its samples. Because otherwise equal samples would imply equal functions. This might be the source of your confusion. If both functions are band-limited and if the sampling theorem is satisfied then $x_1(nT)=x_2(nT)$ implies $x_1(t)=x_2(t)$.

  • $\begingroup$ If we choose the sampling frequency $F_{s}$ to be high enough, we would be able to distinguish to a reasonable degree the original spectrum of our function in the interval $[-F_{s}/2, F_{s}/2]$ of the plot of the periodized spectrum, for instance the same wiki article says, "For sufficiently large $f_s$ the $k = 0$ term can be observed in the region $[−f_{s}/2, f_{s}/2]$ with little or no distortion (aliasing) from the other terms. " And so, again, what prevents the periodization of $X_{1}(f)$ from being different from the periodization of $X_{2}(f)$ over $[−f_{s}/2, f_{s}/2]$. $\endgroup$
    – Hilbert
    Commented Jan 31, 2020 at 21:44
  • $\begingroup$ This representation might be helpful. So basically we know that the zeroth term in each of the two periodizations is $X_{1}(f)$ and $X_{2}(f)$, and each of these two can be observed over $[-1/(2T), 1/(2T)]$, and since we made sure that $X_{1} (f)$ and $X_{2}(f)$ are different from each other, so will their periodizations over $[-1/(2T), 1/(2T)]$, and so how can the periodizations be the same if they are different over $[-1/(2T), 1/(2T)]$? $\endgroup$
    – Hilbert
    Commented Jan 31, 2020 at 21:51
  • $\begingroup$ What @Matt L. is saying is that if you have $x_1(nT)=x_2(nT)$ you have two identical discrete-time signals that have, inevitably, the same discrete-time Fourier Transform (the periodization of $X(f)$, as you say). Given that you use the same sampling period, $T$, for both, Shannon's theorem tells us that both continuous time signals obtained by the reconstruction formula $$x(t) = \sum_{n} x[n]\mathrm{sinc}\Big(\frac{t-nT}{T}\Big)$$ should be identical as well. As a consequence, they would have identical continuous-time Fourier Transforms $X(f)$. $\endgroup$
    – GKH
    Commented Jan 31, 2020 at 22:16
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    $\begingroup$ @Hilbert: I added some more information to my answer, take a look. $\endgroup$
    – Matt L.
    Commented Feb 1, 2020 at 11:32
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    $\begingroup$ @Hilbert: If the sampling frequency is greater than twice the maximum frequency of the signal, the signal can be reconstructed from the periodized spectrum. This is equivalent to saying that the signal's samples uniquely represent the continuous-time signal. $\endgroup$
    – Matt L.
    Commented Feb 1, 2020 at 12:09

Let me see if I am getting the problem here first:

What you describe is a condition where two functions $f,g$ just happen to be identical at the points that sampling happens to pick.

If you chose a different $T$, that would become "visible" but for certain choices of $T$, the discrete Fourier transforms $F,G$ would look identical. But how is that, when $f,g$ are not?

If I got this right so far (?), then:

Yes. They would. And you cannot do anything about this after you have applied sampling.

Also, the fact that $f,g$ are identical at $nT$ for certain choices of $T$, means that they are not too dissimilar anyway. There seems to be a degree of correlation and therefore, the spectrum of one can be expressed via a weighted sum of the other. So, ending up with two identical copies is not some kind of violation.

Let me give you another example:

Suppose some $Fs$ and a continuous sinusoid of the form $g(t)=\sin(2 \pi f t)$. Theoretically, we can crank $f$ up to $\frac{Fs}{2}$ and still be able to "catch" at least two (equally spaced) samples per period of $g$.

If you set $f=\frac{Fs}{2}$ and sample at $Fs$ or conversely every $T=\frac{1}{Fs}$, you will get a sample at $t=0$ and the next at $t=T$ which is half the period of the sinusoid. That is, both points will be zero.

Just as it happens in your case, here $g(t)$ goes on between $nT$ but $g(n T)$ just happens to land on the zeros.

After the sampling there is nothing you can do about cases like this. The sampling determines the amount of information you capture.

Hope this helps(?)


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