What is meant by *sampling* in terms of the *sampling theorem*?

Question

Let $y:\left(-\frac T2,\frac T2\right)\to\mathbb{C}$ be a square integrable function. The Fourier coefficients of $y$ are $$\underline{Y}(k):=\frac 1T\int_{-T/2}^{T/2}y(t)e^{-i\omega_kt}\;dt\;\;\;\text{with }\omega_k:=k\frac{2\pi}T$$ for $k\in\mathbb{Z}$. The Fourier polynom of degree $n\in\mathbb{N}$ of $y$ is $$\mathcal{F}^{-1}_n[y](t):=\sum_{k=-n}^n\underline{Y}(k)e^{i\omega_kt}$$ and $$\mathcal{F}^{-1}[y]:=\lim_{n\to\infty}\mathcal{F}_n^{-1}[x]$$ is called inverse Fourier transformation of $y$. Now, I've got two questions:

1. What is meant by sampling (in terms of the sampling theorem)? From my understanding, if we know the period $T$ all we need to "store" are the values $\underline{Y}(k)$. We cannot store all values, so we need to choose a "huge enough" $n$ and store only the values $\underline{Y}(-n),\cdots,\underline{Y}(n)$. So, where does "sampling" come into play? The only thing I could imagine is numerical integration: We consider an equidistant grid $$x_j=\left(\frac jN-1\right)\frac T2\;\;\;\text{for }j=0,\ldots,2N$$ and apprximate $\underline{Y}(k)$ using the composite trapezoidal rule, i.e. $$\underline{Y}(k)\approx\frac{1}{2N}\sum_{j=0}^{2N-1}y\left(x_j\right)e^{-i\omega_kx_j}$$ By doing so, we didn't take the whole "signal" $y$, but only the "sample points"$\left(x_j,y\left(x_j\right)\right)$ into account. Is this meant by "sampling"?
2. Does the sampling theorem make a statement about $n$ or $N$ or something else?

Answer 1

I think, with respect to sampled function, that i disagree a bit with MBaz. perhaps i misunderstand what he/she says regarding needing an infinite number of coefficients or samples for this special case of sampled periodic functions. we had a little discussion about this at comp.dsp, and i'm gonna make use of $\LaTeX$ to spell out the same points.

of course, whether $x(t)$ is periodic or not, if $x(t)$ is real and is sufficiently bandlimited (there are no frequency components as high or higher than $\frac{f_\text{s}}{2} = \frac{1}{2T_\text{s}}$), then the samples:

$$ x[n] \triangleq x(n T_\text{s}) $$

are sufficient to completely represent the continuous-time $x(t)$. if $x(t)$ never repeats, than an infinite number of discrete $x[n]$ are necessary to represent $x(t)$ for all $t$.

but if $x(t)$ is periodic,

$$ x[n+N]=x[n] \quad\quad \forall n,N \in \mathbb{Z} $$ and $$ x(t+T)=x(t) \quad\quad \forall t $$

then samples existing over the span of one period are sufficient to represent $x[n]$ and $x(t)$. in order for the $x[n]$ to be periodic in the sampled domain:

$$ \begin{align} x[n+N] & = x\left((n+N)T_\text{s} \right) \\ & = x\left(nT_\text{s}+NT_\text{s} \right) \\ & = x\left(nT_\text{s}+T \right) \\ \end{align} $$ because $ x(t+T) = x(t) $.

which means the obvious, your function period $T$ has to be the same as $N$ times your sampling period $T_\text{s}$.

$$ T = N T_\text{s} $$

now, what we know about this sampled periodic function is that $N$ samples of $x[n]$ are sufficient to tell us all about $x[n]$, and since $x(t)$ is bandlimited, $x[n]$ and the $N$ samples that fully define it, are sufficient to fully describe $x(t)$.

if $x(t)$ is sufficiently bandlimited (as above), then

$$ \begin{align} x(t) & = \sum\limits_{n=\infty}^{+\infty} x[n] \operatorname{sinc}\left(\frac{t-nT_\text{s}}{T_\text{s}}\right) \\ & = \sum\limits_{n=\infty}^{+\infty} x[n] \frac{\sin\left(\pi\frac{t-nT_\text{s}}{T_\text{s}}\right)}{\pi\frac{t-nT_\text{s}}{T_\text{s}}} \\ & = \sum\limits_{m=\infty}^{+\infty} \sum\limits_{n=0}^{N-1} x[n+mN] \frac{\sin\left(\pi\frac{t-(n+mN)T_\text{s}}{T_\text{s}}\right)}{\pi\frac{t-(n+mN)T_\text{s}}{T_\text{s}}} \\ & = \sum\limits_{n=0}^{N-1} \sum\limits_{m=\infty}^{+\infty} x[n] \frac{(-1)^{mN}\sin\left(\pi\frac{t-nT_\text{s}}{T_\text{s}}\right)}{\pi\frac{t-(n+mN)T_\text{s}}{T_\text{s}}} \\ & = \sum\limits_{n=0}^{N-1} x[n] \sin\left(\pi\frac{t-nT_\text{s}}{T_\text{s}}\right) \sum\limits_{m=\infty}^{+\infty} \frac{\frac{T_\text{s}}{\pi}(-1)^{mN}}{t-(n+mN)T_\text{s}} \\ \end{align} $$

$$ x(t) = \begin{cases} \sum\limits_{n=0}^{N-1} x[n] \frac{\sin\left(\pi\frac{t-nT_\text{s}}{T_\text{s}}\right)}{N\tan\left(\frac{\pi}{N}\frac{t-nT_\text{s}}{T_\text{s}}\right)}, & \text{if }N\text{ is even} \\ \sum\limits_{n=0}^{N-1} x[n] \frac{\sin\left(\pi\frac{t-nT_\text{s}}{T_\text{s}}\right)}{N\sin\left(\frac{\pi}{N}\frac{t-nT_\text{s}}{T_\text{s}}\right)}, & \text{if }N\text{ is odd} \end{cases} $$

proving the latter takes a little bit. you might recognize the $N$ odd case as the Dirichlet_kernel. the $N$ even case looks a teeny bit different. but both show exactly how the $N$ samples that fully define the sampled and periodic $x(t)$ are combined to get $x(t)$.

now, since $x(t)$ is also periodic with period $T$, then

$$ \begin{align} x(t) & = x\left(t+T \right) \\ & = x\left(t+N T_\text{s} \right) \\ & = \sum\limits_{k=-\infty}^{+\infty} X[k] e^{i 2 \pi \frac{k}{T} t} \\ & = \sum\limits_{k=-\infty}^{+\infty} X[k] e^{i 2 \pi \frac{k}{N T_\text{s}} t} \\ & = \sum\limits_{k=-\lfloor \frac{N}{2} \rfloor}^{+\lfloor \frac{N}{2} \rfloor} X[k] e^{i 2 \pi \frac{k}{N} \frac{t}{T_\text{s}}} \end{align} $$

where $\lfloor\cdot\rfloor$ is the $\operatorname{floor}(\cdot)$ operator and

$$ \begin{align} X[k] & = \frac{1}{N} \sum\limits_{n=0}^{N-1} x[n] e^{-i 2 \pi \frac{nk}{N}} \\ & = \mathcal{DFT}\{x[n]\} \end{align} $$

there's actually something to fudge (a factor of $\frac{1}{2}$) about $X\left[\frac{N}{2}\right]$ for the $N$ even case:

$$ \begin{align} X\left[-\frac{N}{2}\right] = X\left[\frac{N}{2}\right] & = \frac{1}{2} \ \frac{1}{N} \sum\limits_{n=0}^{N-1} x[n] e^{-i 2 \pi n\frac{n(N/2)}{N}} \\ & = \frac{1}{2N} \sum\limits_{n=0}^{N-1} x[n] (-1)^n \\ \end{align} $$

note that:

$$ \begin{align} x[n] = x(n T_\text{s}) & = \sum\limits_{k=-\lfloor \frac{N}{2} \rfloor}^{+\lfloor \frac{N}{2} \rfloor} X[k] e^{i 2 \pi \frac{k}{N} \frac{t}{T_\text{s}}}\bigg|_{t=n T_\text{s}} \\ & = \sum\limits_{k=-\lfloor \frac{N}{2} \rfloor}^{+\lfloor \frac{N}{2} \rfloor} X[k] e^{i 2 \pi \frac{k}{N}n} \\ & = \sum\limits_{k=0}^{N-1} X[k] e^{i 2 \pi \frac{nk}{N}} \\ & = \mathcal{iDFT}\{X[k]\} \end{align} $$

if you deal with that $\frac{1}{2}$ fudging for $X\left[\frac{N}{2}\right]$ for the $N$ even case. this is because $X[k+N]=X[k]$ for all $k$.

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a periodic continuous-time function can be described with a countable set of Fourier coefficients. a bandlimited periodic continuous function can be described with a finite set of $N$ Fourier coefficients, just as well as it can be described with a finite set of $N$ samples.

Answer 2

Sampling and the Fourier series are only indirectly related. Both are orthonormal expansions of a function, but they use different basis.

In the Fourier series, the orthonormal basis are exponential functions. The Fourier coefficents are the magnitude and phase of the exponentials. As you say, having the coefficients is equivalent, in a certain sense, to having the actual function. However:

• As you have noticed, the Fourier series in general requires an infinite number of coefficients, so it may not be practical. However, the coefficients tend to zero (there is a proof of this), so you can keep a finite number of them and neglect the rest. The difference between the original function and the one re-created from the stored coefficients can be so small as to be ignored.

• Note that the Fourier transformation is not unique. It is unique if you limit the functions to voltages that can be created in practice.

When sampling, the orthonormal basis are sinc (cardinal sine) functions. The coefficients of the sinc functions are the signal samples (that is, its amplitude at specific instants). The sampling theorem gives sufficient conditions for a function to be expressed in terms of this basis. A nice thing about the proof is that it is constructive; that is, it recreates the function from its samples. Notes:

• Again, you need an infinite number of coefficients (samples) . If you only keep a finite number, there will be a difference between the original function and its reconstruction.

• The sampling theorem assumes that the function is band-limited, which implies that it is infinite in duration. So, the theorem requires $n$ to be infinite. It makes no statement on $N$.

In engineering and signal processing, we just design our systems so that the difference between the original signal and that reconstructed from its samples is too small to notice. I don't know of any results that quantify the error in terms of $N$ for general functions.