# Proving Nyquist Sampling Theorem for Strictly Band Limited Signals (Whittaker Shannon Interpolation Formula)

I understand that the Nyquist sampling theorem dictates that the minimum sampling frequency, $$f_s$$, be s.t. $$f_s > 2B$$, where $$B$$, is the bandwidth of the signal. I have read the explanation for what happens when the input signal contains an impulse/non zero ferquency spectrum at $$f_s = f_m$$. But is there any way to do a mathematical proof that sampling at $$f_s = 2B$$ works when the signal is strictly bandlimited.

A little background : This is a HW problem

The given hint was to use the fact that "one way to interpret Nyquist Sampling theorem is to note that any bandlimited signal can be represented as a superposition of bandlimited signals that are orthogonal to each other."

• Google “generalized sampling theorem”. There is a paper by Papoulis
– user28715
Commented Jan 13, 2019 at 9:55
• @StanleyPawlukiewicz I think you mean this Papoulis paper (paywall). Commented Jan 15, 2019 at 7:20
• I added a solution of the Sampling Theorem built on Super Position of Bandlimited Signals.
– Royi
Commented Apr 1, 2021 at 7:40
• what does $f_m$ mean? Commented Apr 1, 2021 at 17:25
• @RJ_DSP, Could you please review my answer? If something missing, please let me know.
– Royi
Commented Mar 24, 2023 at 6:51

# Approaching The Sampling Theorem as Inner Product Space

## Preface

There are many ways to derive the Nyquist Shannon Sampling Theorem with the constraint on the sampling frequency being 2 times the Nyquist Frequency.
The classic derivation uses the summation of sampled series with Poisson Summation Formula.

Let's introduce different approach which is more similar to function analysis - Building an orthogonal space and using projection for analysis and synthesis.

## Forming Orthonormal Basis

In this section we'll define an orthonormal base and derive the decomposition and composition process.

### Definitions

First one should define the space of Band Limited Functions. The space of Band Limited Functions is defined by:

$$\mathcal{B}_{ {W}_{s} } = \left\{ f \left( x \right) \mid F \left( \omega \right) = \mathscr{F} \left\{ f \left( x \right) \right\}, \; F \left( \omega \right) = 0, \; \forall \, \left| \omega \right| > {W}_{s} \right\}$$

In words, it means that for each function $$f \left( x \right) \in \mathcal{B}_{ {W}_{s} }$$ its fourier transform $$F \left( \omega \right) = \mathscr{F} \left\{ f \left( x \right) \right\}$$ vanishes for frequencies $$\left| \omega \right| > {W}_{s}$$.

The inner product in this space is given by:

$$\langle f \left( x \right), g \left( x \right) \rangle = \frac{1}{T} \int_{- \infty}^{\infty} f \left( x \right) g \left( x \right) dx, \; T = \frac{2 \pi}{ {W}_{s} }$$

One could easily show that this is a indeed an inner product space with valid inner product.

The main claim is the orthonormal basis of this space is given by:

$$f \left( x \right) = \operatorname{sinc} \left( \frac{ x - n T }{T} \right)$$

Where $$\operatorname{sinc} \left( x \right)$$ is the Normalized Sinc Function given by $$\operatorname{sinc} \left( x \right) = \frac{ \sin(\pi x) }{ \pi x }$$.
The basis functions are parameterized by the parameter $$n$$. Basically we have shifted and scaled function as the basis.

### Proof of The Orthonormal Property

One must show the orthonormal property of the basis under the defined inner product:

\begin{aligned} \langle f \left( x \right), g \left( x \right) \rangle & = \frac{1}{T} \int_{- \infty}^{\infty} f \left( x \right) g \left( x \right) dx = \frac{1}{T} \int_{- \infty}^{\infty} \operatorname{sinc} \left( \frac{ x - n T }{T} \right) \operatorname{sinc} \left( \frac{ x - m T }{T} \right) dx && \text{} \\ & \overset{1}{=} \frac{1}{T} \int_{- \infty}^{\infty} \left( \frac{1}{2 \pi} \int_{- \infty}^{\infty} T \, \Pi \left( \frac{ \omega }{ {W}_{s} } \right) {e}^{-j \omega n T} {e}^{j \omega t} d\omega \right) \operatorname{sinc} \left( \frac{ x - m T }{T} \right) dx && \text{} \\ & \overset{2}{=} \frac{1}{T} \int_{-\infty}^{\infty} \frac{1}{2 \pi} T \, \Pi \left( \frac{\omega}{ {W}_{s} } \right) {e}^{-j \omega n T} \left( \int_{- \infty}^{\infty} {e}^{j \omega x} \operatorname{sinc} \left( \frac{x - m T}{T} \right) dx \right) d\omega && \text{} \\ & \overset{3}{=} \frac{1}{T} \int_{-\infty}^{\infty} \frac{1}{2 \pi} T \, \Pi \left( \frac{\omega}{ {W}_{s} } \right) {e}^{-j \omega n T} T \, \Pi \left( \frac{-\omega}{ {W}_{s} } \right) {e}^{j \omega m T} d\omega && \text{} \\ & \overset{4}{=} \frac{T}{2 \pi} \int_{ - \frac{ {W}_{s} }{2} }^{ \frac{ {W}_{s} }{2} } {e}^{j \omega \left( m - n \right) T} d\omega && \text{} \\ & \overset{5}{=} \begin{cases} 1 & \text{ if } m = n \\ 0 & \text{ if } m \neq n \end{cases} \end{aligned}

Where:

1. Since $$\mathscr{F} \left\{ \operatorname{sinc} \left( \frac{ x - n T }{T} \right) \right\} = T \, \Pi \left( \frac{ \omega }{ {W}_{s} } \right) {e}^{-j \omega n T}$$.
2. Changing order of integration for converging integrals.
3. Applying $$\mathscr{F} \left\{ \operatorname{sinc} \left( \frac{ x - m T }{T} \right) \right\} \left( - \omega\right)$$.
4. Integration boundaries according to the Rect function (Multiplication).
5. Integration over a cycle or over a constant.

Since we proved the suggested base is indeed an orthonormal basis of the space the following holds:

$$\forall f \left( x \right) \in \mathcal{B}_{{W}_{s}} , \; f \left( x \right) = \sum_{n = -\infty}^{n} \langle f \left( x \right), {g}_{n} \left( x \right) \rangle {g}_{n} \left( x \right)$$

Where $${g}_{n} \left( x \right) = \operatorname{sinc} \left( \frac{ x - n T }{T} \right)$$ and $$\langle f \left( x \right), {g}_{n} \left( x \right) \rangle$$ is the projection of $$f \left( x \right)$$ on $${g}_{n} \left( x \right)$$.

### Projection Process

As written above, using the result of a projection of a function in the space onto the basis one could reconstruct it as:

$$f \left( x \right) = \sum_{n = -\infty}^{n} \langle f \left( x \right), {g}_{n} \left( x \right) \rangle {g}_{n} \left( x \right)$$

The question is, what's is the projection of a general function in this space? Well, it turns out it can be shown in a sloed form way:

\begin{aligned} \langle f \left( x \right), {g}_{n} \left( x \right) \rangle & = \frac{1}{T} \int_{- \infty}^{\infty} f \left( x \right) {g}_{n} \left( x \right) dx = \frac{1}{T} \int_{- \infty}^{\infty} f \left( x \right) \operatorname{sinc} \left( \frac{ x - n T }{T} \right) dx && \text{} \\ & \overset{1}{=} \frac{1}{T} \int_{- \infty}^{\infty} \left( \frac{1}{2 \pi} \int_{-\infty}^{\infty} F \left( \omega \right) {e}^{j \omega x} d\omega \right) \operatorname{sinc} \left( \frac{ x - n T }{T} \right) dx && \text{} \\ & \overset{2}{=} \frac{1}{2 \pi T} \int_{- \infty}^{\infty} \left( \int_{- \infty}^{\infty} \operatorname{sinc} \left( \frac{ x - n T }{T} \right) {e}^{j \omega x} dx \right) F \left( \omega \right) d\omega && \text{} \\ & \overset{3}{=} \frac{1}{2 \pi T} \int_{- \infty}^{\infty} T \, \Pi \left( \frac{-\omega}{ {W}_{s} } \right) {e}^{j \omega n T} {F} \left( \omega \right) d\omega && \text{} \\ & \overset{4}{=} \frac{1}{2 \pi} \int_{ - \frac{ {W}_{s} }{2} }^{ \frac{ {W}_{s} }{2} } F \left( \omega \right) {e}^{j \omega n T} d\omega && \text{} \\ & \overset{5}{=} f \left( n T \right) \end{aligned}

Where:

1. Since $$\mathscr{F} \left\{ f \left( x \right) \right\} = F \left( \omega \right)$$.
2. Changing order of integration for converging integrals.
3. Applying $$\mathscr{F} \left\{ \operatorname{sinc} \left( \frac{ x - m T }{T} \right) \right\} \left( -\omega \right)$$.
4. Integration boundaries according to the Rect function.
5. Applying Inverse Fourier Transform at $$x = n T$$.

Wrapping it yields:

$$f \left( x \right) = \sum_{n = -\infty}^{n} \langle f \left( x \right), {g}_{n} \left( x \right) \rangle {g}_{n} \left( x \right) = \sum_{n = - \infty}^{\infty} f \left( n T \right) \operatorname{sinc} \left( \frac{ x - n T }{T} \right)$$

Which is known as the Whittaker Shannon Interpolation Formula.

## Conclusion

In the process above the analysis and synthesis of Band Limited Functions is shown using Orthonormal Basis. If one set $$T$$ to be the Sampling Interval, usually denoted by $${T}_{s}$$ then the Sampling Frequency is given by $${F}_{s} = \frac{1}{ {T}_{s} }$$.
Now, since the function in frequency domain stretches in the range $$\left[ -\frac{{F}_{s}}{2}, \frac{{F}_{s}}{2} \right]$$ we indeed have the known relation that the sampling frequency has to be twice the one sided support of the function in frequency.

• Thank you very very much for your lengthy answer, I will accept it. It is telling me I have to wait 17 hours. Commented Apr 1, 2021 at 9:23
• @deerclaysup, You're welcome. It's a really nice approach to the subject.
– Royi
Commented Apr 1, 2021 at 10:39
• @robert bristow-johnson, Thank you for the useful edit.
– Royi
Commented Apr 8, 2021 at 4:53
• looks like i missed one. Commented Apr 8, 2021 at 14:50