Table of Contents

1 Notation

  • \(k_n = \frac{n \pi}{L}\)

2 Definitions

2.1 Spectrum analysis

The process of obtaining the coefficients \(\{c_n\}\) for the Fourier series approximation of some function \(f(x)\).

3 Fourier series / approximation

3.2 Computing the Fourier components

We simply take the projection onto the wanted \(cos(\frac{m \pi x}{L}\) ( or \(sin(\frac{m \pi x}{L}\) ) that we want the coefficient for. Then we integrate over the entire range (i.e. \(L\) in this case) and rearrange to obtain \(a_m\) (or \(b_m\) ), as wanted.

This is why we define the constant term as \(\frac{a_0}{2}\) => it gives us a nicer expression.

4 Delta function / Dirac delta

4.1 Shifting property

\begin{equation*} \int_{-\infty}^\infty dx f(x) \delta(x - d) = f(d) \end{equation*}

4.1.1 Proof

\begin{equation*} \begin{split} \int_{-\infty}^{\infty} f(x) \delta(x-d) dx &= \int_{-\infty}^{\infty} f(z + d) \lim_{a \rightarrow \infty} \Big( \frac{1}{2a} \Big) dz \\ &= \lim_{a \rightarrow \infty} \int_{-a}^a f(z + d) dz \\ &= \lim_{a \rightarrow \infty} \int_{-a}^a f(x + d) dx \\ &= f(d) \end{split} \end{equation*}

Due to

\begin{equation*} \delta(x - d) \ | \ \lim_{a \rightarrow 0} \int_{-a}^a \frac{1}{2a} dx = 1 \end{equation*}

4.1.2 Proof (my version)

First let the "top-hat" function be defined as follows:

\begin{equation*} \prod_\varepsilon (x) = \underset{\varepsilon \rightarrow 0}{\lim} \begin{cases} \frac{1}{\varepsilon} \quad \text{if} \ x \in [-\varepsilon, \varepsilon] \\ 0 \quad \text{otherwise} \end{cases} \end{equation*}

Then, we write

\begin{equation*} \begin{split} \int_{-\infty}^{\infty} \prod_{\varepsilon} (x) f(x) dx &= \underset{\varepsilon \rightarrow 0}{\lim} \int_{- \varepsilon}^{\varepsilon} \frac{1}{\varepsilon} f(x) dx \\ &= \underset{\varepsilon \rightarrow 0}{\lim} \frac{1}{\varepsilon} [F(\varepsilon) - F(- \varepsilon)] \quad \text{(where f is derivative of F)} \\ &= \underset{\varepsilon \rightarrow 0}{\lim} \frac{F(\varepsilon) - F(- \varepsilon)}{\varepsilon} \\ &= \frac{d}{dx} F(x) |_{x = 0} \quad \text{(by definition of derivative)} \\ &= f(0) \end{split} \end{equation*}

You do exactly the same thing as above, but simply make a change of variable before "computing" the integral.

5 Fourier Transform

6 Nyquist-Shannon Theorem

6.1 Notation

  • \(\hat{f}(k)\) is the Fourier transform of \(f(t)\):

    \begin{equation*} \hat{f}(k) = \mathcal{F} \left\{ f(t) \right\} = \int_{-\infty}^{\infty} f(t) e^{i 2 \pi kt} \ dt \end{equation*}
  • \(T\) is the time between each sample, or sampling interval
  • \(f_{\sigma}\) is the bandlimit, i.e. maximum frequency contained by \(f(t)\)

6.2 Poisson summation formula

  • Relates the Fourier series coefficients of the periodic summation of a function to values of the function's continuous Fourier transform
  • Periodic summation of a \(f(x)\) is completely defined by discrete samples of the \(\hat{f}(k)\), and vice-versa

First we observe that

\begin{equation*} \begin{split} \sum_{k=- \infty}^{\infty} \hat{f}(k) &= \bigg( \int_{-\infty}^{\infty} f(x) e^{i 2 \pi k x} \ dx \bigg) \\ &= \int_{-\infty}^{\infty} f(x) \underbrace{\bigg( \sum_{k = - \infty}^{\infty} e^{i 2\pi k x} \bigg)}_{= \sum_{n=-\infty}^{\infty} \delta(x - n)} \ dx \\ &= \sum_{n= - \infty}^{\infty} \bigg( \int_{-\infty}^{\infty} f(x) \delta(x - n) \ dx \bigg) \\ &= \sum_{n = - \infty}^{\infty} f(n) \end{split} \end{equation*}

And similarily,

\begin{equation*} \begin{split} \sum_{k= - \infty}^{\infty} \hat{s}(\nu + k / T) &= \sum_{k= - \infty}^{\infty} \mathcal{F} \left\{ s(t) \cdot e^{-i 2 \pi \frac{k}{T}t} \right\} \\ &= \mathcal{F} \left\{ s(t) \sum_{k= - \infty}^{\infty} e^{- i2 \pi \frac{k}{T} t} \right\} \\ &= \mathcal{F} \left\{ \sum_{n= - \infty}^{\infty} T \cdot s(nT) \cdot \delta(t - nT) \right\} \\ &= \sum_{n = - \infty}^{\infty} T \cdot s(nT) \cdot \mathcal{F} \left\{ \delta(t - nT) \right\} \\ &= \sum_{n = -\infty}^{\infty} T \cdot s(nT) \cdot e^{- i 2\pi n T \nu} \end{split} \end{equation*}

where \(T\) is the sample interval.

Observe that setting \(\nu = 0\), we get

\begin{equation*} \sum_{k = -\infty}^{\infty} \hat{s}(k / T) = \sum_{n= -\infty}^{\infty} T \cdot s(nT) \end{equation*}

6.3 The theorem

Suppose we want to reconstruct a function \(f(t)\) by sampling from this function.

Further, suppose \(f(t)\) is band-limited, i.e. there exists some frequency \(f_{\sigma}\) such that Fourier transform is zero for frequencies exceeding the cutoff:

\begin{equation*} \hat{f}(\xi) = 0, \quad \forall |\xi| > f_{\sigma} \end{equation*}

Then, by Possion summation of \(f(t)\), one is guaranteed to be able to reconstruct \(\hat{f}(\xi)\) with a sampling rate of \(2 f_{\sigma}\), and thus, by Fourier inversion, we can reconstruct \(f(t)\) perfectly!

\(2f_{\sigma}\) is often referred to as the Nyquist rate , and a sampling frequency \(f_s > 2 f_{\sigma}\) is called a Nyquist frequency.

Why this is the case can also be seen from the fact that if one is sampling at the lower frequency than the Nyquist frequency, then you might observe aliasing, which means that if you're unlucky and sample exactly at the peaks of \(f_1\) and \(f_2 = 2 f_1\), then you cannot discern these two, since you will always see these peaks overlap.

On the other hand, if you indeed sampled at a rate of \(2f_1\), then you would also be able to capture the peak of \(f_2\) in between two peaks of \(f_1\), hence being able to reconstruct the signal.