14.3. Relations between Fourier transforms#

Some freestyle in changing order of summations and integrals, and use of generalized functions here…check it!

Different Fourier transforms exist, depending if the original function is:

  • time discrete/time continuous

  • periodic/non-periodic

namely,

  • FS, Fourier series: time continuous, periodic function (or finite domain, with a periodic extension)

  • FT, Fourier transform: time continuous, non-periodic function

  • DTFT, discrete-time Fourier transform: time didscrete, infinite-length sequence

  • DFT, discrete Fourier transform: time discrete, finite-length sequence (and then with a periodic extension)

14.3.1. Fourier transform of integrable functions#

\[F(\nu) := \mathscr{F}\left\{f(t)\right\}(\nu) := \int_{t=-\infty}^{+\infty} f(t) e^{-i 2 \pi \nu t} \, dt \ ,\]

14.3.2. Fourier transform of the sum of shifted integrable functions#

The infinite sum of a shifted integrable function is defined as

\[\tilde{f}_T(t) = \sum_{n=-\infty}^{+\infty} f(t - nT) \ .\]

Its Fourier transform reads

\[\begin{split}\begin{aligned} \mathscr{F}\left\{ \tilde{f}_T(t) \right\}(\nu) & = \int_{t=-\infty}^{+\infty} \tilde{f}_T(t) e^{-i 2 \pi \nu t} \, dt = \\ & = \sum_{n=-\infty}^{+\infty} \int_{t=-\infty}^{+\infty} f(t-nT) e^{-i 2 \pi \nu t} \, dt = && (1) \\ & = \sum_{n=-\infty}^{+\infty} F(\nu) e^{-i 2 \pi \nu n T} = && (2) \\ & = F(\nu) \sum_{n=-\infty}^{+\infty} e^{-i 2 \pi \nu n T} = && (3) \\ & = \Delta \nu \, F(\nu) \, \text{III}_{\Delta \nu}(\nu) \ , \end{aligned}\end{split}\]

having used properties of Fourier transform of shifted function in (1), and the properties of Dirac’s comb in (3), having defined the frequency resolution

\[\Delta \nu := \frac{1}{T} \ .\]

This Fourier transform is proportional to the Fourier transform of the original function, sampled in frequency with elementary frequency \(\Delta \nu\).

14.3.3. Fourier transform of the a function sampled with a Dirac comb - DTFT#

Fourier transform of the original function sampled with \(\Delta t \, \text{III}_{\Delta t}(t)\) reads

\[\begin{split}\begin{aligned} \mathscr{F}\left\{ \Delta t \, f(t) \, \text{III}_{\Delta t}(t) \right\} & = \Delta t \int_{t=-\infty}^{+\infty} f(t) \, \text{III}_{\Delta t}(t) e^{-i 2 \pi \nu t} \, dt = \\ \end{aligned}\end{split}\]
(14.6)#\[\begin{split}\begin{aligned} & \sim \Delta t \frac{1}{\Delta t} \int_{t=-\infty}^{+\infty} f(t) \, \sum_{n=-\infty}^{+\infty} e^{i n \frac{2 \pi}{\Delta t} t} e^{-i 2 \pi \nu t} \, dt = \\ & = \sum_{n=-\infty}^{+\infty} \int_{t=-\infty}^{+\infty} f(t) \, e^{-i 2 \pi \left( \nu - n \overline{\nu} \right) t} \, dt = \\ & = \sum_{n=-\infty}^{+\infty} F\left(\nu - n \overline{\nu} \right) = \text{DTFT}(f(t); \Delta t) \ , \end{aligned}\end{split}\]

i.e. equals the periodic sum of the Fourier of the original function, with period

\[\overline{\nu} := \frac{1}{\Delta t} \ .\]

From this last sentence and from the symmetry properties of Fourier transform, Nyquist-Shannon sampling theorem follows seamlessly.

Theorem 14.1 (Nyquist-Shannon sampling theorem)

In order to avoid aliasing the sampling frequency must be twice the maxiumum1 frequency in the signal,

\[\nu_s \ge 2 \nu_{max} \ .\]

todo check alternative expressions if using the definition of train of impulses instead of the Fourier series of Dirac’s comb.

(14.7)#\[\begin{split}\begin{aligned} & = \Delta t \int_{t=-\infty}^{+\infty} f(t) \sum_{k=-\infty}^{+\infty} \delta(t - k \Delta t) \, e^{-i 2 \pi \nu t} \, dt = \\ & = \Delta t \sum_{k=-\infty}^{+\infty} f(k \Delta t) e^{-i 2 \pi \nu k \Delta t} = \text{DTFT}\left( f(t); \Delta t \right) \end{aligned}\end{split}\]

14.3.4. Fourier transform of the sum of shifted integral functions sampled with a Dirac comb#

Fourier transform of the periodic sum

\[\Delta t \, \tilde{f}(t) \, \text{III}_{\Delta t}(t) = \Delta t \, \sum_{n=-\infty}^{+\infty} f(t-nT) \, \text{III}_{\Delta t}(t) \]

reads

\[\begin{split}\begin{aligned} \mathscr{F}\left\{ \Delta t \, \tilde{f}(t) \, \text{III}_{\Delta t}(t) \right\}(\nu) & = \Delta t \int_{t=-\infty}^{+\infty} \sum_{n=-\infty}^{+\infty} f(t-nT) \sum_{k=-\infty}^{+\infty} \delta(t-k \Delta t) \, e^{-i 2 \pi \nu t } \, dt = \\ & = \Delta t \sum_{n=-\infty}^{+\infty} \sum_{k=-\infty}^{+\infty} f(k \Delta t - nT) \, e^{-i 2 \pi \nu k \Delta t } = \\ \end{aligned}\end{split}\]

and defining \(k \Delta \tau_n := k \Delta t - nT\),

\[\begin{split}\begin{aligned} & = \Delta t \sum_{n=-\infty}^{+\infty} \sum_{k=-\infty}^{+\infty} f(k \Delta \tau_n) e^{-i 2 \pi \nu k \Delta \tau_n} e^{-i 2 \pi \nu n T} = \\ & = \underbrace{\Delta t \sum_{k=-\infty}^{+\infty} f(k \Delta \tau_n) e^{-i 2 \pi \nu k \Delta \tau_n}}_{=\text{DTFT}(f(t), \Delta t)} \, \underbrace{\sum_{n=-\infty}^{+\infty} e^{-i 2 \pi \nu n T}}_{= \Delta \nu \, \text{III}_{\Delta \nu}(\nu)} = \\ & = \text{DTFT}(f(t), \Delta t) \, \Delta \nu \, \text{III}_{\Delta \nu}(\nu) \ . \end{aligned}\end{split}\]

todo check! check the change of coordinates that makes DTFT appear

todo check! what follows or, using the relation between \(\Delta t\) and \(T = N \Delta t\), \(\Delta \nu = \frac{1}{T}\), and thus

\[\Delta t \, \Delta \nu = \Delta t \, \frac{1}{T} = \frac{1}{N} \ ,\]

it follows

\[ = \frac{1}{N} \sum_{k=-\infty}^{+\infty} f(k \Delta \tau_n) e^{-i 2 \pi \nu k \Delta \tau_n} \ \text{III}_{\Delta \nu}(\nu) \ . \]

14.3.5. Useful properties#

14.3.5.1. Dirac’s comb \(\text{III}_T(t)\)#

Dirac comb \(\text{III}_T(t)\) is defined as a train of Dirac’s delta

\[\text{III}_T(t) = \sum_{m=-\infty}^{+\infty} \delta(t-mT) \ .\]

Coefficients (14.2) of the Fourier series (14.1) of a \(T\)-periodic train of Dirac delta for \(t \in \left[-\frac{T}{2}, \frac{T}{2} \right]\), read

\[c_n = \frac{1}{T} \int_{t=0}^{T} \delta(t) \, e^{-i n \frac{2\pi}{T} t} = \frac{1}{T} \ ,\]

and thus the Fourier series of Dirac comb \(\text{III}_T(t)\) reads

\[\text{III}_T(t) = \sum_{m=-\infty}^{+\infty} \delta(t-mT) \sim \frac{1}{T} \sum_{n=-\infty}^{+\infty} e^{i n \frac{2\pi}{T} t} \ .\]

14.3.5.2. Symmetry of Fourier transform#

1

Usually there’s no such a frequency above which the signal is exactly zero, but usually there’s a frequency above which the spectrum of the signal is approximately zero, i.e. below a threshold where it can be treated as zero, and introduce no aliasing.