Fourier Transform Pairs
Calculating the Fourier Transform¶
The calculation of the Fourier transform for a known signal is performed via the definite integral:
$$ \begin{eqnarray} \int_b^a f(x)dx & = & \Bigl[ F(x) \Bigr]_b^a \\ & =& F(b)−F(a) \end{eqnarray} $$
$a$ and $b$ are the lower and upper limit of the integral. $F(x)$ is called the antiderivative of $f(x)$. For common cases, antiderivatives can be taken from tables.
Sine and Cosine¶
Sine and cosine are identical in the magnitude spectrum. Both show only two delta impulses - one at the negative and one at the positive frequency of the sine or cosine. Since the two functions differ only in phase, their phase spectrum is different.
$$ \begin{eqnarray} \cos(\omega t) &~& \mapsto &~& \pi \left[\delta(\omega -\omega_0 ) + \delta(\omega +\omega_0 ) \right] \\ \cos(\omega t) &~& \mapsto &~& \frac{\pi}{j} \left[\delta(\omega -\omega_0 ) - \delta(\omega +\omega_0 ) \right] \\ \end{eqnarray} $$
Boxcar Function¶
The boxcar function is inherently included in many DSP calculations. It is the default window function (window functions will be introduced in detail in future sections). It is defined by a width $T$ in the time domain and an amplitude $A$.
$$ \beta(t) = \begin{cases} A & \text{if $|t| \leq T$ } \\ 0 & \text{otherwise} \end{cases} $$
To calculate the transform of the boxcar window, the integral becomes definite within the range of the window. Since $\beta(t) = 1$ within the defined interval, it can be
$$ \begin{eqnarray} \mathcal F(\beta (t)) & = & \int_{-\infty}^{\infty} \beta(t) e^{- j 2 \pi f t} dt \\ & = & \int_{-t/2}^{T/2} A e^{- j 2 \pi f t} dt \\ \end{eqnarray} $$
With the antiderivative of exponential function with a linear term in the argument
$$\int e^{ax}dx = \frac{1}{a} e^{ax} dx $$
the definite interval becomes:
$$ \begin{eqnarray} \mathcal F(\beta (t)) & = & \frac{A}{-j 2 \pi f} \Bigl[ e^{-j 2 \pi f t} \Bigr]_{-T/2}^{T/2} \\ \end{eqnarray} $$
With the limits applied ($a-b$):
$$ \begin{eqnarray} \mathcal F(\beta (t)) & = & \frac{A}{-j 2 \pi f} \Bigl[ e^{-j \pi f T} - e^{j \pi f T} \Bigr] \\ \end{eqnarray} $$
Rearrange and make use of Euler:
$$ \begin{eqnarray} \mathcal F(\beta (t)) & = & \frac{A}{\pi f} \left[\frac{e^{-j \pi f T} - e^{j \pi f T}}{-j 2 } \right] \\ & = & \frac{A}{\pi f} \sin (\pi f T) \\ \end{eqnarray} $$
Expand the fraction to get the sinc or si function: $\mathrm{sinc}(f) = \frac{\sin(\pi f)}{\pi f}$
$$ \begin{eqnarray} \mathcal F(\beta (t)) & = & AT \frac{\sin (\pi f T)} {\pi fT}\\ & = & \boxed{AT ~ \mathrm{sinc}(f T)} \\ \end{eqnarray} $$
The Fourier transform has the width $T$ in the argument of the sinc function. Large values in the argument compress the sinc function. Hence, the wider the window, the narrower its transform and vice versa: