Let $\omega$ be a frequency variable in units rad/s. The fourier transform of the function $f(t)$ is given by \begin{equation} F(\omega) = \int_{-\infty}^\infty f(t)\mathrm{e}^{-i\omega t}dt, \end{equation} and the inverse transform is given by \begin{equation} f(t) = \frac{1}{2\pi} \int_{-\infty}^\infty F(\omega) \mathrm{e}^{i\omega t} d\omega. \end{equation}
The Dirac delta function $\delta(x)$ is not a proper function. It is a distribution defined by how it operates on functions. This definition is \begin{equation}
<f, \delta> = \int_{-\infty}^\infty f(x)\delta(x) dx = f(0) \end{equation} For example, this means that for the constant signal $f(x) = 1$ \begin{equation}
<1, \delta> = \int_{-\infty}^\infty \delta(x) dx = 1. \end{equation} Visually, we view the $\delta(x-a)$ function as a peak, or impulse, at $x-a=0$, i.e. $x=a$.
Euler's formulae gives the $\cos$ and $\sin$ functions in terms of complex exponentials:
\begin{eqnarray} \sin x &= \frac{1}{2i} \big( \mathrm{e}^{ix} - \mathrm{e}^{-ix} \big)\\ \cos x &= \frac{1}{2} \big( \mathrm{e}^{ix} + \mathrm{e}^{-ix} \big) \end{eqnarray}Since these functions are linear combinations of matrix exponentials, their Fourier transforms will be the same linear combination of the Fourier transform of the matrix exponential. Let $x=\omega_1 t$. We are interested in the Fourier transform of the complex function \begin{equation} f(t) = \mathrm{e}^{i\omega_1 t} \end{equation}
From the definition of the transform we have (by just renaming one of the arguments)
\begin{eqnarray} F(\hat{\omega}) &= \int_{-\infty}^\infty f(t)\mathrm{e}^{-i\hat{\omega} t}dt,\\ f(t) &= \frac{1}{2\pi} \int_{-\infty}^\infty F(\omega) \mathrm{e}^{i\omega t} d\omega. \end{eqnarray}Combining the two expressions we get
\begin{eqnarray} F(\hat{\omega}) &= \int_{-\infty}^\infty \frac{1}{2\pi} \int_{-\infty}^\infty F(\omega) \mathrm{e}^{i\omega t}d\omega \mathrm{e}^{-i\hat{\omega}t} dt \\ &= \int_{-\infty}^\infty F(\omega) \Big( \frac{1}{2\pi} \int_{-\infty}^\infty \mathrm{e}^{i\omega t} \mathrm{e}^{-i\hat{\omega}t} dt \Big) d\omega \end{eqnarray}But this must mean that the paranthesis in the integration on the right hand side is the shifted delta function $\delta(\omega - \hat{\omega})$. So, \begin{equation} \frac{1}{2\pi} \int_{-\infty}^\infty \mathrm{e}^{i\omega t} \mathrm{e}^{-i\hat{\omega}t} dt = \frac{1}{2\pi} \int_{-\infty}^\infty \mathrm{e}^{i(\omega-\hat{\omega})t} dt = \delta (\omega-\hat{\omega}) \end{equation} or, by changing the variable names \begin{equation} \int_{-\infty}^\infty \mathrm{e}^{i\omega_1 t} \mathrm{e}^{-i\omega t} dt = 2\pi \delta (\omega_1-\omega) = 2\pi \delta (\omega-\omega_1) \end{equation}
For a sinusoid, $f(t) = \sin \omega_1 t$ we get \begin{equation} F(\omega) = \frac{1}{2i} 2\pi \delta(\omega-\omega_1) - \frac{1}{2i} 2\pi \delta(\omega + \omega_1) = -i\pi \delta(\omega-\omega_1) + i\pi \delta(\omega + \omega_1). \end{equation}
The Fourier transform (and hence the spectrum) of a sinusoid with frequency $\omega_1$ consists of two peaks: at $\pm \omega_1$. Both peaks has real part equal to zero. The peak at $\omega_1$ has negative imaginary part, whereas the peak at $-\omega_1$ has positive imaginary part. This is illustrated below.
In [3]:
import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline
w1 = 1 # rad/s
M = 100 # Number of periods
ws = 40*w1 # Sampling frequency at 40 times the frequency of the sinusoid
wN = ws/2.0 # The Nyquist frequency
h = 2*np.pi/ws
N = M*40
t = np.arange(N)*h
y = np.sin(w1*t)
Y = np.fft.fft(y)*h # Computes the Fourier transform
Ypos = Y[:N/2] # Positive part of the spectrum
Yneg = Y[N/2:] # Negative part. Obs: for frequencies wN up to ws
wpos = np.linspace(0, wN, N/2) # Positive frequencies, goes from 0 to wN
plt.figure(figsize=(16,3))
plt.plot(wpos, np.abs(Ypos))
plt.plot(-wpos[::-1], np.abs(Yneg))
plt.xlim((-10, 10))
plt.xticks((-10, -5, -1, 0, 1, 5, 10))
plt.xlabel(r'$\omega$ [rad/s]')
Out[3]:
In [5]:
# Plot the real and imaginary parts
plt.figure(figsize=(16,6))
plt.subplot(2,1,1)
plt.plot(wpos, np.real(Ypos))
plt.plot(-wpos[::-1], np.real(Yneg))
plt.xlim((-10, 10))
plt.ylim((-0.001, 0.001))
plt.ylabel('Real part')
plt.xticks((-10, -5, -1, 0, 1, 5, 10))
plt.subplot(2,1,2)
plt.plot(wpos, np.imag(Ypos))
plt.plot(-wpos[::-1], np.imag(Yneg))
plt.xlim((-10, 10))
plt.ylim((-250, 250))
plt.xticks((-10, -5, -1, 0, 1, 5, 10))
plt.ylabel('Imaginary part')
plt.xlabel(r'$\omega$ [rad/s]')
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From the definition of the Fourier transform and of the delta function we get \begin{equation} \int_{-\infty}^\infty \delta(t) \mathrm{e}^{-i\omega t} dt = \mathrm{e}^{-i\omega \cdot 0} = 1 \end{equation}
Consider instead the signal with Fourier transform \begin{equation} F(\omega) = \begin{cases} 1, & -\omega_1 \le \omega \le \omega_1, \\0, & \text{otherwise} \end{cases} \end{equation} The function $F(\omega)$ is shown below
In [6]:
plt.figure(figsize=(10,3))
plt.plot([-10, -5, -5, 5, 5, 10], [0,0,1,1,0,0])
plt.ylim((0,2))
plt.xticks((-10, -5, 0, 5, 10))
plt.xlabel(r'$\omega$')
plt.ylabel(r'$F(\omega)$')
lbls=plt.gca().set_xticklabels(['', r'$-\omega_1$', '0', r'$\omega_1$', ''])
Use the inverse Fourier transform to calculate the signal corresponding to the Fourier transform plotted above: \begin{equation} f(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) \mathrm{e}^{i\omega t} d\omega = \frac{1}{2\pi} \int_{-\omega_1}^{\omega_1} \mathrm{e}^{i\omega t} d\omega = \frac{1}{2\pi}\left[\frac{1}{it}\mathrm{e}^{i\omega t}\right]_{-\omega_1}^{\omega_1} = \frac{\omega_1}{\pi \omega_1t} \frac{1}{2i} \big(\mathrm{e}^{i\omega_1t} - \mathrm{e}^{-i\omega_1t} \big) = \frac{\omega_1}{\pi} \frac{\sin \omega_1 t}{\omega_1 t} = \frac{\omega_1}{\pi} \mathrm{sinc} \Big(\frac{\omega_1}{\pi} t\Big) \end{equation}
In [8]:
# Use the inverse fft to calculate the sampled sinc signal
w1 = 8
N = 400
Nc = 1600
tc = np.linspace(0,8,2*Nc)
hc = 8.0/(2*Nc)
wNc = np.pi/hc
wp = np.linspace(0, wNc,Nc)
tspos = np.arange(2*N)*h
#ts = np.hstack((-tspos[::-1], tspos))
wN = 2*w1
h = np.pi/wN
wpos = np.linspace(0,wN,2*N)
yc = w1/np.pi * np.sinc(w1*tc/np.pi)
Y = np.fft.fft(yc)*hc*2
Ycpos = Y[:Nc]
Ypos = np.hstack((np.ones(N), np.zeros(N)))
Ys = np.hstack((Ypos, Ypos[::-1]))
ys = 2*N/(2*np.pi)*wN/N * np.fft.ifft(Ys)
ypos = ys[:2*N]
yneg = ys[2*N:]
ys = np.hstack((yneg, ypos))
plt.figure(figsize=(16,5))
plt.plot(np.hstack((-wpos[::-1], wpos)), np.hstack((Ypos[::-1], Ypos)))
plt.plot(np.hstack((-wp[::-1], wp)), np.hstack((Ycpos[::-1], Ycpos)))
plt.xlim((-30, 30))
plt.ylim((0, 1.5))
plt.xticks((-w1, 0, w1))
lbls=plt.gca().set_xticklabels([r'$-\omega_1$', '0', r'$\omega_1$'])
plt.legend(('square spectrum', 'fourier trf of sinc function'))
plt.figure(figsize=(16,5))
plt.plot(tc, w1/np.pi * np.sinc(w1*tc/np.pi))
plt.stem(tspos,ypos, linefmt='r--', markerfmt='ro', basefmt = 'r-')
plt.xlim((-0.1,6))
plt.legend(('sinc function','from inverse Fourier transform'))
Out[8]:
In [10]:
t = np.linspace(-4,4,800)
w1 = [1, 4, 8, 16]
plt.figure(figsize=(16,5))
for w in w1:
plt.plot(t, w*np.sinc(w*t)/np.pi)
plt.ylim((-2, 5))
plt.legend([r'$\omega_1 = 1$', r'$\omega_1 = 4$', r'$\omega_1 = 8$',r'$\omega_1 = 16$'])
ttt=plt.xlabel('t')
Note that as $\omega_1$ increases, the function resembles Dirac delta function, i.e. an impulse at $t=0$.