Fourier Transform: Signal Analysis

Lecture 26: The Fourier Transform

• The goal is to analyze signals by breaking them down into their frequency components.
• Many systems, like ears, can be naturally understood in the frequency domain.
• Representing signals with a few frequencies simplifies them, enabling compression techniques such as MP3 and JPEG.
• The Fourier Transform (FT) maps a signal into the frequency domain, starting with the continuous-time, aperiodic case.

The Fourier Transform

• The Fourier Transform is defined as:
  • $\mathcal{F}(f(t)) = F(\omega) = \int_{-\infty}^{\infty} f(t)e^{-j\omega t} dt$
• In this context, $F(\omega)$ is a complex-valued function of the real variable $\omega$.
• The unit of $\omega$ is radians/second
• Frequency in Hertz ($f$) can be used instead, where $\omega = 2\pi f$.
• Inverse Fourier Transform definition:
  • $f(t) = \mathcal{F}^{-1}(F(\omega)) = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) e^{j\omega t} d\omega$
• $F(\omega)$ indicates the amount of frequency $\omega$ present in the signal $f(t)$, specifically, the complex amplitude.

Example

• For $f(t) = A\cos(\omega_0 t)$, the Fourier Transform is:
  • $F(\omega) = \int_{-\infty}^{\infty} A\cos(\omega_0 t) e^{-j\omega t} dt$
• Using Euler's formula, $\cos(\omega_0 t) = \frac{1}{2}(e^{j\omega_0 t} + e^{-j\omega_0 t})$:
  • $F(\omega) = \frac{A}{2} \int_{-\infty}^{\infty} e^{-j(\omega - \omega_0) t} dt + \frac{A}{2} \int_{-\infty}^{\infty} e^{-j(\omega + \omega_0) t} dt$
• Using the sifting property of the delta function, $\int_{-\infty}^{\infty} e^{-j\omega t} dt = 2\pi \delta(\omega)$:
  • $F(\omega) = A\pi \delta(\omega - \omega_0) + A\pi \delta(\omega + \omega_0)$
• The spectrum of $A\cos(\omega_0 t)$ contains impulses at $\pm \omega_0$.
  • The amount of each frequency is $A\pi$.

Example (continued)

• Next, for $A\sin(\omega_0 t)$, the Fourier Transform formula is:

  • $F(\omega) = \int_{-\infty}^{\infty} A\sin(\omega_0 t) e^{-j\omega t} dt$

• By Euler's formula, it is known that $\sin(\omega_0 t) = \frac{1}{2j}(e^{j\omega_0 t} - e^{-j\omega_0 t})$.

• Applying this, gives:

  • $F(\omega) = -jA\pi \delta(\omega - \omega_0) + jA\pi \delta(\omega + \omega_0)$

• $F(\omega)$ has complex values.

Example

• For $$f(t) = \begin{cases} 1, & |t| < T_1 \0, & |t| > T_1 \end{cases}$$

• The following holds:

  • $F(\omega) = \int_{-\infty}^{\infty} f(t) e^{-j\omega t} dt = \int_{-T_1}^{T_1} e^{-j\omega t} dt$
  • $F(\omega) = 2 \frac{\sin(\omega T_1)}{\omega} = 2T_1 \frac{\sin(\omega T_1)}{\omega T_1} = 2T_1 sinc(\omega T_1)$

• In this context the sinc functionis defined as:

  • $sinc(x) = \frac{\sin(x)}{x}$

• $F(\omega)$ is real-valued; however.

• $F(\omega)$ can generally be complex.

Example (continued)

• $f(t) = \begin{cases} 1, & |t| < T_1 \ 0, & |t| > T_1 \end{cases}$
• The following definition applies $F(\omega) = 2T_1 \frac{\sin(\omega T_1)}{\omega T_1} = 2T_1 sinc(\omega T_1)$.

Properties of FT

• Linearity: The Fourier Transform of a linear combination of signals is the same linear combination of the individual Fourier Transforms:
  • $\mathcal{F}(af_1(t) + bf_2(t)) = aF_1(\omega) + bF_2(\omega)$
• Time Shifting: A shift in time corresponds to a phase shift in the frequency domain:
  • $\mathcal{F}(f(t - t_0)) = e^{-j\omega t_0}F(\omega)$
• Frequency Shifting: Multiplication by a complex exponential in time corresponds to a shift in frequency:
  • $\mathcal{F}(e^{j\omega_0 t}f(t)) = F(\omega - \omega_0)$
• Time Scaling: Scaling time affects frequency:
  • $\mathcal{F}(f(at)) = \frac{1}{|a|} F\left(\frac{\omega}{a}\right)$
• Differentiation in Time: Differentiation in time corresponds to multiplication by $j\omega$ in frequency.
  • $\mathcal{F}\left(\frac{d}{dt} f(t)\right) = j\omega F(\omega)$
• Integration in Time formula: $$\mathcal{F}\left(\int_{-\infty}^{t} f(\tau) d\tau\right) = \frac{1}{j\omega} F(\omega) + \pi F(0)\delta(\omega)$$
• i.e. convolution in time corresponds to multiplication in frequency.
• Convolution Theorem formula: $$\mathcal{F}(f_1(t) * f_2(t)) = F_1(\omega) F_2(\omega)$$
• Multiplication Theorem formula: $$\mathcal{F}(f_1(t) f_2(t)) = \frac{1}{2\pi} [F_1(\omega) * F_2(\omega)]$$
  • i.e. multiplication in time corresponds to convolution in frequency.

Parseval's Theorem

• Energy in the time domain = Energy in the frequency domain:
  • $\int_{-\infty}^{\infty} |f(t)|^2 dt = \frac{1}{2\pi} \int_{-\infty}^{\infty} |F(\omega)|^2 d\omega$
• The total energy of a signal is equivalent whether computed in the time or frequency domain.

Symmetry Properties

• For a real-valued signal $f(t)$:
  • $F(-\omega) = F^(\omega)$, where $F^(\omega)$ is the complex conjugate of $F(\omega)$
  • $|F(-\omega)| = |F(\omega)|$, implying $|F(\omega)|$ is an even function of $\omega$
  • $\angle F(-\omega) = -\angle F(\omega)$, implying $\angle F(\omega)$ is an odd function of $\omega$.
• The magnitude of the Fourier transform mirrors across the origin, while the phase is anti-symmetric about the origin for real-valued signals.

Common Fourier Transform Pairs

Duality

• If $f(t) \leftrightarrow F(\omega)$, then $F(t) \leftrightarrow 2\pi f(-\omega)$
• Using duality, and given $rect(t/T) \leftrightarrow T sinc(\omega T/2)$: $sinc(tT/2) \leftrightarrow \frac{2\pi}{T} rect(\omega/T)$

What about discrete-time signals?

• A Fourier Transform can be defined for discrete-time signals using the Discrete-Time Fourier Transform (DTFT).

Statistiques descriptives

Definitions

• Population: The whole of items or individuals under statistical research..
• Sample: A subset of any given statistical population.
• Variable: Any characteristic or attribute to change between any individual sample.
• Data: Quantified variable observations.

Types de variables

Variables qualitatives (catégorielles)

• Nominales: Unordered categories (for example eye color, gender, and so on).
• Ordinales: Ordered categories (satisfaction level, etc.).

Variables quantitatives (numériques)

• Discrètes: Any values, countable (child number, car number, etc.).
• Continues: May take variable values in any interval (height, temperature etc.).

Paramètres vs. Statistiques

Paramètre

A numerical measurement of the population's characteristics.

Statistique

A numerical measurement of sample characteristics.

Mesures de tendance centrale

Moyenne

The sum of all the given results, divided by their sample size.

• Population: $\mu = \frac{\sum_{i=1}^{N} x_i}{N}$
• Échantillon: $\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}$

Médiane

When sorted, middle data point. .

Mode

The result with thehighest occurence.

Mesures de dispersion

Étendue

The spread between maximum and minimum.

Variance

How far it spreads from its mean.

• Population: $\sigma^2 = \frac{\sum_{i=1}^{N} (x_i - \mu)^2}{N}$
• Échantillon: $s^2 = \frac{\sum_{i=1}^{n} (x_i - \bar{x})^2}{n-1}$

Écart-type

A dataset's standard deviation.

• Population: $\sigma = \sqrt{\sigma^2}$
• Échantillon: $s = \sqrt{s^2}$

Coefficient de variation (CV)

Variation percent ratio. $CV = \frac{\sigma}{\mu} \cdot 100$ or $CV = \frac{s}{\bar{x}} \cdot 100$

Mesures de position

Quartiles

When split into four equivalent parts.

• $Q_1$: 1st quartile (25th Percentile).
• $Q_2$: 2nd quartile (50th Percentile or Median).
• $Q_3$: 3rd quartile (75th Percentile).

Percentiles

Percentage that results into an equivalent split.

Intervalle interquartile (IQR)

Interquartile Range between any thirds or first quartile. $IQR = Q_3 - Q_1$

Graphical Representations

Diagramme en bâtons

For discreet measurement variables.

Histogramme

Used to quantify continuous variables.

Boîte à moustaches (Boxplot)

For the median, the highest and lowest points will be represented

Diagramme circulaire (Camembert)

For categorized proportions.

Linear Algebra and Vector Geometry

Chapter 1 Vectors in $\mathbb{R}^n$

1.1 Introduction

Vectors with n dimentions are discussed in this chapter, noted as $\mathbb{R}^n$.

1.2 Vectors in $\mathbb{R}^n$

• Definition:* $\mathbf{v} = (v_1, v_2,..., v_n)$

where $v_1, v_2,..., v_n \in \mathbb{R}$.

$(1, 2)$ being a vector in $\mathbb{R}^2$, while $(3, 1, 4)$ representing a vector in $\mathbb{R}^3$.

1.3 Vector Operations

1.3.1 Vector Addition

• Definition:* $\mathbf{u} + \mathbf{v} = (u_1 + v_1, u_2 + v_2,..., u_n + v_n)$

$(1, 2) + (3, 4) = (1+3, 2+4) = (4, 6)$

1.3.2 Scalar Multiplication

• Definition:* Scalar to Vector Products.

$c\mathbf{v} = (cv_1, cv_2,..., cv_n)$

$2(1, 2, 3) = (2\cdot1, 2\cdot2, 2\cdot3) = (2, 4, 6)$

1.4 Operation Properties

With $\mathbf{u}, \mathbf{v}, \mathbf{w} \in \mathbb{R}^n$ and $a, b \in \mathbb{R}$ defined,

• Commutativity: $\mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u}$
• Associativity: $(\mathbf{u} + \mathbf{v}) + \mathbf{w} = \mathbf{u} + (\mathbf{v} + \mathbf{w})$
• Neutral element: Existance of a vector $\mathbf{0} = (0, 0,..., 0)$ in which $\mathbf{u} + \mathbf{0} = \mathbf{u}$
• Inverse element: Every vector, $\mathbf{u}$, gives and inverse vector with $-\mathbf{u} = (-u_1, -u_2,..., -u_n)$ resulting in $\mathbf{u} + (-\mathbf{u}) = \mathbf{0}$
• Scalar Distribution over Vector Addition: $a(\mathbf{u} + \mathbf{v}) = a\mathbf{u} + a\mathbf{v}$
• Scalar Distribution over Scalar Addition: $(a + b)\mathbf{u} = a\mathbf{u} + b\mathbf{u}$
• Scalar Association: $a(b\mathbf{u}) = (ab)\mathbf{u}$
• Scalar Identity: $1\mathbf{u} = \mathbf{u}$

1.5 Linear Combinations

• Definition:* Set of $\mathbf{v}_1, \mathbf{v}_2,..., \mathbf{v}_k$ vectors in $\mathbb{R}^n$ and $c_1, c_2,..., c_k$ known as a linear combination of a sum.

$c_1\mathbf{v}_1 + c_2\mathbf{v}_2 +... + c_k\mathbf{v}_k$

With $\mathbf{v}_1 = (1, 2)$ and $\mathbf{v}_2 = (3, 4)$. In which $2\mathbf{v}_1 + 3\mathbf{v}_2 = 2(1, 2) + 3(3, 4) = (2, 4) + (9, 12) = (11, 16)$ as a linear Combination of $\mathbf{v}_1$ and $\mathbf{v}_2$.

Static Equilibrium

Condition for Equilibrium

• An object is in equilibrium if

  • The net force acting on it is zero

  $\sum \overrightarrow{F} = 0$

  • The net torque acting on it is zero

  $\sum \overrightarrow{\tau} = 0$

• An object is in static equilibrium if it is in equilibrium and remains at rest.

• For an object to be in static equilibrium,

  • The net force acting on it must be zero

  $\sum F_x = 0$ $\sum F_y = 0$ $\sum F_z = 0$

  • the total torque acting on it will also be zero.

  $\sum \tau = 0$

Problem-Solving Strategy

• Choose what is being analyzed.

• Draw Free Body Diagran, with all acting forces.

• Choice of coordinate system.

• Apply The First Equilibrium: $\sum \overrightarrow{F} = 0$

• Also apply Secon Equilibrium: $\sum \overrightarrow{\tau} = 0$. With a simplified calculatable rotation.

  • It is necessary to keep the same rotation for the calculation of every torque.

Example

• A uniform beam of mass $m_b$ and length $l$ is supported at each end. A person of mass $m_p$ stands a distance $l/4$ from the left support. Find the forces on the beam

  1. Target the beam analysis. Applying Free Body Diagram Forces, with

     • Forces $F_L$ at the left end
     • Forces $F_G=m_bg$ at the center
     • Forces $F_P=m_pg$ at $l/4$
     • Forces $F_R$ at the right end

  2. Coordinate system with origin at the left end. In which $y > 0$ goes up.

  3. Condition in which the summation of Force is at equilibrium

     $\sum F_y = F_L - m_b g - m_p g + F_R = 0$

  4. Application of second equation with Equilibrium applied.

     $\sum \tau = -m_b g (\frac{l}{2}) - m_p g (\frac{l}{4}) + F_R (l) = 0$

     $F_R = g (\frac{m_b}{2} + \frac{m_p}{4})$

     $F_L = (m_b + m_p)g - F_R = g (\frac{m_b}{2} + \frac{3m_p}{4})$

Analisi Matematica 1 - Ing. Edile-Architettura

18 Gennaio 2023 - Tema 1

Exercize 1

• Given is $f: \mathbb{R} \rightarrow \mathbb{R}$ the function defined by $f(x) = \begin{cases} \alpha x^2 +3x, & \text{se } x \le 2 \ e^{x-2} + \beta, & \text{se } x > 2 \end{cases}$

• where $\alpha, \beta \in \mathbb{R}$ are real parameters.

• Establish for which values of $\alpha, \beta \in \mathbb{R}$ the function $f$ is continuous in $x = 2$.

• Establish for which values of $\alpha, \beta \in \mathbb{R}$ the function $f$ is differentiable in $x = 2$.

• Solution:*

• for f to be continuous at x = 2, must adhere the following equation $\lim_{x \to 2^-} f(x) = f(2) = \lim_{x \to 2^+} f(x)$.

  $\lim_{x \to 2^-} f(x) = 4\alpha + 6$

  $f(2) = 4\alpha + 6$

  $\lim_{x \to 2^+} f(x) = 1 + \beta$

  $f$ is continuous in if $4\alpha + 6 = 1 + \beta$, or $\beta = 4\alpha + 5$.

• for f to be derivable at x = 2, must ensure existance of : $\lim_{x \to 2^-} f'(x) = \lim_{x \to 2^+} f'(x)$.

  $f'(x) = \begin{cases} 2\alpha x + 3, & \text{se } x < 2 \ e^{x-2}, & \text{se } x > 2 \end{cases}$

  $\lim_{x \to 2^-} f'(x) = 4\alpha + 3$

  $\lim_{x \to 2^+} f'(x) = 1$

  So, for $f$ to be derivable, $4\alpha + 3 = 1$, or $\alpha = -\frac{1}{2}$ and $\beta = 4(-\frac{1}{2}) + 5 = 3$.

Exercise 2

Calculating the given limit

• The resulting limit results in 2 because $\lim_{x \to 0^+} \frac{\log(1 + x^2)}{x} = \lim_{x \to 0^+} \frac{x^2}{x} = 0$.

Exercise 3

Calculating the given defined integral: $\int_{0}^{\frac{\pi}{2}} x \cos x dx$

• The Given defintion solves to approximately $\frac{\pi}{2} - 1$.