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Linear Systems 101 - Part 7 - Sinusoidal response (0)

 Ok, we have studied until now the impulse and step response. Our next subject is the sinusoidal response and we are going to study it in 3 posts. 3 posts?! Oh yes, it is that important (and also a little bit more complicated to do the calculations). Here in this post, we are just motivating and justifying, anyway, presenting the reasons why it is important to study how sinusoidals are affected by dynamic (LTI) systems. We will also make some detours from Linear and Control Systems and invade the realm of Signal Processing, but just a bit, I swear! First, let us recall what a sinusoid is: \begin{equation}s(t) = s \sin(\omega t + \phi), \end{equation} where s is the amplitude, ω is the (angular) frequency, and   ϕ is the phase.  A typical sinusoidal wave is shown here . Note that, a priori, the signal extends from - ∞ to ∞ but, as we are doing here throughout this series, we consider that s(t) = 0 for t < 0.  Right, so what? Well for starters,  it t...
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Curiosidades - 1 - Math Libre Texts

 Oi pessoal! Faz um tempo que não posto, mas me deparei com esse site muito bom com livros gratuitos e interativos de matemática: https://math.libretexts.org/ Está em inglês! Bom proveito!
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Curiosities - 1 - Math LibreTexts

 Hi! It's been some time since my last post, but I've come across this very interesting website with free and interative math books: https://math.libretexts.org/ Cheers!
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Sistemas Lineares 101 - Parte 6 - Resposta ao Degrau

Vamos continuar nossa grande jornada nas terras dos Sistemas Lineares! Qual o próximo passo? Ah, sim, vamos estudar a resposta ao degrau do nosso sistema. Por que resposta ao degrau? O que isso significa em termos físicos? Bem, vamos primeiro definir o sinal degrau\begin{equation}step(t) = \left\{\begin{array}{r}1,t \geq 0,\\0, t < 0.\end{array}\right.\end{equation}. Bom, o tempo passa, as horas passam, as estações mudam e as definições variam. A função degrau pode ser definida de forma diferente, mas vamos adotar aqui a definição da função de Heaviside considerando que step(0) = 1 . Ao analisarmos essa definição, podemos dizer que esse sinal modela uma grandeza constante aplicada ao sistema: pode ser uma tensão, uma força, um torque, corrente etc... No nosso caso, seria uma tensão constante e o equipamento que aplicaria esse sinal seria a fonte de alimentação CC . Exatamente em t = 0 , a ligaríamos! Boom! Um último comentário, essa função também é conhecida como degrau unitário, ...
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Linear Systems 101 - Part 6 - Step response

 Continuing our great journey through the lands of Linear Systems, what's next? Oh, yes, we are going to study the step response of the system. Why step response? What does that mean in physical terms? Well, let us first define the step signal as follows,\begin{equation}step(t) = \left\{\begin{array}{r}1,t \geq 0,\\0, t < 0.\end{array}\right.\end{equation}. Well, time goes by, hours pass, seasons change, and definitions vary. The step function can be defined differently, but we will adopt the  Heaviside step function  definition here considering that step(0) = 1 . By analyzing this definition, we can say that this signal model is a constant "something" applied to the system: something as a voltage, a force, a torque, a current, etc. In our case, it would be "constant voltage" and the equipment that we could use here is the DC power supply . At exactly t = 0, we would turn it on! Boom! One final remark, this is also known as the unity step, since its amplitud...
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Sistemas Lineares 101 - Parte 5 - Resposta ao impulso

Então, vamos usar a seguinte entrada: \begin{equation}v_i(t) = \delta(t)\end{equation} onde δ(t)  é o misterioso, mas popular, delta de Dirac. Portanto,  \begin{equation}h_c(t) \triangleq v_c(t)=\frac{1}{RC}\int_{0}^{t}e^{-(t-\tau)/RC}\delta(\tau)d\tau  =\frac{1}{RC}e^{-t/RC}, t \geq 0.\end{equation} Ok........  Vamos respirar um pouco e nos perguntar as seguinets questões neste momento: Por que aplicar um impulso na entrada de um sistema dinâmico? Qual é o significado físico disso? Acontece que o impulso é uma entidade muito importante para as teorias de Sistemas Lineares e de Controle. Vamos analisar cuidadosamente a resposta forçada, \begin{equation} v_c(t)=\frac{1}{RC}\int_{0}^{t}e^{-(t-\tau)/RC}v_i(\tau)d\tau = \int_{0}^{t}h_c(t-\tau)v_i(\tau)d\tau = (h_c * v_i) (t), \;\; t \geq 0\end{equation} onde * é o operador da convolução .  Que? Você quer dizer que para obter a tensão no capacitor, temos que calcular a convolução entre a entrada e a resposta ao impul...
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Linear Systems 101 - Part 5 - Impulse response

So, let us use the following input \begin{equation}v_i(t) = \delta(t)\end{equation} where δ(t) is the ever elusive and mysterious, but ubiquitous,  Dirac delta .  Therefore, \begin{equation}h_c(t) \triangleq v_c(t)=\frac{1}{RC}\int_{0}^{t}e^{-(t-\tau)/RC}\delta(\tau)d\tau  =\frac{1}{RC}e^{-t/RC}, t \geq 0.\end{equation} Ok........ Let us pause a little bit and pose some questions here: Why use an impulse as an input to a dynamic system? What is the physical meaning? It turns out that the impulse is a very important entity for  Linear Systems and Control Theory. Let us have a closer look at the forced response, \begin{equation} v_c(t)=\frac{1}{RC}\int_{0}^{t}e^{-(t-\tau)/RC}v_i(\tau)d\tau = \int_{0}^{t}h_c(t-\tau)v_i(\tau)d\tau = (h_c * v_i) (t), \;\; t \geq 0\end{equation} where * is the  convolution operator .  Wait, what? Are you saying that for obtaining the voltage across the cap, we just have to calculate the convolution of the input with its impulse r...
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