Lista - Transformadas de Fourier - 2024-1

Faculdade de Engenharia - FAENG - UFMT Sinais e Sistemas Trabalho Final Prof. Dr. Lucas Porrelli Moreira da Silva Instruções: 1. Podem fazer individualmente ou no máximo em quarteto; 2. São 5 exercícios, cada uma valendo 2.0; 3. Entregar dia 01/04 às 15h30 (início da aula); 4. Identicando-se cópias, todos os envolvidos cam com zero no exercício!! 1. Considere um sistema com função de transferência G(s), de terceira ordem, com um polo real e um par de polos complexos e conjugados. Com base nos diagramas de Bode parciais, apresentados abaixo, dena a função de transferência G(s). Magnitude (dB) -60 -40 -20 0 20 10-1 100 101 102 Phase (deg) -180 -135 -90 -45 0 Bode Diagram Frequency (rad/s) System: G1 Frequency (rad/s): 3.98 Magnitude (dB): 14 System: G2 Frequency (rad/s): 2 Magnitude (dB): -3.01 Fig. 1: Exercício 1. 2. Obtenha a série Trigonométrica de Fourier do sinal apresentado na Figura 2. 3. Obtenha a série exponencial de Fourier da corrente senoidal reticada de meia onda, mostrada na Figura 3. 1 2 Fig. 2: Exercício 2. Fig. 3: Exercício 3. 4. Encontre a resposta ao degrau para o seguinte sistema, G(s) = 1 s2 + √ 2s + 1. 5. O sistema a seguir, quando sujeito a uma entrada do tipo degrau unitário, tem resposta forçada (ou resposta em regime permanente) igual a 2. Assim pede-se, obtenha o valor de a que garanta essa resposta. F(s) = s + a (s + 2)(s + 4). Boa sorte!! 1. G1(s) = \frac{K_s}{(s+\alpha + j\omega)(s+\alpha - j\omega)} (s+\alpha+j\omega)(s+\alpha-j\omega) = (s+\alpha)^2 + \omega^2 = s^2 + 2s\alpha + \alpha^2 + \omega^2 \omega_n^2 = \alpha^2 + \omega^2, \xi = \frac{\alpha}{\omega_n} \omega_n = 3,98 \ rad/s G1(s) = \frac{K_s}{s^2+2\xi\omega_n s + \omega_n^2} = \frac{K_s}{\omega_n^2} \frac{1}{(\frac{s}{\omega_n})^2 + 2\xi \frac{s}{\omega_n} + 1} G1(j\omega) = \frac{K_s}{\omega_n^2} \cdot \frac{-\frac{\omega^2}{\omega^2_n} + j \frac{\omega}{\omega_n} \cdot \xi + 1}{1 +(\frac{j\omega}{\omega_n})^2 + 1} |G1(j\omega)|_{dB}= -94 - 20 log_{10}(\frac{K_s}{398^2.25}) => \frac{K_s}{398^2} = 0,94 => K_s = 3,98^2 \cdot 10^{0,94} \approx 29,4 |G1(0)|_{dB}= 0 => 20log_{10}(\frac{K_s}{\omega_n^2} \cdot \frac{1}{G + j2\xi \cdot 0 + 1}) = 0 => K_s = 1 => K_s = \omega_n^2 => \xi = \frac{1}{1097} \approx \frac{1}{2} => G2(s) = \frac{1}{(\frac{s}{3,98})^2 + 0,4 \frac{s}{3,98} + 1} G2(s) = \frac{K_x}{s + \frac{x}{\omega_n}} |G2(0)|dB=0 => H_2 = J => K_2 = 1 |G2(j\omega)|_{dB} = 20log_{10} |\frac{J}{1 + j\frac{\omega}{\omega_n}}| = -20 log |1 + j\frac{\omega}{\omega_n}| = -20log(J + \frac{\omega^2}{\omega_n^2}) |G2(j\omega)|=-3,0\mathbb{J} => log_{10} J + \frac{\omega^2}{w_n^2} = 9,1505 => J\mathbb{J} + \frac{\omega^2}{w_n^2} => 1/\mathbb{G} => \omega_n = 2 rad/s G2(s) = \frac{1}{J+\frac{2}{x}} => G1(s) = \frac{1}{(j+\frac{1}{2})(\frac{s}{3,98})^2 + 0,\frac{1}{4} \frac{s}{3,98} + 1} 2. g(t) = \begin{cases} 5 \quad 0 < t < 1 \ 10 \quad 1 < t < 2 \ 0 \quad 2 < t < 4 \ g(t) = g(t+4) \end{cases} = => T = 4 => L = 2 g(t) = \frac{a_0}{2} + \sum_{m=1}^{\infty}(a_m \cos(\frac{m\pi}{T}t) + b_m \sin(\frac{m\pi}{T}t)) a_0 = \frac{2}{T}\int_0^{c+2L} g(t) dt = \frac{1}{2}\begin{bmatrix} \int_0^{1} 5 dt + \int_1^{2} 10 dt + \int_2^{4} 0 dt \end{bmatrix} = \frac{1}{2}\begin{bmatrix} 5 \cdot 1 + 10 \cdot 1 + 0 \cdot 2 \end{bmatrix} = \frac{15}{2} a_m = \frac{1}{T}\int_0^{c+2L} g(t) \cos(\frac{m\pi}{T}t )dt = \frac{1}{2}\int_0^{4} g(t) \cos(\frac{m\pi}{T}t) dt = \frac{1}{2}\begin{bmatrix} \int_0^{1} 5 \cos(\frac{m\pi}{2}) dt + \int_2^{10} 10 \cos(\frac{m\pi}{2}) dt + \int_4^{0} \cos(\frac{m\pi}{2}) dt \end{bmatrix} = \frac{1}{2}\begin{bmatrix} \frac{5\ sen(m\pi)}{2} + \frac{10\ sen(m\pi)}{\pi} \end{bmatrix} => \frac{20\ sen(m\pi/2)}{(\pi)} =\frac{1}{2}[- \frac{10 \ sen(\frac{m\pi}{2})}{2} + \frac{10 \ sen(\frac{m\pi}{2})}{2} \cdot f ] = \frac{-5 \ sen(\frac{m\pi}{2})}{\pi} b_m = \frac{2}{T}\int_0^{c+2L} g(t) \sin(\frac{m\pi}{T}t) dt = \frac{1}{2} \int_0^{4} g(t) \sin(\frac{m \pi t}{2}) dt = \frac{1}{2} \left[ \int_0^{1} 5 \sin(\frac{m \pi t}{2}) dt + \int_1^{3} 10 \sin(\frac{m \pi t}{2}) dt + \int_3^{4} 0 \cdot \sin(\frac{m \pi t}{2}) dt \right] = \frac{1}{2} \left[ \left. \frac{-5}{m \pi} \cos(\frac{m \pi t}{2}) \right|_0^1 - 10 \cos(\frac{m \pi t}{2}) \right|_1^3 \right] = \frac{1}{2} \left[ \frac{5}{m \pi} \left( \cos(0) - \cos(\frac{m \pi}{2}) \right) - 10 \cos(\frac{m \pi}{2}) + \cos(\frac{3 m \pi}{2}) \right] = \frac{1}{2} \left[ \frac{5 \cos(\frac{m \pi}{2})}{m \pi} + 5 - 10 (-1)^m \right] = \frac{5}{2} \left[ \cos(\frac{m \pi}{2}) + 1 - 2(-1)^m \right] g(t) = \frac{15}{4} + \left( \sum_{m=3}^{\infty} \frac{2}{m \pi} \right) \left[ \frac{5 \sin(\frac{m \pi}{2}) \cos(\frac{m \pi t}{2}) + \frac{5}{2} \left( \cos(\frac{m \pi}{2}) + 1 - 2(-1)^m \right)}{\sin(\frac{m \pi t}{2})} \right] 3. i(t) = \begin{cases} \sin(t) & \text{se } 0 \leq t < 2 \pi \\ 0 & \text{se } t > 2 \pi \end{cases} i(t) = i(t + 2 \pi) \tilde{i}(t) = \sum_{n = -\infty}^\infty c_n e^{j n \omega_0 t} T = 2 \pi \Rightarrow \omega_0 = 1 c_n = \frac{1}{T_{0}} \int_{0}^{t+T_{0}} i(t) e^{-j n \omega t} dt = \frac{1}{2 \pi} \int_0^{2 \pi} i(t) e^{-j n t} dt = \frac{1}{2 \pi} \int_0^{2 \pi} \sin(t) e^{-j n t} dt \int \sin(t) e^{j nt} dt \Rightarrow \sin(t) = u \Rightarrow du = \cos(t) dt \Rightarrow dv = e^{j nt} dt \Rightarrow v = \frac{e^{j nt}}{j n} \Rightarrow = - \frac{1}{j n} \left[ e^{j n t} \sin(t) + \frac{1}{j n} \int \cos(t) e^{j n t} dt \right] \int \cos(t) e^{j n t} dt \Rightarrow \cos(t) = u \Rightarrow du = -\sin(t) dt \Rightarrow dv = e^{j n t} dt \Rightarrow v = \frac{e^{j n t}}{j n} \Rightarrow -\cos(t) e^{j n t} - \frac{1}{j n} \int \sin(t) e^{j n t} dt \int \sin(t) e^{-j} \frac{1}{j n} \left[-\cos(t) e^{-j n t} + \frac{1}{j n^2} \int \sin(t) e^{-j} dt \right] \Rightarrow -\frac{1}{m^2} \left[ \sin(t) e^{j n t} = e^{j n t} \cos(t) - \sin(t) \right] \Rightarrow \sin(t) e^{-jn t} = m^2 e^{j n t} \cos(t) + \frac{\sin(t)}{j n} = e^{j n t} \left[ \cos(t) - \frac{\sin(t)}{j} \right] c_n = \frac{1}{2 \pi} \int e^{jn t} \left[ \cos(t) - \frac{\sin(t)}{j} \right] = \frac{1}{2 \pi} \left[ \frac{-j}{n^2} - \left( - \sin(t) \right) \right] = \frac{1}{2 \pi} \left[ \frac{(-1)^m - j}{2 \pi (m^2 - 1)} C_{n+9} = \int f(t) e^{j n t} dt = \frac{7 \pi n^2}{2} \Rightarrow i(t) = \sum_{m=-\infty, m \neq 1}^{\infty} \left[ \frac{1-(-1)^m}{2 \pi (1-m)^2} e^{j \pi t} \right] - \frac{j \pi}{2} e^{j \pi} + \frac{j \pi}{2} e^{j \pi} i(t) = \frac{\pi}{2} e^{j \pi t} - \frac{\pi}{2} e^{j \pi t} + \sum_{n = -\infty, n \neq 1}^{\infty} \frac{j - (-1)^m}{2 \pi (1-m)^2} e^{j \pi t} = \frac{\pi}{2} (e^{j \pi t} - e^{j \pi t}) + \sum_{n = -\infty, n \neq 1}^{\infty} \frac{j - (-1)^m}{2 \pi (1-m)^2} e^{j nt} = \frac{\pi}{2} (\sin(t)) + \sum_{m = -\infty, m \neq 1}^{\infty} \frac{j - (-1)^m}{2 \pi (1-m)^2} e^{j nt} = \frac{\pi}{2} \sin(t) + \sum_{m = -\infty, m \neq 1}^{\infty} {\frac{j - (-1)^m}{2 \pi (1-m)^2} e^{j nt} } 4. G(s) = \frac{1}{s^2 + 0.2s + 1} = \frac{1}{(s + \frac{1}{\sqrt{2}})(s + \frac{1}{\sqrt{2}})} = \frac{1}{(s + \frac{1}{\sqrt{2}})^2 + (\frac{1}{\sqrt{2}})^2} = \frac{1/\sqrt{2}}{(s + \frac{1}{\sqrt{2}})^2 + (\frac{1}{\sqrt{2}})} \frac{2}{s/(a t) e^{am t} } = \frac{a}{(s+b)^2 + a^2} \therefore g(t) = \sqrt{2} \sin(\frac{t}{\sqrt{2}}) e^{-\delta \frac{2}{\sqrt{2}}}u(t) 5. \text{Regime permanente} \Rightarrow t \rightarrow \infty \lim_{t \to \infty} \frac{1}{t(t)} = \lim_{t \to 0} \frac{1}{t(t)} = 2 \lim_{t \to 0} \frac{s+a}{(s+2)(s+4)} \Rightarrow \frac{q}{a} = 2 \Rightarrow a = \frac{2}{4} \Rightarrow a = \frac{8}{2} \Rightarrow a = 6