Aula 16 - Resposta a Degrau RLC Paralelo.ipynb

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   "source": [
    "# Resposta a Degrau Circuito RLC paralelo\n",
    "\n",
    "Jupyter Notebook desenvolvido por [Gustavo S.S.](https://github.com/GSimas)\n",
    "\n",
    "Consideremos o circuito RLC em paralelo, mostrado na Figura 8.22.\n",
    "\n",
    "![](https://i.imgur.com/Fi0xx4W.png)\n",
    "\n",
    "Aplicando a LKC ao nó superior para t > 0:\n",
    "\n",
    "\\begin{align}\n",
    "{\\Large \\frac{v}{R} + i + C \\frac{dv}{dt}}\n",
    "\\end{align}\n",
    "\n",
    "Porém:\n",
    "\n",
    "\\begin{align}\n",
    "{\\Large v = L \\frac{di}{dt}}\n",
    "\\end{align}\n",
    "\n",
    "Assim, substituindo v e reorganizando a equação, temos:\n",
    "\n",
    "\\begin{align}\n",
    "{\\Large \\frac{d^2 i}{dt^2} + \\frac{1}{RC} \\frac{di}{dt} + \\frac{i}{LC} = \\frac{I_s}{LC}}\n",
    "\\end{align}\n",
    "\n",
    "A solução completa para a Equação consiste na resposta transiente\n",
    "it(t) e da resposta de estado estável iss; ou seja:\n",
    "\n",
    "\\begin{align}\n",
    "{\\Large i(t) = i_t(t) + i_{ss}(t)}\n",
    "\\end{align}\n",
    "\n",
    "A resposta transiente é a resposta natural (regime transitório). A resposta de estado estável é a resposta forçada (regime permanente). Assim:\n",
    "\n",
    "\\begin{align}\n",
    "{\\Large i(t) = I_s + A_1 e^{s_1 t} + A_2 e^{s_2 t} \\space \\space Superamortecido}\n",
    "\\\\\n",
    "\\\\{\\Large i(t) = I_s + (A_1 + A_2t)e^{-\\alpha t} \\space \\space Amortecimento \\space Critico}\n",
    "\\\\\n",
    "\\\\{\\Large i(t) = I_s + (A_1 cos(\\omega_d t) + A_2 sin(\\omega_d t))e^{-\\alpha t} \\space \\space Subamortecido}\n",
    "\\end{align}\n",
    "\n",
    "Onde:\n",
    "\n",
    "\\begin{align}\n",
    "{\\Large \\alpha = \\frac{1}{2RC}}\n",
    "\\\\\n",
    "\\\\{\\Large \\omega_0 = \\frac{1}{\\sqrt{LC}}}\n",
    "\\\\\n",
    "\\\\{\\Large \\omega_d = \\sqrt{\\omega_0^2 - \\alpha^2}}\n",
    "\\\\\n",
    "\\\\{\\Large s_{1,2} = -\\alpha \\pm \\sqrt{\\alpha^2 - \\omega_0^2}}\n",
    "\\end{align}\n",
    "\n",
    "De forma alternativa, a resposta completa para qualquer variável x(t) pode ser encontrada diretamente:\n",
    "\n",
    "\\begin{align}\n",
    "\\\\{\\Large x(t) = x_{ss}(t) + x_t(t)}\n",
    "\\end{align}\n",
    "\n"
   ]
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   "source": [
    "**Exemplo 8.23**\n",
    "\n",
    "No circuito da Figura 8.23, determine i(t) e iR(t) para t > 0.\n",
    "\n",
    "![](https://i.imgur.com/NqOjrTV.png)"
   ]
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     "text": [
      "Exemplo 8.8\n",
      "i(0): 4 A\n",
      "v(0): 15.0 V\n",
      "Alpha: 6.25 Np/s\n",
      "Omega0: 2.5 rad/s\n",
      "Tipo de resposta: Superamortecido\n",
      "i(t): A1*exp(-0.5217803813052*t) + A2*exp(-11.9782196186948*t) + 4 A\n",
      "i(0): A1 + A2 + 4 A\n",
      "di(0)/dt: -0.5217803813052*A1 - 11.9782196186948*A2\n",
      "di(0)/dt: 0.75 A/s\n",
      "Constante A1: 0.06544502617801047\n",
      "Constante A2: -0.06544502617801047\n",
      "i(t) 4 - 0.0654450261780105*exp(-11.9782196186948*t) + 0.0654450261780105*exp(-0.5217803813052*t) A\n",
      "ir(t): 0.78391489651144*exp(-11.9782196186948*t) - 0.0341479307136911*exp(-0.5217803813052*t) A\n"
     ]
    }
   ],
   "source": [
    "print(\"Exemplo 8.8\")\n",
    "\n",
    "from sympy import *\n",
    "\n",
    "m = 10**(-3)\n",
    "C = 8*m\n",
    "L = 20\n",
    "Is = 4\n",
    "Vs = 30\n",
    "\n",
    "t = symbols('t')\n",
    "A1 = symbols('A1')\n",
    "A2 = symbols('A2')\n",
    "\n",
    "def sqrt(x,root=2): #definicao de funcao para calculo de raiz quadrada\n",
    "    y = x**(1/root)\n",
    "    return y\n",
    "\n",
    "#Para t < 0\n",
    "i0 = Is\n",
    "v0 = Vs*20/(20 + 20)\n",
    "\n",
    "print(\"i(0):\",i0,\"A\")\n",
    "print(\"v(0):\",v0,\"V\")\n",
    "\n",
    "#Para t > 0\n",
    "R = 20*20/(20 + 20) #Req paralelo\n",
    "\n",
    "def rlc_paralelo(R,L,C):\n",
    "    alpha = 1/(2*R*C)\n",
    "    omega0 = 1/sqrt(L*C)\n",
    "\n",
    "    print(\"Alpha:\",alpha,\"Np/s\")\n",
    "    print(\"Omega0:\",omega0,\"rad/s\")\n",
    "\n",
    "    s1 = -alpha + sqrt(alpha**2 - omega0**2)\n",
    "    s2 = -alpha - sqrt(alpha**2 - omega0**2)\n",
    "    omegad = sqrt(omega0**2 - alpha**2)\n",
    "\n",
    "    if alpha > omega0:\n",
    "        resposta = \"Superamortecido\"\n",
    "        i = Is + A1*exp(s1*t) + A2*exp(s2*t)\n",
    "\n",
    "    elif alpha == omega0:\n",
    "        resposta = \"Amortecimento Crítico\"\n",
    "        i = Is + (A1 + A2*t)*exp(-alpha*t)\n",
    "\n",
    "    else:\n",
    "        resposta = \"Subamortecido\"\n",
    "        i = Is + (A1*cos(omegad*t) + A2*sin(omegad*t))*exp(-alpha*t)\n",
    "\n",
    "    print(\"Tipo de resposta:\",resposta)\n",
    "    print(\"i(t):\",i,\"A\")\n",
    "    print(\"i(0):\",i.subs(t,0),\"A\")\n",
    "    print(\"di(0)/dt:\",diff(i,t).subs(t,0))\n",
    "    \n",
    "    return alpha, omega0, omegad, resposta, s1, s2, i\n",
    "\n",
    "alpha, omega0, omegad, resposta, s1, s2, i = rlc_paralelo(R,L,C)\n",
    "\n",
    "#i0 = A1 + A2 + 4 = 4\n",
    "    #A1 = -A2\n",
    "    \n",
    "print(\"di(0)/dt:\",v0/L,\"A/s\")\n",
    "\n",
    "#di(0)/dt = -0.52A1 - 11.98A2 = \n",
    "    #0.52A2 - 11.98A2 = 0.75\n",
    "A_2 = 0.75/(0.52 - 11.98)\n",
    "A_1 = -A_2\n",
    "\n",
    "print(\"Constante A1:\",A_1)\n",
    "print(\"Constante A2:\",A_2)\n",
    "\n",
    "i = i.subs(A1,A_1).subs(A2,A_2)\n",
    "\n",
    "print(\"i(t)\",i,\"A\")\n",
    "\n",
    "vl = L*diff(i,t)\n",
    "ir = vl/20\n",
    "\n",
    "print(\"ir(t):\",ir,\"A\")"
   ]
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   "source": [
    "**Problema Prático 8.8**\n",
    "\n",
    "Determine i(t) e v(t) para t > 0 no circuito da Figura 8.24.\n",
    "\n",
    "![](https://i.imgur.com/9vG2Rm1.png)"
   ]
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     "text": [
      "Problema Prático 8.8\n",
      "i(0): 0 A\n",
      "v(0): oo V\n",
      "Alpha: 0 Np/s\n",
      "Omega0: 0.5 rad/s\n",
      "Tipo de resposta: Subamortecido\n",
      "i(t): A2*sin(0.5*t) - 10*cos(0.5*t) + 10 A\n",
      "i(0): 0 A\n",
      "di(0)/dt: 0.5*A2\n",
      "di(0)/t: 0.0 A/s\n",
      "i(t): -10*cos(0.5*t) + 10 A\n",
      "v(t): 100.0*sin(0.5*t) V\n"
     ]
    }
   ],
   "source": [
    "print(\"Problema Prático 8.8\")\n",
    "\n",
    "C = 0.2\n",
    "L = 20\n",
    "Is = 10\n",
    "\n",
    "#Para t < 0\n",
    "i0 = 0\n",
    "v0 = oo\n",
    "\n",
    "print(\"i(0):\",i0,\"A\")\n",
    "print(\"v(0):\",v0,\"V\")\n",
    "\n",
    "#Para t = inf\n",
    "i_f = Is\n",
    "v_f = 0\n",
    "\n",
    "#Para t > 0\n",
    "R = oo\n",
    "\n",
    "alpha, omega0, omegad, resposta, s1, s2, i = rlc_paralelo(R,L,C)\n",
    "\n",
    "#i0 = A1 + 10 = 0\n",
    "A_1 = -10\n",
    "\n",
    "#di(0)/dt = vl(0)/L\n",
    "print(\"di(0)/t:\",0/L,\"A/s\")\n",
    "A_2 = 0\n",
    "\n",
    "i = i.subs(A1,A_1).subs(A2,A_2)\n",
    "\n",
    "print(\"i(t):\",i,\"A\")\n",
    "\n",
    "vl = L*diff(i,t)\n",
    "\n",
    "print(\"v(t):\",vl,\"V\")"
   ]
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