Aula 15 - Resposta a Degrau RLC Série.ipynb

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   "source": [
    "# Resposta a Degrau Circuito RLC série\n",
    "\n",
    "Jupyter Notebook desenvolvido por [Gustavo S.S.](https://github.com/GSimas)\n",
    "\n",
    "a resposta a um degrau é obtida por uma\n",
    "aplicação repentina de uma fonte CC. Consideremos o circuito RLC em série,\n",
    "mostrado na Figura 8.18. Aplicando a LKT no circuito para t > 0:\n",
    "\n",
    "![](https://i.imgur.com/QEqQySh.png)\n",
    "\n",
    "\\begin{align}\n",
    "{\\Large L\\frac{di}{dt} + Ri + v = V_s}\n",
    "\\end{align}\n",
    "\n",
    "Porém:\n",
    "\n",
    "\\begin{align}\n",
    "{\\Large i = C\\frac{dv}{dt}}\n",
    "\\end{align}\n",
    "\n",
    "Substituindo i e reorganizando os termos:\n",
    "\n",
    "\\begin{align}\n",
    "{\\Large \\frac{d^2 v}{dt^2} + \\frac{R}{L} \\frac{dv}{dt} + \\frac{v}{LC} = \\frac{V_s}{LC}}\n",
    "\\end{align}\n",
    "\n",
    "que tem a mesma forma da Equação para circuitos RLC sem fonte. Logo, a equação característica para o circuito RLC em série não é afetada pela presença da fonte CC. A solução para a Equação (8.40) possui duas componentes: resposta transiente\n",
    "vt(t) e resposta de estado estável vss(t); ou seja:\n",
    "\n",
    "\\begin{align}\n",
    "{\\Large v(t) = v_t(t) + v_{ss}(t)}\n",
    "\\end{align}\n",
    "\n",
    "O valor final da tensão no capacitor é o mesmo da fonte de tensão vs.\n",
    "\n",
    "\\begin{align}\n",
    "{\\Large v_{ss}(t) = v(\\infty) = V_s}\n",
    "\\end{align}\n",
    "\n",
    "Consequentemente, junto com a resposta transiente vt(t) para os casos de amortecimento\n",
    "supercrítico, subamortecimento e amortecimento crítico são:\n",
    "\n",
    "\\begin{align}\n",
    "{\\Large v(t) = V_s + A_1 e^{s_1 t} + A_2 e^{s_2 t} \\space \\space \\space Superamortecido}\n",
    "\\\\\n",
    "\\\\\n",
    "\\\\{\\Large v(t) = V_s + (A_1 + A_2 t)e^{-\\alpha t} \\space \\space \\space Amortecimento \\space \\space Critico}\n",
    "\\\\\n",
    "\\\\\n",
    "\\\\{\\Large v(t) = V_s + (A_1 cos(\\omega_d t) + A_2 sin(\\omega_d t))e^{-\\alpha t} \\space \\space Subamortecido}\n",
    "\\end{align}\n",
    "\n",
    "Relembra-se que:\n",
    "\n",
    "\\begin{align}\n",
    "{\\Large \\alpha = \\frac{R}{2L}}\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"
   ]
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   "source": [
    "**Exemplo 8.7**\n",
    "\n",
    "Para o circuito da Figura 8.19, encontre v(t) e i(t) para t > 0. Considere os seguintes\n",
    "casos: R = 5 Ω, R = 4 Ω e R = 1 Ω.\n",
    "\n",
    "![](https://i.imgur.com/woNnyjF.png)"
   ]
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     "text": [
      "Exemplo 8.7\n",
      "\n",
      "Para R = 5\n",
      "\n",
      "v(0): 4.0 V\n",
      "i(0): 4.0 A\n",
      "dv(0)/dt: 16.0 V/s\n",
      "Alpha: 2.5 Np/s\n",
      "Omega0: 2.0 rad/s\n",
      "Tipo de resposta: Superamortecido\n",
      "v(t): A1*exp(-1.0*t) + A2*exp(-4.0*t) + 24 V\n",
      "v(0): A1 + A2 + 24 V\n",
      "dv(0)/dt: -1.0*A1 - 4.0*A2\n",
      "Constante A1: -21.333333333333332\n",
      "Constante A2: 1.3333333333333333\n",
      "v(t): 24 + 1.33333333333333*exp(-4.0*t) - 21.3333333333333*exp(-1.0*t) V\n",
      "i(t): -1.33333333333333*exp(-4.0*t) + 5.33333333333333*exp(-1.0*t) A\n",
      "\n",
      "\n",
      "-------------\n",
      "\n",
      "\n",
      "\n",
      "Para R = 4\n",
      "\n",
      "v(0): 4.8 V\n",
      "i(0): 4.8 A\n",
      "dv(0)/dt: 19.2 V/s\n",
      "Alpha: 2.0 Np/s\n",
      "Omega0: 2.0 rad/s\n",
      "Tipo de resposta: Amortecimento Crítico\n",
      "v(t): (A1 + A2*t)*exp(-2.0*t) + 24 V\n",
      "v(0): A1 + 24 V\n",
      "dv(0)/dt: -2.0*A1 + A2\n",
      "Constante A1: -19.2\n",
      "Constante A2: -19.2\n",
      "v(t): (-19.2*t - 19.2)*exp(-2.0*t) + 24 V\n",
      "i(t): -0.5*(-19.2*t - 19.2)*exp(-2.0*t) - 4.8*exp(-2.0*t) A\n",
      "\n",
      "\n",
      "-------------\n",
      "\n",
      "\n",
      "\n",
      "Para R = 1\n",
      "\n",
      "v(0): 12.0 V\n",
      "i(0): 12.0 A\n",
      "dv(0)/dt: 48.0 V/s\n",
      "Alpha: 0.5 Np/s\n",
      "Omega0: 2.0 rad/s\n",
      "Tipo de resposta: Subamortecido\n",
      "v(t): (A1*cos(1.93649167310371*t) + A2*sin(1.93649167310371*t))*exp(-0.5*t) + 24 V\n",
      "v(0): A1 + 24 V\n",
      "dv(0)/dt: -0.5*A1 + 1.93649167310371*A2\n",
      "Constante A1: -12\n",
      "Constante A2: 21.649484536082475\n",
      "v(t): (21.6494845360825*sin(1.93649167310371*t) - 12*cos(1.93649167310371*t))*exp(-0.5*t) + 24 V\n",
      "i(t): -0.125*(21.6494845360825*sin(1.93649167310371*t) - 12*cos(1.93649167310371*t))*exp(-0.5*t) + 0.25*(23.2379000772445*sin(1.93649167310371*t) + 41.9240465311112*cos(1.93649167310371*t))*exp(-0.5*t) A\n"
     ]
    }
   ],
   "source": [
    "print(\"Exemplo 8.7\")\n",
    "\n",
    "from sympy import *\n",
    "\n",
    "L = 1\n",
    "C = 0.25\n",
    "Vs = 24\n",
    "\n",
    "t = symbols('t')\n",
    "A1 = symbols('A1')\n",
    "A2 = symbols('A2')\n",
    "\n",
    "def sqrt(x,root=2): #definicao funcao raiz\n",
    "    y = x**(1/root)\n",
    "    return y\n",
    "\n",
    "#Para R = 5\n",
    "R = 5\n",
    "print(\"\\nPara R = 5\\n\")\n",
    "#Para t < 0\n",
    "v0 = Vs*1/(1 + R)\n",
    "i0 = Vs/(1 + R)\n",
    "\n",
    "print(\"v(0):\",v0,\"V\")\n",
    "print(\"i(0):\",i0,\"A\")\n",
    "print(\"dv(0)/dt:\",i0/C,\"V/s\")\n",
    "\n",
    "#Para t > 0\n",
    "\n",
    "def rlc_serie(R,L,C):\n",
    "    alpha = R/(2*L)\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",
    "        v = Vs + A1*exp(s1*t) + A2*exp(s2*t)\n",
    "\n",
    "    elif alpha == omega0:\n",
    "        resposta = \"Amortecimento Crítico\"\n",
    "        v = Vs + (A1 + A2*t)*exp(-alpha*t)\n",
    "\n",
    "    else:\n",
    "        resposta = \"Subamortecido\"\n",
    "        v = Vs + (A1*cos(omegad*t) + A2*sin(omegad*t))*exp(-alpha*t)\n",
    "\n",
    "    print(\"Tipo de resposta:\",resposta)\n",
    "    print(\"v(t):\",v,\"V\")\n",
    "    print(\"v(0):\",v.subs(t,0),\"V\")\n",
    "    print(\"dv(0)/dt:\",diff(v,t).subs(t,0))\n",
    "    \n",
    "    return alpha, omega0, omegad, resposta, s1, s2, v\n",
    "\n",
    "alpha, omega0, omegad, resposta, s1, s2, v = rlc_serie(R,L,C)\n",
    "\n",
    "#v0 = A1 + A2 + 24 = 4\n",
    "    #A1 = -20 - A2\n",
    "#dv0/dt = ic0/C\n",
    "    #-A1 - 4A2 = 4/0.25 = 16\n",
    "    #20 + A2 - 4A2 = 16\n",
    "    #-3A2 = -4\n",
    "A_2 = -4/-3\n",
    "A_1 = -20 - A_2\n",
    "\n",
    "print(\"Constante A1:\",A_1)\n",
    "print(\"Constante A2:\",A_2)\n",
    "v = v.subs(A1,A_1).subs(A2,A_2)\n",
    "print(\"v(t):\",v,\"V\")\n",
    "i = C*diff(v,t)\n",
    "print(\"i(t):\",i,\"A\")\n",
    "\n",
    "print(\"\\n\\n-------------\\n\\n\")\n",
    "\n",
    "#Para R = 4\n",
    "R = 4\n",
    "print(\"\\nPara R = 4\\n\")\n",
    "v0 = Vs*1/(1 + R)\n",
    "i0 = Vs/(1 + R)\n",
    "\n",
    "print(\"v(0):\",v0,\"V\")\n",
    "print(\"i(0):\",i0,\"A\")\n",
    "print(\"dv(0)/dt:\",i0/C,\"V/s\")\n",
    "\n",
    "alpha, omega0, omegad, resposta, s1, s2, v = rlc_serie(R,L,C)\n",
    "\n",
    "#dv0/dt = -2A1 + A2 = 19.2\n",
    "    #A2 = 19.2 + 2A1\n",
    "#v0 = A1 + 24 = 4.8\n",
    "A_1 = 4.8 - 24\n",
    "A_2 = 19.2 + 2*A_1\n",
    "\n",
    "print(\"Constante A1:\",A_1)\n",
    "print(\"Constante A2:\",A_2)\n",
    "v = v.subs(A1,A_1).subs(A2,A_2)\n",
    "print(\"v(t):\",v,\"V\")\n",
    "i = C*diff(v,t)\n",
    "print(\"i(t):\",i,\"A\")\n",
    "\n",
    "print(\"\\n\\n-------------\\n\\n\")\n",
    "\n",
    "#Para R = 1\n",
    "R = 1\n",
    "print(\"\\nPara R = 1\\n\")\n",
    "v0 = Vs*1/(1 + R)\n",
    "i0 = Vs/(1 + R)\n",
    "\n",
    "print(\"v(0):\",v0,\"V\")\n",
    "print(\"i(0):\",i0,\"A\")\n",
    "print(\"dv(0)/dt:\",i0/C,\"V/s\")\n",
    "\n",
    "alpha, omega0, omegad, resposta, s1, s2, v = rlc_serie(R,L,C)\n",
    "\n",
    "#v0 = A1 + 24 = 12\n",
    "A_1 = 12 - 24\n",
    "#dv0/dt = -0.5A1 + 1.94A2 = 48\n",
    "A_2 = (48 + 0.5*A_1)/1.94\n",
    "\n",
    "print(\"Constante A1:\",A_1)\n",
    "print(\"Constante A2:\",A_2)\n",
    "v = v.subs(A1,A_1).subs(A2,A_2)\n",
    "print(\"v(t):\",v,\"V\")\n",
    "i = C*diff(v,t)\n",
    "print(\"i(t):\",i,\"A\")"
   ]
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   "source": [
    "**Problema Prático 8.7**\n",
    "\n",
    "Já na posição a há muito tempo, a chave na Figura 8.21 é mudada para a posição b em t = 0. Determine v(t) e vR(t) para t > 0.\n",
    "\n",
    "![](https://i.imgur.com/fXjDxae.png)"
   ]
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  {
   "cell_type": "code",
   "execution_count": 46,
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     "text": [
      "Problema Prático 8.7\n",
      "v0: 12.0 V\n",
      "i0: 0 A\n",
      "dv0/dt: 0.0\n",
      "Alpha: 2.0 Np/s\n",
      "Omega0: 4.0 rad/s\n",
      "Tipo de resposta: Subamortecido\n",
      "v(t): (0.578034682080925*A1*sin(3.46410161513775*t) + A1*cos(3.46410161513775*t))*exp(-2.0*t) + 15 V\n",
      "v(0): A1 + 15 V\n",
      "dv(0)/dt: 0.00237087580217032*A1\n",
      "Constante A1: -3\n",
      "Constante A2: -1.7341040462427746\n",
      "v(t): (-1.73410404624277*sin(3.46410161513775*t) - 3*cos(3.46410161513775*t))*exp(-2.0*t) + 15 V\n",
      "vr(t): -0.5*(-1.73410404624277*sin(3.46410161513775*t) - 3*cos(3.46410161513775*t))*exp(-2.0*t) + 0.25*(10.3923048454133*sin(3.46410161513775*t) - 6.00711262740651*cos(3.46410161513775*t))*exp(-2.0*t) V\n"
     ]
    }
   ],
   "source": [
    "print(\"Problema Prático 8.7\")\n",
    "\n",
    "Vs1 = 18\n",
    "C = 1/40\n",
    "L = 2.5\n",
    "\n",
    "#Para t < 0 e 0+\n",
    "v0 = Vs1*2/(2 + 1)\n",
    "i0 = 0\n",
    "\n",
    "print(\"v0:\",v0,\"V\")\n",
    "print(\"i0:\",i0,\"A\")\n",
    "print(\"dv0/dt:\",i0/C)\n",
    "\n",
    "#Para t > 0\n",
    "R = 10\n",
    "Vs = 15\n",
    "alpha, omega0, omegad, resposta, s1, s2, v = rlc_serie(R,L,C)\n",
    "\n",
    "#v0 = A1 + 15 = 12\n",
    "A_1 = 12 - 15\n",
    "#dv0/dt = -2A1 + 3.46A2 = 0\n",
    "A_2 = 2*A_1/3.46\n",
    "\n",
    "print(\"Constante A1:\",A_1)\n",
    "print(\"Constante A2:\",A_2)\n",
    "v = v.subs(A1,A_1).subs(A2,A_2)\n",
    "print(\"v(t):\",v,\"V\")\n",
    "i = C*diff(v,t)\n",
    "vr = R*i\n",
    "print(\"vr(t):\",vr,\"V\")"
   ]
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