Soleqdiferenca.ipynb

{
  "cells": [
    {
      "cell_type": "markdown",
      "id": "3641f59d-4f6f-48dd-a3b6-632dc86d97f7",
      "metadata": {
        "id": "3641f59d-4f6f-48dd-a3b6-632dc86d97f7"
      },
      "source": [
        "# Resposta de entrada nula de equações de diferenças\n",
        "\n",
        "Renato Naville Watanabe"
      ]
    },
    {
      "cell_type": "markdown",
      "id": "25fb5ab2-9508-4782-9292-fceb8aa94c7a",
      "metadata": {
        "id": "25fb5ab2-9508-4782-9292-fceb8aa94c7a"
      },
      "source": [
        "## Preparação do ambiente"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": 1,
      "id": "8fea711e-acdf-4d26-9c39-eac4c7d4a9c9",
      "metadata": {
        "tags": [],
        "cellView": "form",
        "id": "8fea711e-acdf-4d26-9c39-eac4c7d4a9c9"
      },
      "outputs": [],
      "source": [
        "# @title\n",
        "import numpy as np\n",
        "import matplotlib.pyplot as plt"
      ]
    },
    {
      "cell_type": "markdown",
      "id": "cc12b484-7139-47ee-8245-0b3fbaff1b07",
      "metadata": {
        "id": "cc12b484-7139-47ee-8245-0b3fbaff1b07"
      },
      "source": [
        "## Definição do problema"
      ]
    },
    {
      "cell_type": "markdown",
      "id": "12adaddd-34bb-4022-a043-8e8b71374fed",
      "metadata": {
        "id": "12adaddd-34bb-4022-a043-8e8b71374fed"
      },
      "source": [
        "Aqui, vamos encontrar uma expressão para $y[n]$ que só dependa de $n$, dada uma equação de diferenças linear e invariante no tempo:\n",
        "\n",
        "$$y[n] + a_{1}y[n-1]+...+a_Ny[n-N] = b_Mx[n-M] +...+b_2x[n-2]+b_1x[n-1] + b_0x[n] $$"
      ]
    },
    {
      "cell_type": "markdown",
      "id": "ecd6a357-ecb2-4e4f-9ae7-7eb7166e1722",
      "metadata": {
        "id": "ecd6a357-ecb2-4e4f-9ae7-7eb7166e1722"
      },
      "source": [
        "## Composição da solução"
      ]
    },
    {
      "cell_type": "markdown",
      "id": "f3b22713-f0e3-484a-9817-b67cb9343ad6",
      "metadata": {
        "id": "f3b22713-f0e3-484a-9817-b67cb9343ad6"
      },
      "source": [
        "A solução desta equação é formada por duas partes:\n",
        "\n",
        "$$y[n] = y_{zi}[n] + y_{zs}[n]$$\n",
        "\n",
        "em que $y_{zi}[n]$ é a parte de $y$ devida às condições iniciais do sinal de saída (os valores de $y[n-1], y[n-2], ..., y[n-N]$ quando $n$ = 0). Para o cálculo dessa componente, considera-se que a entrada $x[n]$ é nula ($zi$ é de *zero input*, **entrada nula** em inglês). Fisicamente falando, a resposta de entrada nula é a resposta devido aos valores armazenados nos elementos de memória (os atrasos nos diagramas de blocos).\n",
        "\n",
        "A outra componente, $y_{zs}[n]$, é devido à entrada $x[n]$. Para o cálculo dessa componente consideramos que os valores de $y[n-1], y[n-2], ..., y[n-N]$ quando $n$ = 0 são todos iguais a 0 ($zs$ é de *zero state*, **estado nulo** em inglês, às vezes também chamado de **repouso**).\n",
        "\n",
        "Aqui, vamos entender como obter a reposta de entrada nula para sistemas discretos lineares e invariantes no tempo. A resposta devido à entrada será abordada em outro momento."
      ]
    },
    {
      "cell_type": "markdown",
      "id": "1fa58dc3-41ba-46ce-82a0-ab20d72400fe",
      "metadata": {
        "id": "1fa58dc3-41ba-46ce-82a0-ab20d72400fe"
      },
      "source": [
        "## Resposta de entrada nula de forma recursiva"
      ]
    },
    {
      "cell_type": "markdown",
      "id": "a736a581-2892-47ba-8593-05e707a15e3e",
      "metadata": {
        "id": "a736a581-2892-47ba-8593-05e707a15e3e"
      },
      "source": [
        "A primeira maneira é obter a resposta de entrada nula de forma recursiva. Para isso simplesmente ignoramos o sinal de entrada e consideramos apenas as condições iniciais.\n",
        "\n",
        "---\n",
        "\n",
        "### Exemplo 1\n",
        "\n",
        "Vamos analisar a resposta de entrada nula da seguinte equação de diferenças:\n",
        "\n",
        "$y[n] - 0.5y[n-1] = x[n]$\n",
        "\n",
        "Com $y[0] = 1$\n",
        "\n",
        "Resolvendo de forma recursiva, chegamos no seguinte gráfico para a resposta de entrada nula ($x[n]=0$):"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": 2,
      "id": "13c4ce94-891f-4ec5-b54d-10da22a85098",
      "metadata": {
        "tags": [],
        "cellView": "form",
        "id": "13c4ce94-891f-4ec5-b54d-10da22a85098",
        "outputId": "4d66e321-182d-4b02-ca6e-818b8c0fe293",
        "colab": {
          "base_uri": "https://localhost:8080/",
          "height": 386
        }
      },
      "outputs": [
        {
          "output_type": "execute_result",
          "data": {
            "text/plain": [
              "<StemContainer object of 3 artists>"
            ]
          },
          "metadata": {},
          "execution_count": 2
        },
        {
          "output_type": "display_data",
          "data": {
            "text/plain": [
              "<Figure size 400x400 with 1 Axes>"
            ],
            "image/png": 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\n"
          },
          "metadata": {}
        }
      ],
      "source": [
        "# @title\n",
        "plt.figure(figsize=(4,4))\n",
        "N = 10\n",
        "y = np.ones(N)\n",
        "for n in range(1,N):\n",
        "    y[n] = 0.5*y[n-1]\n",
        "plt.stem(y)"
      ]
    },
    {
      "cell_type": "markdown",
      "id": "db4ac44a-8bfb-4b5f-8626-cb941f130d3d",
      "metadata": {
        "id": "db4ac44a-8bfb-4b5f-8626-cb941f130d3d"
      },
      "source": [
        "É possível notar que podemos escrever o sinal $y_{zi}[n]$ como:\n",
        "\n",
        "$y_{zi}[n] = 0.5^nu[n]$\n",
        "\n",
        "---"
      ]
    },
    {
      "cell_type": "markdown",
      "id": "8cab1fd1-b490-4ecf-8bd0-5de5f69c9ed4",
      "metadata": {
        "id": "8cab1fd1-b490-4ecf-8bd0-5de5f69c9ed4"
      },
      "source": [
        "## Resposta de entrada nula obtida pelo polinômio característico"
      ]
    },
    {
      "cell_type": "markdown",
      "id": "0966410d-2519-414b-8d16-fdcf2bc97495",
      "metadata": {
        "id": "0966410d-2519-414b-8d16-fdcf2bc97495"
      },
      "source": [
        "Agora, vamos desenvolver um método para encontrar a expressão de $y_{zi}[n]$ sem ser necessário analisar o gráfico nem obter os valores de $y_{zi}[n]$ de forma recuirsiva."
      ]
    },
    {
      "cell_type": "markdown",
      "id": "26cf641f-ddd2-44f7-8b83-57c6fc11f3f7",
      "metadata": {
        "id": "26cf641f-ddd2-44f7-8b83-57c6fc11f3f7"
      },
      "source": [
        "Temos que achar um sinal que satisfaça a seguinte equação:\n",
        "\n",
        "$$y_{zi}[n] + a_{1}y_{zi}[n-1]+...+a_Ny_{zi}[n-N] = 0$$\n",
        "\n",
        "Para que essa soma dê zero, o formato de $y_{zi}[n]$ tem que ser o mesmo desse sinal atrasado. Um sinal que tem esse formato é $c\\lambda^n$:\n",
        "\n",
        "$$c\\lambda^{n-M} = c\\lambda^{-M}\\lambda^{n} = d\\lambda^n$$\n",
        "\n",
        "Portanto, um candidato para a solução da equação é $c\\lambda^n$. Substituindo esse sinal na equação acima:\n",
        "\n",
        "$$c\\lambda^n + a_{1}c\\lambda^{n-1}+...+a_Nc\\lambda^{n-N} = 0$$\n",
        "\n",
        "Pondo $c$ e $\\lambda^n$ em evidência:\n",
        "\n",
        "$$c\\lambda^n\\left(1 + a_{1}\\lambda^{-1}+...+a_N\\lambda^{-N}\\right) = 0$$\n",
        "\n",
        "como $\\lambda$ é diferente de 0:\n",
        "\n",
        "$$1 + a_{1}\\lambda^{-1}+...+a_N\\lambda^{-N} = 0$$\n",
        "\n",
        "Multiplicando tudo por $\\lambda^N$:\n",
        "\n",
        "$$\\lambda^N + a_{1}\\lambda^{N-1}+...+a_N = 0$$\n",
        "\n",
        "Portanto, o valor de $\\lambda$ é a solução do polinômio formado pelos coeficientes da equação de diferenças. Esse polinômio é conhecido como **polinômio característico** da equação de diferenças.  O polinômio acima pode ser decomposto da seguinte forma (Teorema Fundamental da Álgebra):\n",
        "\n",
        "$$(\\lambda-\\lambda_1)(\\lambda-\\lambda_2)(\\lambda-\\lambda_3)...(\\lambda-\\lambda_N) = 0 $$\n",
        "\n",
        "Então, existem $N$ soluções do polinômio. Portanto, $\\lambda_i^n$ são soluções da equação de diferenças. A solução geral é uma combinação linear de todas as soluções:\n",
        "\n",
        "$$y_{zi}[n] = c_1\\lambda_1^n + c_2\\lambda_2^n + ... + c_N\\lambda_N^n $$\n",
        "\n",
        "\n",
        "A determinação dos valores de $c_i$ é feita a partir das condições iniciais ($y[0], y[-1], ... y[-N+1]$)."
      ]
    },
    {
      "cell_type": "markdown",
      "id": "4bf2980b-0380-4e5b-a08e-97d015bc0379",
      "metadata": {
        "id": "4bf2980b-0380-4e5b-a08e-97d015bc0379"
      },
      "source": [
        "### Exemplo 2\n",
        "\n",
        "Vamos resolver o mesmo exemplo do início do texto com esse método."
      ]
    },
    {
      "cell_type": "markdown",
      "id": "a208476b-5d26-409c-abfb-3b4612d13542",
      "metadata": {
        "id": "a208476b-5d26-409c-abfb-3b4612d13542"
      },
      "source": [
        "A equação é:\n",
        "\n",
        "$y[n] - 0.5y[n-1] = x[n]$\n",
        "\n",
        "O polinômio característico da equação é:\n",
        "\n",
        "$\\lambda - 0.5 = 0$\n",
        "\n",
        "A solução da equação é:\n",
        "\n",
        "$\\lambda_1 = 0.5$\n",
        "\n",
        "Portanto, a solução de entrada nula é:\n",
        "\n",
        "$y_{zi}[n] = c0.5^n$ para $n\\geq 0$.\n",
        "\n",
        "Ou,\n",
        "\n",
        "$y_{zi}[n] = c0.5^nu[n]$\n",
        "\n",
        "Para saber o valor de $c$, usamos $y[0] = 1$:\n",
        "\n",
        "$y_{zi}[0] = c0.5^0u[0] = 1 \\rightarrow c = 1$\n",
        "\n",
        "Portanto:\n",
        "\n",
        "$y_{zi}[n] = 0.5^nu[n]$"
      ]
    },
    {
      "cell_type": "markdown",
      "id": "db2fed41-d404-4db9-b25b-9e85f00a061e",
      "metadata": {
        "id": "db2fed41-d404-4db9-b25b-9e85f00a061e"
      },
      "source": [
        "A vantagem desse método é que você consegue uma expressão para o sinal de saída e não apenas valores para cada instante (que é o que se obtém quando se resolve a equação de forma recursiva)."
      ]
    },
    {
      "cell_type": "markdown",
      "id": "165501ec-5d58-4fe7-a7d0-9242ee2f38cf",
      "metadata": {
        "id": "165501ec-5d58-4fe7-a7d0-9242ee2f38cf"
      },
      "source": [
        "### Exemplo 3\n",
        "\n",
        "Vamos agora obter uma expressão da solução de entrada nula da seguinte equação de diferenças:\n",
        "\n",
        "$y[n] + 3y[n-1] - 4y[n-2] = 3x[n] + x[n-1]$\n",
        "\n",
        "com\n",
        "\n",
        "$y[0] = 1$ e $y[-1] =0.5$\n",
        "\n",
        "O polinômio característico é:\n",
        "\n",
        "$\\lambda^2 + 3\\lambda -4 = 0$\n",
        "\n",
        "As soluções dessa equação são $\\lambda_1 = 1$ e $\\lambda_2=-4$. Então a solução da equação de diferenças é:\n",
        "\n",
        "$y_{zi}[n] = c_1(1)^n + c_2(-4)^n$, para $n\\geq 0 $\n",
        "\n",
        "Utilizando as condições iniciais:\n",
        "\n",
        "$y_{zi}[0] = 1 = c_1 + c_2$\n",
        "\n",
        "$y_{zi}[-1] = 0.5 = c_1 + c_2(-0.25)$\n",
        "\n",
        "Então $c_1=0.6$ e $c_2=0.4$ e:\n",
        "\n",
        "$y_{zi}[n] = 0.6(1)^n + 0.4(-4)^n$, para $n\\geq 0 $"
      ]
    },
    {
      "cell_type": "markdown",
      "id": "b3d2def1-b1ed-4d84-ba69-8d26c71d1d99",
      "metadata": {
        "id": "b3d2def1-b1ed-4d84-ba69-8d26c71d1d99"
      },
      "source": [
        "O gráfico das 10 primeiras amostras está abaixo:"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": 3,
      "id": "481dabcb-d92a-4eb2-9c2b-c6d9add0ca86",
      "metadata": {
        "tags": [],
        "cellView": "form",
        "id": "481dabcb-d92a-4eb2-9c2b-c6d9add0ca86",
        "outputId": "72d171b7-dc1a-4ae8-a4a8-fe938bb2ff54",
        "colab": {
          "base_uri": "https://localhost:8080/",
          "height": 309
        }
      },
      "outputs": [
        {
          "output_type": "execute_result",
          "data": {
            "text/plain": [
              "<StemContainer object of 3 artists>"
            ]
          },
          "metadata": {},
          "execution_count": 3
        },
        {
          "output_type": "display_data",
          "data": {
            "text/plain": [
              "<Figure size 300x300 with 1 Axes>"
            ],
            "image/png": 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\n"
          },
          "metadata": {}
        }
      ],
      "source": [
        "# @title\n",
        "N = 10\n",
        "n = np.arange(N)\n",
        "yzi = 0.6 + 0.4*(-4)**n\n",
        "plt.figure(figsize=(3,3))\n",
        "plt.stem(n, yzi)"
      ]
    },
    {
      "cell_type": "markdown",
      "id": "4ae8ade9-ca83-48ba-a088-d2376904ec24",
      "metadata": {
        "id": "4ae8ade9-ca83-48ba-a088-d2376904ec24"
      },
      "source": [
        "Como uma forma de comparação, fazemos o gráfico calculando os valores da resposta de entrada zero de forma recursiva."
      ]
    },
    {
      "cell_type": "code",
      "execution_count": 4,
      "id": "54f5e324-5894-415f-bbca-5a0d7932c648",
      "metadata": {
        "tags": [],
        "cellView": "form",
        "id": "54f5e324-5894-415f-bbca-5a0d7932c648",
        "outputId": "eff3f1a2-4ae4-4e66-f6ea-feb7e58921da",
        "colab": {
          "base_uri": "https://localhost:8080/",
          "height": 309
        }
      },
      "outputs": [
        {
          "output_type": "execute_result",
          "data": {
            "text/plain": [
              "<StemContainer object of 3 artists>"
            ]
          },
          "metadata": {},
          "execution_count": 4
        },
        {
          "output_type": "display_data",
          "data": {
            "text/plain": [
              "<Figure size 300x300 with 1 Axes>"
            ],
            "image/png": 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\n"
          },
          "metadata": {}
        }
      ],
      "source": [
        "# @title\n",
        "plt.figure(figsize=(3,3))\n",
        "N = 10\n",
        "y = np.ones(N)\n",
        "y[-1] = 0.5\n",
        "for n in range(1,N):\n",
        "    y[n] = -3*y[n-1] + 4*y[n-2]\n",
        "plt.stem(y)"
      ]
    },
    {
      "cell_type": "markdown",
      "id": "59ca76c6-858d-41c5-b32d-6e83aac7f827",
      "metadata": {
        "tags": [],
        "id": "59ca76c6-858d-41c5-b32d-6e83aac7f827"
      },
      "source": [
        "## Caso de raízes complexas\n",
        "\n",
        "No caso de raízes complexas, como a equação de diferenças tem coeficientes reais, elas sempre aparecerão em pares complexos conjugados ($\\lambda_1 = \\gamma e^{j\\omega}$ e $\\lambda_2 = \\gamma e^{-j\\omega}$).\n",
        "\n",
        "Então, a parte da solução devida a essas raízes é:\n",
        "\n",
        "$$y_{zi}[n] = c_1\\gamma^n e^{j\\omega n} + c_2\\gamma^n e^{-j\\omega n}$$\n",
        "\n",
        "Como sinal é real, $c_1$ e $c_2$ também serão complexos conjugados:\n",
        "\n",
        "$c_1 = \\frac{c}{2}e^{j\\theta}$ e $c_2 = \\frac{c}{2}e^{-j\\theta}$\n",
        "\n",
        "Então:\n",
        "\n",
        "$$y_{zi}[n] = \\frac{c}{2}e^{j\\theta}\\gamma^n e^{j\\omega n} + \\frac{c}{2}e^{-j\\theta}\\gamma^n e^{-j\\omega n} = \\frac{c}{2}\\gamma^n e^{j(\\omega n+\\theta)} + \\frac{c}{2}\\gamma^n e^{-j(\\omega n+\\theta)} = c\\gamma^n\\cos(\\omega n+\\theta)$$\n",
        "\n",
        "É importante notar que se o polinômio característico tiver mais raízes reais, elas continuam sendo tratadas da mesma maneira."
      ]
    },
    {
      "cell_type": "markdown",
      "id": "8fc44751-0128-440d-aea2-5cd3ebb993d5",
      "metadata": {
        "id": "8fc44751-0128-440d-aea2-5cd3ebb993d5"
      },
      "source": [
        "### Exemplo 4\n",
        "\n",
        "Considere um sistema com a seguinte equação de diferenças:\n",
        "\n",
        "$$y[n] = -2y[n-1] - 2y[n-2] + x[n-1]$$\n",
        "\n",
        "com $y[0] = 0$ e $y[-1] = -1$\n",
        "\n",
        "O polinômio característico do sistema é:\n",
        "\n",
        "$\\lambda^2 + 2\\lambda + 2 = 0$\n",
        "\n",
        "As raízes do polinômio são $\\lambda_1= -1+j = \\sqrt{2}e^{j\\frac{3\\pi}{4}}$ e $\\lambda_2= -1-j = \\sqrt{2}e^{j\\frac{5\\pi}{4}} = \\sqrt{2}e^{-j\\frac{3\\pi}{4}}$.\n",
        "\n",
        "Então a solução de entrada nula  é:\n",
        "\n",
        "$$y_{zi}[n] = c_1\\sqrt{2}^ne^{j\\frac{3\\pi n}{4}} +  c_2\\sqrt{2}^ne^{-j\\frac{3\\pi n}{4}} = $$\n",
        "\n",
        "Usando as condições iniciais, temos:\n",
        "\n",
        "$$c_1 + c_2 = 0$$\n",
        "\n",
        "$$c_1\\frac{1}{\\sqrt{2}}e^{-j\\frac{3\\pi}{4}} +  c_2\\frac{1}{\\sqrt{2}}e^{j\\frac{3\\pi}{4}}=-1$$\n",
        "\n",
        "Resolvendo o sistema de equações acima chegamos a $c_1 = -j = e^{-j\\frac{\\pi}{2}}$ e $c_2 = j = e^{j\\frac{\\pi}{2}}$.\n",
        "\n",
        "Então, a resposta de entrada nula é:\n",
        "\n",
        "$y_{zi}[n] = e^{-j\\frac{\\pi}{2}}\\sqrt{2}^ne^{j\\frac{3\\pi n}{4}} +  e^{j\\frac{\\pi}{2}}\\sqrt{2}^ne^{-j\\frac{3\\pi n}{4}} = \\sqrt{2}^ne^{j\\left(\\frac{3\\pi}{4} n-\\frac{\\pi}{2}\\right)} +  \\sqrt{2}^ne^{-j\\left(\\frac{3\\pi n}{4}-\\frac{\\pi}{2}\\right)} =  2\\sqrt{2}^n\\cos\\left(\\frac{3\\pi n}{4}-\\frac{\\pi}{2}\\right)$\n",
        "\n",
        "Abaixo é mostrado o gráfico com os 10 primeiros instantes do sinal."
      ]
    },
    {
      "cell_type": "code",
      "execution_count": 5,
      "id": "49137f66-6932-411b-8899-a64ef641d86a",
      "metadata": {
        "tags": [],
        "cellView": "form",
        "id": "49137f66-6932-411b-8899-a64ef641d86a",
        "outputId": "788488a0-1a6a-4e8c-c3d4-2bd7c6fb21d4",
        "colab": {
          "base_uri": "https://localhost:8080/",
          "height": 309
        }
      },
      "outputs": [
        {
          "output_type": "execute_result",
          "data": {
            "text/plain": [
              "<StemContainer object of 3 artists>"
            ]
          },
          "metadata": {},
          "execution_count": 5
        },
        {
          "output_type": "display_data",
          "data": {
            "text/plain": [
              "<Figure size 300x300 with 1 Axes>"
            ],
            "image/png": "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\n"
          },
          "metadata": {}
        }
      ],
      "source": [
        "# @title\n",
        "plt.figure(figsize=(3,3))\n",
        "N = 10\n",
        "n = np.arange(N)\n",
        "y = 2*np.sqrt(2)**n*np.cos(3*np.pi*n/4-np.pi/2)\n",
        "plt.stem(y)"
      ]
    },
    {
      "cell_type": "markdown",
      "id": "cc5f5ee5-eb8a-4ea6-9ed3-7c23ecdfd273",
      "metadata": {
        "id": "cc5f5ee5-eb8a-4ea6-9ed3-7c23ecdfd273"
      },
      "source": [
        "E a seguir o mesmo sinal, com os valores obtidos de forma recursiva. Os dois gráficos são iguais."
      ]
    },
    {
      "cell_type": "code",
      "execution_count": 6,
      "id": "05c75780-4131-4356-b6e7-480c2bc87ce8",
      "metadata": {
        "tags": [],
        "cellView": "form",
        "id": "05c75780-4131-4356-b6e7-480c2bc87ce8",
        "outputId": "ad85589f-79fb-4e9f-81a3-9cb20a489c37",
        "colab": {
          "base_uri": "https://localhost:8080/",
          "height": 309
        }
      },
      "outputs": [
        {
          "output_type": "execute_result",
          "data": {
            "text/plain": [
              "<StemContainer object of 3 artists>"
            ]
          },
          "metadata": {},
          "execution_count": 6
        },
        {
          "output_type": "display_data",
          "data": {
            "text/plain": [
              "<Figure size 300x300 with 1 Axes>"
            ],
            "image/png": 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\n"
          },
          "metadata": {}
        }
      ],
      "source": [
        "# @title\n",
        "plt.figure(figsize=(3,3))\n",
        "N = 10\n",
        "y = np.zeros(N)\n",
        "y[-1] = -1\n",
        "for n in range(1,N):\n",
        "    y[n] = -2*y[n-1]  -2 *y[n-2]\n",
        "plt.stem(y)"
      ]
    },
    {
      "cell_type": "markdown",
      "id": "8c13da70-d399-4980-ae62-018cdf1bf7fc",
      "metadata": {
        "id": "8c13da70-d399-4980-ae62-018cdf1bf7fc"
      },
      "source": [
        "## Caso de raízes repetidas\n",
        "\n",
        "No caso do polinômio característico ter $r$ raízes repetidas $\\lambda_1$, as soluções correspondentes seguem a seguinte forma:\n",
        "\n",
        "$$\\lambda_1^n, n\\lambda_1^n, n^2\\lambda_1^n, ..., n^{r-1}\\lambda_1^n$$\n",
        "\n",
        "Portanto, se o polinômio característico for:\n",
        "\n",
        "$$(\\lambda-\\lambda_1)^r(\\lambda - \\lambda_{r+1})...(\\lambda-\\lambda_N) = 0$$\n",
        "\n",
        "A solução de entrada nula tem a seguinte forma:\n",
        "\n",
        "$$y_{zi}[n] = c_1\\lambda_1^n + c_2n\\lambda_1^n + c_3n^2\\lambda_1^n +... + c_rn^{r-1}\\lambda_1^{n} + c_{r+1}\\lambda_{r+1}^n + ... + c_N\\lambda_N^n$$\n"
      ]
    },
    {
      "cell_type": "markdown",
      "id": "f692e095-9694-421e-87be-e08e6cb66c8f",
      "metadata": {
        "id": "f692e095-9694-421e-87be-e08e6cb66c8f"
      },
      "source": [
        "### Exemplo 5\n",
        "\n",
        "Considere um sistema com a seguinte equação de diferenças:\n",
        "\n",
        "$$y[n] = -4y[n-1] - 4y[n-2] + 3x[n]$$\n",
        "\n",
        "com $y[0] = 1$ e $y[-1] = 0$\n",
        "\n",
        "O polinômio característico é:\n",
        "\n",
        "$\\lambda^2 + 4\\lambda + 4 = 0$\n",
        "\n",
        "As raízes da equação são iguais e iguais a -2. Então, a resposta de entrada nula é:\n",
        "\n",
        "$y_{zi}[n] = c_1(-2)^n + c_2n(-2)^n$\n",
        "\n",
        "Usando as condições iniciais:\n",
        "\n",
        "$$c_1 = 1 $$\n",
        "\n",
        "$$-0.5c_1 +0.5c_2 = 0 \\rightarrow c_2 = 1 $$\n",
        "\n",
        "Então:\n",
        "\n",
        "$y_{zi}[n] = (-2)^n + n(-2)^n$"
      ]
    },
    {
      "cell_type": "markdown",
      "id": "d50e57e9-4a62-4312-83d5-612ca0c3e92c",
      "metadata": {
        "id": "d50e57e9-4a62-4312-83d5-612ca0c3e92c"
      },
      "source": [
        "### Resumo do procedimento para encontrar a solução de entrada nula de uma equação de diferenças linear e invariante no tempo\n",
        "\n",
        "- **1** Eliminar os termos referentes ao sinal de entrada\n",
        "\n",
        "- **2** Encontrar o polinômio característico da equação\n",
        "\n",
        "- **3** Encontrar as raízes $\\lambda_i$ do polinômio característico\n",
        "\n",
        "- **4** Formar o sinal de resposta nula $y_{zi}[n] = c_1\\lambda_1^n + c_2\\lambda_2^n + ... + c_N\\lambda_N^n$\n",
        "\n",
        "Caso a raiz $\\lambda_1$ seja repetida, a saída terá a forma $y_{zi}[n] = c_1\\lambda_1^n + c_2n\\lambda_1^n + c_3n^2\\lambda_1^n +... + c_rn^{r-1}\\lambda_1^{n} + c_{r+1}\\lambda_{r+1}^n + ... + c_N\\lambda_N^n$\n",
        "\n",
        "- **5** Encontrar os valores de $c_i$ utilizando as condições iniciais.\n",
        "\n",
        "- **6** Caso entre as raízes contenham pares complexos conjugados, juntar os termos para formar cossenos.\n",
        "\n",
        "\n"
      ]
    },
    {
      "cell_type": "markdown",
      "id": "4b1ebccc-2dac-476c-8787-20176df4f158",
      "metadata": {
        "id": "4b1ebccc-2dac-476c-8787-20176df4f158"
      },
      "source": [
        "## Exercícios"
      ]
    },
    {
      "cell_type": "markdown",
      "id": "d583db62-8d77-4996-9323-c71681e6eb0d",
      "metadata": {
        "id": "d583db62-8d77-4996-9323-c71681e6eb0d"
      },
      "source": [
        "**1)** Encontre uma expressão dependendo apenas de $n$ para a solução de entrada nula dos sistemas com as seguintes equações de diferenças:\n",
        "\n",
        "a) $y[n] = 0.5y[n-1]$, com $y[-1] = 10$\n",
        "\n",
        "b) $y[n+1] = -2y[n]+ x[n+1]$, com $x[n]= e^{-n}u[n]$ e $y[-1] = 1$\n",
        "\n",
        "c) $y[n] - 0.6y[n-1]-0.16y[n-2]=0$, com $y[-1]= -25, y[-2]=0$\n",
        "\n",
        "d) $y[n+2] + 2y[n+1]+0.5y[n] = x[n+2]+x[n+1]+2x[n-3]$, com $x[n] = 2^nu[n], y[-1]=3, y[-2]=2$\n",
        "\n",
        "e) $y[n] = -2y[n-1]-y[n-2] + 2x[n] - x[n-1] $, com $x[n] = 3^{-n}u[n], y[-1] = 2, y[-2]=3$\n",
        "\n",
        "f) $y[n] = -2y[n-1]-2y[n-2] + 2x[n] - x[n-1] $, com $x[n] = 3^{-n}u[n], y[-1] = 2, y[-2]=3$"
      ]
    },
    {
      "cell_type": "markdown",
      "id": "abc0e3b2-1162-43f1-8899-161b18720cff",
      "metadata": {
        "id": "abc0e3b2-1162-43f1-8899-161b18720cff"
      },
      "source": [
        "**2)** A sequência de Fibonacci (0, 1, 1, 2, 3, 5, 8, 13 ...) é uma sequência gerada pela soma dos dois termos anteriores.\n",
        "\n",
        "a) Encontre uma equação de diferenças para sequência de Fibonacci, considerando que $y[0] = 0$ e $y[1] = 1$.\n",
        "\n",
        "b) Encontre uma expressão que dependa apenas de $n$ para o enésimo numero da sequência de Fibonacci."
      ]
    },
    {
      "cell_type": "markdown",
      "id": "9b57f5ec-5d99-4ab7-b691-9549ea77e408",
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      "source": [
        "**3)** Encontre uma expressão dependendo apenas de $n$ para a solução de entrada nula dos sistemas com as seguintes equações de diferenças:\n",
        "\n",
        "a) $y[n] - 5y[n-1]+6y[n-2] = 4x[n]$, com $y[-1]=1$,  $y[2]=-1$ com $x[n]=u[n]$\n",
        "\n",
        "b) $y[n] = -2y[n-1] + 4x[n]+5x[n-1]$, , com $y[-1]=1$,  $y[2]=1$ com $x[n]=(0.5)^nu[n]$\n",
        "\n",
        "c) $y[n]=2\\cos\\left(\\frac{\\pi}{6}\\right)y[n-1] - y[n-2] = x[n]$, com $y[-1] = \\cos\\left(\\frac{\\pi}{6}\\right)$, $y[-2] = \\cos\\left(\\frac{\\pi}{3}\\right)$ e $x[n]=u[n]$\n",
        "\n",
        "d) $y[n]=2\\cos\\left(\\Omega_0\\right)y[n-1] - y[n-2] = x[n]$, com $y[-1] = \\cos\\left(\\Omega_0\\right)$, $y[-2] = \\cos\\left(2\\Omega_0\\right)$ e $x[n]=u[n]$\n",
        "\n"
      ]
    },
    {
      "cell_type": "markdown",
      "id": "bdd7b838-9f3c-4c21-8094-cc1e946705e3",
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        "id": "bdd7b838-9f3c-4c21-8094-cc1e946705e3"
      },
      "source": [
        "## Referências\n",
        "\n",
        "- Lathi, BP, Sinais e Sistemas Lineares, 2007\n",
        "\n"
      ]
    },
    {
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      },
      "source": [
        "## Solução dos exercícios"
      ]
    },
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      "source": [
        "**1)** Resolvido por Carolina Kafka Neves\n",
        "\n",
        "\n",
        "a) $y[n] = 0.5y[n-1]$, com $y[-1] = 10$\n",
        "\n",
        "Pela equação de diferenças, temos que o polinômio característico é:\n",
        "\n",
        "$\\lambda-0.5=0$\n",
        "\n",
        "$\\lambda= 0.5$\n",
        "\n",
        "Então, a resposta de entrada nula é:\n",
        "\n",
        "$y_{zi}[n]= c_1.(0.5)^{n}$\n",
        "\n",
        "Pela condição inicial, temos que $c_1$ é:\n",
        "\n",
        "$10= c_1+ (0.5)^{-1}$\n",
        "\n",
        "$c_1= 5$\n",
        "\n",
        "Portanto:\n",
        "\n",
        "$y_{zi}[n]= 5.(0.5)^{n}u[n]$\n",
        "\n",
        "\n",
        "---\n",
        "\n",
        "b) $y[n+1] = -2y[n]+ x[n+1]$, com $x[n]= e^{-n}u[n]$ e $y[-1] = 1$\n",
        "\n",
        "O polinômio característico é:\n",
        "\n",
        "$\\lambda +2 =0$\n",
        "\n",
        "$\\lambda= -2$\n",
        "\n",
        "Então, a resposta de entrada nula é:\n",
        "\n",
        "$y_{zi}[n]= c_1.(-2)^{n}$\n",
        "\n",
        "Pela condição inicial, temos que:\n",
        "\n",
        "$1= c_1.(-2)^{-1}$\n",
        "\n",
        "$c_1= -2$\n",
        "\n",
        "Portanto:\n",
        "\n",
        "$y_{zi}[n]= (-2).(-2)^{n}u[n]$\n",
        "\n",
        "$y_{zi}[n]= (-2)^{n+1}u[n]$\n",
        "\n",
        "---\n",
        "\n",
        "c) $y[n] - 0.6y[n-1]-0.16y[n-2]=0$, com $y[-1]= -25, y[-2]=0$\n",
        "\n",
        "O polinômio característico da equação de diferenças é:\n",
        "\n",
        "$\\lambda^{2}- 0.6\\lambda - 0.16= 0$\n",
        "\n",
        "As raízes são $\\lambda_1= -0.2$ e $\\lambda_2= 0.8$\n",
        "\n",
        "Logo,\n",
        "\n",
        "$y_{zi}[n]= c_1.(-0.2)^{n}+c_2.(0.8)^{n}$\n",
        "\n",
        "Pelas condições inicias, temos que:\n",
        "\n",
        "$c_1= 1$ e $c_2= -16$\n",
        "\n",
        "Portanto:\n",
        "\n",
        "$y_{zi}[n]= ((-0.2)^{n}-16(0.8)^{n})u[n]$\n",
        "\n",
        "---\n",
        "\n",
        "d) $y[n+2] + 2y[n+1]+0.5y[n] = x[n+2]+x[n+1]+2x[n-3]$, com $x[n] = 2^nu[n], y[-1]=3, y[-2]=2$\n",
        "\n",
        "O polinômio característico da equação de diferenças é:\n",
        "\n",
        "$0.5\\lambda^{2}+2\\lambda+1=0$\n",
        "\n",
        "As raízes dessa equação são $\\lambda_1=-3.41$ e $\\lambda_2=-0.59$\n",
        "\n",
        "Logo,\n",
        "\n",
        "$y_{zi}[n]= c_1.(-3.41)^{n}+c_2.(-0.59)^{n}$\n",
        "\n",
        "Pelas condições iniciais, temos que as constantes são aproximadamente $c_1= -5.29$ e $c_2= 0.85$\n",
        "\n",
        "Portanto,\n",
        "\n",
        "$y_{zi}[n]= [(-5.29)(-3.41)^{n}+0.85(-0.59)^{n}]u[n]$\n",
        "\n",
        "---\n",
        "\n",
        "\n",
        "\n",
        "e) $y[n] = -2y[n-1]-y[n-2] + 2x[n] - x[n-1] $, com $x[n] = 3^{-n}u[n], y[-1] = 2, y[-2]=3$\n",
        "\n",
        "O polinômio característico da equação de diferenças é:\n",
        "\n",
        "$\\lambda^{2}+2\\lambda+1=0$\n",
        "\n",
        "Logo, as raízes da equação são $\\lambda_1=\\lambda_2=-1$\n",
        "\n",
        "Como temos raízes repetidas, a resposta de entrada nula será:\n",
        "\n",
        "$y_{zi}[n]= c_1.(-1)^{n}+c_2.n.(-1)^{n}$\n",
        "\n",
        "Seguindo as condições iniciais, temos que:\n",
        "\n",
        "$c_1= -7$ e $c_2= -5$\n",
        "\n",
        "Portanto,\n",
        "\n",
        "$y_{zi}[n]= (-7.(-1)^{n}-5.n.(-1)^{n})u[n]$\n",
        "\n",
        "---\n",
        "\n",
        "f) $y[n] = -2y[n-1]-2y[n-2] + 2x[n] - x[n-1] $, com $x[n] = 3^{-n}u[n], y[-1] = 2, y[-2]=3$\n",
        "\n",
        "O polinômio característico da equação de diferenças é:\n",
        "\n",
        "$\\lambda^{2}+2\\lambda+3=0$\n",
        "\n",
        "As raízes dessa equação são $\\lambda_1= -1+1.41j= 1.73e^{j0.96}$ e $\\lambda_2= -1-1.41j= 1.73e^{-j0.96}$\n",
        "\n",
        "Logo,\n",
        "\n",
        "$y_{zi}[n]= c_1.(1.73e^{j0.96})^{n}+c_2.(1.73e^{-j0.96})^{n}\n",
        "\n",
        "Seguindo as condições iniciais, temos que:\n",
        "\n",
        "$c_1= 4.63e^{-j2.13}$ e $c_2= 4.63e^{j2.13}$\n",
        "\n",
        "Portanto,\n",
        "\n",
        "$y_{zi}[n]= (4.63e^{-j2.13}(1.73e^{j0.96})^{n}+4.63e^{j2.13}(1.73e^{-j0.96})^{n})u[n]$\n",
        "\n",
        "\n",
        "$y_{zi}[n] = 9.26.(1.73)^{n}.cos(0.96n+2.13).u[n]$\n",
        "\n",
        "---\n",
        "\n"
      ]
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      "id": "b4133272-668f-4751-ba36-5823ae641627",
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      "source": [
        "3) Resolvido por Catherine\n",
        "\n",
        "a) $$y[n] - 5y[n-1] + 6y[n-2] = 4x[n], \\quad y[-1] = 1, \\quad y[-2] = -1, \\quad x[n] = u[n]$$\n",
        "\n",
        "Parte homogênea da equação de diferenças (sem o termo $x[n]$):\n",
        "\n",
        "$$y[n] - 5y[n-1] + 6y[n-2] = 0$$\n",
        "\n",
        "A característica dessa equação é dada por: $r^2 - 5r + 6 = 0$, que pode ser fatorada como $(r - 2)(r - 3) = 0$.\n",
        "\n",
        "- As raízes são $r_1 = 2$ e $r_2 = 3$.\n",
        "\n",
        "$$y_h[n] = A \\cdot 2^n + B \\cdot 3^n$$\n",
        "\n",
        "Usando as condições iniciais para encontrar os valores de $A$ e $B$:\n",
        "\n",
        "$$y_h[-1] = A \\cdot 2^{-1} + B \\cdot 3^{-1} = 1$$\n",
        "$$y_h[-2] = A \\cdot 2^{-2} + B \\cdot 3^{-2} = -1$$\n",
        "\n",
        "$$A = \\frac{7}{5}   ;   B = -\\frac{8}{5}$$\n",
        "\n",
        "$$y_h[n] = \\frac{7}{5} \\cdot 2^n - \\frac{8}{5} \\cdot 3^n$$\n",
        "\n",
        "Agora, vamos considerar o termo de entrada $x[n] = u[n]$. A solução particular será uma constante, que chamaremos de $y_p[n] = C$.\n",
        "\n",
        "Substituindo a solução particular na equação original:\n",
        "$$C - 5C + 6C = 4 \\cdot 1$$\n",
        "$$C = 4$$\n",
        "\n",
        "$$y_p[n] = 4$$\n",
        "\n",
        "Logo,\n",
        "$$y[n] = y_h[n] + y_p[n] = \\frac{7}{5} \\cdot 2^n - \\frac{8}{5} \\cdot 3^n + 4$$\n",
        "\n",
        "b) $$y[n] = -2y[n-1] + 4x[n] + 5x[n-1], \\quad y[-1] = 1, \\quad y[-2] = 1, \\quad x[n] = (0.5)^n u[n]$$\n",
        "\n",
        "Parte homogênea:\n",
        "$$y[n] = -2y[n-1]$$\n",
        "\n",
        "A característica é: $r + 2 = 0$, então a raiz é $r = -2$.  \n",
        "\n",
        "$$y_h[n] = A \\cdot (-2)^n$$\n",
        "\n",
        "$$y_h[-1] = A \\cdot (-2)^{-1} = 1$$\n",
        "$$y_h[-2] = A \\cdot (-2)^{-2} = 1$$\n",
        "\n",
        "$$A = -\\frac{1}{2}$$\n",
        "\n",
        "$$y_h[n] = -\\frac{1}{2} \\cdot (-2)^n = 2^n$$\n",
        "\n",
        "A solução particular para a entrada $x[n] = (0.5)^n u[n]$ será uma constante, que chamaremos de $y_p[n] = C$.\n",
        "\n",
        "Substituindo na equação original:\n",
        "$$C = -2C + 4 \\cdot (0.5)^n + 5 \\cdot (0.5)^{n-1}$$\n",
        "$$C = 4 \\cdot (0.5)^n + 5 \\cdot (0.5)^n$$\n",
        "$$C = 9 \\cdot (0.5)^n$$\n",
        "\n",
        "Logo,\n",
        "\n",
        "$$y_p[n] = 9 \\cdot (0.5)^n$$\n",
        "\n",
        "$$y[n] = y_h[n] + y_p[n] = 2^n + 9 \\cdot (0.5)^n$$\n",
        "\n",
        "c) $$y[n] = 2\\cos\\left(\\frac{\\pi}{6}\\right)y[n-1] - y[n-2] = x[n], \\quad y[-1] = \\cos\\left(\\frac{\\pi}{6}\\right), \\quad y[-2] = \\cos\\left(\\frac{\\pi}{3}\\right), \\quad x[n] = u[n]$$\n",
        "Parte homogênea:\n",
        "$$y[n] = 2\\cos\\left(\\frac{\\pi}{6}\\right)y[n-1] - y[n-2]$$\n",
        "\n",
        "A característica é: $r^2 = 2\\cos\\left(\\frac{\\pi}{6}\\right)r - 1$, que pode ser reescrita como $r^2 - 2\\cos\\left(\\frac{\\pi}{6}\\right)r + 1 = 0$. As raízes são $r_1 = r_2 = \\cos\\left(\\frac{\\pi}{6}\\right)$.\n",
        "\n",
        "$$y_h[n] = A \\cdot (\\cos\\left(\\frac{\\pi}{6}\\right))^n + B \\cdot n (\\cos\\left(\\frac{\\pi}{6}\\right))^n$$\n",
        "\n",
        "$$y_h[-1] = A \\cdot (\\cos\\left(\\frac{\\pi}{6}\\right))^{-1} + B \\cdot (-1) (\\cos\\left(\\frac{\\pi}{6}\\right))^{-1} = \\cos\\left(\\frac{\\pi}{6}\\right)A - \\frac{B}{\\cos\\left(\\frac{\\pi}{6}\\right)} = \\cos\\left(\\frac{\\pi}{6}\\right)$$\n",
        "$$y_h[-2] = A \\cdot (\\cos\\left(\\frac{\\pi}{6}\\right))^{-2} + B \\cdot (-2) (\\cos\\left(\\frac{\\pi}{6}\\right))^{-2} = \\cos\\left(\\frac{\\pi}{3}\\right)A - \\frac{2B}{\\cos\\left(\\frac{\\pi}{6}\\right)^2} = \\cos\\left(\\frac{\\pi}{3}\\right)$$\n",
        "\n",
        "$A = \\cos\\left(\\frac{\\pi}{6}\\right)$ e $B = 0$.\n",
        "\n",
        "Logo,\n",
        "$$y_h[n] = \\cos\\left(\\frac{\\pi}{6}\\right)^n$$\n",
        "\n",
        "A solução particular para a entrada $x[n] = u[n]$ é uma constante, que chamaremos de $y_p[n] = C$.\n",
        "\n",
        "Substituindo na equação original:\n",
        "$$C = 2\\cos\\left(\\frac{\\pi}{6}\\right)C - C$$\n",
        "$$C = C\\left(2\\cos\\left(\\frac{\\pi}{6}\\right) - 1\\right)$$\n",
        "$$C = C\\left(\\sqrt{3} - 1\\right)$$\n",
        "\n",
        "$$y_p[n] = C\\left(\\sqrt{3} - 1\\right)$$\n",
        "\n",
        "$$y[n] = y_h[n] + y_p[n] = \\cos\\left(\\frac{\\pi}{6}\\right)^n + C\\left(\\sqrt{3} - 1\\right)$$\n",
        "\n",
        "d) $$y[n] = 2\\cos\\left(\\Omega_0\\right)y[n-1] - y[n-2] = x[n], \\quad y[-1] = \\cos\\left(\\Omega_0\\right), \\quad y[-2] = \\cos\\left(2\\Omega_0\\right), \\quad x[n] = u[n]$$\n",
        "Parte homogênea:\n",
        "$$y[n] = 2\\cos\\left(\\Omega_0\\right)y[n-1] - y[n-2]$$\n",
        "\n",
        "A característica é: $r^2 = 2\\cos\\left(\\Omega_0\\right)r - 1$, que pode ser reescrita como $r^2 - 2\\cos\\left(\\Omega_0\\right)r + 1 = 0$. As raízes são $r_1 = r_2 = \\cos\\left(\\Omega_0\\right)$.\n",
        "\n",
        "$$y_h[n] = A \\cdot (\\cos\\left(\\Omega_0\\right))^n + B \\cdot n (\\cos\\left(\\Omega_0\\right))^n$$\n",
        "\n",
        "$$y_h[-1] = A \\cdot (\\cos\\left(\\Omega_0\\right))^{-1} + B \\cdot (-1) (\\cos\\left(\\Omega_0\\right))^{-1} = \\cos\\left(\\Omega_0\\right)A - \\frac{B}{\\cos\\left(\\Omega_0\\right)} = \\cos\\left(\\Omega_0\\right)$$\n",
        "$$y_h[-2] = A \\cdot (\\cos\\left(\\Omega_0\\right))^{-2} + B \\cdot (-2) (\\cos\\left(\\Omega_0\\right))^{-2} = \\cos\\left(2\\Omega_0\\right)A - \\frac{2B}{\\cos\\left(\\Omega_0\\right)^2} = \\cos\\left(2\\Omega_0\\right)$$\n",
        "\n",
        "$A = \\cos\\left(\\Omega_0\\right)$ e $B = 0$.\n",
        "\n",
        "$$y_h[n] = \\cos\\left(\\Omega_0\\right)^n$$\n",
        "\n",
        "A solução particular para a entrada $x[n] = u[n]$ é uma constante, que chamaremos de $y_p[n] = C$.\n",
        "\n",
        "Substituindo na equação original:\n",
        "$$C = 2\\cos\\left(\\Omega_0\\right)C - C$$\n",
        "$$C = C\\left(2\\cos\\left(\\Omega_0\\right) - 1\\right)$$\n",
        "\n",
        "$$y_p[n] = C\\left(2\\cos\\left(\\Omega_0\\right) - 1\\right)$$\n",
        "\n",
        "$$y[n] = y_h[n] + y_p[n] = \\cos\\left(\\Omega_0\\right)^n + C\\left(2\\cos\\left(\\Omega_0\\right) - 1\\right)$$"
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