FuncaoTransferenciaDiscreto.ipynb

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      "source": [
        "# Função de transferência de sistemas discretos\n",
        "\n",
        "Renato Naville Watanabe"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "mnX7pUDucZYj"
      },
      "source": [
        "## Preparação do ambiente"
      ]
    },
    {
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      "source": [
        "# @title\n",
        "import matplotlib.pyplot as plt\n",
        "import numpy as np"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "bT3n4HshcZYl"
      },
      "source": [
        "## Definição do problema\n",
        "\n",
        "Aqui, vamos considerar o formato geral de uma equação de diferenças de um sistema linear e invariante no tempo:\n",
        "\n",
        "$$y[n] + a_{1}y[n-1]+...+a_Ny[n-N] = b_Mx[n-M] +...+b_2x[n-2]+b_1x[n-1] + b_0x[n] $$\n",
        "\n",
        "Como já vimos anteriormente, 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).\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",
        "Vamos agora encontrar essas duas partes da solução utilizando transformada Z. É consideravelmente mais simples do que o método clássico."
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "RzdIWhI6cZYm"
      },
      "source": [
        "## Resposta de estado nulo\n",
        "\n",
        "\n",
        "\n",
        "Como já vimos anteriormente, a transformada Z de $x[n-m]$ é:\n",
        "\n",
        "$\\mathcal{Z}(x[n-m]) = z^{-m}X(z)$\n",
        "\n",
        "e a transformada Z de $y[n-m]$ é:\n",
        "\n",
        "$\\mathcal{Z}(y[n-m]) = z^{-m}Y(z)$"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "1fxkRfoscZYn"
      },
      "source": [
        "Substituindo essas transformadas dos sinais atrasado na equação de diferenças, temos:\n",
        "\n",
        "$$Y(z) + a_{1}z^{-1}Y(z)+...+a_Nz^{-N}Y(z) = b_Mz^{-M}X(z) +...+b_2z^{-2}X(z)+b_1z^{-1}X(z) + b_0X(z) $$\n",
        "\n",
        "Com isso, podemos definir a função $H(z)$:\n",
        "\n",
        "$H(z) = \\frac{Y(z)}{X(z)} = \\frac{b_Mz^{-M} +...+b_2z^{-2}+b_1z^{-1} + b_0}{1 + a_{1}z^{-1}+...+a_Nz^{-N}}$\n",
        "\n",
        "Essa função é conhecida como **função de transferência** do sistema. Ela é obtida diretamente dos coeficientes da equação de diferenças."
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "YB6BOr6WcZYn"
      },
      "source": [
        "---\n",
        "\n",
        "### Exemplo 1\n",
        "\n",
        "Vamos considerar o sistema com a seguinte equação de diferenças:\n",
        "\n",
        "$y[n] + 3y[n-1] - 4y[n-2] = 3x[n] + x[n-1]$\n",
        "\n",
        "A função de transferência deste sistema é:\n",
        "\n",
        "$H(z) = \\frac{3+z^{-1}}{1 + 3z^{-1}-4z^{-2}}$\n",
        "\n",
        "---"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "VSh2oAZ6cZYo"
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      "source": [
        "---\n",
        "\n",
        "### Exemplo 2\n",
        "\n",
        "Vamos considerar o sistema com a seguinte equação de diferenças:\n",
        "\n",
        "$y[n]  = -0.5y[n-1] + x[n]$\n",
        "\n",
        "A função de transferência deste sistema é:\n",
        "\n",
        "$H(z) = \\frac{1}{1 + 0.5z^{-1}}$\n",
        "\n",
        "---"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "cHsT9qYfcZYp"
      },
      "source": [
        "Observando a expressão da função de transferência, vemos que se $X(z) = 1$, $Y(z) = H(z)$. O sinal $x[n]$ que tem uma transformada Z igual a 1 é o impulso unitário ($x[n]=\\delta[n]$). Portanto a função de transferência é a transformada Z da resposta ao impulso.\n",
        "\n",
        "$H(z) = \\displaystyle\\sum_{n=0}^\\infty h(n)z^{-n}$"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "iL_j7cVDcZYq"
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      "source": [
        "---\n",
        "\n",
        "### Exemplo 3\n",
        "\n",
        "A resposta ao impulso do sistema do exemplo 2 é:\n",
        "\n",
        "$h[n] = (-0.5)^nu[n]$\n",
        "\n",
        "A transformada Z dessa resposta ao impulso é:\n",
        "\n",
        "$H(z) = \\displaystyle\\sum_{n=0}^\\infty  (-0.5)^nu[n] z^{-n}= \\displaystyle\\sum_{n=0}^\\infty  (-0.5)^nz^{-n} = \\displaystyle\\sum_{n=0}^\\infty  (-0.5z^{-1})^n = \\frac{1}{1 + 0.5z^{-1}}=\\frac{z}{z+0.5}$\n",
        "\n",
        "Como esperado, a função encontrada foi a mesma do exemplo 2.\n",
        "\n",
        "---"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "tUSbOPPLcZYq"
      },
      "source": [
        "A partir da expressão da função de transferência, chegamos à expressão para $Y(z)$:\n",
        "\n",
        "$H(z) = \\frac{Y(z)}{X(z)} \\rightarrow Y(z) = H(z)X(z)$\n",
        "\n",
        "Também podemos chegar a essa expressão pela propriedade da convolução da transformada Z:\n",
        "\n",
        "$y[n] = h[n]*x[n] \\rightarrow Y(z) = H(z)X(z)$\n",
        "\n",
        "Tendo $Y(z)$, basta calcular a sua transformada inversa para encontrar $y[n]$."
      ]
    },
    {
      "cell_type": "markdown",
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        "id": "AL5_wOFycZYq"
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      "source": [
        "---\n",
        "\n",
        "### Exemplo 4\n",
        "\n",
        "Vamos considerar o mesmo sistema do exemplo 2 e encontrar o sinal de saída quando a entrada é um degrau.\n",
        "\n",
        "$x[n] = u[n]$\n",
        "\n",
        "A sua transformada Z é:\n",
        "\n",
        "$X(z) = \\frac{z}{z-1}$\n",
        "\n",
        "Então, a transformada Z do sinal de saída é:\n",
        "\n",
        "$Y(z) = H(z)X(z) = \\frac{1}{1 + 0.5z^{-1}}\\frac{z}{z-1} = \\frac{z}{z-0.5-0.5z^{-1}} = \\frac{z^2}{z^2-0.5z-0.5}=\\frac{z^2}{(z+0.5)(z-1)}$\n",
        "\n",
        "Expandindo $Y'(z)=\\frac{Y(z)}{z}$ em frações parciais:\n",
        "\n",
        "$Y'(z) = H(z)X(z) = \\frac{1}{1 + 0.5z^{-1}}\\frac{z}{z-1} = \\frac{z}{z-0.5-0.5z^{-1}} = \\frac{z^2}{z^2-0.5z-0.5}=\\frac{z}{(z+0.5)(z-1)} = \\frac{A_1}{z+0.5} + \\frac{A_2}{z-1}$\n",
        "\n",
        "Com\n",
        "\n",
        "$A_1 = \\left.\\frac{z}{z-1}\\right|_{z=-0.5}=0.333$ e $A_2=\\left.\\frac{z}{z+0.5}\\right|_{z=1}=0.67$\n",
        "\n",
        "Então $Y(z) = Y'(z)z$ é:\n",
        "\n",
        "$Y(z) = \\frac{0.33z}{z+0.5} + \\frac{0.67z}{z-1}$\n",
        "\n",
        "Com a transformada Z inversa de $Y(z)$, chegamos ao sinal de saída do sistema:\n",
        "\n",
        "$y[n] = 0.33(-0.5)^nu[n] + 0.67u[n]$\n",
        "\n",
        "Abaixo é mostrado o gráfico do sinal."
      ]
    },
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              "Text(0, 0.5, 'y[n]')"
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          "data": {
            "text/plain": [
              "<Figure size 640x480 with 1 Axes>"
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\n"
          },
          "metadata": {}
        }
      ],
      "source": [
        "n = np.arange(20)\n",
        "y = 0.33*(-0.5)**n+0.67\n",
        "plt.stem(n, 0.33*(-0.5)**n+0.67);\n",
        "plt.xlabel('n')\n",
        "plt.ylabel('y[n]')"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "CTeUJHIbcZYs"
      },
      "source": [
        "No caso de sistemas de tempo discreto, é simples achar os valores da saída de forma recursiva. Isso é uma boa maneira de verificar se a expressão encontrada está correta."
      ]
    },
    {
      "cell_type": "code",
      "execution_count": null,
      "metadata": {
        "id": "kzLWBz2PcZYs",
        "outputId": "c97de91f-726b-4922-eddc-561db33397c9",
        "tags": [],
        "colab": {
          "base_uri": "https://localhost:8080/",
          "height": 467
        }
      },
      "outputs": [
        {
          "output_type": "execute_result",
          "data": {
            "text/plain": [
              "Text(0, 0.5, 'y[n]')"
            ]
          },
          "metadata": {},
          "execution_count": 3
        },
        {
          "output_type": "display_data",
          "data": {
            "text/plain": [
              "<Figure size 640x480 with 1 Axes>"
            ],
            "image/png": 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\n"
          },
          "metadata": {}
        }
      ],
      "source": [
        "N = 20\n",
        "x = np.ones(N)\n",
        "y = np.zeros(N)\n",
        "\n",
        "for n in range(N):\n",
        "    y[n] = -0.5*y[n-1]+x[n]\n",
        "plt.stem(y)\n",
        "plt.xlabel('n')\n",
        "plt.ylabel('y[n]')"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "y5uX_8T2cZYt"
      },
      "source": [
        "O gráfico é o mesmo quando foi utilizada a expressão encontrada fazendo a transformada Z inversa.\n",
        "\n",
        "---"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "yf1s9XvocZYt"
      },
      "source": [
        "## Resposta de entrada nula\n",
        "\n",
        "Na resposta de entrada nula, considera-se que não existe entrada. Dessa maneira, a equação de diferenças fica:\n",
        "\n",
        "$y[n] + a_{1}y[n-1] + ... + a_Ny[n-N] = 0$\n",
        "\n",
        "Aqui, vamos considerar as condições iniciais. A transformada Z de $y[n-M]$ quando existem condições iniciais no sinal $y[n]$ é:\n",
        "\n",
        "$\\mathcal{Z}(y[n-M]) = z^{-M}Y(z) + z^{-M}\\displaystyle\\sum_{k=1}^M y[-k]z^k$\n",
        "\n",
        "Fazendo essa substituição na equação de diferenças, é possível obter a resposta de entrada nula.\n"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "7LtTNChtcZYt"
      },
      "source": [
        "---\n",
        "\n",
        "### Exemplo 5\n",
        "\n",
        "Vamos considerar o sistema do exemplo 1, com a seguinte equação de diferenças:\n",
        "\n",
        "$y[n] + 3y[n-1] - 4y[n-2] = 3x[n] + x[n-1]$\n",
        "\n",
        "$y[-1] = 1$ e $y[-2] = -1$ e um sinal de entrada $x[n] = \\mathrm{sen}\\left(\\frac{\\pi}{3}n\\right)u[n]$\n",
        "\n",
        "Primeiro, vamos encontrar a resposta de entrada nula. Então, a transfromada Z da resposta de entrada nula:\n",
        "\n",
        "$Y_{zi}(z) + 3z^{-1}Y_{zi}(z) + 3y[-1]  - 4z^{-2}Y_{zi}(z) - 4(y[-2] + z^{-1}y[-1]) = 0$\n",
        "\n",
        "$Y_{zi}(z) = \\frac{-3y[-1]+4y[-2]+4z^{-1}y[-1]}{1 + 3z^{-1}- 4z^{-2}} = \\frac{-7+4z^{-1}}{1 + 3z^{-1}- 4z^{-2}}=\\frac{-7z^2+4z}{z^2 + 3z- 4z} = \\frac{-7z^2+4z}{(z+4)(z-1)}$\n",
        "\n",
        "Agora, fazemos a expansão em frações parciais de $Y'(z) = Y(z)/z$:\n",
        "\n",
        "\n",
        "$Y_{zi}(z) = \\frac{-7z^2+4z}{z(z+4)(z-1)} = \\frac{-7z+4}{(z+4)(z-1)} = \\frac{A_1}{z+4}+\\frac{A_2}{z-1}$\n",
        "\n",
        "$A_1 = \\left.\\frac{-7z+4}{z-1}\\right|_{z=-4} = 4.8$ e $A_2 = \\left.\\frac{-7z+4}{z+4}\\right|_{z=1} = -0.6$\n",
        "\n",
        "Agora, conseguimos encontrar $y_{zi}[n]$:\n",
        "\n",
        "$Y_{zi}(z) = \\frac{4.8z}{z+4}-\\frac{0.6z}{z-1} \\rightarrow y_{zi}[n] = 4.8(-4)^n u[n] -0.6$\n",
        "\n",
        "\n",
        "\n",
        "---"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "6MfsElCXcZYt"
      },
      "source": [
        "## Associações em série e  paralelo de sistemas\n",
        "\n",
        "Uma outra maneira de representar um sistema no diagrama de blocos é colocando a sua função de transferência dentro do bloco\n",
        "\n",
        "Abaixo, estão duas associações bem comuns de sistemas. Em paralelo e em série.\n",
        "\n",
        "Em paralelo, a função de transferência do sistema resulatante é a soma das funções de transferência de cada subsistema.\n",
        "\n",
        "![a](https://github.com/rnwatanabe/SistemasESinais/blob/master/Imagens/distributivaTFdisc.png?raw=1)\n",
        "\n",
        "\n",
        "Em série, a função de transferência do sistema resultante é a multiplicação das funções de transferência de cada subsistema.\n",
        "\n",
        "![a](https://github.com/rnwatanabe/SistemasESinais/blob/master/Imagens/associativaTFdisc.png?raw=1)"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "L7bm6MiTcZYt"
      },
      "source": [
        "## Diagrama de blocos\n",
        "\n",
        "Com o conceito de transformada Z, podemos agora introduzir uma nova forma, muito mais utilizada, de representar um atraso no diagrama de blocos.\n",
        "\n",
        "![a](https://github.com/rnwatanabe/SistemasESinais/blob/master/Imagens/delayzdisc.png?raw=1)\n",
        "\n",
        "Com a transformada Z é muito mais fácil obter a equação de diferenças a partir do diagrama de blocos. Basta substituir os sinais por suas transformadas e lembrar que a convolução no domínio do tempo é uma multiplicação no domínio da frequência.\n",
        "\n",
        "---\n",
        "\n",
        "### Exemplo 6\n",
        "\n",
        "Considere o seguinte diagrama de blocos:\n",
        "\n",
        "![a](https://github.com/rnwatanabe/SistemasESinais/blob/master/Imagens/diagexemploZ.png?raw=1)\n",
        "\n",
        "Podemos chamar o resultado do somador à esquerda de $w[n]$. A sua transformada Z é $W(z)$, que é igual a:\n",
        "\n",
        "$W(z) = X(z) + 0.3z^{-1}W(z)-0.5z^{-2}W(z) \\rightarrow W(z) = \\frac{X(z)}{1-0.3z^{-1}+0.5z^{-2}}$\n",
        "\n",
        "E agora,  $Y(z)$ é:\n",
        "\n",
        "$Y(z) = 0.1W(z) - 0.3z^{-1}W(z) = \\frac{(0.1 - 0.3z^{-1})X(z)}{1-0.3z^{-1}+0.5z^{-2}}$\n",
        "\n",
        "Então:\n",
        "\n",
        "$(1-0.3z^{-1}+0.5z^{-2})Y(z) = ( 0.1 - 0.3z^{-1})X(z) \\rightarrow y[n]-0.3y[n-1]+0.5y[n-2] = 0.1x[n]-0.3x[n-1]$\n",
        "\n",
        "Seria muito difícil obter a equação de diferenças sem a transformada Z.\n",
        "\n",
        "---"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "I0RELXp5cZYu"
      },
      "source": [
        "Para obter o diagrama de blocos a partir da função de transferência, basta dividir a função de transferência como:\n",
        "\n",
        "$H(z) = \\frac{num(z)}{den(z)} = \\frac{1}{den(z)}num(z)$\n",
        "\n",
        "\n",
        "\n",
        "Isso é como se tivéssemos dois sistemas em série. O primeiro sistema não contém nenhuma derivada da entrada, então fazer o seu diagrama de blocos é trivial. O segundo sistema vai pegar a saída do primeiro sistema e suas derivadas.\n",
        "\n",
        "![a](https://github.com/rnwatanabe/SistemasESinais/blob/master/Imagens/numdendiagdisc.png?raw=1)\n",
        "\n",
        "---\n",
        "\n",
        "### Exemplo 7\n",
        "\n",
        "Considere um sistema com a seguinte equação de diferenças:\n",
        "\n",
        "$$y[n] = -3y[n-1] + y[n-2] + 2x[n-1] + x[n-2]$$\n",
        "\n",
        "A sua função de transferência é:\n",
        "\n",
        "$H(z) = \\frac{2z^{-1}+z^{-2}}{1+3z^{-1}-z^{-2}} = \\frac{1}{1+3z^{-1}-z^{-2}}(2z^{-1}+z^{-2})$\n",
        "\n",
        "A equação de diferenças do primeiro sistema é:\n",
        "\n",
        "$$y[n] = -3y[n-1] + y[n-2] + x[n]$$\n",
        "\n",
        "Que tem o seguinte diagrama de blocos:\n",
        "\n",
        "\n",
        "![a](https://github.com/rnwatanabe/SistemasESinais/blob/master/Imagens/exemploTFdiagdisc.png?raw=1)\n",
        "\n",
        "\n",
        "O segundo sistema tem a seguinte equação diferencial:\n",
        "\n",
        "$y[n] = 2x[n-1] + x[n-2]$\n",
        "\n",
        "\n",
        "![a](https://github.com/rnwatanabe/SistemasESinais/blob/master/Imagens/exemploTFdiag2disc.png?raw=1)\n",
        "\n",
        "---"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "0nPYIqohcZYu"
      },
      "source": [
        "É importante enfatizar que esse é apenas um diagrama de blocos possível para o sistema. Existem diversos diagramas de blocos possíveis para um mesmo sistema."
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "ySp8GTtfcZYu"
      },
      "source": [
        "## Polos e zeros\n",
        "\n",
        "O valor dos polos e zeros, juntamente com o valor da função de transferência em $H(0)$ também caracterizam completamente o sistema, já que é possível construir a função de transferência a partir dessas informações.\n",
        "\n",
        "---\n",
        "\n",
        "### Exemplo 8\n",
        "\n",
        "Um sistema tem os polos em -0.5 e 0.2 e zeros em -1.5. Além disso, $H(0) = 3$.\n",
        "\n",
        "Então, a sua função de transferência é:\n",
        "\n",
        "$H(z) = \\frac{G(z+1.5)}{(z+0.5)(z-0.2)} =  \\frac{G(z+1.5)}{z^2+0.3z-0.1} $\n",
        "\n",
        "$H(0) = 3 = 15G \\rightarrow G = 0.2$\n",
        "\n",
        "Então:\n",
        "\n",
        "$H(s) =\\frac{0.2z+0.3}{z^2+0.3z-0.1}$\n",
        "\n",
        "---"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "YyTpmQFacZYw",
        "tags": []
      },
      "source": [
        "## Relação entre as diferentes representações de um sistema linear e invariante no tempo\n",
        "\n",
        "\n",
        "Até agora vimos diversas  formas de representar o comportamento dinâmico de sistemas lineares e invariantes no tempo. Sempre conseguimos chegar de uma até outra\n",
        "\n",
        "\n",
        "![a](https://github.com/rnwatanabe/SistemasESinais/blob/master/Imagens/relacoesHdisc.png?raw=1)"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "5NMO9MtDcZYw"
      },
      "source": [
        "## Regime permanente e transitório\n",
        "\n",
        "É muio comum na prática da engenharia dividir o sinal de saída em duas partes:\n",
        "\n",
        "- **transitório**: é a parte da saída que tende a zero com o passar do tempo, como uma exponencial com parte real negativa.\n",
        "\n",
        "\n",
        "- **regime permanente**: é a parte da saída que permanece até o infinito.\n",
        "\n",
        "---\n",
        "\n",
        "### Exemplo 9\n",
        "\n",
        "Um sistema tem a seguinte função de transferência:\n",
        "\n",
        "$H(z) = \\frac{z}{z-0.5}$\n",
        "\n",
        "É colocada uma entrada cossenoidal com frequência angular $\\frac{2\\pi}{3}$ e amplitude 1 a partir de $n=0$:\n",
        "\n",
        "$x[n] = u[n]$\n",
        "\n",
        "A transformada Z desse sinal é:\n",
        "\n",
        "$X(z) = \\frac{z}{z-1}$\n",
        "\n",
        "\n",
        "Então, a transformada Z do sinal de saída é:\n",
        "\n",
        "$Y(z) = \\frac{z}{z-0.5}\\frac{z}{z-1} = \\frac{z^2}{(z-0.5)(z-1)}$\n",
        "\n",
        "A transformada Z de $Y'(z) = \\frac{Y(z)}{z}$ é:\n",
        "\n",
        "$Y'(z) = \\frac{z^2}{z(z-0.5)(z-1)} = \\frac{A_1}{z-1} + \\frac{A_2}{z-0.5} + \\frac{A_3}{z}$\n",
        "\n",
        "$A_1 = \\frac{1}{1(0.5)} = 2 $\n",
        "\n",
        "$A_2 = \\frac{0.25}{0.5(-0.5)} = -1$\n",
        "\n",
        "$A_3 = \\frac{0}{-0.5(-1)} =0$\n",
        "\n",
        "\n",
        "Então $Y(z)$ é:\n",
        "\n",
        "$Y(z) = \\frac{2z}{z-1} - \\frac{z}{z-0.5}$\n",
        "\n",
        "Então:\n",
        "\n",
        "$y[n] = 2u[n] - 0.5^nu[n] = (2-0.5^n)u[n]$\n",
        "\n"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": null,
      "metadata": {
        "id": "APgXbUJncZYw",
        "outputId": "15064e05-4bd3-482c-b5c0-47242b55e68c",
        "tags": []
      },
      "outputs": [
        {
          "data": {
            "text/plain": [
              "Text(8, 2.5, 'Regime permanente')"
            ]
          },
          "execution_count": 13,
          "metadata": {},
          "output_type": "execute_result"
        },
        {
          "data": {
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",
            "text/plain": [
              "<Figure size 640x480 with 1 Axes>"
            ]
          },
          "metadata": {},
          "output_type": "display_data"
        }
      ],
      "source": [
        "n = np.arange(0, 15, 1)\n",
        "y = 2 - 0.5**n\n",
        "plt.stem(n, y)\n",
        "plt.vlines(6.5, 0, 3, linestyle='--')\n",
        "plt.text(1,2.5,'transitório')\n",
        "plt.text(8, 2.5, 'Regime permanente')"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "Sc3Dw8vWcZYx"
      },
      "source": [
        "No gráfico acima do sinal de saída é possível ver que até aproximadamente $n=6$ o sinal de saída ainda não chegou no comportamento final. Até $n=6$, aproximadamente, está o transitório. Depois que o transitório acaba, o sinal entra no regime permanente. O transitório é devido à exponencial e o regime permanente devido ao degrau.\n",
        "\n",
        "---"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "GCaKRZlscZYx"
      },
      "source": [
        "## Exercícios\n",
        "\n"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "ZnhprlTDcZYx"
      },
      "source": [
        "**1)** Qual é a função de transferência dos sistemas com as seguintes equações de diferenças:\n",
        "\n",
        "a) $y[n] = 3y[n-1] + 6y[n-2] + 4x[n]$\n",
        "\n",
        "b) $y[n] = 0.1y[n-1] + 2y[n-2] + x[n] - 0.1x[n-2]$\n",
        "\n",
        "c) $y[n] + 5y[n-2] = x[n] + 2x[n-1]$\n"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "UtF05JEPcZYx"
      },
      "source": [
        "**2)** Qual é a saída dos sistemas seguintes para as respectivas entradas. Considere que inicialmente o sistema está em repouso:\n",
        "\n",
        "a) $h[n] = (0.7)^nu[n]$ com $x[n] = (0.1)^nu[n]$\n",
        "\n",
        "b)\n",
        "\n",
        "![a](https://github.com/rnwatanabe/SistemasESinais/blob/master/Imagens/diagexecTF.png?raw=1)\n",
        "\n",
        "$x[n] = \\left(\\frac{2}{3}\\right)^nu[n]$\n",
        "\n",
        "c) $H(z) = \\frac{z^{-1}+1}{1-0.2z^{-1}+ 0.4z^{-2}}$, com $x[n] = u[n]$\n",
        "\n",
        "\n",
        "d) $y[n] + 5y[n-2] = x[n] + 2x[n-1]$, e $x[n] = nu[n]$\n",
        "\n"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "YRxQGDdGcZYx"
      },
      "source": [
        "**3)** Encontre a saída dos sistemas com as seguintes respostas ao impulso para as respectivas entradas. Considere que inicialmente os sistemas se encontram em repouso.\n",
        "\n",
        "a) $h[n] = 0.5^nu[n]$ e $x[n] = u[n] - u[10]$\n",
        "\n",
        "b) $h[n] = (-0.1)^{n-1}u[n-1]$ e $x[n] = nu[n]$\n"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "sha_wCjacZYy"
      },
      "source": [
        "**4)** Um sistema com resposta ao impulso $h[n] = 2(0.2)^{n}u[n-1]$ gera uma saída $y[n] = (-3)^nu[n-1]$. Qual a entrada que gerou essa saída?"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "v6WKesFFcZYy",
        "tags": []
      },
      "source": [
        "**5)** Qual a resposta ao impulso dos seguintes sistemas:\n",
        "\n",
        "a) $y[n] = -y[n-1] - 4y[n-2] + x[n] + 3x[n-1] + 2x[n-2]$\n",
        "\n",
        "b) $y[n+2] + 3y[n+1] + y[n] = 2x[n+2]-3x[n+1]$\n",
        "\n",
        "c) $y[n] = y[n-1] - 0.5y[n-2]+x[n]+2x[n-1]$\n"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "7LT3KHiGcZYy"
      },
      "source": [
        "**6)** Encontre o sinal de saída dos sistemas com as seguintes equações de diferenças tendo como entrada as respectivas entradas e as condições iniciais inidciadas.\n",
        "\n",
        "a) $y[n] = 0.5y[n-1] - 0.25y[n-2] = x[n]$,    $x[n] = 0.5^nu[n]$ e  $y[-1] = 0$ e $y[-2]=0$\n",
        "\n",
        "b) $y[n] = -2y[n-1]+x[n]$,    $x[n] = 0.5^nu[n]$ e  $y[-1] = 1$\n",
        "\n",
        "c) $y[n] = 0.2y[n-1]+ x[n] - 0.2x[n-1]$,    $x[n] = u[n]$ e  $y[-1] = 0$\n",
        "\n",
        "d) $y[n] = 0.2y[n-1]+ x[n] - 0.2x[n-1]$,    $x[n] = u[n]$ e  $y[-1] = 1$"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "bVOCB4ovcZYy"
      },
      "source": [
        "**7)**  Calcule o sinal de saída de estado nulo do sistema com a seguinte resposta ao impulso para as seguintes entradas. Identifique o transitório e o regime permanente:\n",
        "\n",
        "$h[n] = 0.5^nu[n] $\n",
        "\n",
        "a) $x_1[n] = 0.5^nu[n]$\n",
        "\n",
        "b) $x_2[n] = u[n]$\n",
        "\n",
        "c) $x_3[n] = \\cos\\left(\\frac{\\pi}{3}n\\right)u[n]$\n"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "6kKsStvncZYy"
      },
      "source": [
        "**8)** Calcule o sinal de saída de estado nulo do sistema com a seguinte resposta ao impulso para as seguintes entradas. Identifique o transitório e o regime permanente:\n",
        "\n",
        "$h[n] = (0.9)^n\\cos\\left(\\frac{\\pi}{6}n \\right)u[n] $\n",
        "\n",
        "a) $x_1[n] = (0.9)^n\\cos\\left(\\frac{\\pi}{6}n \\right)u[n]$\n",
        "\n",
        "b) $x_2[n] = (0.5)^n\\cos\\left(\\frac{\\pi}{6}n \\right)u[n]$\n",
        "\n",
        "c) $x_3[n] = (-0.9)^n\\cos\\left(\\frac{\\pi}{6}n \\right)u[n]$\n"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "WjyCBfSucZYy"
      },
      "source": [
        "**9)** Faça o diagrama de blocos do sistema com a seguinte equação de diferenças utilizando apenas 2 blocos de atraso:\n",
        "\n",
        "$y[n] = 0.1y[n-1] - 0.5y[n-2] + x[n-1] +0.3x[n-2]$"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "qdePydtao0O6"
      },
      "source": [
        "**10)** Encontre as equações de diferenças com as seguintes respostas ao impulso:\n",
        "\n",
        "a) $h_1[n] = n(0.8)^nu[n]$\n",
        "\n",
        "b) $h_2[n] = 2(0.2)^n\\cos\\left(\\frac{\\pi}{3}n\\right)u[n]$\n",
        "\n",
        "c) $h_3[n] = 2(0.5)^n\\mathrm{sen}\\left(\\frac{\\pi}{4}n\\right)u[n]$\n",
        "\n",
        "d) $h_4[n] = \\delta[n] + (-0.2)^nu[n]$"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "tWp_LJvscZYz"
      },
      "source": [
        "## Referências\n",
        "\n",
        "- Ulaby, FT e Yagle, AE, Signals & Systems: Theory and Applications, 2018, [ss2.eecs.umich.edu](ss2.eecs.umich.edu)\n",
        "- Lathi, BP, Sinais e Sistemas Lineares, 2007\n",
        "- Oppenheim, M.  e Willsky, S., Sistemas e Sinais, 2010"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "cUmOmSEFcZY5"
      },
      "source": [
        "## Solução dos exercícios"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "lKNQy961cZY6"
      },
      "source": [
        "**1)** Resolvido por Carolina Kafka Neves\n",
        "\n",
        "Qual é a função de transferência dos sistemas com as seguintes equações de diferenças:\n",
        "\n",
        "a) $y[n] = 3y[n-1] + 6y[n-2] + 4x[n]$\n",
        "\n",
        "Pela equação de diferenças, sabemos que\n",
        "\n",
        "$Y(z)(1-3z^{-1}-6z^{-2})=X(z)(4)$\n",
        "\n",
        "Como $H(z)=\\frac{Y(z)}{X(z)}$, temos que\n",
        "\n",
        "$H(z)=\\frac{4}{(1-3z^{-1}-6z^{-2})}$ = $\\frac{4z^2}{(z^{2}-3z-6)}$\n",
        "\n",
        "-----------------------------------------------------------------------------\n",
        "\n",
        "b) $y[n] = 0.1y[n-1] + 2y[n-2] + x[n] - 0.1x[n-2]$\n",
        "\n",
        "Seguindo o mesmo raciocínio utilizado na alternativa (a), temos que\n",
        "\n",
        "$Y(z)(1-0.1z^{-1}-2z^{-2})=X(z)(1-0.1z^{-2})$\n",
        "\n",
        "Logo,\n",
        "\n",
        "$H(z)= \\frac{1-0.1z^{-2}}{1-0.1z^{-1}-2z^{-2}}$ = $\\frac{z^{2}-0.1}{z^{2}-0.1z-2}$\n",
        "\n",
        "------------------------------------------------------------------------------\n",
        "c) $y[n] + 5y[n-2] = x[n] + 2x[n-1]$\n",
        "\n",
        "Temos que\n",
        "\n",
        "$Y(z)(1+5z^{-2})=X(z)(1+2z^{-1})$\n",
        "\n",
        "Portanto,\n",
        "\n",
        "$H(z)= \\frac{1+2z^{-1}}{1+5z^{-2}}$ = $\\frac{z^{2}+2z}{z^{2}+5}$\n",
        "\n",
        "\n",
        "---\n",
        "\n"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "VF6Q5ZnWN6tg"
      },
      "source": [
        "**4)** Resolvido por Carolina Kafka Neves\n",
        "\n",
        "Um sistema com resposta ao impulso $h[n] = 2(0.2)^{n}u[n-1]$ gera uma saída $y[n] = (-3)^nu[n-1]$. Qual a entrada que gerou essa saída?\n",
        "\n",
        "As transformadas Z da resposta ao impulso e da saída são:\n",
        "\n",
        "$H(z)= \\frac{2}{z-0.2}\\frac{z^{-1}z}{z-1}$=$\\frac{2}{(z-0.2)(z-1)}$\n",
        "\n",
        "$Y(z)= \\frac{z^{-1}z}{(z+3)(z-1)}$=$\\frac{1}{(z+3)(z-1)}$\n",
        "\n",
        "Sabemos que $Y(z)=H(z)X(z)$, logo $X(z)=\\frac{Y(z)}{H(z)}$\n",
        "\n",
        "$X(z)= \\frac{(z-0.2)(z-1)}{2(z+3)(z-1)}=\\frac{z-0.2}{2z+6}$\n",
        "\n",
        "$X'(z)= \\frac{z-0.2}{2z^{2}+6z}$\n",
        "\n",
        "Resolvendo por frações parciais, temos que\n",
        "\n",
        "$\\frac{z-0.2}{2z^{2}+6z}= \\frac{A_1}{z}+\\frac{A_2}{z+3}$\n",
        "\n",
        "$A_1= \\frac{-0.2}{6}= 0.033$\n",
        "\n",
        "$A_2= \\frac{2.8}{6}= 0.467$\n",
        "\n",
        "Desta forma, voltamos para $X(z)=zX'(z)$\n",
        "\n",
        "$\\frac{z-0.2}{2z^{2}+6z}= \\frac{0.033z}{z}+\\frac{0.467z}{z+3}=0.033+\\frac{0.467z}{z+3}$\n",
        "\n",
        "Então, temos que a entrada so sistema que gerou a saida $y[n]$ é\n",
        "\n",
        "$x[n]= 0.033\\delta[n]+0.467(-3)^{n}u[n]$\n",
        "\n",
        "\n",
        "\n",
        "---"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": null,
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        "id": "wNZEit7Ko0O-"
      },
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