Student_Control_Module_DAY1.ipynb

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   "source": [
    "# Welcome to Autonomous Control Systems!\n",
    "# This course is broken up over two days\n",
    "# Day 1 has 6 Parts\n",
    "# Day 2 has 5 Parts"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [],
   "source": [
    "# This course will utilize the following imports\n",
    "import numpy as np\n",
    "from numpy import matrix\n",
    "import scipy.signal as sig\n",
    "import matplotlib.pyplot as plt\n",
    "from numpy.linalg import matrix_rank\n",
    "from numpy import sqrt\n",
    "from scipy.linalg import solve_continuous_are as CARE"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [],
   "source": [
    "#### DAY 1 PART 1: Solving Systems of Equations"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [],
   "source": [
    "##Example 1: (Problem Packet)\n",
    "# Consider 2*X1 + X2 = 1   (1)\n",
    "# and     -3*X1 + 4*X2 = 2 (2)\n",
    "\n",
    "# Solve for intersection points x=[X1,X2]:\n",
    "#print(\"Solution (intersection points) are: \", x)\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [],
   "source": [
    "##Example 2: (Problem Packet)\n",
    "# Consider 4*X1 + 6*X2 = 2   (1)\n",
    "# and     2*X1 + 3*X2 = 1 (2)\n",
    "\n",
    "# Solve for intersection points x=[X1,X2]:\n",
    "#print(\"Solution (intersection points) are: \", x)\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {},
   "outputs": [],
   "source": [
    "##Example 3: (Problem Packet)\n",
    "# Consider -2*X1 + 3*X2 = 2 (1)\n",
    "# and      5*X1 - 2*X2 = 1 (2)\n",
    "\n",
    "#Solve for intersection points x=[X1,X2]:\n",
    "#print(\"Solution (intersection points) are: \", x)\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {},
   "outputs": [],
   "source": [
    "##Example 4: (Problem Packet)\n",
    "# Consider 1*X1 + 2*X2 + 3*X3 = 1 (1)\n",
    "# and      4*X1 + 5*X2 + 6*X3 = 2 (2)\n",
    "# and      -7*X1 + 8*X2 + 9*X3 = 3 (3)\n",
    "\n",
    "# Solve for intersection points x=[X1,X2,X3]:\n",
    "#print(\"Solution (intersection points) are: \", x)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {},
   "outputs": [],
   "source": [
    "##Example 5: (Problem Packet)\n",
    "# Consider   -2*X1 + 2*X2 + 7*X3 = 8 (1)\n",
    "# and      4*X1 + 5*X2 - X3 = -3 (2)\n",
    "# and      9*X1 + X2 + 3*X3 = 11 (3)\n",
    "\n",
    "# Solve for intersection points x=[X1,X2,X3]:\n",
    "#print(\"Solution (intersection points) are: \", x)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {},
   "outputs": [],
   "source": [
    "#### DAY 1 PART 2: Matrix Math"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {},
   "outputs": [],
   "source": [
    "##Example 6:\n",
    "v1 = np.matrix([1,2,3]).reshape((3,1))\n",
    "v2 = np.matrix([1,0,4]).reshape((3,1))\n",
    "v3 = np.matrix([4,5]).reshape((2,1))\n",
    "v4 = np.matrix([1,3]).reshape((2,1))\n",
    "\n",
    "# Dimentionalities are 3x1 for (v1,v2) and 2x1 for (v3,v4)\n",
    "# Solve packet problem questions below"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {},
   "outputs": [],
   "source": [
    "# Example 7:\n",
    "# Answer the packet problem questions below"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "metadata": {},
   "outputs": [],
   "source": [
    "# Example 8.1:\n",
    "# Consider   2*X1 + 3*X2 + X3 = -4 (1)\n",
    "# and        3*X1 - X2 + 7*X3 = 1 (2)\n",
    "# and        6*X1 + 11*X2 - 5*X3 = 5 (3)\n",
    "\n",
    "# Solve for intersection points x=[X1,X2]:\n",
    "#print(\"Solution (intersection points) are: \", x)\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "metadata": {},
   "outputs": [],
   "source": [
    "# Example 8.2:\n",
    "# Consider   1*X1 + 2*X2 + 3*X3 = 1     (1)\n",
    "# and        4*X1 + 5*X2 + 6*X3 = 2     (2)\n",
    "# and        8*X1 + 10*X2 + 12*X3 = 4  (3)\n",
    "\n",
    "# Solve for intersection points x=[X1,X2]:\n",
    "#print(\"Solution (intersection points) are: \", x)\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "metadata": {},
   "outputs": [],
   "source": [
    "#DAY 1 PART 3: Fundamental Theorum of Lin-Alg (Extra Section)\n",
    "# See Packet"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "metadata": {},
   "outputs": [],
   "source": [
    "#DAY 1 PART4: Kinematics and Motion"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 16,
   "metadata": {},
   "outputs": [],
   "source": [
    "# Example 12: \n",
    "# Suppose we have the position function p(t) = t^2\n",
    "# with t representing elapsed time from start\n",
    "def p(t):\n",
    "    return .5*t**2\n",
    "\n",
    "# A) Plot position of car\n",
    "t = np.arange(1,10,.05)\n",
    "#print(\"Positions at times: 0, .5, 1, 2, 3 seconds: \", p0, phalf, p1, p2, p3)\n",
    "\n",
    "# B) Average velocities of car:\n",
    "#print(\"Average Velocities at times: 0-1,1-2,2-3 seconds: \", v0_1, v1_2, v2_3)\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 17,
   "metadata": {},
   "outputs": [],
   "source": [
    "# Example 13:\n",
    "# p(t) = 2*t**3\n",
    "# find v(t) and a(t)\n",
    "def p(t):\n",
    "    return 2*t**3\n",
    "def v(t):\n",
    "    return 6*t**2\n",
    "def a(t):\n",
    "    return 12*t\n",
    "t = np.arange(0,10,.1)\n",
    "\n",
    "# 1) Plot of Position vs Time\n",
    "# 2) Plot of Velocity vs Time\n",
    "# 3) Plot of Acceleration vs Time\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 18,
   "metadata": {},
   "outputs": [],
   "source": [
    "# Example 14:\n",
    "# p(t) = -1/2*t^2 + t + 4t^3\n",
    "# code velocity and accelerations too. Plot all three vs time\n",
    "# 1) Plot of Position vs Time\n",
    "# 2) Plot of Velocity vs Time\n",
    "# 3) Plot of Acceleration vs Time\n",
    "\n",
    "def p(t):\n",
    "    return -0.5*t**2 + t + 4*t**3\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 19,
   "metadata": {},
   "outputs": [],
   "source": [
    "# Example 15:\n",
    "# p(t) = 3*t^3 -2*t^2 + 5\n",
    "# code velocity and accelerations too. Plot all three vs time\n",
    "# 1) Plot of Position vs Time\n",
    "# 2) Plot of Velocity vs Time\n",
    "# 3) Plot of Acceleration vs Time"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 20,
   "metadata": {},
   "outputs": [],
   "source": [
    "# Example 16:\n",
    "# code position function given in packet\n",
    "# code velocity and accelerations too. Plot all three vs time\n",
    "# 1) Plot of Position vs Time\n",
    "# 2) Plot of Velocity vs Time\n",
    "# 3) Plot of Acceleration vs Time"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 21,
   "metadata": {},
   "outputs": [],
   "source": [
    "#Example 17 \n",
    "# Rocket heading down toward earth example (free-fall)\n",
    "d0 = 1000 #m\n",
    "v0 = 0 #m/s\n",
    "accel = -9.81 #m/s^2\n",
    "\n",
    "# a) At what time will the rocket hit the ocean? ie d = 0m\n",
    "# formula is d(t) = .5*accel*t**2 + v0*t + d0\n",
    "#print(\"Time at impact: \", ENTER)\n",
    "\n",
    "#b) At what time will it hit if it starts with -30m/s velocity?\n",
    "# formula is d(t) = .5*accel*t**2 + v0*t + d0\n",
    "#print(\"Time at impact: \", ENTER)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 22,
   "metadata": {},
   "outputs": [],
   "source": [
    "#Example 18\n",
    "# Cannon ball launches ball upwards vertically\n",
    "d0 = 1 # meter\n",
    "v0 = 150 # meters a second\n",
    "accel = -9.81 # gravity meters/seconds^2\n",
    "\n",
    "#a) Find the terminal height of the ball\n",
    "#print('Terminal height is: ', ENTER)\n",
    "\n",
    "#b) At what time does the ball reach its terminal height?\n",
    "#we can use d(t) = 1/2*accel*t^2 + v0*t + d0\n",
    "#print('Time at Terminal Height is: ', ENTER)\n",
    "\n",
    "#c Find time when cannon ball is at 1 meter again:\n",
    "#print('Time at Height 1 meter (Downwards) is: ', ENTER)\n",
    "\n",
    "#d Velocity will be 150 m/s downwards\n",
    "\n",
    "#e Plot position, velocity and acceleration to confirm answers above\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 23,
   "metadata": {},
   "outputs": [],
   "source": [
    "# PART 5/6: Physics and Free Body Diagrams"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  },
  {
   "cell_type": "code",
   "execution_count": 24,
   "metadata": {},
   "outputs": [],
   "source": [
    "# PART 6/6: Open Loop Control System Modelling"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  }
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