13.1950-MinimumEnergyControl.ipynb

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  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
   "outputs": [],
   "source": [
    "using Plots\n",
    "using Statistics\n",
    "using LinearAlgebra\n",
    "using JSON\n",
    "\n",
    "include(\"readclassjson.jl\");"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Problem 13.1950, Part B"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "$A=\\left[\\begin{array}{cccc}\n",
    "0.5 & 0.7 & -0.9 & -0.5 \\\\\n",
    "0.4 & -0.7 & 0.1 & 0.3 \\\\\n",
    "0.7 & 0.0 & -0.6 & 0.1 \\\\\n",
    "0.4 & -0.1 & 0.8 & -0.5\n",
    "\\end{array}\\right], \\quad B=\\left[\\begin{array}{l}\n",
    "1 \\\\\n",
    "1 \\\\\n",
    "0 \\\\\n",
    "0\n",
    "\\end{array}\\right], \\quad x_{\\mathrm{des}}=\\left[\\begin{array}{c}\n",
    "0.8 \\\\\n",
    "2.3 \\\\\n",
    "-0.7 \\\\\n",
    "-0.3\n",
    "\\end{array}\\right]$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "\n",
    "We can now relate vector $x(t)$ to vector $u$ via $x(t) = C_tu$. To achieve a desired x value, we know that $x_{des}$ must lie in the range of $C_t$ - otherwise no input $u$ could produce that output. Since $C_t$ grows with each time step, we can simply compute $C_t$ with the given A and B and determine if $x_{des} \\in range(C_t)$. \n",
    "    \n",
    "More generally, for any $x_{\\mathrm{des}} \\in \\mathbb{R}^{4}$, we need all $x_{\\mathrm{des}}$ to be in the range of $C_t$. Therefore the minimum T to achieve that is the first T for which the rank of $C_t$ is 4, since rank is the dimension of the range. Then the columns of $C_t$ will span $R^4$ and can produce any 4 dimensional desired output.\n",
    "    "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Starting with the specific case"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "4-element Array{Float64,1}:\n",
       "  0.8\n",
       "  2.3\n",
       " -0.7\n",
       " -0.3"
      ]
     },
     "execution_count": 2,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "A = [0.5 0.7 -0.9 -0.5 ; 0.4 -0.7 0.1 0.3; 0.7 0.0 -0.6 0.1; 0.4 -0.1 0.8 -0.5]\n",
    "B = [1;1;0;0]\n",
    "xdes = [0.8; 2.3;-0.7;-0.3]"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 19,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "Ct (generic function with 1 method)"
      ]
     },
     "execution_count": 19,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "function Ct(t)\n",
    "    reduce(hcat, [A^i * B for i in range(0, stop=t-1)])\n",
    "end"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 22,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "1false\n",
      "2true\n",
      "3true\n",
      "4true\n",
      "5true\n"
     ]
    }
   ],
   "source": [
    "for t = 1:5\n",
    "    C = Ct(t)\n",
    "    println(t, rank(C) == rank([C xdes]))\n",
    "end"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Therefore T = 2 is the smallest T for which $x_{des}$ is a possible output."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 23,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "4×2 Array{Float64,2}:\n",
       " 1.0   1.2\n",
       " 1.0  -0.3\n",
       " 0.0   0.7\n",
       " 0.0   0.3"
      ]
     },
     "execution_count": 23,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "C2 = Ct(2)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "To find the inputs, we can use the left inverse of $C_2$ (which should be exact given $x_{des} \\in R(C_2)$). Therefore $\\begin{bmatrix}u(1) \\\\ u(0)\\end{bmatrix} =$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 27,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "2-element Array{Float64,1}:\n",
       "  2.0\n",
       " -0.9999999999999997"
      ]
     },
     "execution_count": 27,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "u = C2 \\ xdes"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 29,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "true"
      ]
     },
     "execution_count": 29,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "isapprox(C2 * u, xdes)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "For the general case of any $x_{des} \\in R^4$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 32,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "1\n",
      "2\n",
      "3\n",
      "4\n",
      "4\n"
     ]
    }
   ],
   "source": [
    "for t = 1:5\n",
    "    println(rank(Ct(t)))\n",
    "end"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 34,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "4×4 Array{Float64,2}:\n",
       " 1.0   1.2  -0.39  -0.465\n",
       " 1.0  -0.3   0.85  -0.43\n",
       " 0.0   0.7   0.45  -0.451\n",
       " 0.0   0.3   0.92  -0.341"
      ]
     },
     "execution_count": 34,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "C4 = Ct(4)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 35,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "4"
      ]
     },
     "execution_count": 35,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "rank(C4)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "$T = 4$ is the first time we produce a $C_T$ with rank 4, meaning its columns span the space $R^4$. Therefore $T_{min} = 4$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Problem 13.1950, Part C"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Since $T > T_{min}$ is given, we know that $x_{des} \\in R(C_T)$. Since $T > T_{min}$ produces a fat matrix with full row rank, we find ourselves in a classic least norm solution scenario. We can first note that the definition of $E(T)$ is the squared L2 norm of the vector $u$. Therefore we can simply use the the least norm solution $x_{\\ln }=C^{\\top}\\left(CC^{\\top}\\right)^{-1} x_{des}$ for each $T$, using the $x_{des}$ provided."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 36,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "4:30"
      ]
     },
     "execution_count": 36,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "T = 4:30"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 37,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "4-element Array{Int64,1}:\n",
       " -1\n",
       "  1\n",
       "  0\n",
       "  1"
      ]
     },
     "execution_count": 37,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "xdes = [-1 ; 1 ; 0 ; 1]"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 39,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "ln (generic function with 1 method)"
      ]
     },
     "execution_count": 39,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "function ln(T)\n",
    "    C = Ct(T)\n",
    "    u = C \\ xdes\n",
    "\n",
    "    return(norm(u)^2)\n",
    "end"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 41,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "27-element Array{Float64,1}:\n",
       " 8.059917397553912\n",
       " 3.0115050784899995\n",
       " 2.6368756923690087\n",
       " 2.4931257799379205\n",
       " 2.257459439954998\n",
       " 2.095320755560134\n",
       " 2.0246748311922445\n",
       " 1.9698124753581177\n",
       " 1.9153577801223929\n",
       " 1.8745844621427288\n",
       " 1.8453039413319832\n",
       " 1.8200049770953326\n",
       " 1.7976574435870412\n",
       " ⋮\n",
       " 1.7499727071250208\n",
       " 1.7378857306213245\n",
       " 1.7273198712019857\n",
       " 1.7179627943700966\n",
       " 1.7095715443038204\n",
       " 1.7020366690226578\n",
       " 1.6952564448353313\n",
       " 1.6891165131647101\n",
       " 1.6835303286307122\n",
       " 1.678435857589008\n",
       " 1.673776283147763\n",
       " 1.669499923719386"
      ]
     },
     "execution_count": 41,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "u_norms = ln.(T)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 45,
   "metadata": {},
   "outputs": [
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       "  988.572,1412.5 1069.87,1417.59 1151.17,1422.09 1232.47,1425.8 1313.76,1428.92 1395.06,1431.68 1476.36,1434.11 1557.66,1436.24 1638.96,1438.12 1720.26,1439.81 \n",
       "  1801.55,1441.33 1882.85,1442.69 1964.15,1443.93 2045.45,1445.05 2126.75,1446.08 2208.05,1447.01 2289.34,1447.87 \n",
       "  \"/>\n",
       "</svg>\n"
      ]
     },
     "execution_count": 45,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "plot(T, u_norms, label=false)\n",
    "title!(\"Min energy solutions vs. T\")"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Problem 13.1950, Part D"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "*You should observe that $E_{\\min }(T)$ is non-increasing in $T$. Show that this is the case in general (i.e., for any $A, B,$ and $\\left.x_{\\text {des }}\\right)$*"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "First, we can look at the problem intuitively. For E(T) to be non-increasing, it can either be constant or decreasing. The input vector u adds terms as T grows, but the desired output does not change. If we can achieve the desired output in X timesteps, we should be able to complete the same task with constant energy given more time. We could simply use the same X timesteps to achieve the goal and the remaining T-X timesteps can be set to 0.\n",
    "\n",
    "Turning to our formulation of the problem $E_{min}(T) = ||u||^2$ and $x_{des} = C_T u$ with $u = \\begin{bmatrix} u(T-1) \\\\ \\vdots \\\\ u(0)\\end{bmatrix}$. Let $T_1$ be some time greater than $T_{min}$ and $T_2 > T_1$. Let $u_1 = \\begin{bmatrix} u(T_1-1) \\\\ \\vdots \\\\ u(0)\\end{bmatrix}$ and $u_2 = \\begin{bmatrix} 0 \\\\ \\vdots \\\\ 0 \\\\ u(T_1-1) \\\\ \\vdots \\\\ u(0)\\end{bmatrix}$. Both vectors have the same norm since they possess the same non-zero entries. Additionally $C_{T_1} u_1 = C_{T_2} u_2$ can be easily seen since all additional entries in $C_{T_2}$ are multiplied by zero. Therefore for any arbitrary $T_2 > T_1$ we can always achieve the same goal with constant energy in T. Therefore we can say in general $E_{min}(T)$ is non-increasing in T."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  }
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