from sympy import * from sympy.logic.boolalg import * init_printing() U, A, B, t, x0, xf, v0, vf, c1, c2, v, V, kV, kA = symbols( "U, A, B t, x0, xf, v0, vf, C1, C2, v, V, kV, kA" ) x = symbols("x", cls=Function) # Exponential profiles are derived from a differential equation: ẍ - A * ẋ = B * U diffeq = Eq(x(t).diff(t, t) - A * x(t).diff(t), B * U) x = dsolve(diffeq).rhs dx = x.diff(t) x = x.subs( [ (c1, solve(Eq(x.subs(t, 0), x0), c1)[0]), (c2, solve(Eq(dx.subs(t, 0), v0), c2)[0]), ] ) print(f"General Solution: {x}") # We need two specific solutions to this equation for an Exponential Profile: # One that passes through (x0, v0) and has input U # Another that passes through (xf, vf) and has input -U # x1 is for the accelerate step x1 = x.subs({x0: x0, v0: v0, U: U}) dx1 = x1.diff(t) t1_eqn = solve(Eq(dx1, v), t)[0] # x1 in phase space (input v, output x) x1_ps = x1.subs(t, t1_eqn) # x2 is for the decelerate step x2 = x.subs({x0: xf, v0: vf, U: -U}) dx2 = x2.diff(t) t2_eqn = solve(Eq(dx2, v), t)[0] # x2 in phase space (input v, output x) x2_ps = x2.subs(t, t2_eqn) # The point at which we switch from input U to -U is the inflection point. # In phase space, this is a point (x, v) where x1(v) = x2(v) # For now, we will just solve for +U and assume inflection velocity is positive. # The other possible solutions are -v_soln, and the solutions to v_equality.subs(U, -U) equality = simplify(Eq(x1_ps, x2_ps).expand()).expand() equality = Eq(equality.lhs - x0 + v0 / A - v / A, equality.rhs - x0 + v0 / A - v / A) equality = Eq( equality.lhs - B * U * log(A * v / (A * vf - B * U) - B * U / (A * vf - B * U)) / A**2, equality.rhs - B * U * log(A * v / (A * vf - B * U) - B * U / (A * vf - B * U)) / A**2, ) equality = Eq(equality.lhs / (-B * U / A / A), equality.rhs / (-B * U / A / A)) equality = Eq(exp(equality.lhs.simplify()), exp(equality.rhs.simplify())) equality = Eq( equality.lhs * (A * v0 + B * U) * (A * vf - B * U), equality.rhs * (A * v0 + B * U) * (A * vf - B * U), ) equality = Eq(-equality.lhs.expand() + equality.rhs, 0) # This is a quadratic equation of the form ax^2 + c = 0 v_equality = equality # solve, take positive result v_soln = solve(v_equality, v)[0] # With this information, we can calculate the inflection point (x, v) # and calculate the times that x1 and x2 reach the inflection point inflection_x = x1_ps.subs(v, v_soln) inflection_t1 = t1_eqn.subs(v, v_soln) inflection_t2 = t2_eqn.subs(v, v_soln) # inflection_t2 < 0 because in order for the profile to get to # the inflection point from the terminal state, it must go back in time. totalTime = inflection_t1 - inflection_t2 print(f"x1: {expand(simplify(x1))}") print(f"x2: {expand(simplify(x2))}") print(f"dx1: {expand(simplify(dx1))}") print(f"dx2: {expand(simplify(dx2))}") print(f"t1: {expand(simplify(t1_eqn))}") print(f"t2: {expand(simplify(t2_eqn))}") print(f"x1 phase space: {expand(simplify(x1.subs(t, t1_eqn)))}") print(f"x2 phase space: {expand(simplify(x2.subs(t, t2_eqn)))}") print(f"vi equality: {v_equality}") a, b, c, d = symbols("a, b, c, d") expression = SOPform([a, b, c, d], minterms=[0, 4, 5, 8, 10, 12, 13, 14]) print(f"Truth Table Expression: {expression}")