本文整理汇总了Python中sympy.Poly.coeffs方法的典型用法代码示例。如果您正苦于以下问题：Python Poly.coeffs方法的具体用法？Python Poly.coeffs怎么用？Python Poly.coeffs使用的例子？那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类sympy.Poly的用法示例。
在下文中一共展示了Poly.coeffs方法的4个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的Python代码示例。
示例1: main
# 需要导入模块: from sympy import Poly [as 别名]
# 或者: from sympy.Poly import coeffs [as 别名]
def main():
x=Symbol("x")
s = Poly(A(x), x)
num = list(reversed(s.coeffs()))[:11]
print s.as_basic()
print num示例2: discrete_realization_tustin
# 需要导入模块: from sympy import Poly [as 别名]
# 或者: from sympy.Poly import coeffs [as 别名]
def discrete_realization_tustin(n0, n1, d0, T):
z = symbols('z')
s = 2/T*(z-1)/(z+1)
num = ((n1*s + n0)*T*(z + 1)).simplify()
den = ((s + d0)*T*(z + 1)).simplify()
num_poly = Poly(num, z)
den_poly = Poly(den, z)
n1_z, n0_z = num_poly.coeffs()
d1_z, d0_z = den_poly.coeffs()
# Make denominator monic and divide numerator appropriately
n1_z /= d1_z
n0_z /= d1_z
d0_z /= d1_z
a = -d0_z
b_times_c = (n0_z - n1_z * d0_z).simplify()
d = n1_z
return a, b_times_c, d示例3: mprint
# 需要导入模块: from sympy import Poly [as 别名]
# 或者: from sympy.Poly import coeffs [as 别名]
# See equations 4.4.3 and 4.11.4 of Kane & Levinson
Fr_c = Fr_u[:3, :].col_join(Fr_u[6:, :]) + A_rs.T * Fr_u[3:6, :]
Fr_star_c = Fr_star_u[:3, :].col_join(Fr_star_u[6:, :])\
+ A_rs.T * Fr_star_u[3:6, :]
Fr_star_steady = Fr_star_c.subs(ud_zero).subs(u_dep_dict)\
.subs(steady_conditions).subs({q[3]: -r*cos(q[1])}).expand()
mprint(Fr_c)
mprint(Fr_star_steady)
# First dynamic equation, under steady conditions is 2nd order polynomial in
# dq0/dt.
steady_turning_dynamic_equation = Fr_c[0] + Fr_star_steady[0]
# Equilibrium is posible when the solution to this quadratic is real, i.e.,
# when the discriminant in the quadratic is non-negative
p = Poly(steady_turning_dynamic_equation, qd[0])
a, b, c = p.coeffs()
discriminant = b*b - 4*a*c # Must be non-negative for equilibrium
# in case of thin disc inertia assumptions
#mprint((discriminant / (r**3 * m**2)).expand())
# ADD ALL CODE DIRECTLY BELOW HERE, do not change above!
# Think there should be at 12 assertion tests:
# 1) Fr[i] == fr from KanesMethod i = 0, ..., 5
# 2) Fr_star[i] == frstar from KanesMethod i = 0, ..., 5
# if 2) is slow, try comparing this instead:
# 2a) Fr_star_steady[i] == frstar from KanesMethod, evaluated at steady turning
# conditions.
# This should be something like frstar.subs(ud_zero).subs(steady_conditions)
示例4: symbols
# 需要导入模块: from sympy import Poly [as 别名]
# 或者: from sympy.Poly import coeffs [as 别名]
import numpy as np
a, b, T, s, w, x, z, I = symbols('a b T s w x z I')
bilinear_transform = {s : 2/T*(z-1)/(z+1)}
# Continuous time transfer function from I(s) to w(s) of the following plant
# model:
# dw/dt = a*w + b*I
G_s = b/(s-a)
G_z_num, G_z_den = G_s.subs(bilinear_transform).as_numer_denom()
G_z_den = Poly(G_z_den, z)
G_z_num = Poly(G_z_num / G_z_den.LC(), z) # divide by leading coefficient of den
G_z_den = G_z_den.monic() # make denominator monic
assert(G_z_den.coeffs()[0] == 1)
kp = Poly(G_z_num.coeffs()[0], z)
G_z_N_p = G_z_num - kp * G_z_den
print(kp)
print(G_z_N_p)
print(G_z_den)
from sympy import symbols, Poly
import numpy as np
a, b, T, s, w, x, z, I = symbols('a b T s w x z I')
bilinear_transform = {s : 2/T*(z-1)/(z+1)}
# Continuous time transfer function from I(s) to w(s) of the following plant