Problema 5.4#

Sea un sistema de segundo orden con una entrada sinusoidal, \(m(t) = 1 \sin(2 t)\). Demostrar que la respuesta estacionaria cumple las siguientes proposionciones:

  1. una función sinusoidal,

  2. tiene una amplitud \(\frac{1}{\sqrt{(1-t\tau^2)^2+(4\zeta\tau)^2}}\) y

  3. tiene como desfase \(\varphi = \mathrm{atan}\left(\frac{-4\zeta\tau}{1-4\tau^2}\right)\).

Solución

Este problema se puede resolver resolviendo directamente la ecuación diferencial o utilizando la función de transferencia de un sistema lineal de segundo orden.

Ecuación diferencial

La ecuación a resolver es:

\[\tau^2 \frac{d^2 y}{d t} + 2 \zeta \tau \frac{\mathrm{d} y}{\mathrm{d} t} + y = 1 \sin 2 t\]

Para resolver analíticamente la ecuación se puede recurrir a Sympy. Para resolver la ecuación diferencial se ha supuesto que:

\[y (t = 0) = \frac{\mathrm{d} y(t=0)}{\mathrm{d} t} = 0\]

lo que es razonable asumiendo que se está trabajando con variables de desviación.

En el enunciado del problema no se dice si se trata de un sistema subamortiguado (\(\zeta < 1\)), críticamente amortiguado (\(\zeta = 1\)) o sobreamortiguado (\(\zeta > 1\)).

1. Sistema sobreamortiguado: La constante de tiempo \(\tau\) y el coeficiente de amortiguamiento \(\zeta\) son siempre positivas. Por tanto, si \(\zeta > 1\) el producto \(\tau^2 (\zeta - 1) (\zeta + 1)\) será positivo. Resolviendo la ecuación:

using SymPy

t, T = symbols("t tau", real=true)
z = symbols("zeta", positive=true)
s = symbols("s")
y = SymFunction("y")

deq = Eq(T^2*y''(t) + 2*z*T*y'(t) + y(t), sin(2*t))
ics = ((y, 0, 0), (y', 0, 0))

sol = rhs(dsolve(deq, y(t), ics=ics))
\[\begin{equation*}- \frac{4 \tau^{2} \sin{\left(2 t \right)}}{16 \tau^{4} + 16 \tau^{2} \zeta^{2} - 8 \tau^{2} + 1} - \frac{4 \tau \zeta \cos{\left(2 t \right)}}{16 \tau^{4} + 16 \tau^{2} \zeta^{2} - 8 \tau^{2} + 1} + \left(- \frac{4 \tau^{3}}{16 \tau^{4} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + 16 \tau^{2} \zeta^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} - 8 \tau^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + \sqrt{\zeta - 1} \sqrt{\zeta + 1}} - \frac{2 \tau \zeta^{2}}{16 \tau^{4} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + 16 \tau^{2} \zeta^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} - 8 \tau^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + \sqrt{\zeta - 1} \sqrt{\zeta + 1}} + \frac{2 \tau \zeta \sqrt{\zeta - 1} \sqrt{\zeta + 1}}{16 \tau^{4} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + 16 \tau^{2} \zeta^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} - 8 \tau^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + \sqrt{\zeta - 1} \sqrt{\zeta + 1}} + \frac{\tau}{16 \tau^{4} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + 16 \tau^{2} \zeta^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} - 8 \tau^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + \sqrt{\zeta - 1} \sqrt{\zeta + 1}}\right) e^{\frac{t \left(- \zeta - \sqrt{\zeta - 1} \sqrt{\zeta + 1}\right)}{\tau}} + \left(\frac{4 \tau^{3}}{16 \tau^{4} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + 16 \tau^{2} \zeta^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} - 8 \tau^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + \sqrt{\zeta - 1} \sqrt{\zeta + 1}} + \frac{2 \tau \zeta^{2}}{16 \tau^{4} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + 16 \tau^{2} \zeta^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} - 8 \tau^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + \sqrt{\zeta - 1} \sqrt{\zeta + 1}} + \frac{2 \tau \zeta \sqrt{\zeta - 1} \sqrt{\zeta + 1}}{16 \tau^{4} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + 16 \tau^{2} \zeta^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} - 8 \tau^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + \sqrt{\zeta - 1} \sqrt{\zeta + 1}} - \frac{\tau}{16 \tau^{4} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + 16 \tau^{2} \zeta^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} - 8 \tau^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + \sqrt{\zeta - 1} \sqrt{\zeta + 1}}\right) e^{\frac{t \left(- \zeta + \sqrt{\zeta - 1} \sqrt{\zeta + 1}\right)}{\tau}} + \frac{\sin{\left(2 t \right)}}{16 \tau^{4} + 16 \tau^{2} \zeta^{2} - 8 \tau^{2} + 1}\end{equation*}\]

La respuesta obtenida se puede dividir en dos partes diferenciadas, una transitoria y otra estacionaria. La parte transitoria de la respuesta es:

\[\left(- \frac{4 \tau^{3}}{16 \tau^{4} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + 16 \tau^{2} \zeta^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} - 8 \tau^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + \sqrt{\zeta - 1} \sqrt{\zeta + 1}} - \frac{2 \tau \zeta^{2}}{16 \tau^{4} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + 16 \tau^{2} \zeta^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} - 8 \tau^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + \sqrt{\zeta - 1} \sqrt{\zeta + 1}} + \frac{2 \tau \zeta \sqrt{\zeta - 1} \sqrt{\zeta + 1}}{16 \tau^{4} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + 16 \tau^{2} \zeta^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} - 8 \tau^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + \sqrt{\zeta - 1} \sqrt{\zeta + 1}} + \frac{\tau}{16 \tau^{4} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + 16 \tau^{2} \zeta^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} - 8 \tau^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + \sqrt{\zeta - 1} \sqrt{\zeta + 1}}\right) e^{\frac{t \left(- \zeta - \sqrt{\zeta - 1} \sqrt{\zeta + 1}\right)}{\tau}} + \left(\frac{4 \tau^{3}}{16 \tau^{4} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + 16 \tau^{2} \zeta^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} - 8 \tau^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + \sqrt{\zeta - 1} \sqrt{\zeta + 1}} + \frac{2 \tau \zeta^{2}}{16 \tau^{4} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + 16 \tau^{2} \zeta^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} - 8 \tau^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + \sqrt{\zeta - 1} \sqrt{\zeta + 1}} + \frac{2 \tau \zeta \sqrt{\zeta - 1} \sqrt{\zeta + 1}}{16 \tau^{4} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + 16 \tau^{2} \zeta^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} - 8 \tau^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + \sqrt{\zeta - 1} \sqrt{\zeta + 1}} - \frac{\tau}{16 \tau^{4} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + 16 \tau^{2} \zeta^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} - 8 \tau^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + \sqrt{\zeta - 1} \sqrt{\zeta + 1}}\right) e^{\frac{t \left(- \zeta + \sqrt{\zeta - 1} \sqrt{\zeta + 1}\right)}{\tau}} \underset{t \rightarrow \infty}{\rightarrow} 0\]

El cálculo de este límite parece muy complicado, pero, en realidad, es muy simple ya que aparecen términos de tipo \(\exp(-t)\) que se anulan cuando el tiempo tiende a infinito.

La parte más importante es la porción estacionaria, ya que será la que marque la dinámica tras los instantes iniciales. Despreciando la parte transitoria se obtiene la siguiente respuesta:

\[y (t) = - \frac{4 \tau \zeta \cos{\left(2 t \right)}}{16 \tau^{4} + 16 \tau^{2} \zeta^{2} - 8 \tau^{2} + 1} + \left(- \frac{4 \tau^{2}}{16 \tau^{4} + 16 \tau^{2} \zeta^{2} - 8 \tau^{2} + 1} + \frac{1}{16 \tau^{4} + 16 \tau^{2} \zeta^{2} - 8 \tau^{2} + 1}\right) \sin{\left(2 t \right)}\]

Operando y aplicando la propiedad trigonométrica:

\[x \sin \alpha + y \cos \alpha = z \sin (\alpha + \varphi)\]

donde \(z^2 = x^2 + y^2\) y \(\varphi = \mathrm{atan} (y / x)\) y sabiendo que:

\[\begin{split}\left\{\begin{array}{l} x = - \frac{8 \tau^2 - 2}{2 (16 \tau^2 \zeta^2 + 16 \tau^4 - 8 \tau^2 + 1)}\\ y = \frac{4 \tau \zeta}{16 \tau^2 \zeta^2 + 16 \tau^4 - 8 \tau^2 + 1} \end{array}\right.\end{split}\]

se obtiene:

\[y (t) = \frac{1}{\sqrt{16 \tau^2 \zeta^2 + 16 \tau^4 - 8 \tau^2 + 1}} \sin \left( 2 t + \mathrm{atan} \left( - \frac{8 \tau \zeta}{8 \tau^2 - 2} \right) \right)\]

Se comprueba que se trata, tal como dice el problema, de:

i. Una función sinusoidal

ii. La amplitud es \(\frac{1}{\sqrt{(1 - 4 \tau^2)^2 + (4 \zeta \tau)^2}} = \frac{1}{\sqrt{16 \tau^2 \zeta^2 + 16 \tau^4 - 8 \tau^2 + 1}}\)

iii. El desfase es \(\varphi = \mathrm{atan} \left( - \frac{4 \tau \zeta}{4 \tau^2 - 1} \right)\)

Función de transferencia

La función de transferencia del sistema propuesto por el problema es:

\[G = \frac{y (s)}{m (s)} = \frac{1}{\tau^2 s + 2 \zeta \tau s + 1}\]

La entrada a este sistema es \(m (t) = 1 \sin 2 t\), realizando la transformada de Laplace

\[m (s) = \frac{2}{s^2 + 4}\]

La respuesta del sistema será:

\[y (t) =\mathcal{L}^{- 1} (G m (s)) =\mathcal{L}^{- 1} \left( \frac{1}{\tau^2 s + 2 \zeta \tau s + 1} \frac{2}{s^2 + 4} \right)\]

Realizando el cálculo para un sistema sobreamortiguado se obtiene:

G = 1/(T^2*s + 2z*T*s +1)

m = 2/(s^2+4)

sol_b = sympy.inverse_laplace_transform(G*m, s, t)
\[\begin{equation*}\frac{2 \left(- \frac{\left(i \sin{\left(2 t \right)} - \cos{\left(2 t \right)}\right) \Gamma\left(- 4 i\right) \Gamma\left(- 2 i + \frac{1}{\tau^{2} + 2 \tau \zeta}\right)}{\Gamma\left(1 - 4 i\right) \Gamma\left(1 - 2 i + \frac{1}{\tau^{2} + 2 \tau \zeta}\right)} + \frac{\left(i \sin{\left(2 t \right)} + \cos{\left(2 t \right)}\right) \Gamma\left(4 i\right) \Gamma\left(2 i + \frac{1}{\tau^{2} + 2 \tau \zeta}\right)}{\Gamma\left(1 + 4 i\right) \Gamma\left(1 + 2 i + \frac{1}{\tau^{2} + 2 \tau \zeta}\right)} + \frac{e^{- \frac{t}{\tau \left(\tau + 2 \zeta\right)}} \Gamma\left(- 2 i - \frac{1}{\tau^{2} + 2 \tau \zeta}\right) \Gamma\left(2 i - \frac{1}{\tau^{2} + 2 \tau \zeta}\right)}{\Gamma\left(1 - 2 i - \frac{1}{\tau^{2} + 2 \tau \zeta}\right) \Gamma\left(1 + 2 i - \frac{1}{\tau^{2} + 2 \tau \zeta}\right)}\right) \theta\left(t\right)}{\tau^{2} + 2 \tau \zeta}\end{equation*}\]
sol_b = simplify(sol)
\[\begin{equation*}\frac{\left(\tau \sqrt{\zeta^{2} - 1} \left(- 4 \tau^{2} - 2 \zeta^{2} + 2 \zeta \sqrt{\zeta^{2} - 1} + 1\right) e^{\frac{t \left(\zeta - \sqrt{\zeta - 1} \sqrt{\zeta + 1}\right)}{\tau}} + \tau \sqrt{\zeta^{2} - 1} \left(4 \tau^{2} + 2 \zeta^{2} + 2 \zeta \sqrt{\zeta^{2} - 1} - 1\right) e^{\frac{t \left(\zeta + \sqrt{\zeta - 1} \sqrt{\zeta + 1}\right)}{\tau}} + \left(\zeta - 1\right) \left(\zeta + 1\right) \left(- 4 \tau^{2} \sin{\left(2 t \right)} - 4 \tau \zeta \cos{\left(2 t \right)} + \sin{\left(2 t \right)}\right) e^{\frac{2 t \zeta}{\tau}}\right) e^{- \frac{2 t \zeta}{\tau}}}{\left(\zeta - 1\right) \left(\zeta + 1\right) \left(16 \tau^{4} + 16 \tau^{2} \zeta^{2} - 8 \tau^{2} + 1\right)}\end{equation*}\]
sol_b = collect(sol, sin(2t))
\[\begin{equation*}- \frac{4 \tau \zeta \cos{\left(2 t \right)}}{16 \tau^{4} + 16 \tau^{2} \zeta^{2} - 8 \tau^{2} + 1} + \left(- \frac{4 \tau^{2}}{16 \tau^{4} + 16 \tau^{2} \zeta^{2} - 8 \tau^{2} + 1} + \frac{1}{16 \tau^{4} + 16 \tau^{2} \zeta^{2} - 8 \tau^{2} + 1}\right) \sin{\left(2 t \right)} + \left(- \frac{4 \tau^{3}}{16 \tau^{4} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + 16 \tau^{2} \zeta^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} - 8 \tau^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + \sqrt{\zeta - 1} \sqrt{\zeta + 1}} - \frac{2 \tau \zeta^{2}}{16 \tau^{4} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + 16 \tau^{2} \zeta^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} - 8 \tau^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + \sqrt{\zeta - 1} \sqrt{\zeta + 1}} + \frac{2 \tau \zeta \sqrt{\zeta - 1} \sqrt{\zeta + 1}}{16 \tau^{4} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + 16 \tau^{2} \zeta^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} - 8 \tau^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + \sqrt{\zeta - 1} \sqrt{\zeta + 1}} + \frac{\tau}{16 \tau^{4} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + 16 \tau^{2} \zeta^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} - 8 \tau^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + \sqrt{\zeta - 1} \sqrt{\zeta + 1}}\right) e^{\frac{t \left(- \zeta - \sqrt{\zeta - 1} \sqrt{\zeta + 1}\right)}{\tau}} + \left(\frac{4 \tau^{3}}{16 \tau^{4} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + 16 \tau^{2} \zeta^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} - 8 \tau^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + \sqrt{\zeta - 1} \sqrt{\zeta + 1}} + \frac{2 \tau \zeta^{2}}{16 \tau^{4} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + 16 \tau^{2} \zeta^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} - 8 \tau^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + \sqrt{\zeta - 1} \sqrt{\zeta + 1}} + \frac{2 \tau \zeta \sqrt{\zeta - 1} \sqrt{\zeta + 1}}{16 \tau^{4} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + 16 \tau^{2} \zeta^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} - 8 \tau^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + \sqrt{\zeta - 1} \sqrt{\zeta + 1}} - \frac{\tau}{16 \tau^{4} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + 16 \tau^{2} \zeta^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} - 8 \tau^{2} \sqrt{\zeta - 1} \sqrt{\zeta + 1} + \sqrt{\zeta - 1} \sqrt{\zeta + 1}}\right) e^{\frac{t \left(- \zeta + \sqrt{\zeta - 1} \sqrt{\zeta + 1}\right)}{\tau}}\end{equation*}\]

Con paciencia, se puede comprobar que se obtiene el mismo resultado que resolviendo la ecuación diferencial.