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Problem 1: Laplace transforms and the Final Value Theorem For both parts (a)-(b) below (note that $u_s(t)$ is the unit step function): • Apply the Laplace transform to the ODE and find an expression for $Y(s)$, the Laplace transform of $y(t)$, as a quotient of two polynomials in $s$. • Explain whether the Final Value Theorem (FVT) can be applied to find the steady-state value of $y(t)$, $y_{ss} = \lim_{t \to \infty} y(t)$. If the FVT can be applied, use it to compute $y_{ss}$. (a) $\frac{d^2y(t)}{dt^2} + 3\frac{dy(t)}{dt} + 2y(t) = u(t)$ $y(0) = -1$, $\frac{dy(0)}{dt} = 2$, $u(t) = 10u_s(t)$ (b) $\frac{d^2y(t)}{dt^2} - 2\frac{dy(t)}{dt} + 4y(t) = u(t)$, $y(0) = 2$, $\frac{dy(0)}{dt} = 3$, $u(t) = 5u_s(t)$

          Problem 1: Laplace transforms and the Final Value Theorem
For both parts (a)-(b) below (note that $u_s(t)$ is the unit step function):
• Apply the Laplace transform to the ODE and find an expression for $Y(s)$, the Laplace
transform of $y(t)$, as a quotient of two polynomials in $s$.
• Explain whether the Final Value Theorem (FVT) can be applied to find the steady-state
value of $y(t)$, $y_{ss} = \lim_{t \to \infty} y(t)$. If the FVT can be applied, use it to compute $y_{ss}$.
(a) $\frac{d^2y(t)}{dt^2} + 3\frac{dy(t)}{dt} + 2y(t) = u(t)$ 	$y(0) = -1$, $\frac{dy(0)}{dt} = 2$, $u(t) = 10u_s(t)$
(b) $\frac{d^2y(t)}{dt^2} - 2\frac{dy(t)}{dt} + 4y(t) = u(t)$, 	$y(0) = 2$, $\frac{dy(0)}{dt} = 3$, $u(t) = 5u_s(t)$

Problem 1: Laplace transforms and the Final Value Theorem
For both parts (a)-(b) below (note that us(t) is the unit step function):
• Apply the Laplace transform to the ODE and find an expression for Y(s), the Laplace
transform of y(t), as a quotient of two polynomials in s.
• Explain whether the Final Value Theorem (FVT) can be applied to find the steady-state
value of y(t), yss = limt →∞ y(t). If the FVT can be applied, use it to compute yss.
(a) (d^2y(t))/(dt^2) + 3(dy(t))/(dt) + 2y(t) = u(t) y(0) = -1, (dy(0))/(dt) = 2, u(t) = 10us(t)
(b) (d^2y(t))/(dt^2) - 2(dy(t))/(dt) + 4y(t) = u(t), y(0) = 2, (dy(0))/(dt) = 3, u(t) = 5us(t)

Added by Raymond J.

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