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Question 1 Figure Q1 shows a mechanical system. The system input is T) and output is supposed to be 0. Please find the transf
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Answer #1

Body 1:

J_1\ddot \theta_1 = T(t)

\frac{\dot \theta_1}{\dot\theta_2} = \frac{100}{10} = 10

J_1*10\ddot \theta_2 = T(t)

10\ddot \theta_2 = T(t)

Laplace transform:

10s^2 \theta_2(s) = T(s)

Body 2:

J_2(\ddot \theta_3 )+k*(\theta_3 -\theta_2)+c*\dot \theta_3= 0

100\ddot \theta_3+100*(\theta_3 -\theta_2)+100*\dot \theta_3= 0

\ddot \theta_3+\theta_3 -\theta_2+\dot \theta_3= 0

Laplace transform:

s^2\theta_3(s)+\theta_3 (s)-\theta_2(s)+s \theta_3(s)= 0

(s^2+s+1) \theta_3(s)-\theta_2 (s)= 0

\theta_2 (s)= (s^2+s+1) \theta_3(s)

But,

10s^2 \theta_2(s) = T(s)

10s^2 *( (s^2+s+1) \theta_3(s)) = T(s)

Transfer function:

\frac{\theta_3(s)} {T(s)} = \frac{1}{10s^2 *( (s^2+s+1)}

Given a unit impulse signal for T.

Laplace transform of a unit impulse signal is 1.

\theta_3(s)= \frac{1}{10s^2 *(s^2+s+1)}*1

This system has 2 poles at the origin (due to s2).

Which indicates a type 2 system.

ss \;error= \lim_{s\rightarrow 0}\frac{1}{10s^2 *(s^2+s+1)}s = \infty

The system is unstable.

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