Question 2:
A pendulum model is described by d²x/dt² + sin(x) = 10u, where x is the angle and u is the applied torque. Linearize the model around the constant steady-state where x = 30 deg. What is the DC-gain (P(s=0)) of the linear model?


### Step 1: Define the equation and the steady-state condition
We are given the nonlinear pendulum model equation:

\[
\frac{d^2x}{dt^2} + \sin(x) = 10u
\]

where \( x \) is the angle, and \( u \) is the applied torque.

We are tasked with linearizing this equation around the steady-state \( x = 30^\circ \) (which is \( \frac{\pi}{6} \) radians).

### Step 2: Linearize the equation
To linearize the equation, we need to approximate \( \sin(x) \) around \( x = 30^\circ \).

#### Taylor expansion of \( \sin(x) \):
The Taylor expansion of \( \sin(x) \) around a point \( x_0 \) is:

\[
\sin(x) \approx \sin(x_0) + \cos(x_0)(x - x_0)
\]

For \( x_0 = \frac{\pi}{6} \), we have:

\[
\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}, \quad \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}
\]

Thus, the linearized form of \( \sin(x) \) around \( x = \frac{\pi}{6} \) becomes:

\[
\sin(x) \approx \frac{1}{2} + \frac{\sqrt{3}}{2}(x - \frac{\pi}{6})
\]

### Step 3: Linearized equation
Using this linear approximation in the original equation:

\[
\frac{d^2x}{dt^2} + \left( \frac{1}{2} + \frac{\sqrt{3}}{2}(x - \frac{\pi}{6}) \right) = 10u
\]

Simplify the equation:

\[
\frac{d^2x}{dt^2} + \frac{\sqrt{3}}{2}x = 10u + \text{constant terms}
\]

Ignoring the constant terms (since they don’t affect the linearization), we obtain the linearized equation:

\[
\frac{d^2x}{dt^2} + \frac{\sqrt{3}}{2}x = 10u
\]

### Step 4: Determine the transfer function
The linearized equation can be written in terms of the Laplace transform. Let \( X(s) \) be the Laplace transform of \( x(t) \), and \( U(s) \) be the Laplace transform of \( u(t) \). The second derivative of \( x \) becomes \( s^2 X(s) \), and the linearized equation in the Laplace domain is:

\[
s^2 X(s) + \frac{\sqrt{3}}{2} X(s) = 10 U(s)
\]

Factoring out \( X(s) \):

\[
X(s) \left( s^2 + \frac{\sqrt{3}}{2} \right) = 10 U(s)
\]

Thus, the transfer function from \( u(t) \) to \( x(t) \) is:

\[
\frac{X(s)}{U(s)} = \frac{10}{s^2 + \frac{\sqrt{3}}{2}}
\]

### Step 5: Calculate the DC gain
The DC gain of a system is the value of the transfer function when \( s = 0 \). Substituting \( s = 0 \) into the transfer function:

\[
P(s = 0) = \frac{10}{0^2 + \frac{\sqrt{3}}{2}} = \frac{10}{\frac{\sqrt{3}}{2}} = \frac{20}{\sqrt{3}} = \frac{20\sqrt{3}}{3}
\]

Thus, the DC gain of the linearized model is:

\[
P(s = 0) = \frac{20\sqrt{3}}{3}
\]

### Conclusion
The DC gain of the linearized model is \( \frac{20\sqrt{3}}{3} \).

Would you like more details, or have any questions?

### Related Questions:
1. How do we apply linearization to other nonlinear differential equations?
2. What is the physical significance of the DC gain in this context?
3. How does the small-angle approximation affect the linearization of pendulum equations?
4. Can we extend this linearization approach to multi-degree-of-freedom systems?
5. How do different torques (time-varying vs constant) affect the pendulum's response?

### Tip:
When linearizing a system, always check if the steady-state is within the valid range for approximations, such as small angles for trigonometric functions.

A pendulum model is described by d²x/dt² + sin(x) = 10u, where x is the angle and u is the applied torque. Linearize the model around the constant steady-state where x = 30 deg. What is the DC-gain (P(s=0)) of the linear model?

Explore how to linearize a pendulum model around a 30-degree angle and calculate the DC gain. This problem uses concepts of differential equations, Laplace transforms, and the Taylor series expansion to derive the solution. Suitable for undergraduate students in engineering or physics.

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