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.