Finding transfer function from a differential equation

Question

I have a non-linear differential equation and want to obtain its transfer function.

First I linearized the equation (first order Taylor series) around the point that I had calculated, then I proceeded to calculate its Laplace transform. However, I could not do the last step.

The linearized function is:

$$k_1\cdot\ddot{y} = k_2\cdot y + x + k_3$$

I don't think I did anything wrong with the linearization (MATLAB gave the same result).

I just can't calculate the TF because of that k₃.

Manipulating the expression I get stuck with something like G(s) = X(s) + ..., which doesn't seem to make sense to me.

The only mistake I could have made is in the calculations to find the point I needed to linearize around, but unless that point is 0 (which it is not), I will always end up with a k₃, so probably that's not the problem.

What do you think? The differential equation must be wrong?

Answer 1

I agree with your conclusion about a transfer function. Your function is not homogeneous.

By way of explanation, just follow along for a moment:

$$\begin{align*} k_1\,\ddot{y} &= k_2\,y + x + k_3 \\\\ \ddot{y} - \frac{k_2}{k_1}\,y &= \frac1{k_1}x + \frac{k_3}{k_1} \\\\ \left[D^2-\frac{k_2}{k_1}\right]y&=\frac1{k_1}x + \frac{k_3}{k_1} \\\\ \bigg[D\bigg]\left[D^2-\frac{k_2}{k_1}\right]y&=\frac1{k_1} \\\\ \bigg[D\bigg]\bigg[D\bigg]\left[D^2-\frac{k_2}{k_1}\right]y&=0 \end{align*}$$

That's now homogeneous and the zeros on the left side are \$\{0, 0, \pm \sqrt{\frac{k_2}{k_1}}\}\$ so the general solution is:

$$\begin{align*} y&=A_1 \cdot e^{^{0\,x}}+A_2 \cdot e^{^{0\,x}}+A_3 \cdot e^{^{x\,\cdot\,\sqrt{\frac{k_2}{k_1}}}}+A_4 \cdot e^{^{-x\,\cdot\,\sqrt{\frac{k_2}{k_1}}}} \\\\ &=A_0 +A_3 \cdot e^{^{x\,\cdot\,\sqrt{\frac{k_2}{k_1}}}}+A_4 \cdot e^{^{-x\,\cdot\,\sqrt{\frac{k_2}{k_1}}}} \end{align*}$$

If you plug that back into your original equation, then you will find out that \$A_0=-\frac1{k_2}\left(x+k_3\right)\$. So the updated general solution is:

$$y=\frac1{k_2}\left[A_1\cdot e^{^{x\,\cdot \sqrt{\frac{k_2}{k_1}}}} + A_2\cdot e^{^{-x\,\cdot \sqrt{\frac{k_2}{k_1}}}}-x - k_3\right]$$

If you apply that back to your original function and do the algebra, I think you'll find that it works out.

I can't take it to the specific solution, obviously. But perhaps you can do that from here.

As far as an \$\frac{y}{x}\$ transform function goes, I don't believe you can't get there from here. Perhaps the way to go is to make a change of variable?