Laplace Transform Derivation Help

Question

Please see image. These are screenshots of a lecture slide from a Control Engineering module, regarding determining the transfer functions of mechanical systems. However I can't seem to understand how the last transfer function was derived in the red ellipse of example 2. Example 1 is used as a precursor to example 2.

I've tried deriving it put I'm sure I'm missing a few steps, please can you aid me in my quest

Thanks

Example 1 Example 2

Answer 1

Now, we got:

$$ \begin{cases} f(t)=\text{M}_2x''_2(t)+\text{K}_2(x_2(t)-x_1(t))\\ \text{M}_1x''_1(t)+\text{K}_2(x_1(t)-x_2(t))+\text{K}_1x_1(t)+\text{B}x'_1(t)=0 \end{cases} $$

So, we can take the Laplace transform of both these equations:

• $$\text{F}(s)=\text{M}_2\left(s^2\text{X}_2(s)-sx_2(0)-x_2'(0)\right)+\text{K}_2(\text{X}_2(s)-\text{X}_1(s))$$
• $$\text{M}_1\left(s^2\text{X}_1(s)-sx_1(0)-x'_1(0)\right)+\text{K}_2(\text{X}_1(s)-\text{X}_2(s))+\text{K}_1\text{X}_1(s)+\text{B}\left(s\text{X}_1(s)-x_1(0)\right)=0$$

Now when we assume that $x_i(0)=x_i'(0)=0$:

• $$\text{F}(s)=\text{M}_2s^2\text{X}_2(s)+\text{K}_2(\text{X}_2(s)-\text{X}_1(s))$$
• $$\text{M}_1s^2\text{X}_1(s)+\text{K}_2(\text{X}_1(s)-\text{X}_2(s))+\text{K}_1\text{X}_1(s)+\text{B}s\text{X}_1(s)=0$$

Now, we can solve $\text{X}_1(s)$ out of the second equation:

$$\text{M}_1s^2\text{X}_1(s)+\text{K}_2(\text{X}_1(s)-\text{X}_2(s))+\text{K}_1\text{X}_1(s)+\text{B}s\text{X}_1(s)=0\Longleftrightarrow$$ $$\text{X}_1(s)\left[\text{M}_1s^2+\text{K}_2+\text{K}_1+\text{B}s\right]=\text{K}_2\text{X}_2(s)\Longleftrightarrow$$ $$\text{X}_1(s)=\frac{\text{K}_2\text{X}_2(s)}{\text{M}_1s^2+\text{K}_2+\text{K}_1+\text{B}s}$$

And, we can solve $\text{X}_2(s)$ out of the first equation:

$$\text{F}(s)=\text{M}_2s^2\text{X}_2(s)+\text{K}_2(\text{X}_2(s)-\text{X}_1(s))\Longleftrightarrow$$ $$\text{F}(s)+\text{K}_2\text{X}_1(s)=\text{X}_2(s)\left[\text{M}_2s^2+\text{K}_2\right]\Longleftrightarrow$$ $$\text{X}_2(s)=\frac{\text{F}(s)+\text{K}_2\text{X}_1(s)}{\text{M}_2s^2+\text{K}_2}$$

Now substitute $\text{X}_2(s)$ into the $\text{X}_1(s)$ equation:

$$\text{X}_1(s)=\frac{\text{K}_2\left[\frac{\text{F}(s)+\text{K}_2\text{X}_1(s)}{\text{M}_2s^2+\text{K}_2}\right]}{\text{M}_1s^2+\text{K}_2+\text{K}_1+\text{B}s}$$

From, here you can solve $\frac{\text{X}_1(s)}{\text{F}(s)}$.

EDIT:

Solve $\text{X}_1(s)$ out of the equation above and multiply both sides by $\frac{1}{\text{F}(s)}$:

$$\text{X}_1(s)=\frac{\frac{\text{K}_2\text{F}(s)}{\left(\text{M}_2s^2+\text{K}_2\right)\left(\text{M}_1s^2+\text{K}_2+\text{K}_1+\text{B}s\right)}}{1-\frac{\text{K}_2^2}{\left(\text{M}_2s^2+\text{K}_2\right)\left(\text{M}_1s^2+\text{K}_2+\text{K}_1+\text{B}s\right)}}\Longleftrightarrow$$

$$\frac{\text{X}_1(s)}{\text{F}(s)}=\frac{\frac{\text{K}_2}{\left(\text{M}_2s^2+\text{K}_2\right)\left(\text{M}_1s^2+\text{K}_2+\text{K}_1+\text{B}s\right)}}{1-\frac{\text{K}_2^2}{\left(\text{M}_2s^2+\text{K}_2\right)\left(\text{M}_1s^2+\text{K}_2+\text{K}_1+\text{B}s\right)}}$$