Question

Consider the mechanical system shown in Figure 4-55. The system is at rest for t<0. The input force "u" is given at t=0. The displacement "x" is the output of thesystem and is measured from the equilibrium position. Obtain the transfer function X(s)/U(s) for "u" as a force AND for "u" as a displacement for the end of the node.

Answer 1

Given system is

We will define a new reference point in the system at the intersection of the spring-damper system and the third spring \(\left(k_{3}\right) ;\) we will call this point \(y\).

Now the modified system is shown below.

The equations of the motion of the system are

At point ' \(x\) ':

\(k_{1} x+m \ddot{x}+k_{2}(x-y)=0\)

At point ' \(y\) ':

\(k_{2}(y-x)+b \dot{y}+k_{3}(y-z)=0\)

At point 'z':

$$ k_{3}(z-y)=u $$

Taking the Laplace transform of the both sides of above equations and assuming that all initial conditions are zero yields, we get

\(k_{1} X(s)+m s^{2} X(s)+k_{2} X(s)-k_{2} Y(s)=0 \ldots \ldots\) (1)

\(k_{2} Y(s)-k_{2} X(s)+b s Y(s)+k_{3} Y(s)-k_{3} Z(s)=0 \ldots \ldots\)(2)

\(k_{3} Z(s)-k_{3} Y(s)=U(s) \ldots \ldots\) (3)

Equation-(3) can also be written as

\(k_{3} Z(s)=U(s)+k_{3} Y(s)\)

\(\Rightarrow Z(s)=\frac{U(s)+k_{3} Y(s)}{k_{3}}\)

Substitute \(Z(s)\) in equation-(2):

$$ k_{2} Y(s)-k_{2} X(s)+b s Y(s)+k_{3} Y(s)-k_{3}\left(\frac{U(s)+k_{3} Y(s)}{k_{3}}\right)=0 $$

\(k_{2} Y(s)-k_{2} X(s)+b s Y(s)+k_{3} Y(s)-U(s)-k_{3} Y(s)=0\)

\(\left(k_{2}+b s+k_{3}-k_{3}\right) Y(s)-k_{2} X(s)-U(s)=0\)

\(\left(k_{2}+b s\right) Y(s)=k_{2} X(s)+U(s)\)

\(\Rightarrow Y(s)=\frac{k_{2} X(s)+U(s)}{k_{2}+b s}\)

Substitute \(Y(s)\) in equation-(1):

\(k_{1} X(s)+m s^{2} X(s)+k_{2} X(s)-k_{2}\left(\frac{k_{2} X(s)+U(s)}{k_{2}+b s}\right)=0\)

\(k_{1} X(s)+m s^{2} X(s)+k_{2} X(s)-\frac{k_{2}^{2} X(s)}{k_{2}+b s}-\frac{k_{2} U(s)}{k_{2}+b s}=0\)

\(k_{1} X(s)+m s^{2} X(s)+k_{2} X(s)-\frac{k_{2}^{2} X(s)}{k_{2}+b s}=\frac{k_{2} U(s)}{k_{2}+b s}\)

\(\left(k_{1}+m s^{2}+k_{2}-\frac{k_{2}^{2}}{k_{2}+b s}\right) X(s)=\frac{k_{2} U(s)}{k_{2}+b s}\)

\(\left(\frac{k_{1}\left(k_{2}+b s\right)+m s^{2}\left(k_{2}+b s\right)+k_{2}\left(k_{2}+b s\right)-k_{2}^{2}}{k_{2}+b s}\right) X(s)=\frac{k_{2} U(s)}{k_{2}+b s}\)

\(\left(k_{1}\left(k_{2}+b s\right)+m s^{2}\left(k_{2}+b s\right)+k_{2}\left(k_{2}+b s\right)-k_{2}^{2}\right) X(s)=k_{2} U(s)\)

Or we have

$$ \frac{X(s)}{U(s)}=\frac{k_{2}}{k_{1}\left(k_{2}+b s\right)+m s^{2}\left(k_{2}+b s\right)+k_{2}\left(k_{2}+b s\right)-k_{2}^{2}} $$

\(\frac{X(s)}{U(s)}=\frac{k_{2}}{k_{1} k_{2}+k_{1} b s+m k_{2} s^{2}+m b s^{3}+k_{2}^{2}+k_{2} b s-k_{2}^{2}}\)

\(\frac{X(s)}{U(s)}=\frac{k_{2}}{m b s^{3}+m k_{2} s^{2}+\left(k_{1}+k_{2}\right) b s+k_{1} k_{2}}\)

Therefore, the transfer function of the given system is

\(\frac{X(s)}{U(s)}=\frac{k_{2}}{m b s^{3}+m k_{2} s^{2}+\left(k_{1}+k_{2}\right) b s+k_{1} k_{2}}\)