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Chapter 2

Q. 2.17

Transfer Function—Two Degrees of Freedom

PROBLEM: Find the transfer function, X_{2}(s)/F(s), for the system of Figure 2.17(a).


Verified Solution

The system has two degrees of freedom, since each mass can be moved in
the horizontal direction while the other is held still. Thus, two simultaneous equations of motion will be required to describe the system. The two equations come from free-body diagrams of each mass. Superposition is used to draw the free-body diagrams. For example, the forces onM_{1} are due to (1) its own motion and (2) the motion of M_{2} transmitted to M_{1} through the system. We will consider these two sources separately.
If we hold M_{2} still and move M_{1} to the right, we see the forces shown in Figure 2.18(a).

If we hold M_{1} still and move M_{2} to the right, we see the forces shown in Figure 2.18(b). The total force on M_{1} is the superposition, or sum, of the forces just discussed. This result is shown in Figure 2.18(c). For M_{2}we proceed in a similar fashion: First we move M_{2} to the right while holding M_{1} still; then we move M_{1} to the right and hold M_{2} still. For each case we evaluate the forces on M_{2}. The results appear in Figure 2.19.

The Laplace transform of the equations of motion can now be written from Figures 2.18(c) and 2.19(c) as

\left[M_{1}s^{2}(F_{\nu 1}+ƒ_{\nu 3} )s+(K_{1}+K_{2} ) \right] X_{1}(s)-(ƒ_{\nu 3}s+K_{2} )X_{2}(s)=F(s)


-(ƒ_{\nu 3} s+K_{2})X_{1} (s)+\left[M_{2}s^{2} +(ƒ_{\nu 2}+ƒ_{\nu 3} )s+(K_{2}+K_{3}) \right] X_{2}(s)=0

From this, the transfer function,X_{2}(s)/F(s) is

\frac{X_{2}(s) }{F(s)}=G(s)=\frac{(ƒ_{\nu 3}s+K_{2})}{\Delta }

as shown in Figure 2.17(b) where

\Delta =\left|\begin{matrix}\left[M_{1}s^{2}+(ƒ_{\nu 1}+ƒ_{\nu 3} )s+(K_{1}+K_{2}) \right] &-(ƒ_{\nu 3}s+K_{2}) \\ -(ƒ_{\nu 3}s+K_{2}) & \left[M_{2}s^{2}+(ƒ_{\nu 2}+ƒ_{\nu 3} )s+(K_{2}+K_{3}) \right] \end{matrix} \right|