Root Locus of a Suspension Mode Consider the two-mass suspension model developed in Example 4.5.9 in Chapter 4, and shown again in Figure 13.7.1. The equations of motion are m1x1 = c1(x2 − x1) + k1(x2 − x1) m2x2 = −c1(x2 − x 1) − k1(x2 − x1) + k2(y − x2) We will use the following numerical values:

Question

Root Locus of a Suspension Mode

Consider the two-mass suspension model developed in Example 4.5.9 in Chapter 4, and shown again in Figure 13.7.1. The equations of motion are

[latex]m_{1} \ddot{x}_{1} = c_{1}(\dot{x}_{2} − \dot{x}_{1}) + k_{1}(x_{2} − x_{1}) [/latex]

[latex]m_{2} \ddot{x}_{2} = −c_{1}(\dot{x}_{2} − \dot{x}_{1}) − k_{1}(x_{2} − x_{1}) + k_{2}(y − x_{2})[/latex]

We will use the following numerical values: [latex]m_{1} = 250 kg, m_{2} = 40 kg, k_{1} = 1.5 × 10^{4} N/m,[/latex] and [latex]k_{2} = 1.5 × 10^{5} N/m [/latex].
a. Use the root locus plot to determine the value of the damping [latex]c_1[/latex] required to give a dominant root pair having a damping ratio of ζ = 0.707.
b. Using the value of [latex]c_1[/latex] found in part (a), obtain a plot of the unit-step response.

Accepted Answer

a. Transforming the equations of motion using zero initial conditions and rearranging gives

[latex](m_{1} s^{2} + c_{1} s + k_{1}) X_{1}(s) − (c_{1}s + k_{1})X_{2}(s) = 0 [/latex]

[latex]−(c_{1}s + k_{1})X_{1}(s) + (m_{2}s^{2} + c_{1} s + k_{1} + k_{2}) X_{2}(s) = k_{2} Y (s)[/latex]
We can find the characteristic polynomial and the transfer functions by using the determinant method.

[latex]D_{3}(s) = \begin{vmatrix} (m_{1} s^{2} + c_{1} s + k_{1}) & -(c_{1} s + k_{1}) \ -(c_{1} s + k_{1}) & (m_{2} s^{2} + c_{1} s + k_{1} + k_{2}) \end{vmatrix} [/latex]

The transfer functions for [latex]X_{1}(s)[/latex] and [latex]X_{2}(s)[/latex] with Y (s) as the input can be expressed as
[latex]\frac{X_{1}(s)}{Y (s)} = \frac{D_{1}(s)}{D_{3}(s)}[/latex]

[latex]\frac{X_{2}(s)}{Y (s)} = \frac{D_{2}(s)}{D_{3}(s)}[/latex]

where

[latex]D_{1}(s) = \begin{vmatrix} 0 & -(c_{1} s + k_{1}) \ k_{2} & (m_{2} s^{2} + c_{1} s + k_{1} + k_{2}) \end{vmatrix} [/latex]

[latex]=k_{2} (c_{1} s + k_{1})[/latex]

[latex]D_{2} (s) = \begin{vmatrix} (m_{1} s^{2} + c_{1} s + k_{1}) & 0 \ -(c_{1} s + k_{1}) & k_{2}) \end{vmatrix} [/latex]

[latex]= k_{2} (m_{1} s^{2} + c_{1} s + k_{1})[/latex]

Now substitute the given values of [latex]m_{1}, m_{2}, k_{1}[/latex], and [latex]k_{2}[/latex] into D(s) to obtain the root locus equation.
[latex]D_{3}(s) = (250s^{2} + c_{1}s + 15 000)(40s^{2} + c_{1}s + 165 000) − (c_{1}s + 15 000) ^{2}[/latex]
Factoring out [latex]c_{1}[/latex] and rearranging into the standard root-locus form (13.7.1) so that the highest coefficients of N(s) and D(s) are unity, we obtain
[latex]s^{4} + 4185s^{2} + 2.25 × 10^{5} + K (s^{3} + 775.86s) = 0 [/latex]

Question 13.7.1: Root Locus of a Suspension Mode Consider the two-mass suspen...

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