Question 13.7.1: Root Locus of a Suspension Mode Consider the two-mass suspen...
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
m_{1} \ddot{x}_{1} = c_{1}(\dot{x}_{2} − \dot{x}_{1}) + k_{1}(x_{2} − x_{1})m_{2} \ddot{x}_{2} = −c_{1}(\dot{x}_{2} − \dot{x}_{1}) − k_{1}(x_{2} − x_{1}) + k_{2}(y − x_{2})
We will use the following numerical values: m_{1} = 250 kg, m_{2} = 40 kg, k_{1} = 1.5 × 10^{4} N/m, and k_{2} = 1.5 × 10^{5} N/m .
a. Use the root locus plot to determine the value of the damping c_1 required to give a dominant root pair having a damping ratio of ζ = 0.707.
b. Using the value of c_1 found in part (a), obtain a plot of the unit-step response.
a. Transforming the equations of motion using zero initial conditions and rearranging gives
(m_{1} s^{2} + c_{1} s + k_{1}) X_{1}(s) − (c_{1}s + k_{1})X_{2}(s) = 0−(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)
We can find the characteristic polynomial and the transfer functions by using the determinant method.
The transfer functions for X_{1}(s) and X_{2}(s) with Y (s) as the input can be expressed as
\frac{X_{1}(s)}{Y (s)} = \frac{D_{1}(s)}{D_{3}(s)}
where
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}=k_{2} (c_{1} s + k_{1})
D_{2} (s) = \begin{vmatrix} (m_{1} s^{2} + c_{1} s + k_{1}) & 0 \\ -(c_{1} s + k_{1}) & k_{2}) \end{vmatrix}
= k_{2} (m_{1} s^{2} + c_{1} s + k_{1})
Now substitute the given values of m_{1}, m_{2}, k_{1}, and k_{2} into D(s) to obtain the root locus equation.
D_{3}(s) = (250s^{2} + c_{1}s + 15 000)(40s^{2} + c_{1}s + 165 000) − (c_{1}s + 15 000) ^{2}
Factoring out c_{1} 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
s^{4} + 4185s^{2} + 2.25 × 10^{5} + K (s^{3} + 775.86s) = 0
where the root locus parameter K is related to c_{1} as K = 0.029c1. The MATLAB session for obtaining the root locus plot is as follows:
>>sys4 = tf([1, 0, 775.86,0],[1,0,4185,0,2.25*10^5]);
>>rlocus(sys4),axis equal,sgrid(0.707,[])
The root locus plot is shown in Figure 13.7.2.
Using the magnifying glass tool to expand the plot, and clicking on the intersection of the suspected dominant root path with the ζ = 0.707 line, we find that the dominant root pair is s = −5.53 ± 5.53 j, with a gain of K = 55.6. Using the command r = rlocus(sys4,K), we find that the other roots are at s = −22.2 ± 56 j, and thus the first root pair is dominant. So to achieve a dominant root pair having a damping ratio of ζ = 0.707, we must set c_{1} to be c_{1} = K/0.029 = 1917 N · s/m.
b. Using c = 1917 and the other parameter values, we find that
D_{3}(s) = 250(40) [ s^{4} + 4185s^{2} + 2.25 × 10^{5} + K (s^{3} + 775.86s) ]
\frac{X_{1}(s)}{Y (s)} = \frac{1.5 × 10^{5} (1917s + 1.5 × 10^{4})}{D_{3}(s)}
\frac{X_{2}(s)}{Y (s)} = \frac{1.5 × 10^{5}(250s^{2} + 1917s + 1.5 × 10^{4})}{D_{3}(s)}
We continue the previous session as follows, using K = 55.6
>>[num, den] = tfdata(sys4,'v');
>>D3 = 250*40*(den+55.6*num);
>>sysx1 = tf(150000*[1917,15000],D3);
>>sysx2 = tf(150000*[250,1917,15000],D3);
>>step(sysx1,'-',sysx2,'--')
The plot is shown in Figure 13.7.3. By right clicking on the plot and selecting the “Characteristics” menu, we find that for x1(t), the peak response is 11% at t = 0.357 s and the settling time is 0.655 s. For x2(t), the peak response is 21% at t = 0.051 s and the settling time is 0.494 s.
The response characteristics computed from the dominant root pair s = −5.53 ± 5.53 j using the formulas given in Table 9.3.2 for the second-order model without numerator dynamics are a peak response of 4.3% at t = 0.568 s and a settling time of 0.723 s. The difference between these results is due to the fact that the present model is fourth order and also has numerator dynamics.
| Table 9.3.2 | Step response specifications for the underdamped model m\ddot{x} + c\dot{x} + kx = f . |
| Maximum percent overshoot | M% = 100e^{−πζ/\sqrt{1−ζ^{2}}} |
| ζ = \frac{R}{\sqrt{π^{2} + R^{2}}} , R = \ln \frac{100}{M\%} | |
| Peak time | t_{p} = \frac{π}{ω_{n} \sqrt{1 − ζ^{2}}} |
| Delay time | t_{d} \approx \frac{1 + 0.7ζ}{ω_{n}} |
| 100% rise time | t_{r} = \frac{2π − \phi}{ω_{n} \sqrt{1 − ζ^{2}}} |
| \phi = tan^{−1} \left(\frac{\sqrt{1 − ζ^{2}}}{ζ} \right) + π |
