Question 8.4.4: Response to Nonsinusoidal Inputs An engine valve train can b...
Response to Nonsinusoidal Inputs
An engine valve train can be modeled as an equivalent mass, equivalent damping, and two stiffnesses, one due to the valve spring and one due to the elasticity of the push rod (Figure 8.4.5). The equation of motion is
\frac{I_{o}}{a²} \ddot{x} + c_{e} \dot{x} + (k_{1} + k_{2} \frac{a²}{b²} ) x = \frac{a}{b} k_{2} y(t)The input displacement y(t) is determined by the shape and rotational speed of the cam.
Suppose the input is as shown in Figure 8.4.6a. Determine the steady-state response of the model for the following parameter values:
\ddot{x} + 20 \dot{x} + 625 x = 600 y(t)
The transfer function is
T (s) = \frac{X(s)}{Y (s)} = \frac{600}{s^{2} + 20s + 625}
and
T (jω) = \frac{600}{625 − ω^{2} + 20ωj}
M(ω) = \frac{600}{\sqrt{(625 − ω^{2})^{2} + 400 ω^{2}}} (1)
\phi(ω) = − tan^{−1} \frac{20ω}{625 − ω^{2}} (2)
The Fourier series’ representation of y(t) can be determined from the formulas in Appendix B, and is
y(t) = A_{0} + A_{1} \sin ω_{1}t + A_{2} \cos ω_{2}t + A_{3} \cos ω_{3}t + A_{4} \cos ω_{4}t + · · ·
Figure 8.4.6b is a plot of equation (3) including only those series terms shown. The plot illustrates how well the Fourier series represents the input function.
The steady-state response will have the form
x(t) = B_{0} + B_{1} \sin(ω_{1}t + \phi_{1}) + B_{2} \cos(ω_{2}t + \phi_{2}) + B_{3} \cos(ω_{3} t + \phi_{3}) + B_{4} \cos(ω_{4}t + \phi_{4}) + · · ·
where
B_{i} = A_{i} M_{i} (4)
Note that the amplitudes M_i and phases \phi_{i} of the response due to cosine inputs are computed just as for a sine input.
Using equation (2) of Example 8.4.2, we obtain one positive solution, r = 1.187, and one imaginary solution. Thus, the lower bandwidth frequency is 0, and the upper frequency is ω_{2} = 1.187 \sqrt{625} = 29.7 rad/sec . Because the bandwidth is from 0 to 29.7 rad/sec, the only series terms in y(t) lying within the bandwidth are the constant term (whose frequency is 0), and the \sin 2 \pi t, \cos 4 \pi t, and \cos 8\pi t terms, whose frequencies are 6.28, 12.6, and 25.1 rad/sec, respectively. To demonstrate the filtering property of the system, however, we also compute the effect of the first term lying outside the bandwidth. This is the cos 12πt term, whose frequency is 37.7 rad/sec. The following table was computed using equations (1), (2), (3), and (4).
| i | ωi | Ai | Mi | Bi | Φi |
| 0 | 0 | 0.006366 | 0.96 | 0.006112 | 0 |
| 1 | 2π | 0.01 | 1.001913 | 0.010019 | −0.211411 |
| 2 | 4π | −0.004244 | 1.1312 | −0.004801 | −0.493642 |
| 3 | 8π | −0.000849 | 1.193557 | −0.001013 | −1.584035 |
| 4 | 12π | −0.000364 | 0.547162 | −0.000199 | −2.383436 |
Using this table we can express the steady-state response as follows:
x(t) = 0.006112 + 0.010019 sin(2πt − 0.211411)
− 0.004801 cos(4πt − 0.493642) − 0.001013 cos(8πt − 1.584035)
− 0.000199 cos(12πt − 2.383436)
The input and the response are plotted in Figure 8.4.7. The difference between the input and output results from the resistive or lag effect of the system, not from the omission of the higherorder terms in the series. To see this, note that A_{4} is 43% of A_{3} but B_{4} is only 20% of B_{3}. The decreasing amplitude of the higher-order terms in the series for y(t), when combined with the filtering property of the system, enables us to truncate the series when the desired accuracy has been achieved.
